Jan Dangerfield

Stuart Haring
. Series Editor: Julian Gilbey
Cambridge International
AS & A Level Mathematics:

Mechanics
Coursebook

""':::'w CAMBRIDGE
UNIVERSITY PRESS
 Contents

Contents
Series introduction vi

Acknowledgements viii

How to use this book ix

1 Velocity and acceleration 1

1.1 Displacement and velocity 2
1.2 Acceleration 9
1.3 Equations of constant acceleration 11
1.4 Displacement–time graphs and multi-stage problems 15
1.5 Velocity–time graphs and multi-stage problems 22
1.6 Graphs with discontinuities 27
End-of-chapter review exercise 1 31

2 Force and motion in one dimension 35

2.1 Newton’s first law and relation between force and acceleration 36 iii

2.2 Combinations of forces 39

2.3 Weight and motion due to gravity 42
2.4 Normal contact force and motion in a vertical line 47
End-of-chapter review exercise 2 51

3 Forces in two dimensions 53

3.1 Resolving forces in horizontal and vertical directions in
equilibrium problems 54
3.2 Resolving forces at other angles in equilibrium problems 59
3.3 The triangle of forces and Lami’s theorem for three-force
equilibrium problems 62
3.4 Non-equilibrium problems for objects on slopes and known
directions of acceleration 66
3.5 Non-equilibrium problems and finding resultant forces and
directions of acceleration 71
End-of-chapter review exercise 3 77
Cross-topic revision exercise 1 80
 Cambridge International AS & A Level Mathematics: Mechanics

4 Friction 82
4.1 Friction as part of the contact force 83
4.2 Limit of friction 90
4.3 Change of direction of friction in different stages of motion 95
4.4 Angle of friction 100
End-of-chapter review exercise 4 105

5 Connected particles 108

5.1 Newton’s third law 109
5.2 Objects connected by rods 110
5.3 Objects connected by strings 114
5.4 Objects in moving lifts 121
End-of-chapter review exercise 5 125

6 General motion in a straight line 128

6.1 Velocity as the derivative of displacement with
respect to time 129
iv 6.2 Acceleration as the derivative of velocity with
respect to time 134
6.3 Displacement as the integral of velocity with
respect to time 139
6.4 Velocity as the integral of acceleration with
respect to time 148
End-of-chapter review exercise 6 153
Cross-topic revision exercise 2 155

7 Momentum 157
7.1 Momentum 159
7.2 Collisions and conservation of momentum 161
End-of-chapter review exercise 7 167

8 Work and energy 169

8.1 Work done by a force 171
8.2 Kinetic energy 177
8.3 Gravitational potential energy 179
End-of-chapter review exercise 8 183
 Contents

9 The work–energy principle and power 185

9.1 The work–energy principle 186
9.2 Conservation of energy in a system of conservative forces 193
9.3 Conservation of energy in a system with
non-conservative forces 196
9.4 Power 201
End-of-chapter review exercise 9 206
Cross-topic revision exercise 3 208

Practice exam-style paper 210

Answers 212

Glossary 232

v
 1

Chapter 1
Velocity and acceleration
In this chapter you will learn how to:

■ work with scalar and vector quantities for distance and speed
■ use equations of constant acceleration
■ sketch and read displacement–time graphs and velocity–time graphs
■ solve problems with multiple stages of motion.
 Cambridge International AS & A Level Mathematics: Mechanics

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills
IGCSE / O Level
level Solve quadratics by factorising or using 1 Solve the following equations.
Mathematics the quadratic formula.
a x 2 – 2 x – 15 = 0
b 2x2 + x – 3 = 0
c 3x 2 – 5x – 7 = 0
IGCSE / O Level
level Solve linear simultaneous equations. 2 Solve the following pairs of
Mathematics simultaneous equations.
a 2 x + 3 y = 8 and 5x – 2 y = 1
b 3x + 2 y = 9 and y = 4x – 1

What is Mechanics about?

How far should the driver of a car stay behind another car to be able to stop safely in an
emergency? How long should the fuse on a firework be so the firework goes off at the
highest point? How quickly should you roll a ball so it stops as near as possible to a target?
How strong does a building have to be to survive a hurricane? Mechanics is the study of
questions such as these. By modelling situations mathematically and making suitable
assumptions you can find answers to these questions.
2
In this chapter, you will study the motion of objects and learn how to work out where an
object is and how it is moving at different times. This area of Mechanics is known as
‘dynamics’. Solving problems with objects that do not move is called ‘statics’; you will study
this later in the course.

1.1 Displacement and velocity

An old English nursery rhyme goes like this:

The Grand Old Duke of York,

He had ten thousand men,

He marched them up to the top of the hill,

And he marched them down again.

His men had clearly marched some distance, but they ended up exactly where they started,
so you cannot work out how far they travelled simply by measuring how far their finishing
point is from their starting point.

You can use two different measures when thinking about how far something has travelled.
These are distance and displacement.

Distance is a scalar quantity and is used to measure the total length of path travelled. In
the rhyme, if the distance covered up the hill were 100 m, the total distance in marching up
the hill and then down again would be 100 m + 100 m = 200 m.
 Chapter 1: Velocity and acceleration

Displacement is a vector quantity and gives the location of an object relative to a fixed
TIP
reference point or origin. In this course, you will be considering dynamics problems in only
one dimension. To define the displacement you need to define one direction as positive. In A scalar quantity, such
as distance, has only a
the rhyme, if you take the origin to be the bottom of the hill and the positive direction to
magnitude. A vector
be up the hill, then the displacement at the end is 0 m, since the men are in the same
quantity, such as
location as they started. You can also reach this answer through a calculation. If you displacement, has
assume that they are marching in a straight line then marching up the hill is an increase in magnitude and
displacement and marching down the hill is a decrease in displacement, so the total direction. When you
displacement is ( +100 m) + ( −100 m) = 0 m. are asked for a vector
quantity such as
Since you will be working in only one dimension, you will often refer to the displacement displacement or
as just a number, with positive meaning a displacement in one direction from the origin velocity, make sure you
and negative meaning a displacement in the other direction. Sometimes the direction and state the direction as
origin will be stated in the problem. In other cases, you will need to choose these yourself. well as the magnitude.
In many cases the origin will simply be the starting position of an object and the positive
direction will be the direction the object is initially moving in.

KEY POINT 1.1 WEB LINK

Displacement is a measure of location from a fixed origin or starting point. It is a vector and so has Try the Discussing
both magnitude and direction. If you take displacement in a given direction to be positive, then distance resource at the
displacement in the opposite direction is negative. Introducing calculus
station on the
Underground 3
You also have two ways to measure how quickly an object is moving: speed and velocity.
Mathematics website
Speed is a scalar quantity, so has only a magnitude. Velocity is a vector quantity, so has
(www.underground
both magnitude and direction. mathematics.org).
For an object moving at constant speed, if you know the distance travelled in a given time
you can work out the speed of the object.

KEY POINT 1.2

For an object moving at constant speed:

distance covered
speed =
time taken

This is valid only for objects moving at constant speed. For objects moving at non-
constant speed you can consider the average speed.

KEY POINT 1.3

total distance covered

average speed =
total time taken

Velocity measures how quickly the displacement of an object changes. You can write an
equation similar to the one for speed.
 Cambridge International AS & A Level Mathematics: Mechanics

KEY POINT 1.4

For an object moving at constant velocity:

change in displacement
velocity =
time taken

Let’s see what this means in practice.

Suppose a man is doing a fitness test. In each stage of the test he runs backwards and forwards
along the length of a small football pitch. He starts at the centre spot, runs to one end of the
pitch, changes direction and runs to the other end, changes direction and runs back to the
centre spot, as shown in the diagrams. He runs at 4 m s −1 and the pitch is 40 m long.

To define displacement and velocity you will need to define the origin and the direction you
will call positive. Let’s call the centre spot the origin and to the right as positive.

In the first diagram, he has travelled a distance of positive

10 m. Because he is 10 m in the positive direction, his
displacement is 10 m. His speed is 4 m s −1. Because he
is moving in the positive direction, his velocity is
also 4 m s −1.
10 m
In the second diagram, he has travelled a total 40 m

distance of 30 m, but he is only 10 m from the centre

4 spot, so his displacement is 10 m. His speed is still
4 m s −1 but he is moving in the negative direction so
his velocity is −4 m s −1.

In the third diagram, he has travelled a total distance 10 m

of 50 m, but he is now 10 m from the centre spot in the
negative direction, so his displacement is −10 m. His
speed is still 4 m s −1 and he is still moving in the
negative direction so his velocity is still −4 m s −1.

In the fourth diagram, he has travelled a total distance 10 m

of 70 m, but his displacement is still −10 m. His speed
is still 4 m s −1 and he is moving in the positive direction
again so his velocity is also 4 m s −1.

The magnitude of the velocity of an object is its

speed. Speed can never be negative. For example,
an object moving with a velocity of +10 m s −1 and an 10 m TIP
object moving with a velocity of −10 m s −1 both have a speed of 10 m s −1.
We use vertical lines to
As with speed, for objects moving at non-constant velocity you can consider the average velocity. indicate magnitude of
a vector.
KEY POINT 1.5
So, speed = velocity
net change in displacement
average velocity =
total time taken
 Chapter 1: Velocity and acceleration

In the previous example, the man’s average speed is 4 m s −1 but his average velocity is 0 m s −1. WEB LINK
We can rearrange the equation for velocity to deduce that for an object moving at constant Try the Speed vs
velocity v for time t, the change in displacement s (in the same direction as the velocity) is velocity resource at the
given by: Introducing calculus
station on the
s = vt Underground
The standard units used for distance and displacement are metres (m) and for time are Mathematics website.

seconds (s). Therefore, the units for speed and velocity are metres per second (usually
written in mathematics and science as m s −1, though you may also come across the notation
m/s). These units are those specified by the Système Internationale (SI), which defines the
system of units used by scientists all over the world. Other commonly used units for speed
include kilometres per hour (km/h) and miles per hour (mph).

WORKED EXAMPLE 1.1 TIP

You usually only

A car travels 9 km in 15 minutes at constant speed. Find its speed in m s −1.
include units in the
final answer to a
Answer
problem and not in all
9 km = 9000 m and Convert to units required for the answer which the earlier steps. This is
15 minutes = 900 s are SI units. because it is easy to
confuse units and
s = vt Substitute into the equation for displacement variables. For example, 5
s for displacement can
so 9000 = 900v and solve.
be easily mixed up with
v = 10 m s −1 s for seconds. It is
important to work in
SI units throughout, so
WORKED EXAMPLE 1.2 that the units are
consistent.
A cyclist travels at 5 m s −1 for 30 s then turns back, travelling at 3 m s −1 for 10 s. Find
her displacement in the original direction of motion from her starting position.

Answer

s = vt
So s1 = 5 × 30 Separate the two stages of the journey.
= 150 Remember travelling back means a negative
and s2 = −3 × 10 velocity and a negative displacement.
= −30
So the total displacement is
s = 150 + ( −30)
= 120 m
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 1.3

A cyclist spends some of his journey going downhill at 15 m s −1 and the rest of the time going uphill at 5 m s −1. In
one minute he travels 540 m. Find how long he spent going downhill.

Answer

Let t  be the amount of time spent going downhill. Define the variable.

Then 60 − t is the amount of time spent going uphill. Write an expression for the time spent
travelling uphill.

Total distance = 15t + 5(60 − t ) = 540 Set up an equation for the total distance.
15t + 300 − 5t = 540
10t = 240
t = 24 s

EXPLORE 1.1

Two students are trying to solve this puzzle.

A cyclist cycles from home uphill to the shop at 5 m s −1. He then cycles home and wants
6
to average 10 m s −1 for the total journey. How fast must he cycle on the way home?

Their solutions are shown here. Decide whose logic is correct and try to explain what
WEB LINK
is wrong with the other’s answer.
You may want to have
Student A Student B
a go at the Average
Call the speed on the return journey v. Cycling at 5 m s −1 will take twice as long as it would speed resource at the
The average of 5 m s −1 and v is 10 m s −1 if he were going at 10 m s −1. That means he has used Introducing calculus
so v must be 15 m s −1. up the time required to go there and back in the first
station on the
part of the journey, so it is impossible to average
Underground
10 m s −1 for the total journey.
Mathematics website.

MODELLING ASSUMPTIONS

Throughout this course, there will be questions about how realistic your answers are.
To simplify problems you will make reasonable assumptions about the scenario to
allow you to solve them to a satisfactory degree of accuracy. To improve the
agreement of your model with what happens in the real world, you would need to
refine your model, taking into account factors that you had initially ignored.

In some of the questions so far, you might ask if it is reasonable to assume constant
speed. In real life, speed would always change slightly, but it could be close enough to
constant that it is a reasonable assumption.
 Chapter 1: Velocity and acceleration

With real objects such as bicycles or cars, there is the question of which part of the
object you are referring to. You can be consistent and say it is the front of a vehicle,
but when it is a person the front changes from the left leg to the right leg. You may
choose to consider the position of the torso as the position of the person. In all the
examples in this coursebook, you will consider the object to be a particle, which is
very small, so you do not need to worry about these details. You will assume any
resulting errors in the calculations will be sufficiently small to ignore. This could
cause a problem when you consider the gap between objects, because you may not
have allowed for the length of the object itself, but in our simple models you will
ignore this issue too.

DID YOU KNOW?

Once they have reached top speed, swimmers tend

to move at a fairly constant speed at all points during
the stroke. However, the race ends when the swimmer
touches the end of the pool, so it is important to time
the last two or three strokes to finish with arms
extended. If the stroke finishes early the swimmer
might not do another stroke and instead keep their
arms extended, but this means the swimmer slows
down. In a close race, another swimmer may 7
overtake if that swimmer times their strokes better.
This happened to Michael Phelps when he lost to
Chad Le Clos in the final of the Men’s 200 m
Butterfly in the 2012 Olympics.

EXERCISE 1A

1 A cyclist covers 120 m in 15 s at constant speed. Find her speed.

2 A sprinter runs at constant speed of 9 m s −1 for 7 s. Find the distance covered.

M 3 a A cheetah spots a grazing gazelle 150 m away and runs at a constant 25 m s −1 to catch it. Find how long it
takes to catch the gazelle.
b What assumptions have been made to answer the question?

4 The speed of light is 3.00 × 108 m s −1 to 3 significant figures. The average distance between the Earth and the
Sun is 150 million km to 3 significant figures. Find how long it takes for light from the Sun to reach the Earth
on average. Give the answer in minutes and seconds.

5 The land speed record was set in 1997 at 1223.657 km h −1. Find how long in seconds it took to cover 1km when
the record was set.

M 6 A runner runs at 5 m s −1 for 7 s before increasing the pace to 7 m s −1 for the next 13s.
a Find her average speed.
b What assumptions have been made to answer the question?
 Cambridge International AS & A Level Mathematics: Mechanics

7 A remote control car travels forwards at 6 m s −1 in Drive and backwards at 3 m s −1 in Reverse. The car travels
for 10 s in Drive before travelling for 5 s in Reverse.
a Find its displacement from its starting point.
b Find its average velocity in the direction in which it started driving forwards.
c Find its average speed.

8 A speed skater averages 11m s −1 over the first 5 s of a race. Find the average speed required over the next 10 s
to average 12 m s −1 overall.

9 The speed of sound in wood is 3300 m s −1 and the speed of sound in air is 330 m s −1. A hammer hits one end of
a 33 m long plank of wood. Find the difference in time between the sound waves being detected at the other
end of the plank and the sound being heard through the air.

10 An exercise routine involves a mixture of jogging at 4 m s −1 and sprinting at 7 m s −1. An athlete covers 1km in
3 minutes and 10 seconds. Find how long she spent sprinting.

11 Two cars are racing over the same distance. They start at the same time, but one finishes 8 s before the other.
The faster one averaged 45 m s −1 and the slower one averaged 44 m s −1. Find the length of the race.

12 Two air hockey pucks are 2 m apart. One is struck and moves directly towards the other at 1.3 m s −1. The other
is struck 0.2 s later and moves directly towards the first at 1.7 m s −1. Find how far the first puck has moved
when the collision occurs and how long it has been moving for.

P 13 A motion from point A to point C is split into two parts. The motion from A to B has displacement s1 and
8
takes time t1. The motion from B to C has displacement s2 and takes time t2.
a Prove that if t1 = t2 the average speed from A to C is the same as the average of the speeds from A to B
and from B to C .
b Prove that if s1 = s2 the average speed from A to C is the same as the average of the speeds from A to B
and from B to C if, and only if, t1 = t2.

P 14 The distance from point A to point B is s. In the motion from A to B and back, the speed for the first part of the
motion is v1 and the speed for the return part of the motion is v2 . The average speed for the entire motion is v.
2v1v2
a Prove that v = .
v1 + v2
b Deduce that it is impossible to average twice the speed of the first part of the motion, that is, it is
impossible to have v = 2v1.
 Chapter 1: Velocity and acceleration

1.2 Acceleration
Velocity is not the only measure of the motion of an object. It is useful to know if, and how,
the velocity is changing. We use acceleration to measure how quickly velocity is changing.

KEY POINT 1.6 TIP

For an object moving at constant acceleration, The units of

acceleration are m s −2.
change in velocity
acceleration =
time taken
If an object has constant acceleration, a, initial velocity u and it reaches final velocity v in time t,
then
v−u
a=
t
where u, v and a are all measured in the same direction.

An increase in velocity is a positive acceleration, as shown in the diagram on the left.

A decrease in velocity is a negative acceleration, as shown in the diagram on the right. This
is often termed a deceleration.

positive positive
direction direction

initial final initial final

velocity velocity velocity velocity 9
20 m s−1 30 m s−1 20 m s–1 10 m s–1

EXPLORE 1.2

If the initial velocity is negative, what effect would a positive acceleration have on the
car? Would it be moving more quickly or less quickly?

What effect would a negative acceleration have on the car in this situation? Would it
be moving more quickly or less quickly?

When the acceleration is constant, the average velocity is simply the average of the initial
1
and final velocities, which is given by the formula ( u + v ). This can be used to find
2
displacements using the equation for average velocity from key point 1.5.

KEY POINT 1.7

If an object has constant acceleration, a, initial velocity u and it reaches final velocity v in time t,
then the displacement, s is given by
1
s = ( u + v )t
2
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 1.4

TIP
A parachutist falls from rest to 49 m s −1 over 5 s. Find her acceleration.
‘Rest’ means not
Answer moving, so velocity is
v−u zero.
a=
t
49 − 0
=
5
= 9.8 m s −2 Make sure you use the correct units which
are m s −2 .

WORKED EXAMPLE 1.5

A tractor accelerates from 5 m s −1 to 9 m s −1 at 0.5 m s −2 . Find the distance covered by the tractor over this time.

Answer
v−u v−u
a= Substitute into a = first to find t.
t t
9−5
So 0.5 =
t
10 0.5t = 4
t = 8s

1 1
s= ( u + v )t Substitute into s = ( u + v )t to find s.
2 2
1
= (5 + 9) × 8
2
= 56 m

EXERCISE 1B

1 A car accelerates from 4 m s −1 to 10 m s −1 in 3s at constant acceleration. Find its acceleration.

2 A car accelerates from rest to 10 m s −1 in 4 s at constant acceleration. Find its acceleration.

3 A car accelerates from 3 m s −1 at an acceleration of 6 m s −2 . Find the time taken to reach 12 m s −1.

4 An aeroplane accelerates at a constant rate of 3 m s −2 for 5 s from an initial velocity of 4 m s −1. Find its final
velocity.

5 A speedboat accelerates at a constant rate of 1.5 m s −2 for 4 s reaching a final velocity of 9 m s −1. Find its initial
velocity.

6 A car decelerates at a constant rate of 2 m s −2 for 3s finishing at a velocity of 8 m s −1. Find its initial velocity.

7 A car accelerates from an initial velocity of 4 m s −1 to a final velocity of 8 m s −1 at a constant rate of 0.5 m s −2 .
Find the displacement in that time.
 Chapter 1: Velocity and acceleration

M 8 A sprinter covers 60 m in 10 s accelerating from a jog. Her final velocity is 9 m s −1.

a Calculate her acceleration.
b What assumptions have been made to answer the question?

9 A wagon is accelerating down a hill at constant acceleration. It took 1s more to accelerate from a velocity
of 1m s −1 to a velocity of 5 m s −1 than it took to accelerate from rest to a velocity of 1m s −1. Find the
acceleration.

10 A driver sees a turning 100 m ahead. She lets her car slow at constant deceleration of 0.4 m s −2 and arrives at
the turning 10 s later. Find the velocity she is travelling at when she reaches the turning.

PS 11 A cyclist is travelling at a velocity of 10 m s −1 when he reaches the top of a slope which is 80 m long. There is a
bend at the bottom of the slope which it would be dangerous to go round faster than 11m s −1. Down the slope
he would accelerate at 0.1m s −2 because of gravity if he did not pedal or brake. To go as fast as possible but
still reach the bottom at a safe speed should the cyclist brake, do nothing or pedal?

1.3 Equations of constant acceleration

In Worked example 1.5, you needed two equations to find the required answer. Wherever
possible it is better to go directly from the information given to the required answer using
just one equation because it is more efficient and reduces the number of equations to solve,
and therefore reduces the likelihood of making mistakes.

There are five equations relating the five variables s, u, v, a and t. Each equation relates
11
four of the five variables.

Two of these equations were introduced in Section 1.2, although the first one is normally
given in the rearranged form shown in Key Point 1.8.
TIP
KEY POINT 1.8
In general, these
For an object travelling with constant acceleration a, for time t, with initial velocity u, final velocity equations are only
v and change of displacement s , we have valid if the acceleration
is constant.
v = u + at
1
s = ( u + v )t
2
FAST FORWARD
1
s = ut + at 2
2 In Chapter 6 you will
1 2 consider how acceleration,
s = vt − at speed, distance and
2
time are related when
v 2 = u 2 + 2as
the acceleration is not
These equations are often referred to as the suvat equations. constant.

You will derive these equations in Exercise 1C.

 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 1.6

a A go-kart travels down a slope of length 70 m. It is given a push and starts moving at an initial
velocity of 3 m s −1 and accelerates at a constant rate of 2 m s −2 . Find its velocity at the bottom of
the slope.

b Find the time taken for the go-kart to reach the bottom of the slope.

Answer

a The time, t, is unknown. It is often useful to list what information is

The final velocity, v, is unknown. given and what is unknown.

s = 70
u=3
a=2

v 2 = u2 + 2 as Choose the equation with the known variables

= 32 + 2 × 2 × 70 and the one required.
= 289 In this case, we know u, a and s, and we want
v = ±17 to find v.

We know that v . 3. From the context the velocity is increasing

12 v = 17 ms −1 from 3 m s −1, so only the positive solution is
required.

A negative velocity would indicate movement

in the opposite direction.

1 2
b s = ut + at Use a formula that involves given values rather
2
1 than relying on your calculated values, as this
70 = 3t + × 2t 2 will increase your chances of getting the
2
t 2 + 3t − 70 = 0 correct answer even if your earlier answer was
(t + 10)(t − 7) = 0 wrong.

t = −10 or t = 7 Negative time would refer to time before the

We know that t . 0 so t = 7 s. go-kart started its descent. Only the positive
solution is required.
 Chapter 1: Velocity and acceleration

WORKED EXAMPLE 1.7

A trolley has a constant acceleration. After 2 s it has travelled 8 m and after another 2 s it has travelled a further
20 m. Find its acceleration.

Answer
Let the initial speed be v1. There are unknown velocities at three different
Let the speed after 2 s be v2. times so simply using u and v may be
insufficient and unclear.
Let the speed after 4 s be v3.
Acceleration, a, is unknown. List the information for the first 2 s.
s=8 t=2 There are too many unknowns to be able to
u = v1 v = v2 calculate any of them at this stage.

Acceleration, a, is unknown but the same as for the first 2 s. List the information for the next 2 s.
s = 20 t=2
Note that the speed after 2 s is the final speed
u = v2 v = v3 for the first 2 s but the initial speed for the next
2 s so we can use the same letter to represent it.
1 2 Since v2 and a are the common unknowns in
s = vt − at
2
both stages of the motion, we will write
1 13
8 = 2v2 − × a × 22 equations relating them. First write the
2
2v2 − 2 a = 8 equation for the first stage of the motion.

1 2 We will again write the equation relating v2

s = vt − at
2
and a, but now for the second stage of the
1
20 = 2v2 + × a × 22 motion.
2
2v2 + 2 a = 20

4v2 = 28 Solve simultaneously by adding the equations

v2 = 7 m s −1 and substituting the value for v2 back in to one
of the original equations.
So a = 3 m s −2

or There is an alternative solution by considering

1
s = ut + at 2 the whole 4 s as one motion and creating
2
1 equations involving v1.
8 = 2v1 + × a × 22
2
1
28 = 4v1 + × a × 42
2
giving v1 = 1m s −1
So a = 3 m s −2
 Cambridge International AS & A Level Mathematics: Mechanics

EXERCISE 1C

1 For each part, assuming constant acceleration, write down the equation relating the four variables in the
question and use it to find the missing variable.
a Find s when a = 3 m s −2, u = 2 m s −1 and t = 4 s.
b Find s when a = 2 m s −2, v = 17 m s −1 and t = 8 s.
c Find a when s = 40 m , u = 3 m s −1 and t = 5 s.
d Find a when s = 28 m, v = 13 m s −1 and t = 4 s.
e Find a when s = 24 m, u = 2 m s −1 and v = 14 m s −1.
f Find u when s = 45 m, a = 1.5 m s −2 and t = 6 s.
g Find v when s = 24 m, a = −2.5 m s −2 and t = 4 s.
h Find s when a = 0.75 m s −2 , u = 2 m s −1 and v = 5 m s −1.

2 Assuming constant acceleration, find the first time t, for positive t, at which the following situations occur.
a Find t when a = −2 m s −2, u = 10 m s −1 and s = 24 m.
b Find t when a = 0.5 m s −2, v = 5 m s −1 and s = 21m.
c Find t when a = 1m s −2, u = 3 m s −1 and s = 20 m.

3 Assuming constant acceleration, find v when s = 6 m, u = 5 m s −1 and a = −2 m s −2 if the object has changed
14 direction during the motion.

4 Assuming constant acceleration, find u when s = 60 m, v = 13 m s −1 and a = 1m s −2 if the object has not
changed direction during the motion.

5 a Assuming constant acceleration, find v when s = 18 m, u = 3 m s −1 and a = 2 m s −2.

b Why is it not necessary to specify in this question whether the object has changed direction during the
motion?

6 A car is travelling at a velocity of 20 m s −1 when the driver sees the traffic lights ahead change to red. He
decelerates at a constant rate of 4 m s −2 and comes to a stop at the lights. Find how far away from the lights
the driver started braking.

7 An aeroplane accelerates at a constant rate along a runway from rest until taking off at a velocity of 60 m s −1.
The runway is 400 m long. Find the acceleration of the aeroplane.

8 An aeroplane accelerates from rest along a runway at a constant rate of 4 m s −2 . It needs to reach a velocity of
80 m s −1 to take off. Find how long the runway needs to be.

9 A motorcyclist sees that the traffic lights are red 40 m ahead of her. She is travelling at a velocity of 20 m s −1
and comes to rest at the lights. Find the deceleration she experiences, assuming it is constant.

M 10 A driver sees the traffic lights change to red 240 m away when he is travelling at a velocity of 30 m s −1. To avoid
wasting fuel, he does not brake, but lets the car slow down naturally. The traffic lights change to green after
12 s, at the same time as the driver arrives at the lights.
a Find the speed at which the driver goes past the lights.
b What assumptions have been made to answer the question?
 Chapter 1: Velocity and acceleration

11 In a game of curling, competitors slide stones over the ice at a target 38 m away. A stone is released directly
towards the target with velocity 4.8 m s −1 and decelerates at a constant rate of 0.3 m s −2. Find how far from the
target the stone comes to rest.

12 A golf ball is struck 10 m from a hole and is rolling towards the hole. It has an initial velocity of 2.4 m s −1 when
struck and decelerates at a constant rate of 0.3 m s −2 . Does the ball reach the hole?

PS 13 A driverless car registers the traffic lights change to amber 40 m ahead. The amber light is a 2 s warning
before turning red. The car is travelling at 17 m s −1 and can accelerate at 4 m s −2 or brake safely at 8 m s −2 .
What options does the car have?
1
P 14 The first two equations in Key Point 1.8 are v = u + at and s = ( u + v )t. You can use these to derive the
2
other equations.
1
a By substituting for v in the second equation, derive   s = ut + at 2.
2
1 2
b Derive the remaining two equations, s = vt − at and v 2 = u2 + 2 as, from the original two equations.
2
P 15 Show that an object accelerating with acceleration a from velocity u to velocity v, where 0 , u , v, over a
u+v 1
time t is travelling at a velocity of at time t, that is that at the time halfway through the motion the
2 2
velocity of the object is the mean of the initial and final velocities.

P 16 Show that an object accelerating with acceleration a from velocity u to velocity v, where 0 , u , v, over a
v 2 + u2 1
displacement s is travelling at a speed of at a distance s. Hence prove that when the object does
2 2 15
not change direction the speed at the midpoint of the distance is always greater than the mean of the initial and
final speeds. Deduce also that the mean of the initial and final speeds occurs at a point closer to the start of the
motion than the end.

1.4 Displacement–time graphs and multi-stage problems

It can be useful to show how the position of an object changes over time. You can do this
using a displacement–time graph.

Imagine the following scenario. A girl is meeting a friend 1km down a straight road. She
walks 30 m along the road to a bus stop in 20 s. Then she waits 30 s for a bus, which takes
her to a bus stop 20 m past her friend. The bus does not stop to pick anyone else up or drop
them off. The journey takes 150 s. She walks the 20 m back to meet her friend in 15 s.

The graph would look like the one shown. You always show time on the x-axis and
s
displacement on the y-axis. Notice you are defining the time as being measured from
1020
when the girl starts walking and the displacement from where she starts walking in the 1000
direction of her friend.

Where the graph is horizontal it indicates that the displacement is unchanged and 30
t
therefore the girl is not moving. This was when she was waiting for the bus. If the graph 20 50 200 215

is not horizontal it indicates the position is changing and the steepness of the line
indicates how quickly it is changing.
 Cambridge International AS & A Level Mathematics: Mechanics

A straight line on a displacement–time graph indicates a constant speed, as when the girl
was walking to the bus stop. A curved line indicates a change in speed, for example, when
the bus started moving after picking the girl up and when it slowed down to stop.

Notice that when the girl got off the bus to meet her friend she travelled in the opposite
direction, so the change in her displacement and hence her velocity are negative. On the
graph there is a negative gradient. The speed is the magnitude of the gradient, but the
velocity includes the negative sign to indicate the direction.

Displacement–time graphs can have negative displacements below the x-axis, unlike
distance–time graphs.

KEY POINT 1.9 FAST FORWARD

The gradient of a displacement–time graph is equal to the velocity of the object. In Chapter 6, you will
consider gradients of
curved displacement–
When sketching a graph of the motion of an object, you should show clearly the shape of time graphs.
the graph, and carefully distinguish a straight line from a curve. On a sketch you need to
show only the key points. These include the intercept on the vertical axis, which is the
initial position of the object, and any intercepts on the horizontal time axis, where the
object is at the reference point. If there is more than one stage to the motion, you should
clearly indicate the time and displacement of the object at the change in the motion.

16
WORKED EXAMPLE 1.8

A racing car passes the finish line of a race moving at a constant velocity of 60 m s −1. After 5 s it starts decelerating
at 3 m s −2 until coming to rest. Sketch the displacement–time graph for the motion after the end of the race,
measuring displacement from the finishing line.

Answer
Let t1 be the time from the start of the first stage The first stage of the journey is while the car
and t2 be the time from the start of the second stage. travels at constant velocity.

The second stage is while the car is

Let s1 be the displacement up to time t1 during the decelerating.
first stage and s2 the distance travelled during the second stage.

For the end of the first stage, Find the displacement during the first stage
s = vt because it will be marked on the sketch.

So when t1 = 5
s1 = 60 × 5
= 300 m
For the graph for 0 , t , 5, The graph of the first stage relates the variables
s = 60t s and t and is found using the equation for
constant velocity.
 Chapter 1: Velocity and acceleration

s (m)
The first stage is a straight line graph with
gradient 60 for 5 s.
300
The line starts at the origin because initial
t (s)
5 displacement is 0 m.

After 5 s the displacement is 300 m.

For the end of the second stage, During the deceleration stage use an equation
v = u + at for constant acceleration to find the value of t2
at the end of that stage of the motion.

So, 0 = 60 + (–3) × t2 Use the final velocity from the first stage as the
t2 = 20 initial velocity for the second stage.

v 2 = u2 + 2 as During the deceleration stage use an equation

2 2
0 = 60 + 2 × ( −3) × s2 for constant acceleration to find the value of s2
s2 = 600 at the end of that stage of the motion.

Total time for the journey = t1 + t2 The total time is the value to be marked on the
= 5 + 20 = 25 s sketch.
Total displacement = s1 + s2 The total displacement from the finish line is
= 300 + 600 the sum of both displacements.
= 900 m 17
For 5 , t , 25, The general displacement during the second
s = 300 + s2 stage is the sum of the displacement at the end
of the first stage and the displacement during
and
the second stage.
t = 5 + t2
1 2
s = ut + at
2
1
s2 = 60t2 + × −3 × t22
2
3
= 60t2 − t22
2
For the graph for 5 , t , 25, We can now find the equation of the curve in
3 2 terms of s and t.
s = 300 + 60t2 − t2
2
3
= 300 + 60(t − 5) − (t − 5)2
2
3 2 75
=− t + 75t −
2 2
s (m) The graph for the second stage is a negative
900 quadratic curve, valid for 5 , t , 25 finishing
horizontal at t = 25 (since v = 0).
300
t (s) The join between the graphs at t = 5 is smooth
5 25
with the same gradient on the line before the
join and the curve immediately after the join.
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 1.9

A cyclist is travelling at a velocity of 15 m s −1 when he passes a junction. He then decelerates at a constant rate of
0.6 m s −2 until coming to rest. A second cyclist travels at a constant velocity of 20 m s −1 and passes the junction 4 s
after the first cyclist. Find the time at which the second cyclist passes the first and the displacement from the
junction when that happens.

Answer
Let t1 be the time from the first cyclist reaching The cyclists pass the junction at different
the junction and t2 be the time from the second cyclist times, so it may be useful to define times for
reaching the junction.
each of them separately.

Let s1 be the displacement of the first cyclist from

the junction and s2 be the displacement of the second
cyclist from the junction.

1 2
s = ut + at Find a formula for the displacement of the first
2
1 cyclist.
s1 = 15t1 + × ( −0.6)t12
2

s = vt Find a formula for the displacement of the

s2 = 20t2 second cyclist from the junction, noting that the
18
time is not measured from the same instant.

t2 = t1 − 4 Find how the times are related.

So s2 = 20(t1 − 4)

s1 = s2 One cyclist passes the other when they have the

15t1 −
2
0.3t1 = 20(t1 − 4) same displacement.
3t12 + 50t1 − 800 = 0
(3t1 + 80)(t1 − 10) = 0
80
t1 = 10, −
3
So t1 = 10 s

1
s1 = 15 × 10 + × −3 × 102 Find the displacement from either formula as
2
= 120 m they should give the same answer.

Or s2 = 20(10 − 4)
= 120 m
 Chapter 1: Velocity and acceleration

WORKED EXAMPLE 1.10

Cyclist A is travelling at 16 m s −1 when she sees cyclist B 15 m ahead A B

travelling at a constant velocity of 10 m s −1. Cyclist A then slows initial
at 1.5 m s −2 . Find the minimum gap between the cyclists. position
A B
15 m
at time t
sB
sA
Answer
Let t be the time measured from when the cyclists Find the gap between the cyclists by adding the
are 15 m apart and let the gap between the cyclists at original gap and the change in displacement of
time t be G(t ) .
the leading cyclist, and then subtracting the
G(t ) = G(0) + sB − sA change of displacement of the following cyclist.
G(t ) = 15 + 10t − (16t − 0.75t 2 )

G(t ) = 0.75(t − 4)2 + 3 Complete the square to find the minimum gap
Minimum gap is 3 m at 4 s. and the time at which it occurs.

Or 16 − 1.5t = 10 Alternatively, the closest distance is when the

t=4 cyclists travel at the same speed because once
the cyclist behind slows down the gap will
G(t ) = 15 + 10t − (16t − 0.75t 2 ) = 3 m
increase again.
19

EXERCISE 1D

1 Sketch the displacement–time graphs from the information given. In each case consider north to be the
positive direction and home to be the point from which displacement is measured.
a Bob leaves his home and heads north at a constant speed of 3 m s −1 for 10 s.
b Jenny is 30 m north of home and walks at a constant speed of 1.5 m s −1 until reaching home.
c Ryo is sitting still at a point 10 m south of his home.
d Nina is 300 m north of her home. She drives south at a constant speed of 10 m s −1, passing her home, until
she has travelled a total of 500 m.

2 Sketch the displacement–time graphs from the information given. In each case consider upwards to be the
positive direction and ground level to be the point from which displacement is measured. Remember to
include the values for time and displacement at any points where the motion changes.
a A firework takes off from ground level accelerating upwards for 10 s with constant acceleration 4 m s −2.
b A ball is thrown upwards from a point 1m above the ground with initial speed 5 m s −1. It accelerates
downwards at a constant rate of 10 m s −2 until it stops moving upwards, when it is caught by someone
standing on a ladder.
 Cambridge International AS & A Level Mathematics: Mechanics

c A rocket is falling at 10 m s −1 at a height of 100 m above the ground when its engines turn on to provide a
constant acceleration of 2 m s −2 upwards. The engines remain on until the rocket has reached a height of
175 m above ground level.
d A pebble is thrown upwards from the top of a cliff 18.75 m above the sea. It has initial speed 5 m s −1. It
accelerates downwards at a constant rate of 10 m s −2 until it stops moving upwards, starts falling and
reaches the sea at the bottom of the cliff. Displacement is measured from the top of the cliff.

3 Sketch the displacement–time graphs from the information given. In each case consider forwards to be the
positive direction and the traffic lights to be the point from which displacement is measured. Remember to
include the values for time and displacement at any points where the motion changes.
a A car is waiting at rest at the traffic lights. It accelerates at a constant rate of 3 m s −2 for 5 s then remains at
constant speed for the next 10 s .
b A motorbike passes the traffic lights at a constant speed of 10 m s −1. After 6 s it starts to slow at a constant
rate of 2 m s −2 until it comes to rest.
c A truck is moving at constant speed of 8 m s −1 and is approaching the traffic lights 60 m away. When it is
20 m away it accelerates at a constant rate of 2 m s −2 to get past the lights before they change colour.
d A scooter accelerates from rest 100 m before the traffic lights at a constant rate of 1.5 m s −2 until it reaches
6 m s −1. It then travels at this speed until it reaches a point 50 m beyond the traffic lights. At that point it
starts to slow at a constant rate of 1m s −2 until it stops.

4 The sketch shows a displacement–time graph of the position of a train passing s (m)

a station. The displacement is measured from the entrance of the station to the 100

20 front of the train. Find the equation of the displacement–time graph and hence
the time at which the front of the train reaches the entrance of the station.
0 t (s)
10
−60

5 The sketch shows a displacement–time graph of a car slowing down with s (m)
constant acceleration before coming to rest at a set of traffic lights.
a The equation of the displacement–time graph can be written in the form
s = p(t − q )2 + r. Using the two points marked and the fact that the car is
0 t (s)
stationary at t = 10, find p, q and r. 10

1
b By comparison with the equation s = s0 + ut + at 2, find the initial speed
2 −50
and acceleration of the car.

PS 6 Two cars drive along the same highway. One car starts at junction 1 travelling north at a constant speed of
30 m s −1. The second car starts at junction 2, which is 3 km north of junction 1, travelling south at a constant
speed of 20 m s −1.
a Sketch the two displacement–time graphs on the same set of axes.
b Find the equations of the two displacement–time graphs.
c Solve the equations to find the time at which the cars pass each other and hence find the distance from
junction 1 at which they pass.
 Chapter 1: Velocity and acceleration

7 Two trains travel along the same stretch of track 5 km long. One starts at the southern end travelling north at
a constant speed of 25 m s −1. The second train starts at the northern end 40 s later travelling south at a
constant speed of 15 m s −1.
a Sketch the two displacement–time graphs on the same set of axes.
b Find the time for which the first train has been moving and the distance the first train has travelled when
the trains pass each other.

8 A cyclist is stationary when a second cyclist passes travelling at a constant speed of 8 m s −1. The first cyclist
then accelerates for 5 s at a constant rate of 2 m s −1 before continuing at constant speed until overtaking the
second cyclist. By sketching both graphs, find the equations of the two straight line sections of the graphs and
hence find how long it is before the first cyclist overtakes the second.

M 9 Two rowing boats are completing a 2 km course. The first boat leaves, its crew rowing at a speed of 3.2 m s −1 .
The second boat leaves some time later, its crew rowing at 4 m s −1, and overtakes the first boat after the second
has been travelling for 40 s.
a Find how much earlier the second boat completes the course.
b What assumption has been made in your answer?

10 The leader in a race has 500 m to go and is running at a constant speed of 4 m s −1, but with 100 m to go
increases her speed by a constant acceleration of 0.1m s −2 . The second runner is 100 m behind the leader when
the leader has 500 m to go, and running at 3.8 m s −1 when she starts to accelerate at a constant rate. Find the
minimum acceleration she needs in order to win the race.

PS 11 A van driver wants to pull out from rest onto a road where cars are moving at a constant speed of 20 m s −1. 21
When there is a large enough gap between cars, the van driver pulls out immediately after one car passes. She
then accelerates at a constant rate of 4 m s −2 until moving at 20 m s –1. To do this safely the car behind must
always be at least 10 m away. Find the minimum length of the gap between the cars for the van driver to pull out.

12 A police motorcyclist is stationary when a car passes, driving dangerously at a constant speed of 40 m s −1 . At
the instant the car passes, the motorcyclist gives chase, accelerating at 2.5 m s −2 until reaching a speed of
50 m s −1 before continuing at a constant speed. Show that the motorcyclist has not caught the car by the time
he reaches top speed. Find how long after the car initially passed him the motorcyclist catches the car.

13 The front of a big wave is approaching a beach at a constant speed of 11.6 m s −1. When it is 30 m away from a
boy on the beach the wave starts decelerating at a constant rate of 1.6 m s −2 and the boy walks away from the
sea at a constant speed of 2 m s −1. Show that the wave will not reach the boy and find the minimum distance
between the boy and the wave.

PS 14 Swimmers going down a waterslide 30 m long push themselves off with an initial speed of between 1m s −1 and
2 m s −1. They accelerate down with constant acceleration 0.8 m s −2 for the first 20 m before more water is added
and the acceleration is 1m s −2 for the last 10 m of the slide. For safety there must be at least 5 s between
swimmers arriving at the bottom of the slide. Find the minimum whole number of seconds between swimmers
being allowed to start the slide.

P 15 A ball is projected in the air with initial speed u and goes up and down with acceleration g downwards. A
timer is at a height h. It records the time from the ball being projected until it passes the timer on the way up
as t1 and on the way down as t2. Show that the total of the two times is independent of h and that the initial
g (t1 + t2 )
speed can be calculated as u = . Show also that the difference between the times is given by
2
2 u2 − 2 gh
. Hence find a formula for h in terms of t1, t2 and g .
g
 Cambridge International AS & A Level Mathematics: Mechanics

1.5 Velocity–time graphs and multi-stage problems

As well as using a displacement–time graph, we can show the motion of an object on a
velocity–time graph.

Imagine the following scenario. An athlete goes for a run. He starts at rest and gradually
increases his speed over the first 30 s before maintaining the same speed of 5 m s −1 for 60 s.
Then he gradually reduces his speed until coming to rest another 30 s later. He then
returns to his starting point by increasing his speed quickly at the start and continually
trying to increase his speed for 90 s, but only managing to increase it by smaller and
smaller amounts, peaking at 6 m s −1. He then slows down over 10 s before coming to rest at
his start point.

The graph would look like the one shown here. You always
v (m s−1)
show time on the x-axis and velocity on the y-axis.
5
A horizontal graph line indicates that the velocity is unchanged
and therefore the athlete is moving at constant speed. If the
0 t (s)
graph is not horizontal it indicates the velocity is changing and 30 90 120 210 220
the steepness of the line indicates how quickly it is changing.
Note that when the athlete returned to the start, the velocity
−6
became negative because the direction of motion changed.

KEY POINT 1.10

22
The gradient of a velocity–time graph is equal to the acceleration of the object.

v−u v (m s−1)
From this graph you can see that the gradient is , which is the same as the formula
t
given for acceleration in Section 1.2.

You can use the formula for the area of a trapezium to show that the area under the graph v
1
line is ( u + v )t, which is the same as the formula for the displacement. This rule can be u
2
generalised so that if the motion changes and the velocity–time graph has more than one t (s)
t
line, the area under the graph may be found as the sum of separate areas under the lines.

KEY POINT 1.11

The area under the line of a velocity–time graph is the displacement of the object.

Note that in the previous scenario of the athlete, part of the graph is under the x-axis.
The area below the axis is a ‘negative area’ and it indicates a negative displacement. In this FAST FORWARD
particular example, the athlete started and ended at the same point and so the area above In Chapter 6 you will
the axis should equal the area below the axis to indicate no overall change in displacement. consider gradients of
and areas under curved
Note also that part of the graph is curved. This indicates that the acceleration is not
velocity–time graphs.
constant.
 Chapter 1: Velocity and acceleration

MODELLING ASSUMPTIONS

In the same way as we asked if it is reasonable to assume constant speed, we might ask
if it is reasonable to assume constant acceleration. In many cases it is close enough, but
it is often harder to maintain the same acceleration when moving at high speeds.

In scenarios involving people, we often say that someone is not moving then walks
at a given speed. We assume that the change is instantaneous. In the case of
walking at low speeds the time taken to reach that speed is sufficiently small that it
is not a bad assumption, but for runners there may be some error in making that
assumption.

DID YOU KNOW?

Olympic sprinters take about 60 m to reach top speed. By the end of the 100 m race they are
normally starting to slow down. You might expect that, because runners start to slow down after
about 100 m, race times for 200 m will be more than double the times for 100 m. In fact, for most
of the time since world records were recorded, the 200 m world record has been less than double
the 100 m record because the effect of starting from a stationary position is larger than the effect of
slowing down by a small amount for the second 100 m.

WORKED EXAMPLE 1.11 23

a Arthur travels at a constant speed of 5 m s −1 for 10 s and then decelerates at a constant rate of 0.5 m s −2 until
coming to rest. Sketch the velocity–time graph for his motion.

b Brendan travels at a constant 4 m s −1 starting from the same time and place. Show that Arthur and Brendan
are travelling at the same speed after 12 s and hence find the furthest Arthur gets ahead of Brendan.

1
c Show that for t . 10 the gap between them is given by g(t ) = − t 2 + 6t − 25 and hence find the time when
4
Brendan overtakes Arthur.

Answer
a Let T be the time spent in deceleration. Use an equation of constant acceleration to
v = u + at find the time for the second stage of the
0 = 5 – 0.5T motion.
T = 10 s
t = T + 10
= 20 s
 Cambridge International AS & A Level Mathematics: Mechanics

v (m s−1) Constant velocity means a horizontal line.

5
Deceleration from a positive velocity means a
negative gradient.
t (s)
10 20 The x-axis intercept is at the total time from
the start.

b v (m s−1) Find where the lines cross to solve when

5 velocities are equal.
4

t (s)
10 12 20
v = 5 − 0.5(t − 10) = 4 Time in the second stage is 10 s less than total
t = 12 s time since start.

1
sA = 5 × 10 + × (5 + 4) × 2 = 59 m The largest gap between them is equal to the
2
difference in displacements at the time when
sB = 4 × 12 = 48 m they have the same velocity. After this time
Therefore, the largest gap is 11m . Brendan is travelling faster than Arthur and so
starts to catch up.

Distance travelled is area under graph: a

24
rectangle plus a trapezium for Arthur and a
rectangle for Brendan.

c Starting gap = 0 The gap at time t is the starting gap plus the
At time t: distance covered by the leading person minus
1  1  the distance covered by the following person.
sA = 50 + (t − 10)  5 +  5 − (t − 10)  
2   2 
For t . 10, Arthur is in the second stage of the
1 motion, so the total distance is the distance
= 50 + 5(t − 10) − (t − 10)2
4
covered in the first stage plus the distance
sB = 4t
1 covered in the second stage up to time t.
g(t ) = 0 + 50 + 5(t − 10) − (t − 10)2 − 4t
4
1 2
= − t + 6t − 25
4
Brendan overtakes when the gap is 0, so Solve g(t ) = 0 to find t.
1
− t 2 + 6t − 25 = 0
4
t 2 − 24t + 100 = 0
t = 5.37 or 18.6

Since the equations are valid only for t . 10, Check the context and validity of the equations
t = 18.6 s to determine which is the relevant solution.
 Chapter 1: Velocity and acceleration

EXERCISE 1E

1 Sketch the velocity–time graphs from the information given. In each case take north to be the positive
direction.
a Rinesh starts from rest moving north with a constant acceleration of 3 m s −2 for 5 s.
b Wendi is moving north at 2 m s −1 when she starts to accelerate at a constant rate of 0.5 m s −2 for 6 s.
c Dylon is moving south at a constant speed of 4 m s −1.
d Susan is moving north at 6 m s −1 when she starts decelerating at a constant rate of 0.3 m s −2 until she comes
to rest.

2 Sketch the velocity–time graphs from the information given. In each case take upwards to be the positive direction.
a A ball is thrown up in the air from the surface of a pond with initial velocity 20 m s −1. It accelerates
under gravity with constant acceleration 10 m s −2 downwards. Once it has reached its highest point it
falls until it hits the surface of the pond and goes underwater. Under the water it continues to accelerate
with constant acceleration 1m s −2 for 1s.
b A parachutist falls from a helicopter which is flying at a constant height. She accelerates at a constant
rate of 10 m s −2 downwards for 0.5 s before the parachute opens. She then remains at constant speed for 5 s.
c A hot-air balloon is floating at a constant height before descending to a lower height. It descends with
constant acceleration 5 m s −2 for 6 s, then the burner is turned on and the balloon decelerates at a constant
rate of 2 m s −2 until it is no longer descending.
d A firework takes off from rest and accelerates upwards for 7 s with constant acceleration 5 m s −2 , before 25
decelerating at a constant rate of 10 m s −2 until it explodes at the highest point of its trajectory.

3 The graph shows the motion of a motorcyclist when he starts travelling along a v (m s−1)
highway until reaching top speed. Find the distance covered in reaching that
30
speed.
10
0 t (s)
8

4 The graph shows the motion of a ball when it is thrown upwards in the air until v (m s−1)
it hits the ground. Find the height above the ground from which it was thrown.
10

0 t (s)
1

−15

5 The sketch graph shows the motion of a boat. Find the distance the boat travels v (m s−1)
during the motion.
12

0 t (s)
4 10
 Cambridge International AS & A Level Mathematics: Mechanics

6 The graph shows the journey of a cyclist going in a straight line from home to v (m s−1)
school. Find the distance between her home and the school.
7

0 t (s)
60 120 300

7 A racing car is being tested along a straight 1 km course. It starts from rest accelerating at a constant rate of
10 m s −2 for 5 s. It then travels at a constant speed until a time t s after it started moving. Show that the
distance covered by time t is given by s = 125 + 50(t − 5) . Hence find how long it takes to complete the course.

8 A rowing boat accelerates from rest at a constant rate of 0.4 m s −2 for 5 s. It continues at constant velocity for
some time until decelerating to rest at a constant rate of 0.8 m s −2. In total the boat covers a distance of 30 m.
Find how long was spent at constant speed.

9 A cyclist accelerates at a constant rate from rest over 10 s to a speed v m s −1 . She then remains at that speed for
a further 20 s. At the end of this she has travelled 300 m in total. Find the value of v.

M 10 A boat accelerates from rest at a rate of 0.2 m s −2 to a speed v m s −1. It then remains at that speed for a further
30 s. At the end of this it has travelled 400 m in total.
a Find the value of v.
b What assumptions have been made to answer the question?

11 A crane lifts a block from ground level at a constant speed of v m s −1 . After 5 s the block slips from its shackles
26
and decelerates at 10 m s −2 . It reaches a maximum height of 6 m . Find the value of v.

12 A car is at rest when it accelerates at 5 m s −2 for 4 s. It then continues at a constant velocity. At the instant the
car starts moving, a truck passes it moving, at a constant speed of 22 m s −1 . After 10 s the truck starts slowing
at 1m s −2 until coming to rest.
a Show that the velocities are equal after 12 s and hence find the maximum distance between the car and the
truck respectively.
b Show that the distance covered at a time t s from the start by the car and the truck, for t . 10, are given by
1
40 + 20(t − 4) and 220 + 22(t − 10) − (t − 10)2, respectively. Hence find the time at which the car passes
2
the truck.

13 Two cyclists are having a race along a straight road. Bradley starts 50 m ahead of Chris. Bradley starts from
rest, accelerates to 15 m s −1 in 10 s and remains at this speed for 40 s before decelerating at 0.5 m s −2. Chris
starts 5 s later than Bradley. He starts from rest, accelerates to 16 m s −1 in 8 s and maintains this speed.
a Show that Bradley is still ahead when he starts to slow down, and find how far ahead he is.
b Find the amount of time Bradley has been cycling when he is overtaken by Chris.

PS 14 A driver travelling at 26 m s −1 sees a red traffic light ahead and starts to slow at 3 m s −2 by removing her foot
from the accelerator pedal. A little later she brakes at 5 m s −2 and comes to rest at the lights after 6 s.
a Sketch the velocity−time graph of the motion.
b Find the equations of the two sections of the graph.
c Hence find the time when she needs to start braking.
 Chapter 1: Velocity and acceleration

P 15 A car accelerates from rest to a speed v m s −1 at a constant acceleration. It then immediately decelerates at a
constant deceleration until coming back to rest t s after starting the motion.
a Show that the distance travelled is independent of the values of the acceleration and deceleration.
b Suppose instead the car spends a time T s at speed v m s −1 but still returns to rest after a total of t s after
starting the motion. Show that the distance travelled is independent of the values of the acceleration and
deceleration.

1.6 Graphs with discontinuities

What happens when a ball bounces or is struck by a bat? It would appear that the velocity
instantaneously changes from one value directly to a different value. If this did happen
instantaneously, the acceleration would be infinite. In practice, the change in velocity
happens over a tiny amount of time which it is reasonable to ignore, so we will assume the
change is instantaneous.

The velocity–time graph will have a discontinuity as in the following graph as the velocity
instantaneously changes.
v (m s−1)

0 t (s)
27

The displacement–time graph cannot have a discontinuity, but the gradient will
instantaneously change, so the graph will no longer be smooth at the join between two
stages of the motion. For the velocity–time graph shown, the displacement–time graph will
look like the following.

s (m)

0 t (s)

KEY POINT 1.12

On the velocity–time graph of an object that instantaneously changes velocity by bouncing or being
struck, the change is represented by a vertical dotted line from the velocity before impact to the
velocity after impact.
 Cambridge International AS & A Level Mathematics: Mechanics

MODELLING ASSUMPTIONS DID YOU KNOW?

In practice, the objects may not instantaneously change velocity. In the example of a tennis
ball being hit by a racket, the strings stretch very slightly and spring back into shape. It is
during this time that the ball changes velocity. In the case of a tennis ball striking a solid
wall or a solid object striking the ball, the ball may compress slightly during contact
before springing back into shape. In these cases, the time required to change is so small
that you can ignore it. By modelling the objects as particles, you can assume the objects Golf balls look and
do not lose shape and the time in contact is sufficiently small to be negligible. feel solid, but in the
instant after impact
from a golf club
WORKED EXAMPLE 1.12 moving at around
200 km h −1 the ball
appears to squash so
a A ball is travelling at a constant speed of 10 m s −1 for 2 s until it strikes a wall. that its length is only
It bounces off the wall at 5 m s −1 and maintains that speed until it reaches about 80% of its
where it started. When it passes that point it decelerates at 1m s −2 . Find the original diameter and
times and displacements when each change in the motion occurs. its width increases

b Sketch a velocity–time graph and a displacement–time graph for the motion.

Measure displacements as distances from the starting point and the original
direction of motion as positive.
28 a The distance to the wall is
s = 10 × 2 = 20 m
The time between hitting Note that times are measured from the start
the wall and returning to the of the motion.
starting point is, therefore,
20
t= = 4 s so t = 6 s
5
The time from starting to
decelerate until it stops is
0 − ( −5)
= 5 s so t = 11s
1
The distance covered is Note that displacements are measured from
1
5 × 5 + × ( −1) × 52 = 12.5 m the starting position taking the original
2
so displacement is direction as positive.

s = −12.5 m Although decelerating, the acceleration is

positive because the velocity is negative.
b v (m s−1) Notice the graph is discontinuous at t = 2.
10
Although the ball is decelerating after t = 6,
the gradient is positive because the velocity
0 t (s) is negative.
2 6 11

5
 Chapter 1: Velocity and acceleration

s (m) Notice that at t = 2 the gradient is different

20 on either side of the cusp. This indicates a
discontinuity in the velocity.

0 t (s)
2 6 11

12.5

EXERCISE 1F

1 An ice hockey puck slides along a rink at a constant speed of 10 m s −1. It strikes the boards at the edge of the
rink 20 m away and slides back along the rink at 8 m s −1 until going into the goal 40 m from the board. Sketch
a velocity–time graph and a displacement–time graph for the motion, measuring displacement from the
starting point in the original direction of motion.

2 A bowling ball rolls down an alley with initial speed 8 m s −2 and decelerates at a constant rate of 0.8 m s −2 .
After 2.5 s it strikes a pin and instantly slows down to 2m s −1. It continues to decelerate at the same constant
rate until coming to rest. Sketch a velocity–time graph and a displacement–time graph for the motion.

3 In a game of blind cricket, a ball is rolled towards a player with a bat 20 m away, who tries to hit the ball. On
one occasion, the ball is rolled towards the batsman at a constant speed of 4 m s −1. The batsman hits the ball
back directly where it came from with initial speed 6 m s −1 and decelerating at a constant rate of 0.5 m s −2. 29
Sketch a velocity–time graph and a displacement–time graph for the motion, taking the original starting point
as the origin and the original direction of motion as positive.

4 A ball is dropped from rest 20 m above the ground. It accelerates towards the ground at a constant rate of
10 m s −2. It bounces on the ground and leaves with a speed that is half the speed it struck the ground with. It is
then caught when it reaches the highest point of its bounce. Sketch a velocity–time graph and a displacement–
time graph for the motion, measuring displacement above the ground.

5 A ball is struck towards a wall, which is 5 m away, at 2.25 m s −1. It slows down at a constant rate of 0.2 m s −2
until is strikes the wall. It bounces back at 80% of the speed it struck the wall at. It again slows down at a
constant rate of 0.2 m s −2 until coming to rest. Sketch a velocity–time graph and a displacement–time graph for
the motion, measuring displacement from the wall, taking the direction away from the wall as positive.

M 6 A billiard ball is on the centre spot of a 6 m long table and is struck towards one of the cushions with initial
speed 3.1m s −1. It slows on the table at 0.2 m s −2 . When it bounces off the cushion its speed reduces to 70% of
the speed with which it struck the cushion. The ball is left until it comes to rest.
a Sketch the velocity–time and displacement–time graphs for the ball, taking the centre of the table as the
origin for displacement and the original direction of motion as positive.
b What assumptions have been made in your answer?

P 7 A ball is released from rest 20 m above the ground and accelerates under gravity at 10 m s −2 . When it bounces
its speed halves. If bounce n occurs at time tn the speed after the bounce is vn . Show that vn = 15 − 2.5tn and
deduce that, despite infinitely many bounces, the ball stops bouncing after 6 s.
 Cambridge International AS & A Level Mathematics: Mechanics

Checklist of learning and understanding

● The equations of constant acceleration are:
v = u + at
1
s= ( u + v )t
2
1
s = ut + at 2
2
1 2
s = vt − at
2
v 2 = u 2 + 2 as
● A displacement–time graph shows the position of an object at different times. The gradient is
equal to the velocity.
● A velocity–time graph shows how quickly an object is moving at a given time. The gradient is
equal to the acceleration. The area under the graph is equal to the displacement.

30
 Chapter 1: Velocity and acceleration

END-OF-CHAPTER REVIEW EXERCISE 1

1 A man playing with his young son rolls a ball along the ground. His son runs after the ball to fetch it.
a The ball starts rolling at 10 m s −1 but decelerates at a constant rate of 2 m s −2 . Find the distance covered
when it comes to rest.
b Once the ball has stopped, the boy runs to fetch it. He starts from rest beside his father and accelerates at a
constant rate of 2 m s −2 for 3s before maintaining a constant speed. Find the time taken to reach the ball.

2 A car is travelling at 15 m s −1 when the speed limit increases and it accelerates at a constant rate of 3 m s −2 until
reaching a top speed of 30 m s −1.
a Find the distance covered until reaching top speed.
b Once the car is at top speed, there is a set of traffic lights 600 m away. The car maintains 30 m s −1 until it
starts to decelerate at a constant rate of 5 m s −2 to come to rest at the lights. Find the time taken from
reaching top speed until it comes to rest at the traffic lights.

3 In a race, the lead runner is 60 m ahead of the chaser with 200 m to go and is running at 4 m s −1. The chaser is
running at 5 m s −1.
a Find the minimum constant acceleration required by the chaser to catch the lead runner.
b If the lead runner is actually accelerating at a constant rate of 0.05 m s −2 , find the minimum constant
acceleration required by the chaser to catch the lead runner.

4 A jet aeroplane coming in to land at 100 m s −1 needs 800 m of runway.

31
a Find the deceleration, assumed constant, the aeroplane can produce.
b On an aircraft carrier, the aeroplane has only 150 m to stop. There are hooks on the aeroplane which catch
arresting wires to slow it down. If the aeroplane catches the hook 50 m after landing, find the deceleration
during the last 100 m.

5 The sketch shows a velocity–time graph for a sled going down a slope. Sketch the v (m s−1)
displacement–time graph, marking the displacements at each change in the motion.
18

0 t (s)
6 15

6 The sketch shows a velocity–time graph for rowers in a race. v (m s−1)

Given that the race is 350 m long and finishes at time 50 s, find the 2v
value of v.
v

0 t (s)
10 50
 Cambridge International AS & A Level Mathematics: Mechanics

M 7 A footballer kicks a ball directly towards a wall 10 m away and walks after the ball in the same direction at a
constant 2 m s −1. The ball starts at 4 m s −1 but decelerates at a constant rate of 0.5 m s −2 . When it hits the wall it
rebounds to travel away from the wall at the same speed with which it hit the wall.
a Find the time after the initial kick when the ball returns to the footballer.
b What assumptions have been made in your answer?

PS 8 An entrant enters a model car into a race. The car accelerates from rest at a constant rate of 2 m s −2 down a
slope. When it crosses the finishing line a firework is set off. The sound travels at 340 m s −1 . The time between
the entrant starting and the firework being heard at the start of the course is 12 s.
a Find the length of the course.
b Find the actual time it took for the model car to complete the course.

9 A lion is watching a zebra from 35 m behind it. Both are stationary. The lion then starts chasing by accelerating
at a constant rate of 3 m s −2 for 5 s. Once at top speed the lion decelerates at 0.5 m s −2 . The zebra starts moving
1s after the lion started, accelerating at a constant rate of 2 m s −2 for 7 s before maintaining a constant speed.
a Show that the lion has not caught the zebra after 8 s .
b Show that the gap between them at time t s, for t . 8, after the start of the lion’s motion is given by
1 2 51
t − 5t + and hence determine when the lion catches the zebra, or when the lion gets closest and how
4 2
32 close it gets.

PS 10 A car is behind a tractor on a single-lane straight road. Both are moving at 15 m s −1. The speed limit is 25 m s −1,
so the car wants to overtake. The safe distance between the car and the tractor is 20 m.
a To overtake, the car goes onto the other side of the road and accelerates at a constant rate of 2 m s −2 until
reaching the speed limit, when it continues at constant speed. Show that the distance the car is ahead of the
tractor at time t s after it starts to accelerate is given by t 2 − 20 for 0 , t < 5 , and deduce that the car is not
a safe distance ahead of the tractor before reaching the speed limit.
b The car pulls in ahead of the tractor once it is a safe distance ahead. Find the total time taken from the start
of the overtaking manoeuvre until the car has safely overtaken the tractor.
c To overtake safely on the single-lane road, when the car returns to the correct side of the road in front of the
tractor there must be a gap between his car and oncoming traffic of at least 20 m. Assuming a car travelling
in the opposite direction is moving at the speed limit, find the minimum distance it must be away from the
initial position of the overtaking car at the point at which it starts to overtake.

11 Two hockey players are practising their shots. They are 90 m apart and hit their balls on the ground directly
towards each other. The first player hits his ball at 6 m s −1 and the other hits hers at 4 m s −1. Both balls decelerate
at 0.1m s −2 . Find the distance from the first player when the balls collide.

12 The sketch shows a velocity–time graph for a skier going down a slope. v (m s−1)
Given that the skier covers 80 m during the first stage of acceleration, find the total
2v
distance covered.
v

0 t (s)
t 5t 7t
 Chapter 1: Velocity and acceleration

P 13 Two trains are travelling towards each other, one heading north at a constant speed of u m s −1 and the other
heading south at a constant speed of v m s −1. When the trains are a distance d m apart, a fly leaves the
northbound train at a constant speed of w m s −1 . As soon as it reaches the other train, it instantly turns back
travelling at w m s −1 in the other direction. Show that the fly meets the southbound train having travelled a
wd 2 uwd
distance of and returns to the northbound train when the train has travelled a distance of .
w+v ( w + v )( w + u )

P 14 Two cars are on the same straight road, the first one s m ahead of the second and travelling in the same direction.
The first one is moving at initial speed v m s −1 away from the second one. The second one is moving at initial
speed u m s −1 where u > v. Both cars decelerate at a constant rate of a m s −2 . Show that the second car overtakes
s
at time t = irrespective of the deceleration, provided the cars do not come to rest before the second one
u−v
passes. Show also that the distance from the starting point of the second car to the point where it overtakes
depends on a and find a formula for that distance.

15 A woman walks in a straight line. The woman’s velocity t s after passing through a fixed point A on the line is
v m s −1. The graph of v against t consists of four straight line segments (see diagram).

v (m s−1)

2
1.5

52 60
O t (s) 33
30 40

−2.2

The woman is at the point B when t = 60. Find:

i the woman’s acceleration for 0 , t , 30 and for 30 , t , 40 [3]
ii the distance AB [2]
iii the total distance walked by the woman. [1]
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q1 November 2011

16 A car travels in a straight line from A to B, a distance of 12 km, taking 552 s. The car starts from rest
at A and accelerates for T1 s at 0.3 m s −2, reaching a speed of V m s −1. The car then continues to move at V m s −1
for T2 s. It then decelerates for T3 s at 1m s −2, coming to rest at B.
i Sketch the velocity–time graph for the motion and express T1 and T3 in terms of V . [3]
ii Express the total distance travelled in terms of V and show that 13V 2 − 3312V + 72 000 = 0.
Hence find the value of V . [5]
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q5 November 2013
 Cambridge International AS & A Level Mathematics: Mechanics

17 The diagram shows the velocity–time graph for a particle P which travels on a straight line AB,
where v m s −1 is the velocity of P at time t s . The graph consists of five straight line segments.
The particle starts from rest when t = 0 at a point X on the line between A and B and moves
towards A. The particle comes to rest at A when t = 2.5.

v (m s−1)

O t (s)
2.5 4.5 14.5

i Given that the distance XA is 4 m , find the greatest speed reached by P during this stage of the motion. [2]

In the second stage, P starts from rest at A when t = 2.5 and moves towards B. The distance AB is 48 m. The
particle takes 12 s to travel from A to B and comes to rest at B. For the first 2 s of this stage P accelerates at
3 m s −2, reaching a velocity of V m s −1. Find:
ii the value of V [2]
iii the value of t at which P starts to decelerate during this stage [3]
iv the deceleration of P immediately before it reaches B. [2]
34 Cambridge International AS & A Level Mathematics 9709 Paper 42 Q6 November 2010
 35

Chapter 2
Force and motion in one dimension
In this chapter you will learn how to:
■ relate force to acceleration
■ use combinations of forces to calculate their effect on an object
■ include the force on an object due to gravity in a force diagram and calculations
■ include the contact force on a force diagram and in calculations.
 Cambridge International AS & A Level Mathematics: Mechanics

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills
Chapter 1 Use equations of constant 1 Find t when u = 4, v = 6 and a = 5.
acceleration. 2 Find v when u = 3, a = 4 and s = 2.

What is a force and how does it affect motion? DID YOU KNOW?
When an object is moving, how will it continue to move? What makes it speed up or slow
down? What affects the acceleration or deceleration of an object? These are questions that
have been considered by many philosophers over the course of history. The first person
to publish what is now considered to be the correct philosophy was Isaac Newton (1643–
1727), which is why Mechanics is often referred to as Newtonian Mechanics. Newton used
his theories to explain and accurately predict the movements of planets and the Moon, as
well as to explain tides and the shape of the Earth.

Newton was not the first to consider the idea of a force. Aristotle (384–322 BCE), the Ancient
Greek philosopher, believed that a cause (or force) was the thing that had an effect on an object Newton published his
to make it move. He came up with a theory of how force was related to motion, but he did not work in a book called
define forces clearly and he had no supporting evidence for his claims. His theory led to the Philosophiae Naturalis
conclusion that heavier bodies fall to Earth faster than lighter bodies, which is not true. Principia Mathematica
in 1687, often known
simply as the Principia.
36 2.1 Newton’s first law and relation between force and acceleration
A force is something that can cause a change in the motion of an object. There are many
different types of force that can act on an object. Think of different ways to get an object
on a table to start moving or change speed.

Most forces are caused by objects being in contact with other objects. Pushing or dragging
an object can cause it to accelerate. A force can act through a string under tension; pulling on
the string can cause an attached object to speed up or slow down. A solid rod can act like a
string, but it can also push back when it is under compression and there is a thrust in the rod.

Objects that are in contact with each other may experience friction, which may include air
resistance. This generally slows objects down, but sometimes friction is required to cause
the motion. For example, a car engine gets the car to move by using the friction with the
ground to start the wheels rolling along the ground. In icy or muddy conditions there is not
much friction so cars cannot accelerate quickly.

If you gently push an object off the edge of a table it will accelerate towards the ground
despite nothing (apart from the air) being in contact with it. This is because there is a force
due to gravity. In fact, there is a force due to gravity between all objects, but other than the
gravity pulling objects towards the Earth, the forces are so small as to be negligible.

Gravity is the only force you will consider in this course that acts on an object without
being in contact with the object. Other forces that act in this way, for example magnetic
attraction or repulsion, will not be considered in this course.

Newton progressed from using mathematics to calculate the position, speed and
acceleration of an object with constant acceleration to explaining why objects move as
they do. He produced three laws of motion which are still used today in many situations to
calculate and describe how objects move.
 Chapter 2: Force and motion in one dimension

KEY POINT 2.1

Newton’s first law states that an object remains at rest or continues to move at a constant velocity
unless acted upon by a net force.

This is not immediately obvious because the forces are not visible. You see that objects
DID YOU KNOW?
sliding along the ground slow down and eventually stop without anyone trying to slow
them down and without the object hitting another object. A ball moving through the air Aristotle thought
appears to be changing direction as it falls under gravity, yet it does not touch anything that at every moment
while changing direction. something must be
causing an object to
Newton’s second law expresses how a force relates to the motion of an object. For an object continue to move,
of constant mass, the net force acting on the object is proportional to the product of its so an object flying
mass and acceleration. through the air must
be pushed by the air
F α ma to continue moving.
The force is measured in newtons (symbol N). One newton is defined as the amount Newton was the first
to contradict him.
of force required to accelerate 1kg at 1m s−2 . Using kilograms for the unit of mass,
metres for length and seconds for time, so that acceleration is in ms −2, the constant of
proportionality is 1.

KEY POINT 2.2

Newton’s second law leads to the equation

37
Force = mass × acceleration

Force is a vector quantity, so may be positive or negative depending on which direction is

FAST FORWARD
assigned to be positive.
Newton’s third law
relates forces between
MODELLING ASSUMPTIONS
objects and their effects
on each other. You
In all cases at this stage you will consider objects as particles, so you can ignore any will learn about this in
complexities due to the shape of the object. For many of the problems, it will have Chapter 5.
such a small effect that you can treat it as negligible. This means that the error it
causes in the calculations is small enough to be ignored.

For example, when you consider an object like a bag of sand, if the mass of the bag
is sufficiently small compared to the mass of the sand you say it is negligible and the
bag is termed light.

In some questions, there is a general force called resistance, which acts in the FAST FORWARD
opposite direction to motion. This may be due to friction, air resistance or both. Friction will be
In some situations you will ignore resistance forces altogether. This is a modelling considered in more
assumption that you make to simplify the situation. detail in Chapter 4.
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 2.1

a A cyclist on a bike, with combined mass of 100 kg, accelerates from rest with acceleration 0.2 m s−2. Find the
force the cyclist generates.

b A second cyclist on a bike, with combined mass of 80 kg, accelerates from rest by generating the same force
as the first cyclist. Find her acceleration.

c The first cyclist is travelling at 20 m s−1 when he starts to brake with a force of 500 N. Find the distance
covered while coming to rest.

Answer

a F = ma = 100 × 0.2 = 20 N Use F = ma.

b F = ma Use F = ma and solve the equation.

20 = 80 a
1
a = or 0.25 m s−2
4
c F = ma Notice that the force is in the opposite
−500 = 100 a direction to motion so is negative.
a = −5m s−2
v 2 = u2 + 2 as Use equations of constant acceleration.
38 2
0 = 20 + 2 × −5 × s
s = 40 m

You can see that the same force was exerted by both cyclists, but the acceleration was smaller
for the larger mass. Mass can be said to be a measure of the amount of material present in an
object, but can also be described as the reluctance of an object to change velocity.

EXERCISE 2A

1 Find the horizontal force required to make a car of mass 500 kg accelerate on a horizontal road at 2 m s−2.

2 A wooden block of mass 0.3 kg is being pushed along a horizontal surface by a force of 1.2 N. Find its
acceleration.
3 A gardener drags a roller on horizontal land with a force of 360 N causing it to accelerate at 1.2 m s−2. Find the
mass of the roller.

4 A man pushes a boy in a trolley, with total mass 60 kg, along horizontal land from rest with a constant force
of 42 N for 10 s. Find the distance travelled in this time.

M 5 A snooker ball of mass 0.2 kg is struck so it starts moving at 1.2 m s−1. As it rolls, the table provides a constant
resistance of 0.08 N. It strikes another ball 1m away.
a Find the speed with which it strikes the other ball.
b What assumptions have you made when answering this question?

6 Find the constant force required to accelerate a mass of 5 kg from 3 m s−1 to 7 m s−1 in 8 s on a horizontal surface.
 Chapter 2: Force and motion in one dimension

7 A ship of mass 20 tonnes is moving at 10 m s−1 when its engines stop and it decelerates in the water. It takes
500 m to come to rest. Find the resistance force, which is assumed to be constant, of the water on the ship.

8 A winch provides a constant force of 80 N and causes a block to accelerate on horizontal land from 2 m s−1 to
10 m s−1 in 6 s. Find the mass of the block.

9 A Formula 1 car and its driver have total mass of 800 kg. The driver is travelling at 100 m s−1 along a horizontal
straight when, 100 m before a bend, he starts to brake, slowing to 40 m s−1. Assuming a constant braking force,
find the force of the brakes on the car.

10 A strongman drags a stone ball from rest along a horizontal surface. He moves it 10 m in 4 s by exerting a
constant force of 100 N. Find the mass of the stone ball.

11 A car and driver, of total mass 1350 kg, are moving at 30 m s−1 on a horizontal road when the driver sees
roadworks 400 m ahead. She breaks, decelerating with a constant force of 600 N until arriving at the
roadworks. Find the time elapsed before arriving at the roadworks.

12 A train travelling at 30 m s−1 on a horizontal track starts decelerating 360 m before coming to rest at a station.
The brakes provide a constant resistance of 100 kN. Find the mass of the train.

13 A block is being dragged along a horizontal surface by a constant horizontal force of size 45 N. It covers 8 m
in the first 2 s and 8.5 m in the next 1 s. Find the mass of the block.

2.2 Combinations of forces TIP

Newton’s first and second laws refer to net or resultant forces because there may be more 39
It is normally easier to
than one force acting on an object at any one time.
put all the information
in a question into this
KEY POINT 2.3 equation rather than
working out net force
Newton’s second law is given more generally as
separately.
Net force = mass × acceleration

An object in equilibrium may have several forces acting on it, but their resultant is zero so it
FAST FORWARD
remains at rest or moving at constant velocity.
You will consider
KEY POINT 2.4 problems with objects
in equilibrium in
Objects with a net force of zero acting on them are said to be in equilibrium and have no Chapter 3.
acceleration.

You should use a diagram to work out which forces act in which directions. In this chapter TIP
the diagrams will be simple and only include one or two forces horizontally or vertically.
You should get used to drawing diagrams for simple situations so that you are prepared for Always draw a force
more complicated situations in later chapters. diagram, even in simple
situations, to ensure
Consider an aeroplane flying through the air. The forces acting on it are its weight, lift from you have considered all
the air on the wings, the driving force from the engine and propellers, and air resistance. forces in the problem
Compare the following two illustrations. and added them into
the relevant equations.
 Cambridge International AS & A Level Mathematics: Mechanics

a a

lift lift

TIP
air resistance air resistance thrust
thrust
You do not usually
include units on
diagrams where forces
weight weight are indicated by
unknowns, otherwise
Both illustrations have the same information, but the diagram on the right is simpler to there can be confusion
draw and clearly shows the important information. about whether a letter
refers to an unknown or
Diagrams are not pictures. In Mechanics you normally draw objects as circles or a unit. Remember to use
rectangles and show forces as arrows going out from the object. Objects being pushed S.I. units throughout.
from behind or dragged from in front are both shown as an arrow going forward from the
object. The net force (or resultant force) is not shown on the diagram.

Accelerations are shown beside the diagram using a double arrow.

WORKED EXAMPLE 2.2

a A car of mass 600 kg has a driving force of 500 N and air resistance of 200 N. Find how long it takes to
accelerate from 10 m s−1 to 22 m s−1.
a

40
200 N 500 N

b The car stops providing a driving force and the brakes are applied. It decelerates from 22 m s−1 to rest in
220 m. Find the force of the brakes.

Answer

a a
The diagram is very simple and clear.
The car is shown as a rectangle.
200 500
The forces are arrows going out from the
rectangle.
The acceleration is a double arrow above the
diagram.
No resultant force is marked on the diagram.

F = ma All the horizontal forces make up the net force.

500 − 200 = 600 a Forces are negative if they are in the opposite
a = 0.5 m s−2 direction to the motion.

v = u + at Acceleration is constant so you can use the

22 = 10 + 0.5t relevant equation of constant acceleration to
t = 24 s finish the problem.
 Chapter 2: Force and motion in one dimension

b a It is useful to draw new diagrams whenever the

situation changes.
200 Note that the air resistance is still there, but
F
the driving force is not there. The unknown F
is now the braking force.
Note that acceleration is still in the direction of
motion, although it is clear it will be negative
because all the forces are acting the other way.

v 2 = u2 + 2 as Use an equation of constant acceleration first

2 2
0 = 22 + 2 a × 220 to find the acceleration.
Three of the variables are known.
a = −1.1m s−2
F = ma Use F = ma now that two of the variables in
−200 − F = 600 × ( −1.1 ) this equation are known.
F = 460 N

EXERCISE 2B

1 A boat and the sailor on it have a combined mass of 300 kg. The boat’s engine provides a constant driving
force of 100 N. It maintains a constant speed despite water resistance of 60 N and air resistance. Find the
magnitude of the air resistance. 41

2 A boy and his friends have a tug-of-war with his father. His father pulls on one end of the rope with a force
of 200 N. The boy and three friends each pull with equal force on the other end of the rope. The rope is in
equilibrium. Find the force each child exerts on the rope.

3 A team of sailors pulls a boat over the sand to the sea. Each sailor is capable of providing a force of up to
300 N. The resistance from the sand is 2200 N. Find the minimum number of sailors needed on the team to
maintain a constant speed.

4 A cyclist and her bike have a combined mass of 80 kg. She exerts a driving force of 200 N and experiences air
resistance of 150 N. Find the acceleration of the cyclist.

5 A car of mass 1500 kg experiences air resistance of 450 N. It accelerates at 3 m s−2 on horizontal ground. Find
the driving force exerted by the engine.

6 A boat of mass 2 tonnes has a driving force of 1000 N and accelerates at 0.2 m s−2. Find the resistance that the
water provides.

7 A water-skier of mass 75 kg is towed by a horizontal rope with constant tension of 150 N. There is constant
resistance from the water of 120 N. Find the time taken to reach a speed of 10 m s−1 from rest.

8 A learner driver is being tested on the emergency stop. The car has mass 1250 kg and is moving at 21m s−1.
When the driver presses the brakes there is a braking force of 10 000 N in addition to the air resistance of
500 N. Find the distance covered in coming to rest.
 Cambridge International AS & A Level Mathematics: Mechanics

M 9 A small aircraft is accelerating on a runway. Its engines provide a constant driving force of 20 kN. There is average
air resistance of 1000 N. It starts from rest and leaves the runway, which is 900 m long, with speed 80 m s−1.
a Find the mass of the aircraft.
b What assumptions have been made to answer the question?

10 A drag racing car races over a track of 400 m. The car has mass 600 kg and accelerates from rest with a
constant driving force of 21kN. There is air resistance of 1800 N. Find the speed of the car at the finish line of
the race.

11 A wooden block of mass 6 kg is being dragged along a horizontal surface by a force of 10 N. It accelerates
from 1m s−1 to 3 m s−1 in 4 s. Find the size of the friction force acting on the wooden block.

12 A motorcyclist is travelling at 20 m s−1 on level road when she approaches roadworks and slows down to
10 m s−1 over a distance of 250 m. The combined mass of the motorcyclist and the motorcycle is 360 kg. There is
air resistance of 80 N. Find the braking force of the motorcycle.

13 An aeroplane of mass 8 tonnes is flying horizontally through the air at 240 m s−1. There is air resistance of
20 kN. The pilot reduces the driving force from the engines to slow to 160 m s−1 over 40 s before starting to
descend. Find the magnitude of the reduced driving force.

PS 14 A car of mass 1400 kg slows down from 30 m s−1 to 20 m s−1 when the driver sees a sign for reduced speed limit
400 m ahead. There is air resistance of 1000 N. Determine whether the driver needs to provide a braking force
or just reduce the amount of driving force exerted, and find the size of the force.

42
2.3 Weight and motion due to gravity
If an object falls under gravity it moves with constant acceleration, whatever the mass of
the object. This may seem contradictory, because an object like a feather will fall to Earth
much more slowly than a hammer. However, this is actually because of air resistance.
Commander David Scott on Apollo 15 demonstrated that on the Moon where there is no
atmosphere the two objects do land at the same time.

DID YOU KNOW?

Galileo Galilei (1564–1642) was the first to demonstrate that the mass
does not affect the acceleration in free fall. It was thought he did this by
dropping balls of the same material but different masses from the Leaning
Tower of Pisa to show they land at the same time. However, no account
of this was made by Galileo and it is generally considered to have been
a thought experiment. Actual experiments on inclined planes did verify
Galileo’s theory.

The acceleration in freefall due to gravity on Earth is denoted by the letter g and has a
numerical value of approximately 10 m s−2.

If an object of mass m kg falls under gravity with acceleration g m s−2 then the force on
the object due to gravity must be F = mg. This force is called the weight of an object and
always acts towards the centre of the Earth, or vertically downwards in diagrams.
 Chapter 2: Force and motion in one dimension

KEY POINT 2.5

The weight of an object of mass m kg is given by W = mg. It is the force due to gravity, so is
measured in newtons.

MODELLING ASSUMPTIONS

The value of g is actually closer to 9.8 m s−2 , but even that varies slightly depending
on other factors. Because of the rotation of the Earth, the acceleration of an object
in freefall is lower at the equator than at the poles. Gravity is also weaker at high
altitudes and may even be weaker at depths inside the Earth. There can also be very
slight local variations, for example due to being near large mountains of dense rock.
For the purposes of this course, we will assume that g is 10 m s−2.

Above the surface of the Earth, the force due to gravity decreases. The difference is
negligible for small distances, but this becomes important in space. In deep space,
the gravitational pulls of not only the Earth but also the Sun become negligible.
Under Newton’s first law, objects like Voyager 1 and Voyager 2 in the far reaches of
the Solar System will continue to move with the same velocity until they reach close
enough to another star to feel its gravitational effect.

WORKED EXAMPLE 2.3

43

a A ball of mass 0.2 kg is thrown vertically upwards out of a window 4 m above the ground. The ball is
released with speed 8 m s−1. Assuming there is no air resistance, find how long it takes to hit the ground.

b If instead there is a constant air resistance of 0.1N against the direction of motion, find how long the ball
takes to hit the ground.

Answer
a Taking upwards as positive: Define clearly which direction is positive.

With only gravity acting, the acceleration is – g

if upwards is positive.
a = −g

0.2g N

1 2 On the way up and the way down, there is

s = ut + at
2 no change in the forces, so the whole motion
1 can be dealt with as a single motion with
−4 = 8t + × −10t 2
2 acceleration – g.
5t 2 − 8t − 4 = 0
t = 2 or − 0.4
t is positive so the time to hit the ground is 2 s.
 Cambridge International AS & A Level Mathematics: Mechanics

b On the way up: Separate the way up from the way down
because resistance forces oppose motion, so
a
the forces are different on the way down.

You first need to find how long it takes to reach

0.1 N the highest point and the distance travelled in
this time.

0.2g N
F = ma Use Newton’s second law to find the
−0.2 g − 0.1 = 0.2 a acceleration.
a = −10.5
v = u + at The maximum height is reached when the
0 = 8 + −10.5t1 velocity is reduced to zero.
t1 = 0.762
So the time to reach highest point is 0.762 s
(3 significant figures).

v 2 = u2 + 2 as You will need to know the height reached so you

02 = 82 + 2 × −10.5s1 can think about the motion on the way down.
s1 = 3.05 It is better to use given values rather than
So distance travelled upwards is 3.05 m calculated values as far as possible.
44
(3 significant figures).

On the way down: Draw a new diagram for the new situation.

0.1 N The motion is now downwards and the

resistance force is in the opposite direction to
a the motion, so it now acts upwards.

0.2g N

F = ma The ball is moving down for this stage, so we

0.2 g − 0.1 = 0.2 a can define downwards as positive.
a = 9.5

1 2
s = ut + at Use the equation of constant acceleration with
2
1 the total distance including the maximum
s1 + 4 = 0t2 + × 9.5t22 height s1 found previously.
2
giving t2 = 1.22 (3 significant figures)

t = t1 + t2 = 1.98 s (3 significant figures) Note that answers to previous calculations are

written to 3 significant figures but you should
always use values from the calculator, by using
Answer key or memories, in later calculations to
avoid premature rounding errors.
 Chapter 2: Force and motion in one dimension

WORKED EXAMPLE 2.4

a A ball is dropped from a height of 30 m above the ground. Two seconds later, another ball is thrown upwards
from the ground with a speed of 5 m s−1. They collide at a time t s after the first ball was dropped. Find t.
b They collide at a height h m above the ground. Find h.

Answer
a Taking upwards as positive: The mass is unknown but labelled as m1,
although this does not affect the acceleration
because only gravity acts on the ball.
a

m1 g N
Let t1 be the time after the first ball is dropped and s1 be
its displacement from its starting position.

1 2
s = ut + at Measure height from the ground and time
2
from the time when the first ball is dropped.
1
s1 = 0t1 + × −10t12 45
2
h = 30 − 5t 2

The second ball has a similar diagram, also

with unknown, but possibly different, mass.
a

m2 g N
Let t2 be the time after the second ball is thrown and s2
be its displacement from its starting position.

1 2 Again measure height from the ground and

s = ut + at
2 time from the time when the first ball is
1
s2 = 5t2 + × −10t22 dropped.
2
h = 5(t − 2) − 5(t − 2)2

5(t − 2) − 5(t − 2)2 = 30 − 5t 2 Solve the equations simultaneously.

25t = 60
t = 2.4 s

b h = 30 − 5t 2 = 30 − 5 × 2.42 = 1.2 m Substitute into the equation for height.

 Cambridge International AS & A Level Mathematics: Mechanics

EXERCISE 2C

1 Find the weight of a man of mass 70 kg.

2 A young goat has weight 180 N. Find its mass.

3 A ball is dropped from a height of 20 m. Find the time taken to hit the ground.

4 A water fountain projects water vertically upwards with initial speed 10 m s−1. Find the maximum height the
water reaches.

5 A ball is thrown vertically downwards with speed 5 m s−1 from a height of 10 m. Find the speed with which it
hits the ground.

PS 6 A wrecking ball of mass 1 tonne is dropped onto a concrete surface to crack it. It needs to strike the ground at
5 m s−1 to cause a crack. Find the minimum height from which it must be dropped.

7 A coin of mass 0.05 kg is dropped from the top of the Eiffel Tower, 300 m high. It experiences air resistance of
0.01N. Find the speed with which it hits the ground.

8 A winch lifts a bag of sand of mass 12 kg from the ground with a constant force of
240 N until it reaches a speed of 10 m s−1. Then the winch provides a force to keep the
bag moving at constant speed. Find the time taken to reach a height of 40 m.

9 A firework of mass 0.4 kg is fired vertically upwards with initial speed 40 m s−1.
46
The firework itself provides a force of 2 N upwards. The firework explodes after
6 s. Find the height at which it explodes.

10 A flare of mass 0.5 kg is fired vertically upwards with speed 30 m s−1. The flare
itself provides a force of 0.8 N upwards, even when the flare is falling, to keep
the flare high for as long as possible. The flare is visible over the horizon when it
reaches a height of 25 m.
a Find how long the flare is visible for.
b What assumptions have you made in your answer?

11 A feather of mass 10 g falls from rest from a height of 3 m and takes 2 s to hit the ground. Find the air
resistance on the feather.

12 A ball of mass 0.3 kg is thrown upwards with speed 10 m s−1. It experiences air resistance of 0.15 N. It lands on
the ground 1.2 m below. Find the speed with which it hits the ground.

P 13 A bouncy ball is dropped from a height of 5 m. When it bounces its speed immediately after impact is 80% of
the speed immediately before impact.
a Find the maximum height of the ball after bouncing.
b Show that the height is independent of the value used for g.

14 A parachutist of mass 70 kg falls out of an aeroplane from a height of 2000 m and falls under gravity until
600 m from the ground when he opens his parachute. The parachute provides a resistance of 2330 N. Find
the speed at which he is travelling when he reaches the ground.
 Chapter 2: Force and motion in one dimension

15 A ball is thrown vertically up at 10 m s−1. A second later another ball is thrown vertically up from the same
point at 8 m s−1. Find the height at which they collide.

16 A pebble is dropped from rest into a deep well. At time t later it splashes into the water at the bottom of the
well. Sound travels at 340 m s−1 and is heard at the top of the well 5 s after the pebble was released. Find the
depth of the well.

17 A ball of mass 2 kg is projected up in the air from ground level with speed 20 m s−1. It experiences constant air
resistance R. It returns to ground level with speed 15 m s−1. Find R.

2.4 Normal contact force and motion in a vertical line

When an object rests on a table, why does it not fall? There is a force due to gravity, so
there must be another force in the opposite direction keeping it in equilibrium. This is
called the reaction force.

reaction force

mg

47
The reaction force is the force on an object from the surface it is resting on. It is usually
denoted by the letter R. It is perpendicular to the surface it is in contact with, so sometimes
N is used for the normal contact force.

KEY POINT 2.6

The contact force between the object and the surface it is on is called the reaction force and is
always perpendicular to the surface.

When the object is on a horizontal surface, the contact force is usually the same
magnitude as the weight. It simply prevents the object leaving or falling through the
surface. However, when the surface is tilted with the object on it, or when other forces act
on the object pushing it into the surface or pulling it away from the surface, the normal
contact force is not usually the same magnitude as the weight.

FAST FORWARD

Some people mistakenly think the normal contact force is equal and opposite to the force of gravity.
You will look at Newton’s third law in Chapter 5, which is about forces that are equal and opposite.
There is a force that is equal and opposite to the force of gravity on an object, but it acts on the
Earth, not the object. Because the mass of the Earth is so large, the acceleration caused by the force
is usually negligible. However, when you consider the motion of planets, the effect of gravity on
both the Earth and other planets is important.
 Cambridge International AS & A Level Mathematics: Mechanics

DID YOU KNOW?

Moon

GS
GE
GM
Sun GM
GE
GS

Earth

In 1887, King Oscar II of Sweden and Norway established a prize for anyone who could solve the
three-body problem, which asks what happens to a system of three objects each with gravity acting
on them from each other, like the Sun, the Earth and the Moon as shown in the diagram. Henri
Poincaré (1854–1912) showed that, although we know the equations for the objects, there is no way
to solve them. Moreover, he showed that if there is the slightest change in the initial positions or
velocities of the bodies, the outcome may be entirely different. This led to the development of chaos
theory and this effect became known as the butterfly effect. Another example of this is the weather,
which is why it’s so difficult to predict.

MODELLING ASSUMPTIONS

If an object is on a table you may expect the table top to bend or even break if the
object is heavy enough. You will assume that this is never the case and that the forces
will never cause the surface to bend or break.
48
As an object is lifted off the surface the contact force is reduced. When the force is
reduced to zero, you would expect the object to lose contact with the surface and
be lifted off. However, there are some cases where this does not happen in the real
world. Vacuum suction pads, for example, can provide a force pulling the object
towards the surface, as can electrostatic forces or sticky surfaces. You will ignore
these possibilities in this course.

Therefore, you will assume a normal contact force will be non-negative and there is
no limit to how large it can be.

WORKED EXAMPLE 2.5

a A crane is lifting a pallet on which rests a stone block of mass 5 kg. The motion
is vertically upwards. The crane lifts the pallet from rest to a speed of 3 m s−1 in
6 m. Find the contact force on the stone block during the acceleration.

b If the contact force exceeds 650 N the pallet may break and so this situation
is considered unsafe. Assuming the same acceleration as in part a, find how
many stone blocks the crane can lift safely.
 Chapter 2: Force and motion in one dimension

Answer
a Taking upwards as positive:

R
a

5g N
2 2
v = u + 2 as Use an equation of constant acceleration to
2 2
3 = 0 + 2a × 6 find the acceleration.

a = 0.75 m s−2
F = ma Use Newton’s second law to find the contact
R − 5 g = 5 × 0.75 force.
R = 53.75 N
b Taking upwards as positive:

R
a

We define the number of stone blocks as n, but

still consider the blocks as a single object.
5ng N
49
F = ma
R − 5 ng = 5 n × 0.75
R , 650 The restriction is given as an inequality.
5ng + 5n × 0.75 , 650
n , 12.1
Hence the maximum number of blocks is 12. Note that the question asked for a number of
blocks, so it is the largest integer satisfying the
inequality.

EXPLORE 2.1

An electronic scale and an object can be used to measure the acceleration in an

elevator.

Use the scale to find the mass of the object. The scale works out the mass by
measuring the contact force and dividing by g.

Put the object on the scale on the floor of the elevator. As you go up and down in the
elevator, the reading on the scale should change.

As the elevator goes up, write down the maximum and minimum reading on the scale.
The contact force can be calculated by multiplying by g. By using F = ma for the object
you can now calculate the maximum acceleration and deceleration of the elevator.

Try this again when the elevator is going down.

 Cambridge International AS & A Level Mathematics: Mechanics

EXERCISE 2D

1 An elevator is carrying a man of mass 70 kg upwards, accelerating at constant acceleration from rest to
10 m s−1 in 2 s. Find the size of the contact force on the man.

2 An elevator is carrying a woman of mass 55 kg upwards. It is travelling upwards at 8 m s−1 and starts to slow
down at a constant rate when it is 9 m from where it stops. Find the size of the contact force on the woman.

3 An elevator is carrying a trolley of mass 30 kg downwards, accelerating at a constant rate from rest to 7 m s−1
in 2 s. Find the size of the contact force on the trolley.

4 An elevator is carrying a child of mass 40 kg downwards. It is travelling at 8 m s−1 and starts to slow down at
constant acceleration when it is 6.25 m from where it stops. Find the size of the contact force on the child.

5 A forklift truck carries a wooden pallet. On the pallet is a box of tiles with mass
35 kg. The truck lifts the pallet and tiles with an initial acceleration of 2 m s−2.
Find the contact force on the tiles.

6 A weightlifter is trying to lift a bar with mass 200 kg from the floor. He lifts with
a force of 1800 N but cannot lift it off the floor. Find the size of the contact force
from the floor on the bar while the weightlifter is trying to lift the bar.

7 A plate of mass 0.7 kg is being held on a horizontal tray. The tray is lifted from rest on the floor and
accelerates at a constant rate until it reaches a height of 1.25 m after 5 s. Find the contact force on the plate.
50
8 A man of mass 75 kg is standing in the basket of a hot-air balloon. The balloon is rising at 5 m s−1 and 4 s later
it is descending at 3 m s−1. Assuming constant acceleration, find the contact force on the man.

M 9 A girl of mass 45 kg is sitting in a helicopter. The helicopter rises vertically with constant acceleration from
rest to a speed of 40 m s−1 when it reaches a height of 200 m.
a Find the contact force on the girl.
b What assumptions have been made to answer the question?

PS 10 A drone carries a parcel of mass 4 kg. The parcel is held in place by two pads, on its top and on its bottom.
The drone hovers at a height of 30 m before descending for 2 s to a height of 8 m. Find the contact force acting
on the parcel as it descends and determine whether the force acts from the pad on top or bottom.

Checklist of learning and understanding

● A force is something that influences the motion of an object. Its size is measured in newtons (N).
● Force is related to acceleration by the equation: net force = mass × acceleration.
● Objects acted on only by the force of gravity have an acceleration of g m s−2 .
● The weight of an object is the force on it due to gravity and has magnitude W = mg.
● The reaction force or normal contact force is the force on an object due to being in contact with
another object or surface. It acts perpendicular to the surface and is usually denoted by R (or
sometimes N).
 Chapter 2: Force and motion in one dimension

END-OF-CHAPTER REVIEW EXERCISE 2

1 A cyclist travelling on a horizontal road produces a constant horizontal force of 40 N. The total mass of the
cyclist and the bicycle is 80 kg. Considering other forces to be negligible, find the distance covered as the cyclist
increases her speed from 10 m s−1 to 12 m s−1.
2 A bag of sand of mass 20 kg is lifted on a pallet by a crane. The bag is lifted from rest to a height of 5 m in 8 s at
constant acceleration. Find the contact force on the bag of the sand.
3 A rower starts from rest and accelerates to 4 m s−1 in 20 s. The combined mass of the rower and the boat is
100 kg. The rower provides a constant horizontal driving force of 60 N but is held back by a constant resistance
from the water. Find the size of the resistance force.
4 A stone of mass 0.3 kg is dropped from the top of a cliff to the sea 40 m below. There is constant air resistance
of 0.4 N as it falls.
a Find the speed with which the stone hits the water.
b The sound of the stone hitting the sea travels at 340 m s−1. Find the time between releasing the stone and
hearing the sound at the top of the cliff.
5 A train of mass 9000 kg is on a horizontal track. Its engine provides a constant driving force of 4000 N. There is
constant air resistance of 400 N.
a Find the time taken to reach a speed of 48 m s−1 from rest.
b When travelling at 48 m s−1 the train enters a horizontal tunnel 400 m long. In the tunnel air resistance
increases to 1000 N. Find the speed at which the train leaves the tunnel.
M 6 A submarine has mass 20 000 tonnes. With the engines on full power it can travel at 11m s−1 on the surface and 51
14 m s−1 underwater.
a When at maximum speed on the surface, the engines are turned off and it takes 4 km to come to a stop.
Find the resistance from the water on the submarine.
b Assuming the same resistance from the water, find the distance it would take to stop from maximum
speed underwater when the engines are turned off.
c Why is it not a reasonable assumption that the resistance underwater is the same as the resistance when
the submarine is at the surface?
7 A diver of mass 60 kg dives from a height of 10 m into a swimming pool. Through the air there is resistance of
50 N.
a Find the speed at which the diver enters the water.
b Once in the water, the water provides an upwards force of 2000 N. Find the greatest depth in the water the
diver reaches.
PS 8 A car of mass 400 kg is approaching a junction and needs to stop in 40 m. It is travelling at 15 m s−1 and there is air
resistance of 1200 N. Determine whether the car needs to brake or accelerate and find the size of the relevant force.
9 A ball of mass 0.1kg is projected vertically upwards from ground level with speed 9 m s−1. It reaches a height of
3 m. There is air resistance against the motion.
a Find the size of the air resistance.
b Find the speed with which the ball hits the ground.
10 A car of mass 350 kg is travelling at 30 m s−1 when it starts to slow down, 100 m from a junction. At first, it slows
just using the air resistance of 200 N. Then, at a distance of s m from the junction, it slows using brakes
providing a force of 2000 N as well as the air resistance. Find the distance from the junction at which the brakes
must be applied if the car is to stop at the junction.
 Cambridge International AS & A Level Mathematics: Mechanics

PS 11 A firework of mass 0.3 kg has a charge which provides an upward force of 7 N for 3s.
a Assuming no air resistance, find the maximum height reached by the firework.
b The explosive for the firework has a fuse which burns at a rate of 12 mm per second. Find how long the
fuse should be so the firework explodes at the maximum height.
12 A boy drags a cart of mass 5 kg along a horizontal road with force 10 N. There is air resistance of 2 N. At some
point the boy lets go of the cart and the cart slows down under air resistance until coming to rest. In total, the
cart has travelled 36 m. Find the length of time the boy was dragging the cart.
13 A light pallet is at rest on the ground with a stone of mass 30 kg on top of it but not attached. A crane lifts the
pallet by providing a force of 310 N upwards to a height of 8 m, at which point the pallet instantly stops and the
stone loses contact with it. Find the maximum height reached by the stone.
P 14 An air hockey table is 2 m long. A puck of mass 50 g is on the table at the middle point. A player hits the puck
with initial speed 4 m s−1 directly towards one side. Once it is moving there is air resistance of R N. Every time
the puck hits a side, the speed is reduced by 20%.
32
a Show that if R , the puck returns past the middle point of the table.
205
b Given that the puck does not return to the middle point a second time, find a lower bound for R.
15 A particle P is projected vertically upwards, from a point O, with a velocity of 8 m s−1. The point A is the highest
point reached by P. Find:
i the speed of P when it is at the midpoint of OA [4]
52
ii the time taken for P to reach the midpoint of OA while moving upwards. [2]
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q3 November 2012
16 Particles P and Q are projected vertically upwards, from different points on horizontal ground, with velocities
of 20 m s−1 and 25 m s−1 respectively. Q is projected 0.4 s later than P. Find:
i the time for which P’s height above the ground is greater than 15 m [3]
ii the velocities of P and Q at the instant when the particles are at the same height. [5]
Cambridge International AS & A Level Mathematics 9709 Paper 42 Q5 November 2010
17 A particle of mass 3 kg falls from rest at a point 5 m above the surface of a liquid which is in a container.
There is no instantaneous change in speed of the particle as it enters the liquid. The depth of the liquid in the
container is 4 m. The downward acceleration of the particle while it is moving in the liquid is 5.5 m s−2.
i Find the resistance to motion of the particle while it is moving in the liquid. [2]
ii Sketch the velocity–time graph for the motion of the particle, from the time it starts to move until the
time it reaches the bottom of the container. Show on your sketch the velocity and the time when the
particle enters the liquid, and when the particle reaches the bottom of the container. [7]
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q6 November 2014
In this chapter you will learn how to:
■ resolve forces in two dimensions
■ find resultants of more than one force in two dimensions
■ use F = ma in two directions
■ find directions of motion and accelerations.
 Cambridge International AS & A Level Mathematics: Mechanics 1

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills
IGCSE I O Level Use Pythagoras' theorem. 1 Find the hypotenuse of a right-angled triangle
Mathematics with short sides of length 5 m and 12 m.
IGCSE I O Level Use trigonometry for right-angled 2 A triangle, ABC, has a right angle at B.
Mathematics triangles. Length AC is 8 m and LEA C is 40°. Find
lengths AB and BC.
IGCSE I O Level Use the sine rule and the cosine 3 A triangle, ABC, has length AC 6m, LBAC
Mathematics rule. 40° and length BC 7 m. Find LABC and
length AB.
Pure Mathematics 1 Use the trigonometry identity
4 If sin0 =¾,find cos 0 .
sin 2 0 + cos 2 0 = 1
Pure Mathematics 1 Use the trigonometry identity
5 If sin0 = / , find tan 0.
sin0 3
- - = tan 0
cos0

How do you combine forces that are not acting in the same line?
Imagine two children are playing with a toy. They both pull it with a force of 10 N.
What would be the net force? Before you can answer this question, you need to know the

■
directions in which the forces are acting. If both children want to take the toy to the same
place and their forces act in the same direction, the net force would be 20 N. If they are
trying to take the toy away from each other and their forces act in opposite directions,
there would be no net force . But what if the forces are not parallel? For example, one could
be to the north and one to the east.

This chapter covers how to solve problems with forces in two dimensions.

3.1 Resolving forces in horizontal and vertical directions in equilibrium

problems
A force is a vector quantity. When vectors are added it is the equivalent of joining one
vector on to the end of the other.

This property can be used in reverse by splitting a vector into the sum of two others called
components. You choose the two vectors to be in perpendicular directions to make it
possible to set up equations. The components and the original vector will then always form
a right-angled triangle, so you can find the values of each component using trigonometry
for right-angled triangles or Pythagoras' theorem.
v
ou can use t h e tngonometnc
• .
. re 1at10ns h 1ps
' sm. = -"--"----
opposite an d cos 0 = _ adjacent
_;__ _
I
.1 0 I
I
hypotenuse hypotenuse I
I
to find how the components of a force relate to the original force. In the diagram: :Fy
I
I
Fx = Fcos0 -------------------·
Fx
Fy = Fsin0
Note that if you knew the other angle in this triangle, you would have to use sin to find Fx
and cos to find Fy.
 Chapter 3: Forces in two dimensions

KEY POINT 3.1

The components of a force are at right angles to each other. The original force is the hypotenuse of When drawing force
the triangle. diagrams, you can
draw the triangles
to help work out the
Components are not extra forces. They are the parts of a force already given, which act in components, but it
certain directions. is best not to mark
the components as
Equations are formed by finding the net component horizontally and the net component
separate forces or you
vertically. This is called resolving the forces in each direction.
may count the force
twice.
0 DID YOU KNOW?

The shape that a chain, wire or rope makes when it hangs between two points has a mathematical
formu la. It is called the catenary curve after the Latin word for chain . You can resolve for each link
in the chain, or particle on a wire or rope, to form differential equations. You can then solve them
to get the equation of the curve. The formula for the curve is a hyperbolic function (derived from
the exponential function), but a small part of the curve looks very similar to a parabolic curve, like
those for quadratic graphs.

In equilibrium the net force in both perpendicular directions will be zero.

WORKED EXAMPLE 3.1

■
A particle of mass 4 kg is held in place by a force of magnitude 100 N acting at an angle 0 above the horizontal and
a horizontal force of F N. Find the values of 0 and F .

Answer
IOON
The dashed lines show the horizontal and
vertical directions to allow calculation of the
F:: components of the 100 N force.

'
4gN

I00sin0 = 4g Resolving vertically.

e = 23 .6°
F = 100cos0 Resolving horizontally.
= 91.7 There are no units as this is the value of F
I
'

Cambridge International AS & A Level Mathematics: Mechanics

I

WORKED EXAMPLE 3.2

A boat is held in place by a force of 5 N due east, a force of 10 N due south and a force F N , on a bearing of 0. Find
the values of F and 0 .

Answer

north

F
•
I
To be in equilibrium F must have a component
to the west to cancel out the 5 N force and a
-----,: I component to the north to cancel out the 10 N force .
I
I
I
a• A triangle is drawn to make it easier to work out the
components, but the components are not marked.
---,► 5N
Bearings are always measured clockwise from
north. If the bearing is not acute it is often easier to
mark an acute angle, here a, relative to one of the
ION
' four basic. Here the bearing 0 = 360° - a.

F sin a= 5 Resolving east-west.

F cos a= 10 Resolving north-south.

•
1
tana = Dividing the equations.
2
a= 26.6°
T herefore 0 = 333.4°.
p 2 = 52 + 102 By Pythagoras' theorem.
F = 11.2

EXERCISE 3A

1 Find the components of the forces in the diagrams:

a horizontally, specifying whether it is left or right
b vertically, specifying whether it is upwards or downwards .

15N ii 12N

iv

~
iii

8N

21N
 Chapter 3: Forces in two dimensions

V 13N

vii ,~--r--· viii 120N

s.s N ~
175° \ _ 110°

U _____ _
2 a A force, F, has a horizontal component, Fx, of 10 N and acts at 20° above the rightwards
horizontal, as shown in the diagram. Find F and the vertical component, Fy.

b A force, F, has a vertical component, Fy, of 8 N and acts at 25° to the right of the upwards vertical. Find F
and the horizontal component, F,,.
c A force , F, has a vertical component, Fy , of 8 Nanda horizontal component, F,,, of 10 N. Find F and the
angle, 0, that the force makes with the rightwards horizontal.
d A force of 25 N has a horizontal component, F,.0 of 17 N and acts above the horizontal. Find the vertical
component, Fy, and the angle, 0, above the rightwards horizontal at which the force acts.
e A force of 3.8 N has a vertical component, ~ ,, of 3 N and acts to the left of the vertical. Find the horizontal
component, Fx , and the angle, 0, above the leftwards horizontal at which the force acts. ■
3 A particle in equilibrium has three forces of magnitudes 5 N, 6 N and F N acting on it in the horizontal plane
in the directions shown. Find the values of F and 0.
F

SN

6N

4 A lightshade of mass 2 kg is hung from the ceiling by two strings. One is fixed with tension 8 N at 20° to the
vertical. The other is fixed with tension T N at an angle 0 to the vertical.
a By modelling the lightshade as a particle, draw a force diagram for this situation.
b Resolve horizontally to find a value for T sin 0 and resolve vertically to find a value for T cos 0.
c Hence, find the values of T and 0.
north
5 A ship is being blown by a breeze with a force of 100 N on a bearing of 280° , as
shown in the diagram. It is pulled by a rope attached to the shore with force 50 N • F
on a bearing of 170°. A tugboat holds it in place. Find the size and bearing of the
force F applied by the tugboat.

SON
I
I

Cambridge International AS & A Level Mathematics: Mechanics

I

6 A wooden block of weight 20 N is at rest on a horizontal surface. It is pulled by a

force of 30 N acting at 10° above the horizontal, as shown in the diagram , and
remains at rest because of a horizontal frictional force, F.
a Draw the force diagram for this situation. 77
b Find the size of F and the size of the normal contact force.

7 A winch is dragging a caravan along a horizontal road at constant velocity. The

caravan has mass 750 kg. The winch provides a force of 850 N and acts at angle
0 above the horizontal, as shown in the diagram. There is friction of 700 N.

a Draw the force diagram for this situation.

b Find the value of 0 and size of the normal contact force.

8 A box of weight 50 N is being dragged at constant velocity along a horizontal road by a force, F , acting at 15°
above the horizontal. It experiences friction of 10 N .
a Draw the force diagram for this situation.
b Find F and the normal contact force.

9 A small aeroplane of mass 5000 kg is towed along a runway at constant speed by a rope acting at 20° below
the horizontal. There is friction and air resistance horizontally with total force 4000 N. Find the tension in the
rope and the normal contact force.

8
■
10 A wooden block is held in position by three horizontal forces, as shown in the F
diagram. One acts to the left with force 56 N. One acts with force Fat an angle 0,
where sin0 =¾, above the rightwards horizontal. One acts with force G at an 56N - --

angle (() , where sin(() = ~ . below the rightwards horizontal. Find F and G.
17
G
11 A block with weight 44 N is held in equilibrium by two ropes, one with tension,
Ti, acting at angle sin- 1 ~~ to the upwards vertical and the other with tension, T2 ,
5
acting at angle sin- 1 to the upwards vertical. Find Ti and T2 .
13
12 A box with weight 400 N is at rest on a horizontal surface. A man is pulling on a rope to try to get the box to
move. The force he can exert depends on the angle at which he holds the rope, so that when the rope is at an
angle 0 above the horizontal, the force he exerts is 1600 sin 0 N. He starts by holding the rope horizontally and
gradually increases the angle, thereby increasing the force. Another man tries to prevent this motion of the
box, by pulling horizontally. He can exert a maximum force of 700 N. Find the angle at which the box can no
longer remain on the ground. Hence, determine whether the box lifts off the ground first or slides along the
ground first.

0 13 A particle has three forces acting on it, as shown in the diagram, where sin0 = i.5
Show that F + G = 150./3 by resolving horizontally, and write down another
equation by resolving vertically. Hence, show that G = 75./3 + 100 and find F.

G
 Chapter 3: Forces in two dimensions

0 14 A particle has three horizontal forces acting on it, as shown in the diagram.
/3 an d fi n d an expression
Sh ow t h at cos a = -14 -
- 13--
cos- . 1or
.- sm
. a. U se
15
cos 2 a + sin 2 a = I to get an equation in /3. Hence, find a and /3.

13N

3.2 Resolving forces at other angles in equilibrium problems

Try resolving horizontally and vertically for the forces in equilibrium in this diagram.
R
You should get the two equations:
R cos 65° = T cos 25°
R sin 65° + T sin 25° = 10

ION

If there are two unknown forces and neither of them is vertical or horizontal, resolving
horizontally and vertically will lead to two equations, both of which involve two unknowns .

■
In pro blems that
You can solve these equations simultaneously, but it could be challenging. It would be
invo lve a slope,
easier if one equation involved only one unknown. you should resolve
Sometimes it is easier to resolve forces in directions other than horizontal and vertical. forces parallel and
perpendicular to the
A force has no component in the direction perpendicular to its line of action. This means slope. In other cases,
th at if you resolve perpendicular to an unknown force , the unknown force will not appea r choose directions
in the equation. perpendicular to an
unknown force. Choose
If you resolve in a d irection perpendicular to R in the example illustrated, R will not the directions carefully
appear in the equation so you can so lve directly for T. You will need to find the component so there are as few
of the 10 N force in this direction. unknowns as possible

As an a lternative to drawing a right-angled triangle, it may be easier to consider the angle

between the force and the direction in which you are resolving. When resolving parallel to
in each direction ,
to make solving
equations easier.
thej
a certain direction, as marked by p in the following diagram, the component of the force F
in that direction will be adjacent to the angle 0 between the force and the direction p .
Therefore, the component FP is found using the cosine of the angle.

The component of a force, F , parallel to a given direction, p , can be found by FP = F cos 0,

where 0 is the angle between the force and the direction p.
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 3.3

A boat is held in equilibrium by three forces of 10 N , F N and 20 N, as shown in the diagram. Find the values
of F and 0.

Resolving horizontally and vertically will leave

awkward simultaneous equations in F and 0.

Since F is an unknown force, we resolve

perpendicular to F so it does not appear in
F the equations, to find 0.

10 N Then you can find F .

Answer
20N
To help, dashed lines are added to the force
diagram to create right-angled triangles, with
the forces as the hypotenuses and the other
two sides parallel and perpendicular to F .
F Mark the angle a to compare the 20 N force
with the direction of F.
ION
You can find 0 from a because they add up to 80°.

■ 20 sin a = 10 sin 80°

a = 29.5°
Therefore 0 = 80° - a = 50.5°
Resolve perpendicular to F.

Notice F does not appear in this equation.

F + 10 cos80° = 20 cos a
Resolve parallel to F .
F = 15.7

WORKED EXAMPLE 3.4

A block of mass 10 kg is held in equilibrium on a slope at an angle of 20° to the horizontal

by a force, F , acting at 15° above the slope. Find F and the normal contact force. When you draw a
diagram involving a
Answer slope, make sure the
slope does not look
The normal contact (reaction) force is like it is at 45°, as it
perpendicular to the slope. will make it clearer if
angles at other points
If you draw the weight arrow down to the
in the diagram are the
horizontal line from the bottom of the slope,
same as the angle of
it may make it easier to find missing angles. the slope or not.
You will be resolving perpendicular and parallel
lOgN to the slope, so add dotted lines to form the
right-angled triangles, making sure the forces
are the hypotenuses of the triangles. This allows
you to find components in the directions of the
dotted lines.
 Chapter 3: Forces in two dimensions

WEB LINK
F cos 15° = I Og sin 20° Resolve parallel to slope.
You may want to
F = 35.4N Notice R does not appear in this equation.
have a go at t he
R + Fsinl 5° = 10 g cos20° Resolve perpendicular to slope.
Make it equal
R = 84.8N resource at the Vector
Geometry station on
the Underground
Mathematics website.

Note that you do not

need to be able to use
i - j vector notation
for this Mechanics
syllabus.

1 Find the components of the following forces in the direction of the dashed arrow. Although it might seem
clear from the diagram, make sure you specify whether the component is in the given direction or in the
opposite direction.
a b 14N

~- ----
--- ■
16 1°
C ~''. d
6.sN ~
14
-~ 3.8 ~ -~

2 Find the components of the following forces perpendicular to the direction of the dashed arrow. Make
it clear whether the component is in the perpendicular direction clockwise or anticlockwise from the
direction given.
a 42N b 12.3N

________ .,..

C 0.4 N
◄OI(.----,---..

I~ -.._
d

4.5 N
◄(_
n,,, -
12 1°

___..(____/
,.off

3 A particle has three fo rces acting on it, as shown in the diagram. By resolving
perpendicular to and parallel to F , find F and 0.

4 A boat is held in equilibrium by two tugboats. One pulls with a force of 100 N on
a bearing of 190°. One pulls on a bearing of 340° with tension T . The wind blows
with a force on the boat of Fon a bearing of 50°. By resolving perpendicular to T ,
find T . Find also F.
I

Cambridge International AS & A Level Mathematics: Mechanics

5 A book of mass 3 kg is prevented from sliding down a slope at 15° to the horizontal by friction acting up the
slope and parallel to it. Find the force of friction and the normal contact force.

6 A wooden block of mass 4 kg is held at rest on a slope at angle 0 to the horizontal by a force of 12 N acting
up the slope and parallel to it. Find the slope's angle and the normal contact force.

7 A particle of mass 2 kg is held in equilibrium on a slope at 13° to the horizontal by a force F acting at 10° to
the slope above it. Find F and the normal contact force .

8 A box of mass 12 kg is held in equilibrium on a slope at 18° to the horizontal by a force of size 50 N acting at an
angle 0 above the slope. Find 0 and the normal contact force .

9 A boy is dragging a bag of mass 8 kg up a slope at an angle of 17° to the horizontal and exerts a force of 50 N
parallel to the slope to do this. Air resistance, F, parallel to the slope prevents the boy from increasing his
speed, so he maintains a constant speed. Find the magnitude of the air resistance and the normal contact force.

10 A girl is dragging a sled of mass 20 kg up a slope at angle 14° to the horizontal. She pulls at an angle of 0
above the slope with a force of 70 N. She maintains a constant speed despite friction of 10 N parallel to the
slope. Find 0 and the normal contact force.

11 A particle of mass 4 kg is at rest on a slope at an angle of 49° to the horizontal. There is a frictional force of
10 N acting up the slope and a force F going up the slope acting at 9° above the slope. Find F and the normal
contact force .

~ 12 A heavy box of mass 50 kg is on a slope at angle 25° to the horizontal. There is no friction to prevent it sliding
down the slope, but there are three rods attached, at 40°, 50° and 60° above the slope, for people to drag it. A

■ man and two boys hold the rods to keep the box in equilibrium.
a Show that, if the man pulls with a force of 170 N and each boy can pull with a force of up to 90 N , they can
hold the box in equilibrium.
b If instead the man pulls with force 180 N and each boy can pull with a force up to 70 N, determine whether
or not they can hold the box in equilibrium and state which rod each should hold.

~ 13 A box of mass 20 kg is on a horizontal surface. There are three rods attached on one side, at 10°, 25° and
35° above the horizontal, for people to drag it. Three people are available to pull on these rods and they are
capable of providing forces of 150 N, 200 N and 250 N.
a The box is being pulled in the opposite direction by a horizontal force of 425 N. Show that only two of the
people are required to keep the box in equilibrium. State which of the rods each person holds.
b The horizontal force is increased to 550N. Show that if the box is to be prevented from moving
horizontally, it cannot remain on the ground.

3.3 The triangle of forces and La mi's theorem for three-force

equilibrium problems
The methods in this section are not required by the syllabus. However, they provide neat
and efficient methods for solving some problems. Although the questions can all be solved
using the methods from the previous sections, they may be solved more quickly using
alternative methods involving the triangle of forces or Lami's theorem.

If three forces act on an object to keep it in equilibrium, they will have no resultant. This
means that we can draw them end to end and they will finish where they started and form a
triangle. We can then use trigonometry to solve the problem.
 Chapter 3: Forces in two dimensions

• KEY POINT 3.4

By drawing a triangle of forces, we can use the sine rule or cosine rule directly without resolving
components. The lengths of the sides will be the magnitudes of the forces.

F irst , draw the fo rce diagram as a t riangle of forces.

A The fo rces can be

drawn in a triangle of
fo rces in any order.
Choose an order where
B it is easiest to work out
the angles.
You can add a ngles to the diagrams . You should extend the straight lines in the triangle,
as shown in the diagram.

A B C
Applying the sine rule to the t riangle gives

■
sin(l80° - a) sin(l 80° - /3) sin(l 80° - y) ·
A B C
Since sin 0 = sin(l 80° - 0), t h is leads to
sma sin/3 siny

Lami 's theorem states that for a particle in equilibrium with three forces on it, the ratio of the
magnitude of the force with the sine of the angle between the other two forces is the same for
each force.
A B C
sin a sin /3 sin r

EXPLORE 3.1

Forces of size 5 N, 6 N and 12 N act on an object. Can the object be in equilibrium?

Here are the opinions of two students.

Student A Student B .

There is no way of making two of them equal If the forces were at different angles, it might
to the third, so they cannot cancel out, and be possible for it to be in equilibrium.
the object cannot be in equilibrium.

Is one of the students correct? If the forces were of different sizes , in which
circumstances would each student be correct?
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 3.5

An object is in equilibrium by the action of forces oflO N, 8 N and 9N, as shown in ION
the diagram. Find the values of 0 and <p.

9N

Answer

Redraw the diagram as a triangle of forces .

ION

<p

cos(l80o - ) = 32 + 102 - 92 Use the cosine rule to find the a ngle between
<p 2 x 8 x l0
the 8 N and 10 N forces.
<p = 121.2°

cos(l80o - ) = 92 + 102 - 8"

■
<p 2 x 9 x l0 Use the cosine rule to find the angle between
<p = 130.5° the 10 N and 9 N forces , and use the alternate
angles theorem using the parallel 'north' lines.

WORKED EXAMPLE 3.6

A ship is held in equilibrium by ropes on bearings of 120° and 220°. The wind is blowing due north and exerting a
force of 90 N on the ship. Find the tensions in the two ropes.

Answer
90N

-
90
- - = - Ti~-
sin 100° sin 140° Use Lami's theorem.
Ti=58 .7 N
- 90 T2
- - = --=----
sin100° sin120° Use Lami's theorem again.
T2 = 79 .1N
 Chapter 3: Forces in two dimensions

WORKED EXAMPLE 3.7

A particle is held in equilibrium by three fo rces, as shown in the diagram. 40 N

F ind the size s of F and a .

120°

Answer
Using the method of resolving forces: Resolving perpendicular to F followed by
50 sin(l 80° - a) = 40 sin 60° resolving parallel to F

a = 43.9° or 136.1° Note that a must be bigger than 60° or there

Here a= 136.1 °. would be no component of the forces to the left
of the 40 N force so the particle could not be in
F = 50 cos(l 80° - a) + 40 cos 60° = 56. 1
equilibrium.
40 50
=--- Using Lami's theorem.
sin a sin 120°
a= 43.9° or 136.1°
Here a= 136.1°.

- - --F- - - - = - 50
- -

■
sin(3600 - 120° - a) sin 120°
⇒ F = 56.l
Using the method of the triangle of forces:

40 50 Using the sine rule.

- - - - - = - - - -- -
sin( 180° - a) sin( 180° - I 20°)
a = 43.9° or 136.1 °
Here a= 136.1°.
/3 = 76.1°
F 2 = 40 2 + 50 2 - 2 X 40 X 50 X cos /3 Using the cosine rule.
F = 56.1

1 A particle is held in place by forces of 8 N , 11 N and 12 N, as shown in the

diagram. Find the values of 0 and <p.

2 A mass of 5 kg is held in equilibrium by two ropes with tensions of 30 N and 40 N .

Find the angles that the ropes make with the vertical.
12N
 Cambridge International AS & A Level Mathematics: Mechanics

3 A mass of 7 kg is held in equilibrium by two ropes. One has tension 20 N and acts at 40° to the upwards
vertical. Find the tension in the other rope and the angle that it makes with the upwards vertical.
4 A ship is held in place by two ropes with forces 40 N and 35 N, as shown in the diagram, which prevent the wind
blowing it away. The wind has force F and acts at an angle 0 to the 35 N force, as shown. Find the sizes of 0 and F.
40N

5 Three ropes pull a boat, which remains in equilibrium. The ropes act due north and on bearings of 100°
and 210°. The one acting north has tension 25 N . Find the tensions in the other ropes .
6 A box has two ropes holding it in place. It is pushed by a force of ION. The angles between the force and the
ropes are 120° and 150°. Find the tensions in the ropes.
7 An 8 N force, a 9 N force and a 10 N force on an object result in no net force. Find the angle between the 8 N and
the 9 N forces .
~ 8 A land yacht is a vehicle with a sail that gets blown by the wind, but it moves on solid ground. An adult and
a child are holding ropes attached to the land yacht. The adult is capable of pulling with a force of 300 N.
The child is capable of pulling with a force of 80 N . T hey cannot pull in the same direction or they get in each
other's way, so there needs to be an angle of at least 20° between their ropes.

■ a For what strength of wind can the two of them work together to prevent the yacht from moving?
b For what strength of wind can the adult prevent the child from moving the boat?
c When the wind is blowing with a force of 130 N, the adult pulls the land yacht directly against the wind.
The child can cause the path of the yacht to deviate from the direction in which the adult pulls. Find the
maximum angle of deviation the child can cause.
0 9 A particle is held in equilibrium by three forces. Two of the forces have the same size, F N. Prove that the
third force acts along the line of the angle bisector of the lines of action of the other two forces.
0 10 Four forces on an object, A , B , C and D , result in no net force . If the angle between forces A and Bis a and
the angle between forces C and Dis y, show that A 2 + B 2 + 2AB cos a= C 2 + D 2 + 2CD cosy.

3.4 Non-equilibrium problems for objects on slopes and known

directions of acceleration
You used Newton's
When forces are not in equilibrium, the net force will not be zero, so we can apply Newton's
second law in
second law. The object will accelerate. Chapter 2, Section 2.1.
We calculate the acceleration using F = ma , but we need to resolve the forces into components in a
relevant direction and find the net force in that direction.
We need to choose carefully which directions to resolve in. When an object is on a slope it is clear the
object is not going to fly off the slope or go into the slope, so any acceleration will be parallel to the
slope, either up it or down it. In this situation there will be no net force in the direction perpendicular
to the slope, so we should resolve in directions perpendicular and parallel to the slope.
Alternatively, if a ship is being towed in a straight line by two tugboats, you may be able to see the
direction of motion from the bearing of the ship. There will be no acceleration perpendicular to the
direction of motion, so we should resolve in directions perpendicular and parallel to the motion.
 Chapter 3: Forces in two dimensions

In some situations, as well as the acceleration being unknown, one of the forces or an angle A
is also unknown. For example, suppose two people are pul ling a car with ropes at known angles.
The force from person A is known, but person B is pull ing with enough force to keep the car
following the path marked by the dotted line. Without knowing the size of the force, it is impossible
to work out the acceleration from one equation.
In this case, we must resolve forces in the perpendicular direction to get a second equation. We
know that there is no acceleration in this direction, so this equation is set up in the same way as
with equil ibrium problems.

The net force in the direction perpendicular to the acceleration is zero.

MODELLING ASSUMPTIONS

The scenarios in these questions involve net forces that cause acceleration. Did the forces
instantly appear at those sizes? Forces like gravity will always be there, but someone
pulling on a rope may have to increase the force from zero.
If that happens, why was there not a smaller acceleration while the force was
increasing to the size given? There are different assumptions that may have been made
to model the situation more easily, without significantly affecting the values calculated.
In some cases, the object is said to be held in place. That means there is initially some

■
other force keeping the object in equilibrium. That force is instantaneously removed so
the forces under consideration are already at the values given. In other cases, the time
taken to reach the given force values is considered negligible, and it is modelled as if the
forces are instantly at the values given.
We have also noted earlier that we are ignoring the shape of objects and considering
them all to be particles. In many cases this does not have an impact because the
object slides along a surface like a particle does . However, round objects like balls,
wheels or cylinders can roll. This has an impact on the motion , but at this stage we
will treat them as if they are particles, just sliding.

WORKED EXAMPLE 3.8

A box of mass 25 kg is dragged along the floor by a force of 30 N acting at 20° above the horizontal. Find the
acceleration and the normal contact force .

Answer
R
30N

25gN

F=ma Resolving horizontally.

30 cos 20° = 25a
a= 1.13 ms-2
 Cambridge International AS & A Level Mathematics: Mechanics

R + 30sin20° = 25g Resolving vertically, there must be no resultant

force , otherwise the box would leave the floor or
R = 240N
sink into the floor.

WORKED EXAMPLE 3.9

A boat of mass 40 kg has an engine providing a driving force of 30 Nin an easterly direction. It is also being blown
by the wind with a force T to the north. The boat moves on a bearing of 60°. Find T and the acceleration of the boat.
Answer
T
I
,~ direction of
60° ,,' motion
, ... ,'
_ , '- ----•30N

T sin 60° = 30 sin 30° Resolve perpendicular to the direction of motion first
T = 17.3N because there will be no net force in this direction.

F = ma Resolve in the direction of motion.

T cos 60° + 30 cos 30° = 40a
a = 0.866 m s- 2

■ WORKED EXAMPLE 3.10

A table is sliding down a slope at an angle 20° to the horizontal. There is resistance of
10 N acting up the slope parallel to it. The table takes 5 s to slide 10 m down the slope
from rest. Find the mass of the table.
Answer
R
The table is modelled as a particle, so we do
not worry about its shape for the diagram.
a REWIND

Look back to Chapter 1,

Section 1.3, if you
mg need a reminder of the
equations of constant
s = ut + .!_ at 2 Use information given to find the acceleration
acceleration.
2 first.
10 = .!_ a X 52
2
a= 0.8m s- 2
~ WEB LINK

You may want to have

F = ma Resolve parallel to the slope. a go at the Make it stop
mg sin20° -10 = m x 0.8 resource at the Vector
Here F is the net force and, since we are
m=3 .82kg Ge ometry station on
taking the direction of motion as positive, the
the Underground
10 N force is negative. Mathematics website.
 Chapter 3: Forces in two dimensions

1 A wooden block of mass 5 kg is on a horizontal surface. It is dragged by a force of 20 N acting at 14° above the
horizontal, as shown in the following diagram. Find the acceleration of the block and the normal contact force.

SgN

2 A book of mass 2 kg is dragged along a horizontal surface by a rope at 6° above the horizontal. It accelerates
at 0.3ms-2 .
a Draw the force diagram for this situation.
b Find the tension in the rope and the normal contact force.

3 A box of mass 10 kg is pulled along a horizontal surface by a rope with tension 20 N at an angle 0 above the
horizontal. The box accelerates at 1.2 m s- 2 . Find 0 and the normal contact force.

4 A car of mass 1000 kg is being towed by two people holding ropes. One pulls with a tension of 80 Nat an
angle of 18° to the direction of motion. The other pulls at an angle of 25° to the direction of motion, as

■
shown in the diagram. Find the tension, T, in the second rope and the acceleration of the car.
80N

5
< ➔ direction

T
of motion

A box of mass 20 kg is dragged by a force of 40 N at an angle of 15° to the direction of motion, and a force of
30 N at an angle of 0 to the direction of motion. Find the value of 0 and the acceleration of the box.

6 A truck of mass 15 000 kg is being towed by two ropes. One pulls with a tension of 3000 Nat an angle of 20° to
the direction of motion. The other pulls with a tension, T, at an angle of 10° to the direction of motion. There
is resistance of 500 N against the motion, in the same line as the motion.
a Draw the force diagram for this situation.
b Find T and the acceleration of the truck.

7 A ship of mass 10 000 kg is being towed due north by two tugboats with acceleration 0.1 m s- 2 . One pulls
with a tension of 2000 Non a bearing of 330°. The other pulls with a tension, T, on a bearing of 0. There is
resistance against the motion of 1000 N. Find T and 0.

8 A train of mass 230 tonnes provides a driving force of 300 000 N to accelerate up a slope at an angle of 5° to
the horizontal. The force diagram is shown. Find the acceleration of the train.

~ 230000gN
 Cambridge International AS & A Level Mathematics: Mechanics

9 A log of mass 200 kg is dragged up a slope at an angle of 13° to the horizontal by a rope attached to a truck.
The rope is at an angle of 20° above the slope.
a Draw the force diagram for this situation.
b The log accelerates at 0.3 m s-2 . Find the tension in the rope.

10 A windsurfer and his board have a total mass of 80 kg. They are being pushed by the water with a force
of 20 N westwards. The wind is pushing them northwards with a force F. The windsurfer accelerates on a
bearing of 340°. Find the force F and the acceleration of the windsurfer.

11 A buoy of mass 12 kg is on the surface of a lake. The tide pushes it with a force of 25 N and the wind pushes
it with a force of 15 N, as shown in the diagram. The buoy moves in the direction shown. Find the value
of 0 and the acceleration.
direction
25 N of motion
~
15N

12 A girl pulls a toy car of mass 0.8 kg by a string along a horizontal path. The tension in the string is 3 N and

■
the string is held at an angle of 40° above the horizontal. There is air resistance of 2 N. Find the time taken to
reach a speed of 2 m s- 1 from rest.

13 A shopper drags a trolley of mass 25 kg from rest along horizontal ground. The shopper is pulling the trolley
by a force of 30 N with his arm, which is at 15° above the horizontal. There is friction of 10 N. Find the speed
of the trolley after being pulled a distance of 6 m.

CD 14 A ball of mass 3 kg is rolled with initial speed 4 m s- 1 up a slope at an angle 10° to the horizontal.

a Find the maximum distance up the slope the ball reaches.

b What assumptions have been made to answer the question?

15 A cyclist of mass 70 kg (including her bicycle) arrives at an uphill stretch of road of length 30 m with an angle
9° to the horizontal, travelling at 10 m s- 1• She exerts a force of 15 N parallel to the slope and there is wind
resistance of 5 N against her. Find the time taken to reach the top of the slope.

~ 16 A ball of mass m kg is rolled up a slope at an angle 0 to the horizontal, where sin 0 = ¾- The ball passes a
point A with speed 7 m s-1. A point B is 5 m further up the slope than point A. Find the time between passing
B on the way up and returning to B on the way down.

17 A van of mass 2000 kg is towed from rest by two ropes. One pulls with a tension of 130 Nat 10° to the
direction of motion and the other acts at 15° to the direction of motion. Find the distance covered in 10 s.

18 A ship of mass 15 000 kg is moving due east at 2 m s- 1 when it starts being towed by a tugboat. The wind is
blowing it on a bearing of 60°, so the tugboat exerts a force of 5000 Non a bearing of 100° to make the ship
continue to go east. Find the speed of the ship after 5 s.

19 A box of mass 9 kg is dragged along horizontal ground by a force F acting at 30° above the horizontal. There
is friction of 5 N. The box starts at rest and reaches a speed of 4 m s- 1 in 10 m. Find the size of the force F.
 Chapter 3: Forces in two dimensions

~ 20 A car of mass 1200kg arrives at a steep upwards slope of length 130m at 34° to the horizontal. It is travelling
at 12 m s- 1 and there is air resistance of 100 N. Find the minimum force , assumed constant, the engine must
provide for the car to reach the top of the slope.

3.5 Non-equilibrium problems and finding resultant forces and

directions of acceleration
In the previous section, the direction of acceleration was known or could be worked out
from the situation. In the situation here, with forces A and B, the direction of acceleration
is unknown.

In situations like this , we can work out the single force equivalent to the combination of the
other forces by drawing the vectors end to end, as in the following diagram. This is called
the resultant of the other forces. If these are the only forces in the situation, the resultant is
the net force for use in Newton's second law.
A ......-_ _B.....,,

■
I
I
I
I
I
,' F
I
I
I
I
I
I

We can then use the diagram and trigonometry to work out the magnitude and direction of
the resultant of the forces A and B, which is shown by F .

In the following situation with forces A, Band C , the direction of acceleration is again
unknown.

When there are three forces, if we draw the A

vectors end to end we will get a quadrilateral.
It may not be easy to calculate the resultant
----► B from this diagram.

So, when there are more than two forces , we find the components of
the net force by resolving horizontally and vertically. By adding these
horizontal and vertical components, we can find the horizontal and
vertical components, F,'( and Fy, of the resultant force, F. We can
use the components of the resultant to calculate the magnitude and
direction of the resultant force.
 Cambridge International AS & A Level Mathematics: Mechanics

KEY POINT 3. 7

The magnitude of the resultant force, F, with components Fx horizontally and Fy vertically, can be
calculated using Pythagoras' theorem as F = ✓ F; + F} .

The direction of the resultant force, F , with components F_, horizontally and Fy vertically, can be
calculated using trigonometry as tan 0 = Fy, where 0 is the angle with the x-direction .
F_,

Do not show the resultant force on the force diagram because it is easy to confuse it with a separate force.
Instead, to show the resultant force, draw a second diagram alongside the force diagram.
A

EXPLORE 3.2

■ Two students are discussing the following situation. A heavy stone has three ropes
attached. They are pulled on bearings of 0I0°, 020° and 060°. Three people can pull
with forces of 200 N, 150 N and 100 N. Which person should pull on which rope to
maximise the net force if the direction is unimportant?

Student A StudentB
The total net force will be the same whoever Who pulls each rope will affect both the net
pulls each rope, but the direction may force and direction. We will need to work out
change. each case to decide which gives the largest
net force.
Which one of the students is correct?

If student A is correct, what effect does the arrangement of the people pulling the
ropes have on the direction of motion and why?

If student B is correct, is there a general rule as to who should pull each rope to
maximise the net force and why does it work?

If instead the direction is more important than net force , how can you decide
who should pull each rope so that the net force is as close as possible to a given
direction?
 Chapter 3: Forces in two dimensions

WORKED EXAMPLE 3.11

A boat of mass 100 kg experiences a force of 30 N eastward from the wind and a force of
40 N from the tide on a bearing of 35°, as shown in the diagram . Find the direction of the
subsequent motion and the acceleration.

Answer 30 N

direction of motion ,'

,,~ Draw a diagram with the resultant force to

,, ,
,, , show where the angle is being measured from.
_,,
,,, ,,
_,,/s,
,
,,, ,, Adding vectors is equivalent to drawing them
end to end.
direction of motio~ _,,
,,,, Do not draw the resultant as a separate force
, ,, on the force diagram.

■
,,,'
,,
, , ,' 0

30N
R2 = 30 2 + 40 2 - 2 X 30 X 40cosl25° Use the cosine rule to find the resultant.
R = 62.3N
F= ma
Use Newton's second law in the direction of
R = 100a acceleration.
a = 0.623 m s- 2
40 R Use the sine rule to find the angle.
= - --
sin 0 sin 125°
0 = 31.8°
So the direction of motion is on a bearing of 58°.
 . .. .

Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 3.12

A particle of mass 3 kg is attached to three ropes in the horizontal plane with forces 2N
I~
of 2 N , 4 N and 3 N , as shown in the diagram.

Find the direction of the subsequent motion and the acceleration.

e::-----.---► 4N
~ ,
Answer 3N
direction of motion
,,--,.; Draw a separate diagram showing the
F ,,
,,
'I
,, ,F resultant force.
. ,,:::}?_______ ! y

Fx Do not draw it on the force diagram.

FT = 4 + 3cos30° = 6.60
Find the components of the resultant
F';, = 2 - 3sin30° = 0.50 horizontally and vertically.

F
tan0 = ---1'.. Use trigonometry to find 0.
Fr
0 = 4.33°
so the direction is 4.33° above the positive x -direction.
F2 = F} + F; Use Pythagoras' theorem to find the resultant.
F = 6.62N

■
F= ma
Use Newton's second law in the direction of
F = 3a acceleration.
a= 2.2lms- 2

1 A particle of mass 3 kg is at rest and has two forces acting on it. One has magnitude 5 N and the other has
magnitude 3 N . They act in the directions shown. Find the magnitude and direction of acceleration of the
resulting motion.

2 A mass of 4 kg is held above the ground and released from rest. There is wind blowing it with a force of 20 N
horizontally. Find the angle from the downward vertical at which it initially falls.

3 A boat has its motor running, creating a force of 500 N. The wind is blowing it with a force of 200 N. The
directions of the forces are shown on the diagram. Find the direction of the subsequent motion.
SOON

I
~ 200N
 Chapter 3: Forces in two dimensions

4 A particle of mass 25 kg has two forces acting on it, one of 20 N and one of 35 N, in 20N

the directions shown. Find the magnitude and direction of the resulting acceleration.

5 Three coplanar forces act on a particle, as shown in the following diagram. X has
components ON in the x-direction and 20 Nin the y-direction. Y has components 25 Nin
the x-direction and -10 Nin the y-direction. Z has components -10 Nin the x-direction
a nd -15 Nin they-direction. Find the magnitude and direction of the resultant of the
three forces.
X

y
z
40N

6 Three coplanar forces act on a particle, as shown in the diagram.

a Force F has components of -30 N in the x -direction and -40 Nin the
y-direction. Find the value of a.
b Find the magnitude and direction of the resultant of the three forces.

7 Three coplanar forces act on a particle, as shown in the following diagram. Show that the x component and

■
y component of the resultant are equal. Hence, determine the direction of the resultant force.
40N

20N

8 Three people drag a bag of sand of mass 150 kg, labelled A in the diagram. They pull in the horizontal plane
with forces 40 N, 25 N and 35 N in the directions shown, compared to the direction AB.
a Find the magnitude and direction of acceleration of the resultant motion.
b What assumptions have been made to answer the question?

B 40N

9 A boat of mass 1000 kg is pulled by three tugboats. One pulls due north with force 500 N, one pulls due east
with force 350 N and one pulls on a bearing of 040° with a force of 250 N. Find the bearing and acceleration of
the resultant motion.
 Cambridge International AS & A Level Mathematics: Mechanics

10 In a competition of strength, four people pull a mass with ropes at different angles. Arjun 330N
I
The direction in which the mass moves determines the winner. Arjun wants the mass
to go north, Bob wants it to go east, Chen wants it to go south and David wants it to
go west. The men pull with the forces in the directions shown in the diagram. Find
the direction of the resultant motion and determine who wins.

11 A rowing boat of mass 120 kg is being pulled from rest by three boats. One pulls north
with a force of 100 N, the second pulls on a bearing of 020° with a force of 80 N and the
third pulls on a bearing of 045° with a force of 90 N. There is resistance from the water Chen 300N
of 200 N directly against the motion. Find the bearing and acceleration of the resultant
motion.

12 A hovercraft has an engine providing a force of 150 N on a bearing of 340°. The wind blows on a bearing of
310°, which means the hovercraft accelerates from rest on a bearing of 320°. Find the force of the wind on the
hovercraft.

0 13 The wind is blowing a boat with force F. The motor of the boat can exert a driving force of D N , where D < F.
Show with a diagram that, whatever direction the wind is taking the boat with the motor switched off, the motor
is capable of deflecting the direction by a maximum of sin- 1 ~.

14 A building is unstable after a natural disaster. A car is stuck under the building and needs to be dragged out
as quickly as possible, although the exact direction is less important. Three people can pull ropes, one due
north, one at a bearing of 010° and one at a bearing of 030°. Akhil can pull with a force of 300 N , Ben can
pull with a force of 240 N and Khadijah can pull with a force of 210 N. Find who should pull each rope to

■
maximise the acceleration and what the net force will be.

• A force can be split into components using the idea that force is a vector and can be written as
the sum of other vectors.
• The components are usually found in two perpendicular directions with the force as the
hypotenuse of a right-angled triangle and the other two sides as components.
• Directions chosen are usually horizontally and vertically, parallel and perpendicular to a slope,
or parallel and perpendicular to the direction of motion.
• Resolving perpendicular to an unknown force means the unknown will not appear in the
equation.
• When the direction of acceleration is unknown it is normally best to find components of a
resultant force and use them to find the direction and magnitude of the resultant.
 Chapter 3: Forces in two dimensions

FHi·l·1#i=M4i!ili!Miifiii!Mhii
1 Three forces act on a particle in equilibrium in the horizontal plane, as shown in the diagram. Find the size of
the unknown force F and the angle 0.

!SN

2 Three forces act on a particle in equilibrium in the horizontal plane, as shown in the diagram. By resolving in a
direction perpendicular to F , show that 0 = 47.2° and find F.
44N

3
30N

A girl is dragging a suitcase of mass 18 kg on horizontal ground, using a strap. The strap is at 40° to the
horizontal. She pulls with a force of 15 N. There is air resistance of 5 N.
a Find the magnitude of the normal contact force from the ground on the suitcase.
•
b Find the acceleration of the suitcase.

4 Two people drag a car of mass 1200 kg forward with ropes. One pulls with force 400 N on a bearing of 005°. One
pulls with force 360 Non a bearing of 352° . Find magnitude of the acceleration and its direction to the nearest 0.1°.

5 A boat is in equilibrium held by a rope to the shore. The rope exerts a force T at T
an angle 0 from north. The wind blows the boat with force 40 N in a northwest
direction. The current pushes it south with a force of 50 N. Show that
T sin 0 = 20.J'I and find an expression for T cos 0. Hence, show that
8
tan e = + IO.J'I and find 0 and T.
17
6 A car of mass 300 kg is on a slope, which is at an angle of 5° to the horizontal.
When it is pulled down the slope by a rope parallel to the slope with a force of T, it
accelerates at 2 m s-2. Find the acceleration of the car when it is pulled up the slope SON

by a rope parallel to the slope with a force of T.

~ 7 Three boys are having a strength competition. They hold ropes attached to the same object of mass 10 kg. One
pulls due north with force 32 N and another pulls on a bearing of 200° with force 45 N. The third wants to
make the object accelerate due east and pulls with a force of 24 N.
a Find the bearing at which the third boy should pull.
b Find the resultant acceleration.
 Cambridge International AS & A Level Mathematics: Mechanics

8 A girl can drag a stone block of mass 18 kg up a slope at an angle of 13° to the horizontal with an acceleration
of 0.7 m s-2 . Assuming this is the maximum force she can exert to drag the block, find the mass of the heaviest
stone block she would be able to drag up the slope.
9 A box of mass 8 kg is held at rest at the top of a slope 6 m long at an angle of 12° to the horizontal. Assume air
resistance and friction are negligible.
a The box is released. Find the time taken for the box to reach the bottom of the slope.
b Instead, a boy pushes the box downwards with a force of 20 N parallel to the slope. Find how much
sooner the box reaches the bottom of the slope than under gravity alone.
CD 10 A girl is sitting on a sledge, which her friend drags across the horizontal surface of a frozen lake. The sledge is
initially at rest and then the friend pulls on a rope at an angle of 35° above the horizontal with a force of 8 N for
2 m before releasing the rope. The total mass of the girl and the sledge is 50 kg. There is air resistance of 2.4 N.
a Find the speed of the sledge when the friend releases the rope.
b What assumptions have been made to answer the question?
c After the rope is released, air resistance causes the sledge to slow down until coming to rest. Find the
total distance before the sledge comes to rest.
11 In a test of strength competition, a competitor must get a 10 kg stone as far as they can up a slope. The slope is
at 10° to the horizontal. The competitor can drag the stone for 5 m from rest up the slope and then must release
it. Frictional forces are to be considered negligible.

■ a A competitor drags the stone with a rope at an angle of 16° above the slope and a force of 65 N. Find the
speed at which the stone is released.
b Find how far the stone travels after being released before coming to rest.
12 The four athletes in a bobsleigh team start the race by running along the ice. They push for 40 m on a horizontal
track, providing an average horizontal force of 180 N each. The total mass of the bobsleigh and the four athletes
is 600 kg.
a Find the speed at the end of the horizontal stretch of track.
The athletes then get into the bobsleigh. The track continues with a downhill stretch oflength 1300 m on a
slope at an angle of 5° to the horizontal. There is air resistance ofl 75 N.
b Find the total time to complete the entire track.
0 13 A ball of mass m kg slides down a slope, which is at an angle of 0° to the horizontal. It passes two light
gates x m apart. At the first gate, the speed of the ball is measured as um s- 1, and at the second its speed
is measured as v m s- 1• Assuming the resistance is constant, show the resistance force has a total size of
!!!_ (2xg sin0 + u 2 - v2 ).
2x
0 14 A car of mass m kg is rolling down a slope of length x m, which is at an angle of 30° to the horizontal. It has a
booster that provides a force of mg N over a distance of 1 m , which the driver sets off at a distances m after the
car starts moving. Assuming the booster is used before the end of the slope, show that the speed at the bottom
of the slope is given by v2 = g(x + 2) and deduce that the final speed is independent of when the booster is
applied. (Note that if the booster were applied for a fixed time rather than a fixed distance this would not be
true.)
 Chapter 3: Forces in two dimensions

@ 15 58N

Coplanar forces of magnitudes 58 N, 31 N and 26 N act at a point in the directions shown in the diagram.
Given that tan ex =/ , find the magnitude and direction of the resultant of the three forces. [6)
2
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q2 November 2011
@ 16

A particle P of mass 1.05 kg is attached to one end of each of two light inextensible strings, of lengths 2.6 m
and 1.25 m. The other ends of the strings are attached to fixed points A and B , which are at the same
horizontal level. P hangs in equilibrium at a point 1m below the level of A and B (see diagram). Find
the tensions in the strings. [6)

■
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q3 November 2013
@ 17 SON

A
-4 B

A block of mass 60 kg is pulled up a hill in the line of greatest slope by a force of magnitude 50 N acting
at an angle cx 0 above the hill. The block passes through points A and B with speeds 8.5 m s- 1 and 3.5 m s- 1
respectively (see diagram). The distance AB is 250 m and Bis 17.5 m above the level of A. The resistance
to motion of the block is 6 N . Find the value of ex. [11)
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q7 November 2014
 Cambridge International AS & A Level Mathematics: Mechanics ' .

1 A car of mass 1500 kg is on a straight horizontal road. The car accelerates from 20 m s- 1 to 24 m s- 1 in 10 s.
The car has a constant driving force and there is resistance of 100 N. Find the size of the driving force. [4]
2 A particle starts from rest at a point X and moves in a straight line until 40 s later it reaches a point Y , which
is 145 m from X. For Os < t < 5 s the particle accelerates at 0.8 m s- 2 . For 5 s < t < 30 s it remains at constant
velocity. For 30 s < t < 40 s it decelerates at a constant rate, but does not come to rest.
a Find the velocity at time t = 5 s and t = 40 s. [5]
b Sketch the velocity-time graph. [2]
3 A particle P is released from rest down a slope, which is at an angle of 20° to the horizontal. There is no
friction between the particle and the slope.
a Find the particle's speed after 0. 7 s. [2]
b Find the speed when the particle has travelled 1.2 m. [2]
4 A crate of weight 400 N is lifted by a forklift truck. The truck lifts the crate from rest to a height of 2 m in 5 s.
Assuming constant acceleration, find the normal contact force from the truck on the crate. [4]
5 A force F acts in a horizontal plane and has components 25 N in the x-direction and -17 Nin the y-direction
relative to a set of axes. The force acts at an angle a below the x-axis.
a Find the sizes of F and a. [4]
b Another force has magnitude 29 N and acts at an angle of 70° above the positive x-axis. The resultant of

■
these two forces has magnitude RN and makes an angle of 0 with the positive x-axis. Find the values
~Rmd0. ~

6 The graph shows the velocity of a parachutist as she falls from an aircraft
until she hits the ground 50 s later.
There are four stages to the motion: falling freely under gravity with the
parachute closed; decelerating with the parachute open; falling at constant
speed with the parachute open; and coming to rest instantaneously on hitting
the ground.
a Find the total distance fallen . [2]
b The parachutist has mass 70 kg. Show that the upward force on the parachutist
due to the parachute during the second stage is 1148 N. [5]
7 A particle of mass 6.3 kg is attached to one end of a light inextensible string. x
The other end of the string is attached to a fixed point X . It is held in equilibrium
by a horizontal force F when the string is at an angle a to the vertical,
where tan a = ~~ . Find the tension in the string and the size of F. [4]

8 Two forces, each of size 8 N, have a resultant of 13 N .

a Find the angle between the forces. [2]
b The two given forces of magnitude 8 N act on a particle of mass m kg, which remains at rest on a
horizontal surface with no friction . The normal contact force between the surface and the particle has
magnitude 7 N. Find m and the acute angle that one of the 8 N forces makes with the surface. [3]
 Cross-topic review exercise 1

9 Three coplanar forces of magnitudes 7 N, 10 N and 12 N act at a point A, 7N

as shown in the diagram.
a Find the component of the resultant of the three forces in the direction
AB and perpendicular to the direction AB. [3)
b Hence, find the magnitude and direction of the resultant of the three forces. [3]
ION
10 a A cyclist lets her bike accelerate down a slope with constant gradient, at constant
acceleration. She passes a point A, then 4 s later passes a point B 32 m away. Another 2 s later she
passes a point Ca further 19m away. Find the acceleration of the cyclist. [5)
b Assuming there is no friction or resistance and the cyclist is not pedalling, find the angle that the slope
makes with the horizontal, giving your answer to the nearest 0.1°. [3)
@ 11 A particle P is in equilibrium on a smooth horizontal table under the action
of four horizontal forces of magnitudes 6 N, 5 N, F N and F N acting in the
directions shown. Find the values of a and F. (6)

SN

FN
Cambridge International AS & A Level Mathematics 9709 Paper 42 Q3 November 2010

@ 12 A cyclist starts from rest at point A and moves in a straight line with acceleration 0.5 m s- 2 for a distance of 36 m.
The cyclist then travels at constant speed for 25 s before slowing down, with constant deceleration, to come to rest
at point B. The distance AB is 210 m.
i Find the total time that the cyclist takes to travel from A to B. [5]
■
24s after the cyclist leaves point A, a car starts from rest from point A, with constant acceleration 4 m s-2 ,
towards B. It is given that the car overtakes the cyclist while the cyclist is moving with constant speed.
ii Find the time that it takes from when the cyclist starts until the car overtakes her. (5)

Cambridge International AS & A Level Mathematics 9709 Paper 41 Q7 November 2015

13 A small bead Q can move freely along a smooth horizontal straight FN

wire AB of length 3 m. Three horizontal forces of magnitudes F N, 10 N

20N
and 20 N act on the bead in the directions shown in the diagram. The
magnitude of the resultant of the three forces is RN in the direction
shown in the diagram. ION

Find the values of F and R. [5)

ii Initially the bead is at rest at A. It reaches B with a speed of 11.7 m s- 1. Find the mass of the bead. [3)

Cambridge International AS & A Level Mathematics 9709 Paper 41 Q5 November 2015

@ 14 A particle P of weight 21 N is attached to one end of each of two A
------------- , ---- B --
I
light inextensible strings, S 1 and S2 , of lengths 0.52 m and 0.25 m
: S2
respectively. The other end of S1 is attached to a fixed point A, and the I
I
other end of S2 is attached to a fixed point B at the same horizontal
level as A. The particle P hangs in equilibrium at a point 0.2 m below
the level of AB with both strings taut (see diagram). Find the tension in
S 1 and the tension in S2 • [6)

Cambridge International AS & A Level Mathematics 9709 Paper 43 Q4 November 2012

In this chapter you will learn how to:
■ calculate the size of frictional forces
■ use friction to solve problems in motion
■ determine the direction of motion of an object
■ solve problems where a change in direction of motion changes the direction of friction .
 Chapter 4: Friction

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills

IGCSE / 0 Level Use Pythagoras' theorem. 1 Find the hypotenuse of a right-angled triangle with
Mathematics short sides of length 8 m and 15 m.
I GCSE/ 0 Level Use trigonometry for right- 2 A triangle, LABC, has a right angle at B. Length
Mathematics angled triangles . BC is 7 m and LBA C is 35°. Find length AC.
Chapter 2 / Chapter 3 Resolve forces and use 3 A box of mass 5 kg is on a slope at an angle of
Newton's second law. 10° to the horizontal. It is pulled down a slope
with force 8 N parallel to the slope. Find the
acceleration of the box.

How does friction work?

When a box is at rest on the floor it is in equilibrium with the weight balanced by a contact
force. It does not matter if there is friction or not because no frictional force is required
for the box to stay in equilibrium. However, when you gently push the box horizontally,
it may still remain in equilibrium and not move. This is because friction prevents it. As
you increase the pushing force the box may still not move. This suggests that friction can
change value in order to prevent motion.

■
At some point the pushing force on the box will be large enough to overcome friction and
the box will slide along the floor. What factors affect the point at which this occurs? Does
it depend on the size or shape of the object? It is reasonable to expect the size of the force
will depend on the two surfaces in contact. But what else affects it?

Once the force is large enough to overcome friction, how does friction behave? Does
friction remain fixed or does it change depending on the motion?

All these questions will be considered in this chapter.

4.1 Friction as part of the contact force

Connect a spring balance to a block of wood on a horizontal surface. Increase the

force on the spring balance horizontally until the block starts moving. When the
block is at rest, the friction force takes a large enough value to prevent motion. When
it starts moving, try to keep it moving slowly at a constant speed and read off the
force on the spring balance. This will be equivalent to the frictional force. Try this
on different surfaces and you should see that some surfaces have different amounts
of friction. Try moving the block at different constant speeds. The size of friction
should not be affected by the speed of the object.

Try resting a small mass on top of the block before pulling it horizontally. The
frictional force should be larger now. This suggests that the mass may affect the size of
friction. However, the mass also affects the normal contact force. By adding a mass and
simultaneously lifting the block slightly with another spring balance (quite difficult in
practice), the size of the force of friction goes down again despite the larger mass. This
suggests it is the size of the normal contact force that affects friction, not the mass.
I

Cambridge International AS & A Level Mathematics: Mechanics

If there is friction between two surfaces, the contact is called rough . If there is no friction
the contact is called smooth .

If there is no motion, friction takes whatever value is required to prevent motion. This means that
if an object is at rest on a horizontal surface with no forces other than its weight acting on it, there
will be no friction. When a force acts on the object, but is not strong enough to cause motion,
friction will act in the opposite direction to the force. As the force is increased friction will
increase until the point when the force is large enough to overcome the friction and cause motion.

When the force on the object is large enough that the object is still in equilibrium but any
more force would cause motion, the object is said to be in limiting equilibrium. At this point
friction will take a fixed , maximum value. That value depends on two main factors: how
rough the surfaces are and the normal contact force between them. Each pair of surfaces
has a coefficient of friction , denoted by µ , which gives a numerical value for how rough the
surface is. The size of friction is limited to a value µ times the normal contact force.

0 DID YOU KNOW?

Surprisingly, friction will not noticeably depend on the amount of area in contact between the two
surfaces. A larger area would create more friction, but it also spreads out the normal contact force
over a larger area so has almost no net effect.

• KEY POINT 4.1

Friction can take any value up to its limiting value.

■
F,sµR
If the object is moving relative to the surface, friction will take the limiting value:
F=µR
where R is the normal contact force.

A typical value forµ is between 0.3 and 0. 7, although surfaces that are not as rough may
have a smaller coefficient of friction and surfaces that are extremely rough may have a
larger coefficient of friction , possibly bigger than 1, although this is unusual. A 'smooth'
surface has no friction , which is equivalent toµ and takes the value 0.

Friction depends on the normal contact force between the surfaces, so we say it is 'part
of the contact force', and it acts parallel to the surface whereas the normal contact force
is perpendicular. This means the total contact force is the resultant of the normal contact
force and the friction force. We calculate it in the usual way by considering a right-angled
triangle and using Pythagoras' theorem.

KEY POINT 4.2

The total contact force can be found from C = ,JR 2 + F 2 .

The direction of the total contact force is at an angle of tan- 1 (~)to the normal contact force.

When an object is stationary, but about to move, friction will take the limiting value. We
say the object is in limiting equilibrium, or we can use the phrase 'on the point of slipping'.
This means that any extra force would make the object start moving.

Sometimes it is not clear which way friction acts. Suppose a car is on a rough slope with a
tow rope attached on the end of the car facing up the slope. Tension in the rope is acting
 Chapter 4: Friction

parallel to the line of greatest slope in an upwards direction. We don't know if the tension
is there to try to pull the car up the slope, or to help prevent the car moving down the slope.
Friction will act in different directions depending on the situation.

The two possibilities are shown by the two force diagrams. If the tension in the rope is
large, friction may act down the slope to prevent the car going up the slope (shown in the
left diagram). If the tension is small, friction may act up the slope to prevent the car going
down the slope (shown in the right diagram).

mg mg

In situations where it is not clear which way friction acts, you must make an assumption
about what happens in the subsequent motion. You need to be aware of the significance of
getting a negative value .

• KEY POINT 4.3

If you assume fricti on acts in one direction and then solving the equations gives a negative val ue fo r Forces 'parallel to the
friction , it means your assumption was wro ng. It means th at friction has the same magnitude but in slope' act along the line
the other direction. You should state that the direction is not as marked on your diagram. of greatest slope. Any
other direction parallel
to the surface will not

■
The value of friction given by equations may be too large because friction is limited to µR . be as steep, which is
If that happens, then the object cannot be in equilibrium. why roads up steep
slopes wind up rather
than go straight up.
In this course, fo rces
If the value calculated for friction to keep the object in equilibrium is larger than the limiting value, will generally act along
the object cannot remain in equilibrium. lines of greatest slope.

WORKED EXAMPLE 4.1

A box of mass 5 kg is at rest on horizontal ground. The box is being pulled by a horizontal force of 8 N . Find the
total contact force .
Answer

r7 r 8N
I I

' SgN
R = 5g = SON Resolve vertically to fi nd R .

F = SN Resolve horizontally to find F.

Note that the box is at rest , so friction must be of the

magnitude to prevent motion.

C = ✓ F 2 + R 2 = 50.6 N Use Pythagoras' theorem to find the total contact force.

I

Cambridge International AS & A Level Mathematics: Mechanics

'

WORKED EXAMPLE 4.2

a A book of mass 8 kg is at rest on a rough slope, which is at an angle of 20° to the horizontal. The book is
held in limiting equilibrium by a force of 10 N up the line of greatest slope. F ind the coefficient of friction
and the magnitude of the contact force .
b Find the largest force up the slope for which the book remains at rest.

Answer
a R
The force diagram assumes the book is on the point of

~
slipping down rather than up the slope.

Note that it may not be obvious from the question

20°
8gN
whether the book is on the point of slipping up or down
the slope.

R = 8g cos 20 = 75 .2N Resolve perpendicular to the slope first to find R.

8g sin20 = F+ 10
Resolve parallel to the slope to find F.
F = 17.4N
If you had assumed the book was about to slip up the
slope and marked friction as acting down the slope, you
would have got -17.4 N for friction and realised you
had made the wrong assumption: friction should be the

■
other way.

F=µ R
The book is on the point of slipping, so friction is
F limiting.
µ =-
R
= 0.23 1
C = ✓F 2 + R 2 Use Pythagoras' theorem to find the contact force.
= 77 .2N
b R If the force up the slope is the largest possible to prevent

~
motion, friction must act down the slope.

8gN

R = 8g cos 20 Resolve perpendicular to the slope first to find Ras before.

= 75 .2 N

8g sin 20 + F =P Resolve parallel to the slope to find P.

F =µR Since the book is on the point of slipping, friction is limiting

= 17.4N so we can find F using the value for J1 found in part a.

P = 44.7N
 Chapter 4: Friction

WORKED EXAMPLE 4.3

A waste container of mass 400 kg is in equilibrium on a rough slope at an angle of 18° to the horizontal. The
coefficient of friction between the slope and the skip is 0.3. It is held in equilibrium by a winch with tension TN.
Find the range of possible values for T
Answer

When T is minim al: R Firstly, consider the case where the winch is providing

~
the minimum force to prevent the skip from sliding down

~
the slope.

400gN

R = 400g cos 18 = 3800 N Resolve perpendicular to the slope first to find R.

400gsinl 8=T+F Resolve parallel to the slope next to find T

F = µR = 1140 Since the tension is the minimum possible, friction must

T = 94.8 take the maximum value.

When Tis max imal: R Secondly, consider the case where the winch is providing

~~ the maximum force, which is not enough to make the

skip slide up the slope, so friction acts down the slope in

dr 400gN
this case.

■
Resolve perpendicular to the slope first to find R as
R = 400g cos 18 = 3800N before.

400 g sin 18 + F = T Resolve parallel to the slope next to find T.

F = µR = 1140 Since the tension is the maximum possible, friction must

T = 2380 take the maximum value.

Therefore, the range of values for Tis 94.8 :,;;: T :,;; 2380.

1 A box is at rest on horizontal ground.

a When it is pulled to the right by a force of 40 N, as shown in the diagram, R

find the size and direction of the force of friction.

b When it is instead pushed to the left by a force of 25 N at 20° above the - --40N
horizontal, find the size and direction of the force of friction.
mg
c When there is no sideways force acting on the box, find the size of the force
of friction.
 Cambridge International AS & A Level Mathematics: Mechanics

2 A box of mass 14 kg is at rest on a slope that is at 15° to the horizontal.

a When there is no external force acting on the box, find the size and direction of the force of friction.
b When it is pulled up the slope by a force of 50 N parallel to the line of greatest
slope, as shown in the diagram, find the size and direction of the force of
friction. (Note that friction is not marked. You will have to decide which
direction you think friction is acting and come to a conclusion based on
whether the answer you get is positive or negative.)
14gN
c When the box is dragged down the slope by a force of 20 N at 10° above the line
of greatest slope, find the size and direction of the force of friction.
d When it is pulled up the slope by a force of 15 N at 35° above the horizontal, find the size and direction of
the force of friction .

3 A box of mass 20 kg is at rest on rough horizontal ground. Find the magnitude of the total contact force in
each of these cases.
a The box is pulled horizontally to the right by a force of 40 N.
b The box is pushed to the left by a force of 50 N at 15° above the horizontal,
as shown in the diagram.
~-,----,
SON

c The box is pushed to the left by a force of 50 N at 15° below the horizontal.

4 A book of mass 4 kg is at rest on a rough slope at angle 14° to the horizontal. Find the magnitude of the total
contact force in each of these cases.

■
. a No other force acts on the book .
b The book is pulled down the slope by a force of 5 N parallel to the line of
greatest slope.
c The book is pulled up the slope by a force of 15 N at 9° above the line of
greatest slope, as shown in the diagram.

5 A tin of mass 0.5 kg is on a rough horizontal table with coefficient of friction 0.3. Find the largest horizontal
force that can be exerted on the tin before the tin starts to move.

6 A block of wood of mass 3 kg is on a rough slope, which is at an angle of 25° to the horizontal. The coefficient
of friction between the block and the slope is 0.4. It is held in place by a force, P, going up the line of greatest
slope.
a Find the smallest possible size of P to prevent the block sliding down the slope.
b Given that the block remains in equilibrium, find the largest possible size of P.

7 A chair of mass 6 kg is at rest on a rough horizontal floor with coefficient of friction 0.35. It is pulled
horizontally by a force of 25 N . A boy pushes down on the chair so that the chair is on the point of slipping
but remains at rest. Find the force that the boy exerts on the chair.

8 Two men are trying to drag a bin of mass 100 kg up a rough slope at an angle 20° to the horizontal. The
coefficient of friction is O.25. One man pulls up the slope with a force of 400 N. The other tries to lift the bin
perpendicularly to the slope, providing a force such that the bin is on the point of slipping up the slope. Find
the force exerted by the second man.

9 A sledge of mass 200 kg is being pulled by a woman along rough horizontal ground. She exerts a force of
500 N at I 8° above the horizontal and the sledge is on the point of slipping. Find the coefficient of friction.
 Chapter 4: Friction

4D 10 A gardener is trying to move a heavy roller of mass 150 kg along rough ground at an angle of 5° to the
horizontal. He exerts a force of 200 N down the slope and parallel to it and the roller is on the point of slipping.
a Find the coefficient of friction.
b What assumptions have been made to answer the question?

0 11 In a factory, a machine picks up a box by clamping it on both sides. The box of mass 4 kg is held clamped on
both sides by identical clamps with the contacts horizontal. The machine provides a contact force of 50 N
with each clamp. Find the minimum coefficient of friction between each clamp and the box for the box not
to slip.

12 A box of mass 30 kg is at rest on a rough slope at an angle of 20° to the horizontal. When a girl pushes up the
slope along the line of greatest slope with a force of 25 N, the box does not slip down. Find the range of values
for the coefficient of friction between the box and the slope.

13 A ring of mass 2.5 kg is threaded on to a fixed horizontal wire. It is made of a 60N

rubbery material to give it an extremely high coefficient of friction (above 1) and
prevent it sliding along the wire. When it is at rest, the higher part of the ring is in
contact with the wire, so the normal contact force from the wire is upwards, as
shown in the diagram. The ring is attached to a string, which provides a tension of
60 N at an angle of 50° above the horizontal. The ring is now in limiting 2.5gN
equilibrium. The force diagram for the situation is given in the diagram. Note that
60N
the normal contact force is now acting downwards because there cannot be a
vertical component of acceleration, so the lower part of the ring is now in contact

■
with the wire. Find the coefficient of friction between the ring and the wire.
F

2.5gN

~ 14 A box of mass 50 kg is at rest on a slope, which is at an angle of 26° to the horizontal. The coefficient of
friction is 0.4. The box is held in place by a rope attached to a winch pulling up the slope and parallel to it.
Find the minimum and maximum possible values for the tension, T, which the winch could provide for the
box to remain in equilibrium.

15 A car of mass 1350 kg is at rest on a rough slope at an angle of 7° to the horizontal. A man tries to push it
down the slope, exerting a force of 500 N, but cannot get it to move.
a Find the angle that the total contact force makes with the slope.
b When the man stops pushing, the car remains in equilibrium. Find the angle that the total contact force
makes with the slope.

0 16 A ring of mass 2 kg is held in place at rest on a rough horizontal wire. It is attached to a string that is at an
angle of 40° above the horizontal.
a Explain why once the ring is released it can never be in equilibrium, however high the coefficient of
friction , when the tension in the string satisfies T sin 40 = 2g.
b Show that when the tension is 100 N the coefficient of friction must be at least I. 73 for the ring to be in
equilibrium, but when the tension increases to 200 N the coefficient of friction can be as low as 1.41 with
the ring remaining in equilibrium. Explain why.

17 A ring of mass 3 kg is at rest on a rough horizontal wire. It is attached to a string that is at an angle of 60°
above the horizontal. The coefficient of friction between the ring and the wire is 0. 7. Find the set of values for
the tension, T , which will allow the ring to remain in equilibrium.
 Cambridge International AS & A Level Mathematics: Mechanics I

4.2 Limit of friction

We have seen that when an object is in limiting equilibrium or on the point of slipping, friction
takes the maximum value. When an object is moving, friction will remain at the limiting value.

lf the object is moving relative to the surface, friction will take the value F = µR.
When an object is moving or about to start moving, mark the friction as µR on the force diagram.

When an object moves at constant speed it is in equilibrium. However, when an object on

a surface is accelerating, it will accelerate parallel to the surface. On horizontal ground
the acceleration will be horizontal. On a slope, the acceleration will be along the line of
greatest slope.

If we resolve parallel to the surface to find acceleration, we will not find a solution because
the size of friction is not known. The size of friction will depend on two factors , µ , which
may be given, and R. We will normally need to resolve perpendicular to the surface where
there is no acceleration to calculate the normal contact force first. This will allow us to
find the value of R and, hence, friction. Then we can resolve parallel to the surface using
Newton's second law to find acceleration.

In order to calculate the acceleration in the direction parallel to a rough surface, resolve

■
perpendicular to the surface first to find the normal contact force and, hence, the frictional force.
Then resolve parallel to the surface and calculate acceleration using F = ma.

EXPLORE4.2

Two students, Basma and Bijal, are discussing the best way to drag a heavy box along
a rough horizontal surface. Here are their arguments.

Basma Bijal
I would pull horizontally to get all the force I I would pull at an angle above the horizontal.
can exert on the box working in the direction This would reduce the contact force and
I want to go. therefore reduce the friction.

Discuss which argument is more convincing.

Practical experiments may help you answer the question. Test the situation using a
wooden block and spring balance. Increase the horizontal force until it is just less than
the force required to start the block moving. Try to keep the force the same, but change
the angle at which it acts. Does the block start moving if the force is acting at an angle?

In Section 4.1 we considered the situation where a car is held on a slope, but we didn't
know which way friction was acting. You also need to know how to deal with situations
where it is not known if there is motion nor, if there is, in which direction the motion would
be. Start by assuming the situation that seems likely to be correct, but be ready to spot a
contradiction.

Consider the same example where a car is on a rough slope and there is a rope pulling up
the line of greatest slope, but this time we do not know whether the car remains stationary.
 Chapter 4: Friction

If we assume the car slips down the slope, the friction must be limiting and act up the
slope. However, if we solve the equ ations and get a negative value for acceleration, this
contrad icts th e assumption and su ggests the car does not, in fact , slide down the slope.

If instead we as sume the car is pu lled up the slope, the friction must be limiting and act
down the slope. However, if t h is leads to a negative value for acceleration, this again would
contrad ict the assu mption and suggests the car is not, in fact, pulled up the slope.

~ WEB LINK
T hese two resu lts together would lead to t he conclu sion that the car is in equilibrium and
fr iction may not be limiting.
You may want to
• KEY POINT 4. 7 have a go at the
reso urce Afhctional
It may be necessary to make an assumption about the direction of motion when setting up the force story at the Vector
diagram. If the outcome contradicts the assumption, then you need to change your initial assum ption. Geometry station on
the Underground
Mathematics website.
I
MODELLING ASSUMPTIONS

We have assumed that the limiting value for friction is the same whether the object
is moving or not. In reality, there is a small difference between static friction and
dy namic fric tion. From t he experiment in Explore 4.2 you may have realised that to
start the block moving takes slightly more force than the amount required to keep
it at constant velocity once it is already moving. The difference is slight and for the
purposes of this course we will ignore it and assume they are both the same.

■
Once the object moves, the exact point on the surface in contact with the object
is always ch a nging, so each part of the contact may have a different value for the
coefficient of fr iction. We will assume that the difference in the values ofµ across a
broadly simila r surface is negligible. If the surface changes significantly, this will be
stated in the question and we wil l use a different value for µ for the different surface.
Awkward shap es may make it difficu lt for a n object to slide smoothly along a surface.
For example, a hook shape m ay lodge itself in the surface. However, in this course we
a re treating obj ects as particles so, whatever the size and shape of the actual object,
the size of fri ction will not be affected by those factors.

DID YOU KNOW?

Frederick the G reat, King of Prussia fro m 1740 until 1786, wanted to build a foun tain 30 m
tall fo r his gardens at Sanssouci. He asked Leonhard Euler (1707- 1783), one of the greatest
mathematicians of the age, to help calculate how to get the water from the river under enough
pressure to create the fountain. Euler did his calculations assuming no friction, but advised the
engineers that he wo uld need to do experiments to see if the calculations were valid.

The engineers did not take his advice and the fountains were built according to theory alone. The
pipes burst and the water never made it to the fountain. Frederick blamed Euler, despite Euler's
warnings.

Euler was the first to create equations modelling frictionless fluids, but it took more than a century
to work out how to add friction to the model of fluid dynamics in equations known as the Navier-
Stokes equations. These are still not full y understood and there is a $1million prize for so lving other
aspects of these equation s.
!

Cambridge International AS & A Level Mathematics: Mechanics

I

WORKED EXAMPLE 4.4

a A curling player tries to slide a curling stone of mass 20 kg along a horizontal ice rink to stop on top of a
target that is 46 m away from where it was released. The coefficient of friction between the ice and the stone
is 0.05. The player releases the stone with a speed of 6.5 m s- 1• Find how far from the target it stops .
b In a game of curling there are sweepers who sweep the ice to polish it and reduce the coefficient of friction.
Assuming they lower the coefficient equally along the entire path, find the reduced coefficient of friction
required to get the stone to land on the target.

Answer
a
a It is useful to add the direction of motion to
direction of the diagram and show the acceleration in that
motion _ _ __.
direction, even though the acceleration will be
,R
negative.

' 20g N

R = 20g Resolve vertically first to find R.

= 200N

■
F=ma Resolve horizontally to find a.
- µR = 20a
a= - 0.5ms- 2
v2 = u 2 + 2as Use an equation of motion for constant
02 = 6.52 - 2 X - 0.5s acceleration to find the distance.
s = 42.25
D istance from target = 46 - s Make sure you answer the question.
= 3.75m
b v2
= u 2 + 2as Use an equation of motion for constant
02 = 6.5 2 - 2a X 46 acceleration to find the acceleration.
a = - 0.459 m s-2
As before, Solve the equations to find µ.
R = 20g
= 200
and
-µR = 20a
µ = 0.0459
 Chapter 4: Friction

WORKED EXAMPLE 4.5

A woman drags a box of mass 20 kg up a rough slope. The slope is at an angle of 10° to the horizontal and the
coefficient of friction between the box and the slope is 0.45. The woman pulls the box using a rope held at an angle
of 20° above the slope, with a tension of 120 N. Find whether the force is large enough to create motion and, if it is,
find the acceleration.

Answer

Draw a force diagram, making the 20° angle

L µR
I o

20gN
-- 20°---- with a dotted line parallel to the slope.
This diagram assumes that there will be
motion up the slope, so friction will be limiting
down the slope.

R+ 120 sin 20 = 20g cos 10 Resolve perpendicular first to find R.

R = 156N
F= ma Resolve parallel to the slope to find a.
120 cos 20 - 20g sin 10 - µ R = 20a
a = 0.393 m s- 2 This is positive, so consistent with the
assumption that there is motion up the slope.

■
1 A box of mass 14kg is on horizontal ground. It is dragged by a horizontal force of21 N. The surface is rough
and the coefficient of friction between the surface and the box is 0.1.
a Resolve vertically to find the size of the normal contact force.
b Find the size of the frictional force.
c Find the acceleration of the box.

2 A skip of mass 3000 kg is held at rest by a winch on a slope at an angle of 15° to the horizontal. The slope is
rough and the coefficient of friction between the slope and the skip is 0.25. When the winch is removed the
skip starts to slide down the slope.
a Resolve perpendicular to the slope to find the size of the normal contact force.
b Find the size of the frictional force.
c Find the acceleration of the skip.

3 A boy is dragging a box of mass 20 kg up a rough slope at an angle of 12° to the horizontal. The coefficient of
friction is 0.28 . He provides a force of 100 N parallel to the slope. Find the acceleration of the box.
 Cambridge International AS & A Level Mathematics: Mechanics

4 A gardener is pulling a wheelbarrow of mass 8 kg from rest along rough horizontal 50N
ground. The coefficient of friction between the wheelbarrow and the ground is 0.6.
The gardener provides a force of 50 N at an angle of 30° above the horizontal, as
..-----I
~
shown in the diagram.
a Find the acceleration of the wheelbarrow.
b What happens when the wheelbarrow has 20 kg of soil in it and the gardener exerts the same force at the
same angle?

5 A ski-plane has skis to land and take-off on snow. It has a mass of 3000 kg and has a propeller providing a
force of 20 000 N horizontally. It accelerates from rest on horizontal ground at 2.2 m s-2 . Find the coefficient
of friction between the ground and the ski-plane.

6 A bin of mass 16 kg is held on a downhill slope at an angle of 20°. When the bin is released, it slides down the
slope with acceleration 1.2 m s- 2 . Find the coefficient of friction between the bin and the ground.

7 A ring of mass 2 kg is on a fixed rough horizontal wire with coefficient of friction 0.4. It is pulled by a rope
with tension 15 N at an angle of 5° above the horizontal. Find the acceleration of the ring.

8 A ring of mass 3 kg is on a fixed rough horizontal wire. It is pulled by a rope with tension 20 N at an angle of
10° above the horizontal and accelerates at 2 m s-2 . Find the coefficient of friction between the ring and the
wire.

9 A ski-plane of mass 5000 kg accelerates from rest along a rough horizontal runway of length 600 m. It needs
to reach a speed of 25 m s- 1 by the end of the runway to take off. The propeller provides a horizontal force of

■ 16 000 N. Find the maximum coefficient of friction to allow the ski plane to take off.

10 A downhill skier of mass 80 kg is accelerating down a rough slope of length 400 m at 22° to the horizontal.
There is air resistance of 50 N and the coefficient of friction between the snow and the skis is 0.3. The skier is
moving at 20 m s- 1 at the top of the slope. Find the speed of the skier at the bottom of the slope.

11 A bag of sand of mass 200 kg is being winched up a slope of length 10 m, which is at an angle of 6° to the
horizontal. The slope is rough and the coefficient of friction is 0.4. The winch provides a force of 1000N
parallel to the slope. At the bottom of the slope the bag is moving at 2 m s- 1. Find the distance it has moved
when its speed has reduced to 1.5 m s- 1.

12 A man wants to drag a block of wood of mass 50 kg along horizontal rough ground, where the coefficient
of friction is 0.45. If he pushes it he can generate a force of 250 N horizontally. Alternatively, he can pull
via a string with a force of only 230 N at an angle of 25° above the horizontal. Which would give the larger
acceleration?

13 Two men are pushing a palette of bricks of mass 120 kg along rough horizontal ground. The first man pushes
horizontally with a force of 150N. The second man pulls via a rope at an angle of20° above the horizontal
with a force of 140N. They maintain a constant velocity.
a Find the coefficient of friction between the palette and the ground.
b The second man no longer pulls the rope. By first finding the new normal contact force, find the
deceleration of the palette.

14 A snooker ball of mass 0.4kg is struck towards a cushion from 0.8m away with speed 3ms- 1. The surface
of the snooker table has a coefficient of friction of 0.3. When the ball bounces from the cushion its speed is
reduced by 20%. Find how far from the cushion it stops.
 Chapter 4: Friction

15 A wooden block of mass 10 kg is on rough horizontal ground with coefficient of friction 0.6. It is dragged by a
force of 80 N acting at 15° to the horizontal.
a Find the acceleration if the force is above the horizontal.
b Find the acceleration if the force is below the horizontal.

C, 16 A box of mass 50 kg is slowing down from 10 m s- 1 on rough horizontal ground. The coefficient of friction
between the box and the ground is 0.3. To start with, the box is being slowed by a string providing a tension
of 25 N horizontally. Then the string breaks and the box comes to a halt under friction alone after a total
distance of 14.5 m.
a Find how far the box travelled before the string broke.
b What assumptions have been made to answer the question?

4.3 Change of direction of friction in different stages of motion

A shopper is pushing a shopping trolley, but rather than just pushing it, the shopper gives it
a shove, lets go and walks after it. After a few metres, the trolley stops because of friction.

When the shopper does the same thing up a slope, friction also causes the trolley to stop,
but once the trolley has stopped, friction then acts in the opposite direction to prevent the
trolley falling back down the slope.

When the shopper does the same thing up a steeper slope, the trolley may start moving
back towards the shopper. In this situation, friction will be limiting to start with and act
down the slope to stop the trolley moving up the slope. Once the trolley comes to rest,
friction will act up the slope to try to prevent the trolley moving back down the slope. If
the force due to gravity is large enough, the trolley will start moving back down the slope
and friction will again become limiting, but will now act up the slope .
■
• KEY POINT 4.8

When the motion of an object can be split into different stages, you need to draw a different force
diagram for each stage and deal with the stages separately. The direction of the frictional force will
be different if the object changes direction.

EXPLORE4.3

Two students, Nina and Jon, are discussing the problem of a ball rolling up a slope
and then back down the slope.

Nina says she can save a lot of time in working out how long it takes to return to
the starting point, by working out how long it takes to reach the highest point and
doubling it. She says the speed when it reaches the starting point on the way down
will be the same as when it started on the way up.
Jon says that's not true . The uphill stage and downhill stage have to be worked out
separately. He says that the downhill bit will take longer and the speed will be lower
because friction has slowed down the ball.
I

Cambridge International AS & A Level Mathematics: Mechanics

Nina says that's nonsense. Of course friction will slow it down more quickly when
going uphill, but that just means it stops after a smaller distance and a smaller time
than without friction. It will still return to the starting point after twice the time it
took to reach the highest point.

Who is correct?

MODELLING ASSUMPTIONS

When an object is placed on a slope, it may topple over rather than slide down the
slope. However, for this course, we are considering all objects as particles, so they
have no shape and cannot fall over.
When a ball is placed on a slope, the centre of the ball is always above a point lower
on the slope than the point where the ball touches the slope. Therefore, the ball will
always roll down the slope, regardless of how much friction there is. Rolling is also
different from sliding. However, because we are considering all objects as particles,
objects like balls or cylinders, which may roll, are treated as particles that are sliding
and we will ignore any differences this might give.

WORKED EXAMPLE 4.6

■
a A box of mass 10 kg is pushed from rest along rough horizontal ground by a
horizontal force of size 50 N for 3 s. The coefficient of friction is 0.45 . Find the
speed when it stops being pushed.
b The box then slows down because of the friction. Find the total distance the
box has moved.

Answer
a R Draw the force diagram with friction limiting
because we know the box will move.
µR----t-+--+----50N

IOg N
R = 10g Resolve vertically first to find R .
=lOON
F= ma
Resolve horizontally, taking the direction of
50- µR = 10a motion as positive, to find a.
a=0.5m s-2
V = U + at Use an equation of constant acceleration to
= 0 + 0.5 X 3 find the velocity.
= 1.5 m s- 1

b s = ut + -1 at 2 Use an equation of constant acceleration to

2 1 find the displacement for the first stage.
So s, = 0 X 3+- X 0.5 X 32
2
= 2.25
 Chapter 4: Friction

Draw a new force diagram for the second

a REWIND

Look back to
stage of the motion because the situation has Chapter 1, Section 1.3,
changed. if you need a reminder
of the equations of
constant acceleration .
"'!OgN

R = 10g = lOON Resolve vertically to find the new value for R,

which in this case is the same as the old value.

-µR = l0a Resolve horizontally, taking the direction of

a= -4.5ms-2 motion as positive, to find the value for a for
the second stage.

v2 = u 2 + 2as Use an equation of constant acceleration to

2
So 0 = 1.5 2 + 2 x -4.5s2 find the displacement for the second stage.
S2 = 0.25

Hence s = s1 + s2 Find the total distance for the two stages of the
= 2.25 + 0.25 motion.
= 2.5m

WORKED EXAMPLE 4.7 ■

a A ball of mass 3 kg rolls up a slope with initial speed 10 m s- 1• The slope is at an angle of 20° to the
horizontal and the coefficient of friction is 0.3. By modelling the ball as a particle, find the distance up the
slope when the ball comes to rest.
b Show that after coming to rest the ball starts to roll down the slope.
c Find the speed of the ball when it returns to its starting point.

Answer
a R
Draw the force diagram with friction acting

~
µR ~ ----
down the slope against the direction of motion.

3gN

R = 3g cos 20 Resolve perpendicular first to find R.

= 28.2 N
F=ma Resolve parallel, assigning up the slope as
- µR - 3g sin 20 = 3a positive, to find a.
a = - 6.24 m s- 2
v2 = u 2 + 2as Use an equation of constant acceleration to
0 2
= 10 2
- 2 X 6.24 XS find the distance.
s = 8.0lm
 Cambridge International AS & A Level Mathematics: Mechanics

b R
Draw a new force diagram because the

~
situation has changed and friction now acts up

~
the slope to prevent motion down the slope.
Friction is limiting because we are assuming
3gN
there will be motion down the slope.

R = 3g cos 20 Resolve perpendicular to find the new value

= 28.2N for R, which in this case is the same as the old
value.
3g sin 20 - µR = 3a Resolve parallel, assigning down the slope as
a= 0.601 m s-2 positive, to find a.

This is positive, so the ball will roll back down the slope. The acceleration down the slope should
come out as positive to be consistent with the
assumption that there is motion down the
slope.
Alternatively, suppose the ball is in equilibrium. This could also be done by calling the friction
Then, F = 3g sin 20 F and finding the size of F required to prevent
= 10.3 motion and showing F > µR .
But µR = 8.46 < F , which is not possible.
c v2 = u2 + 2as Use an equation of constant acceleration to

■
2
= 0 + 2 X 0.601 X 8.01 find the speed.
v = 3.I0ms- 1

iiiiliHiii
1 At the end of a downhill run, a skier of mass 80 kg slides up a rough slope at an angle of 10° to the horizontal,
to slow down. He arrives at the upward slope with an initial speed of 12 m s- 1. The coefficient of friction
between the skier and the slope is 0.4. Find how far up the slope he comes to rest, and show that he remains at
rest there without falling back down the slope.

2 A ball of mass 3 kg rolls with initial speed 8 m s- 1 up a rough slope at an angle of 15° to the horizontal. The
coefficient of friction between the ball and the slope is 0.6.
a By modelling the ball as a particle, find how long it takes for the ball to come to rest and show that the ball
remains at rest there.
b Why is this model different from reality?

3 A book of mass 3 kg is at rest on a rough slope at an angle of 15° to the horizontal. It takes a force of 20 N
parallel to the slope to break equilibrium and drag it up the slope.
a Find the coefficient of friction between the slope and the book.
b Find the acceleration of the book down the slope if the 20 N force is applied down the slope.

4 A box of mass 12 kg is at rest on a rough slope at an angle of 18° to the horizontal. The coefficient of friction
between the slope and the box is 0.4.
a Find the force it takes parallel to the upwards slope to break equilibrium and drag the box up the slope.
b If the force were applied down the slope and parallel to it, find the acceleration.
. Chapter 4: Friction

~ 5 A car of mass 1250 kg is at rest on a rough slope at an angle of 35° to the horizontal. It takes a force of
13 000 N to move it up the slope. Show that without any force the car would slide down the slope, and find the
minimum force to prevent it moving down.

6 A bin of mass 10 kg is at rest on a rough slope at an angle of 32° to the horizontal. It is held on the point of
moving up the slope by a force of 90 N parallel to the slope. Show that when the force is removed the bin
would slide down the slope, and find its acceleration.

7 A trolley of mass 5 kg is rolling up a rough slope, which is at an angle of 25° to the horizontal. The coefficient
of friction between the trolley and the slope is 0.4. It passes a point A with speed 12 m s- 1. Find its speed when
it passes A on its way back down the slope.

8 A ball of mass 1.5 kg is sliding up a slope, which is at 30° to the horizontal. The coefficient of friction between
the ball and the slope is 0.45. It passes a point A at 10 m s- 1• By modelling the ball as a particle, find the time
taken to return to A.

C, 9 A pinball game involves hitting a ball up a slope whenever it reaches the bottom of the slope. The pinball has
mass 0.2 kg and rolls down a rough slope of length 1.2 mat angle 12° to the horizontal and with coefficient of
friction 0.1. The ball starts at the top of the slope at rest. When it reaches the bottom of the slope it is hit back
up and its speed is increased by 50%.
a Find the maximum height up the slope the pinball reaches after it has been hit back up the slope.
b What assumptions have been made to answer the question?

10 A wooden block of mass 3.5 kg is sliding up a rough slope and passes a point A with speed 20 m s- 1• The slope

■
is at 29° to the horizontal. The block comes to rest 25 m up the slope. Find its speed as it passes point A on the
way down.

11 A boy drags a sledge of mass 4 kg from rest down a rough slope at an angle of 18° to the horizontal. He pulls
it with a force of 8 N for 3 s by a rope that is angled at 10° above the parallel down the slope. After 3 s the rope
becomes detached from the sledge. The coefficient of friction between the slope and the sledge is 0.4 . Find the
total distance the sledge has moved down the slope from when the boy started dragging it until it comes to
rest.

12 A particle slides up a slope at angle 34° to the horizontal with coefficient of friction 0.4. It passes a point Pon
the way up the slope with speed 3 m s- 1 and passes it on the way down the slope with speed 2 m s- 1. Find the
coefficient of friction between the particle and the slope.

0 13 A particle slides up a slope at angle 0 to the horizontal with coefficient of friction µ. It passes a point A
on the way up the slope at speed um s- 1 and passes it on the way down the slope with speed v m s- 1• Prove that:

v2 = u2 ( sin 0 - µ cos 0 )
sine+µ cos e
so vis independent of the mass of the particle and the value of g. Deduce also that the speed on the way down
is always smaller than the speed on the way up.

14 A toy car of mass 80 g rolls from rest 80 cm down a rough slope at an angle of 16° to the horizontal. When it
hits a rubber barrier at the bottom of the slope it bounces back up the slope with its speed halved, and reaches
a height of 10 cm. Find the coefficient of friction between the car and the slope.
 Cambridge International AS & A Level Mathematics: Mechanics

4.4 Angle of friction

The concept of the angle of friction is not required by the syllabus. However, an
understanding of the angle of friction can make some problems on the syllabus easier to
solve and can help give alternative methods to solve problems on topics beyond the syllabus.

If a box is being pulled horizontally by a rope with tension T and is on the point of
slipping, the force diagram would look like the first diagram. This diagram has four
forces , but we can draw a simpler diagram with only three forces if we combine the normal
contact force and friction into a single contact force , C, as shown in the second diagram.
R C

µR _..-+-t-t---•T

mg
mg

The angle of friction ?i, is the angle between the normal contact force and the
total contact force when friction is limiting.
R
By drawing the components of the total contact force in a right-angled
triangle it can be seen that tan ?i, = µ: = µ.

■ The angle of friction is related to the coefficient of friction by A= tan- 1µ .

By considering the total contact force as a single force rather than two forces (the normal
contact force and friction), problems like the previous one with four forces can be reduced Cl REWIND

to problems with three forces . This means that you can use methods involving the triangle Look back to
of forces or Lami's theorem. Chapter 3, Section 3.3,
if you need a reminder
In the simplest case of an object on a slope in limiting equilibrium under gravity, a problem
of the triangle of
with three forces becomes a problem with two forces .
forces and Lami's
R C theorem.

mg
mg

It is now easy to determine the total contact force. The contact force and the weight must
be equal in magnitude and act in opposite directions, so the contact force will be vertical.

The angle between the normal contact force and the total contact force is the angle
between the normal contact force and the vertical. This is also equal to the angle between
the slope and the horizontal. If the object is in limiting equilibrium, then tan ?i, = µ and we
have the coefficient of friction .
 Chapter 4: Friction

• KEY POINT 4.10

The angle of friction is the steepest slope on which an object can remain at rest without slipping
under gravity.

EXPLORE4.4

The coefficient of friction can be found by experiment using the angle of friction.
Two people will be required to find it safely.

To find the coefficient of friction between an object and a table, place the object on
the table. Then gradually lift one side of the table so the surface is at an angle to the
horizontal. At the point where the object starts to slip down the table, remove the
object from the table and measure the angle between the table and the horizontal.
Use the equation tan ,1 =µto find the coefficient of friction.

WORKED EXAMPLE 4.8

A man tries to drag a suitcase of mass 18 kg along a rough horizontal surface. He drags it with a rope at an angle
of 20° above the horizontal. The coefficient of friction between the ground and the suitcase is 0.4. The suitcase is
in limiting equilibrium. Find the tension in the rope.

■
Answer
C
Mark the contact force as a single force so the
' I
problem now has only three forces.
ra"~ I _2.9: - - T There is limiting equilibrium, so the angle
between C and the normal is tan- 1 µ.

18gN

T = 18g
We can now apply Lami's theorem.
sin(l80-tan- 1µ) sin(70 + tan- 1µ)

T = 66.9N

WORKED EXAMPLE 4.9

a A 2 kg brick is at rest on a plank. The plank is lifted at one end to make an angle of 20° with the
horizontal and the brick remains stationary on the plank. Find the total contact force between the brick
and the plank.
b The plank is lifted further to an angle of 25° and the brick is on the point of slipping down the slope. Find
the coefficient of friction between the plank and the brick.
c The plank is lifted further to an angle of 35° and the brick is held in place by a force at an angle of 15° above
the angle of the upwards slope. Find the size of the force .
 I

Cambridge International AS & A Level Mathematics: Mechanics I

Answer
a C
The only two forces are the total contact force
and the weight, so they must cancel each other
out by being equal in magnitude but opposite
in direction.

2gN

C = 2g Resolving vertically gives C immediately.

= 20N

b C
As before, the total contact force must be
vertically upwards, so the angle between the
normal and the total contact force is 25°.

2g N
µ=tan 25 Because the brick is now in limiting
= 0.466 equilibrium, the coefficient of friction is found
from the tan of the angle between the contact
force and the perpendicular.

■
C C
Using the total contact force, the force diagram
now has only three forces in it and the triangle
of forces can be used.
Since there is limiting friction, the angle of
friction between C and the perpendicular to
2gN the slope will be the same as in the previous
part, which was also limiting.

, ,
, ,, Angles in the triangle can often be found more
,
,, easily by extending the sides if required or by
,, comparing with vertical or horizontal lines on
140° the original diagram.

T ----2:L_ Use the sine rule to find T .

sin 10 sin 130
T = 2g sin 10
sin 130
= 4.53N
 Chapter 4: Friction

1 A boy is trying to drag a box along a rough horizontal surface by pulling horizontally. The box has mass
12 kg . The coefficient of friction between the box and the surface is 0.4 . The box is on the point of slipping.
Find the size of the force exerted by the boy.

2 A girl tries to drag the box of mass 12 kg along the rough horizontal surface with coefficient of friction 0.4.
She exerts a force at an angle of 10° above the horizontal and the box is on the point of slipping. Find the size
of the force exerted by the girl.

3 A builder tries dragging a sack of sand of mass 25 kg along a rough horizontal surface with coefficient of
friction 0.5. He pulls at 12° above the horizontal and the sack is on the point of slipping. Find the size of the
total contact force.

4 A bench has mass 17 kg and is at rest on horizontal ground . A woman tries to move it by pulling it with a force
of 80 N at 15° above the horizontal and the bench is on the point of slipping.
a Find the angle of friction.
b Hence, find the coefficient of friction between the bench and the ground.

5 A trailer has mass 120 kg. A winch pulls the trailer with a force of 500 N at an angle 0 above the horizontal.
The trailer is in limiting equilibrium on horizontal ground with coefficient of friction 0.45. Find 0.

6 A metal block of mass 20 kg is on a rough slope at an angle of 12° to the horizontal. The coefficient of friction

■
between the book and the slope is 0.4. A boy is trying to move the block up the slope by pushing parallel to
the slope. He increases the force until equilibrium breaks. Find the maximum size of the force the boy pushes
with before the block slips.

7 A girl drags a sledge up a rough slope, which has an angle ofl0° to the horizontal. The sledge has mass 8 kg and
the coefficient of friction between the slope and the sledge is 0.3. She pulls the sledge with a rope at an angle of12°
to the slope and increases the tension until equilibrium is broken. Find the tension in the rope when this happens.

8 A car is towed down a rough slope, which is at an angle of 5° to the horizontal. The coefficient of friction
between the car and the slope is 0.35. The car is towed using a rope at an angle of 13° to the slope. Equilibrium
is broken when the tension in the rope is 4000 N. Find the mass of the car.

~ 9 A box of mass 12kg is at rest on a rough horizontal surface with coefficient of friction 0.6. A force is exerted
on it at an angle 0 above the horizontal so that the force required to break equilibrium is minimised. Show
that 0 is the angle of friction and find the size of the force required to break equilibrium.

10 A box has mass 40 kg and is on a rough slope with coefficient of friction 0.3. It is pulled up the slope by a
force of 300 N at 10° above the slope and is in limiting equilibrium. Find the angle that the slope makes with
the horizontal.

0 11 A ring of mass m kg is at rest on a fixed rough horizontal wire with coefficient of frictionµ. It is attached to a
string that is at an angle of a above the horizontal. Show that when T < m~ sm a and 0 = tan- 1 µ the ring
cos a - 0)
will be in equilibrium.
I.
Show further that if a + 0 ,;;; 90° and T > mg sm a the ring will always move, but if a + 0 > 90° and
cos(a-0)
T > mg sin °
sin(a + 0 - 90)
the ring will remain in equilibrium.
 , . 'y ~·· .J • ~~1 ·~:- '~~J .~ , ' -~~ :::~:~~

Cambridge International AS & A Level Mathematics: Mechanics · ·. . , .. :-,{) \

. · . · ~li.....~~.k~-~~~
' , .2~·,~

0 12 A particle of weight W is at rest on a rough slope, which makes an angle of a to the horizontal. The coefficient
of friction between the particle and the slope is µ . Assuming 0 + a < 90°, where 0 = tan- 1 µ, show that the
minimum force F required to break equilibrium and make the particle slide up the slope is F = W sin(0 + a)
and that F makes an angle 0 to the slope above the particle.

Show further that in the case where a < 0, the minimum force F required to break equilibrium and make the
particle slide down the slope is F = W sin( 0 - a) and that F makes an angle 0 to the slope below the particle.

• Friction can take any value up to the limiting value, which depends on the normal contact force,
R, and the coefficient of friction, µ.
• F~µR
• If there is motion, or the object is on the point of slipping or in limiting equilibrium, friction
will take the maximum possible value.
• The total contact force is the combination of the normal contact force and the friction.
• If a situation becomes different because a force changes or the direction of motion switches,
the normal contact force may be affected, so the friction may change. It is best to draw a new
diagram every time a different situation arises.
 I
j

Chapter 4: Friction

IUi·#•j#!=MiiiiliB'hitiiiiMhii
1 A horizontal force, T, acts on a particle of mass 12kg, which is on a rough horizontal plane. Given that the
particle is on the point of slipping and that the coefficient of friction is 0.35, find the size of T.
2 A particle of mass 15 kg is on a slope at an angle 25° to the horizontal. The coefficient of friction between the
particle and the slope is 0.3. A force, P, acts up the slope along the line of greatest slope. Find the set of values
for P for the particle to be in equilibrium.
3 A bowler rolls a ten-pin bowling ball of mass 4 kg along a horizontal lane. The ball is released with a speed of
9ms- 1 • The coefficient of friction between the ball and the lane is 0.04. The first pin is 18.5m away. Find the
speed at which the ball hits the pin.
4 A brick of mass 4.3 kg is being pushed up a slope by a force of 40 N parallel to the slope. The slope is at 13° to
the horizontal and the coefficient of friction between the brick and the slope is 0.55. Find the acceleration of
the brick.
5 A boat of mass 5 tonnes is being launched from rest into the sea by sliding it down a ramp. The ramp is at 5° to
the horizontal and is lubricated so the coefficient of friction is only 0.08. The ramp is 40 m long before the boat
enters the sea. Find the speed with which the boat enters the sea.

6 A bag of mass 49 kg is on rough horizontal ground with coefficient of friction 0.3. A force T acts at 0 above the
horizontal, where sin 0 =i and the bag is in limiting equilibrium. Show that R = ST, where R is the normal
5 3
contact force, and find another equation relating Rand T. Hence, find Rand T.
7 A book of mass 1.3 kg is on a plank of wood, which is held at an angle of 16° to the horizontal. The coefficient
of friction between the book and the plank is 0.45.
a
b
Show that the book remains at rest and find the size of the frictional force.
The book is held stationary while the plank is raised to make an angle of 27° with the horizontal. Show
that when the book is released it accelerates down the slope, and find the size of the acceleration.
•
~ 8 Two boys are arguing over who gets to play with a toy. The toy has mass 3 kg and is at rest on rough horizontal
ground with a coefficient of friction of 0.3. The older boy pulls with a force of 26 N at an angle of 39° above
the horizontal. The younger boy pulls in the opposite direction with a force of 24 N at an angle of 9° above the
horizontal. Determine whether the toy moves. If it accelerates, find the size of the acceleration and direction. If
not, find the size of the friction.
9 A mass of 6kg is on a slope at an angle of 14° to the horizontal. The coefficient of friction between the slope
and the mass is 0.4. There is a force of 5 N acting down the slope and parallel to it.
a Show that the force is not great enough to overcome friction, and find the magnitude of the total contact
force between the mass and the slope.
b When the force of 5 N is removed, find the total contact force and the angle it makes with the slope.
10 A box of mass 9 kg rests on a slope, which is at an angle of 34° to the horizontal. It is held in place by a
horizontal force of 20 N.
a By considering the total contact force as a single force , or otherwise, find the size of the total contact force .
b Given that friction is limiting, find the coefficient of friction between the box and the slope. !
I·
11 A particle of mass 6 kg is on a slope at an angle of 20° to the horizontal. The coefficient of friction between the particle f
and the slope is 0.1. The particle is 5 m from the bottom of the slope. It is projected up the slope with speed 4m s- 1.
a Find the distance travelled up the slope from the starting point until the particle comes to rest.
b Find the time until the particle reaches the bottom of the slope.
I
I

Cambridge International AS & A Level Mathematics: Mechanics

I

12 A particle of mass 8 kg is at rest on a slope at angle 15° to the horizontal. The

coefficient of friction between the particle and the slope is 0.05 . The particle is pulled
up the slope by a rope with tension 30 N at an angle of 20° above the line of the slope.
a Find the acceleration of the particle.
After travelling 10 m the string is cut and there is no tension.
b Find the speed of the particle when the string is cut.
The particle slows down until coming to rest.
c Find how far the particle has travelled in total when it reaches its highest point on the slope.
d Find the total time until it reaches that point.
0 13 A particle of mass m is on rough, horizontal ground with coefficient of friction µ 1. It is initially moving at speed
um s- 1• After a distance xm the surface changes to another surface with coefficient of friction µ 2 . The particle
2
2
comes to rest, having travelled a distance of y m on this surface. Show that µ 2 =u - µ,gx_
2gy
0 14 A mass of m is at rest on a plank of wood on level ground with coefficient of friction µ 1 • One end of the plank
is lifted until the mass starts to slip. The angle at which this happens is a .
a Show that µ 1 = tan a .

The angle of the plank is then raised to an angle /3 and the mass is held in place. The mass is then released and

■
travels a distance x down the slope. At the end of the slope the particle slides along the level ground, slowing
down under friction where the coefficient of friction is µ 2 , until coming to rest at a distance y from the bottom
of the slope. You may assume the mass starts sliding along the floor at the same speed as it has when it reaches
the end of the slope.
x(sin f3 - tan a cos /3)
b Sh ow t h at µ 2 = ----'---'----------'--'-.
y

@ 15 A particle moves up a line of greatest slope of a rough plane inclined at an angle a to the horizontal, where
sin a = 0.28. The coefficient of friction between the particle and the plane is ½-
Show that the acceleration of the particle is -6 m s- 2 . [3)
ii Given that the particle's initial speed is 5.4 m s- 1, find the distance that the particle
travels up the plane. [2)

Cambridge International AS & A Level Mathematics 9709 Paper 43 QI November 2013

 Chapter 4: Friction

@ 16 7.2N

Fig. 1 Fig. 2
A block of weight 7.5 N is at rest on a plane which is inclined to the horizontal at angle a, where tan a= J__
24
The coefficient of friction between the block and the plane is µ. A force of magnitude 7.2 N acting parallel to a
line of greatest slope is applied to the block. When the force acts up the plane (see Fig. I) the block remains at
rest.
17
Show that µ ;,,, . (4]
24
When the force acts down the plane (see Fig. 2) the block slides downwards.
31
ii Show thatµ< . (2]
24
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q3 November 2014
@ 17

The diagram shows a particle of mass 0.6 kg on a plane inclined at 25° to the horizontal. The particle is acted ■
on by a force of magnitude P N directed up the plane parallel to a line of greatest slope. The coefficient of
friction between the particle and the plane is 0.36. Given that the particle is in equilibrium, find the set of
possible values of P. (91
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q6 November 2012
In this chapter you will learn how to:

■ use Newton's third law for objects that are in contact

■ calculate the motion or equilibrium of objects connected by rods
■ calculate the motion or equilibrium of objects connected by strings
■ calculate the motion or equilibrium of objects that are moving in elevators.
 Chapter 5: Connected particles

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills

Chapter 1 Use the suvat equations 1 A ball is thrown vertically upwards with initial
for motion with constant speed 5 m s- 1•
acceleration. a How high does it rise?
b How long does it take to reach the maximum
height?
Chapter 2 Use Newton's second law, know 2 A box of mass 20 kg is at rest on a smooth surface.
that weight = mg and know
a Work out the normal contact force.
about normal contact forces.
b What assumptions have you made in
answering part a?
Chapter 3 Resolve forces in equilibrium 3 A box is sitting on a slope that makes an angle
and deal with non-equilibrium 30° with the horizontal. The box has weight 4 N.
problems. a Work out the frictional force that is preventing
the box from slipping down the slope.

The angle that the slope makes with the

horizontal is increased to 0. The box starts
from rest and slides with constant acceleration
2.5 m s- 2 down the slope.
b Work out the frictional force in this situation.

How is the motion of an object affected by it being attached to

something else?
When a car tows a trailer, the motion of the car is altered by the trailer. Mostly this is due to
the extra weight of the trailer, although there may be additional resistance forces on the trailer.
Would the motion of the car be the same if the trailer and its contents could be put inside the car?

In this chapter you will study the forces acting on different types of connected objects and
look at how these forces affect or prevent motion. In particular, you will consider objects
connected by rigid rods (such as a car towing a trailer), objects connected by strings (such
as masses hanging on the ends of a rope that passes over a pulley) and objects in moving
lifts (elevators). You will not be considering objects such as planets that affect each other
'remotely' using gravitational attraction.

You will use Newton's second law to calculate the acceleration of moving systems and
Newton's third law to calculate normal contact forces (normal reaction forces).

5.1 Newton's third law

ewton -' third law states that for every action there is an equal and opposite reaction. This
means that in every interaction there is a pair of forces that have the same magnitude but
act in opposing directions.

For example, when a boy uses a rope to pull a box, the force
with which the box pulls on the boy is equal and opposite to
the force with which the boy pulls the box. In both cases, the
force is the tension in the rope.
. - - - .d
 Cambridge International AS & A Level Mathematics: Mechanics

When two objects are in contact, each pushes on the other with an equal and opposite
normal contact force .

A box resting on the floor pushes down on the floor with a vertical contact force and the
floor pushes up on the box with an equal and opposite contact force. If no other forces act,
these contact forces are each equal to the weight of the box.

A box resting on a slope pushes into the slope with a contact normal contact force
force and the slope pushes back with an equal and opposite friction
contact force . When you draw a force diagram, you usually show
only the forces acting on the box, so here you show the normal
contact force from the slope on the box. weight

In Chapters 3 and 4, you studied the forces acting on an object and the motion or equilibrium
of that object. You now do the same thing but for systems made up of connected objects.

5.2 Objects connected by rods

A I' d is anything that can be modelled as a rigid connector with no mass. Examples of
objects connected by rods include a car towing a caravan, a truck pulling a trailer, and a
train made up of an engine pulling some carriages. In each of these situations you will only
consider motion in a straight line; that is, in one dimension.
You can analyse the forces and the motion in these systems using Newton's second law.

■
In a connected system, you can apply Newton's second law to the entire system or to the individual
components of the system.

When you consider the individual components in a system of two objects connected by a
rod such as a tow-bar, you need to include a tensi n force in the connecting rod or tow-
bar. In some situations this tension may turn out to be negative. This means that the rod is
under compression and the force is a tbr st.
T T
□-----•--- (--□
T
..
T

tension (pull ing inwards)

□i----..-(----JJ),-----1□
thrust (pushing outwards)

WORKED EXAMPLE 5.1

A car towing a trailer travels along a horizontal straight road. The car has mass 1500 kg and the trailer has mass
500 kg. The resistance to motion is 80 N on the car and 20 N on the trailer. The driving force produced by the
engine of the car is 360 N. Find the tension in the tow-bar.
Answer
trailer and car together:
normal contact force It is always a good idea to start with a diagram
R showing the forces.
res1~stance driving force
360N You can treat the system as a single entity
IOON to fi nd the acceleration. This is because the
weight 20 000 N
internal tensions and thrusts cancel.

Newton's second law for the system:

There is no motion vertically, so the vertical
360 - 100 = 2000a so a= 0.13ms- 2 components cancel out.
 Chapter 5: Connected particles

trailer: car:
normal contact force normal contact force The components must be treated separately
-~ Rt resistance30 N?Re
+driving
when the internal tensions or thrusts are
resistance T T force required.
20 N 360
tension in tow-bar N Draw separate diagrams to show the forces on
weight 5000 N weight 15000N
the trailer and the forces on the car.
Note that the force pulling the trailer forwards
is the tension in the tow-bar.
Newton's second law fo r the trai ler and car separately:
The car and trailer have the same acceleration.
T - 20 = 500a and 360 - 80 - T = 1500a
Hence, T = 85N . Either eliminate a or substitute a = 0.13.

WORKED EXAMPLE 5.2

A car towing a trailer travels along a horizontal straight road. The car has mass 1500 kg and the trailer has mass
500 kg. The resistance to motion is 80 N on the car and 20 N on the trailer. The driver applies the brakes, so the
driving force is replaced by a braking force of 100 N opposing the forward motion.
a Find the force in the tow-bar.
The car then descends a hill at 3° to the horizontal. The resistances and braking force are unchanged.

Answer
b Find the new force in the tow-bar.
■
a trailer: car: Draw the force in the tow-bar as a tension
normal reaction Rt normal reaction Re
unless you know that it is a compression.
80 N----,-+-,
+--tc=::!:::::::i--•T
20N t
weight 5000N
1o;F-t--J
weight 15000N
N ew ton's second law for the sys tem:
The resu ltant horizontal forces on the system
- I00 - 80 - 20 = 2000a so a = - 0.1m s- 2 are the braking force and the resistances.

New ton's second law for the trailer and car separately: The car and trailer have the same acceleration.
T - 20 = 500a and - 100 - 80- T = 1500a

Hence, T = - 30 N. Either eliminate a or substitute a = -0.1.

The fo rce in the tow-bar is a thru st of 30 N .

b Draw a new force diagram for the new situation.

The forces labelled R1, Re and Twill not
necessarily have the same values as in part a.
 ·' :· , . ,' ·:., :' -.. -.,_ ··,;t·:~:_(1
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Newton's second law for the system (parallel to the slope): The acceleration is parallel to the slope
15 000 sin 3 + 5000 sin 3 - 100 - 80 - 20 = 2000a (down the slope). The angle between the
so a= 0.423ms- 2, vertical and the normal to the slope is 3°. The
Newton 's second law for the trailer and car separately: weight of the car has a component 15 000 sin 3
T + 5000sin3 - 20 = 500a down the slope. The component normal to
the slope (15 000 cos 3) is balanced by Re.
and 15000sin 3 - 100 - 80 -T = 1500a Similarly for the weight of the trailer.
Hence, T = - 30 N.

. In this case, the force in the tow-bar is still a thrust of 30 N .

WORKED EXAMPLE 5.3

A model train consists of an engine (locomotive) coupled to a chain of four trucks. The coupling between the
engine and the first truck and each coupling between trucks are modelled as rigid rods . The train is moving on a
straight horizontal track. The engine has mass 0.8 kg and each truck has mass 0.2 kg when empty. The resistance
to motion is 0.06 N on the engine and 0.01 N on each truck. The driving force produced by the engine is 3 N.
a A mass of 0.1 kg is placed in each truck. Find the tension in each coupling.
b Find the tension in each coupling if, instead, the 0.4kg is all placed in the last truck.

Answer
a 3N 3N 3N 3N 8N

■
o.q_uu_, o.~ o.~ o.q_uu_, o.06~
Draw a diagram to show the forces acting on
the engine and each truck.
-~ T4 ~ T3~ T2~ T1 ~ .
3N 3N 3N 3N SN

Newton's second law for t he system: The resultant horizontal forces on the system
3 _ (0.06 + 0.01 + 0.01 + 0.0l + 0.0 l) = (0. 8 + 4 x 0. l + 4 x 0 _2 )a are the driving force and the resistances.
3 - 0.1=2a
a= 1.45 m s- 2

Newton's second law for the engine and each truck separately:
The acceleration is the same for the engine
3 - 0.06 - Ti = 0.8a T1 = 1.78N and for each truck.
Ti - 0.01- T2 = 0.3a T2 = 1.335N
There are five equations and four unknowns.
T2 - 0.01- T3 = 0.3a T3 = 0.89N
The 'spare' equation can be used to check the
T3 - 0.01 - T4 = 0.3a T4 = 0.445N values.
and T4 - 0.01 = 0.3a
b
6N 2N 2N 2N 8N
O.? O·~ O,~ 0.9 0.06 ~ N
T4 T3 ~ T2 T1 ~ •
Draw a new force diagram.
6N 2N 2N 2N SN
Newton's second law for the system: The resultant horizontal forces on the system
are the driving force and the resistances.
3 - 0.1 = 2a
a = l.45m s- 2
 Chapter 5: Connected particles

Newton's second law for the engine and each truck separately: The acceleration is the same for the engine
3 - 0.06 - Ti = 0.8a T1= 1.78N and for each truck.
T1 - 0.01 - T2 = 0.2(/ T2 = 1.48 N
T2 - 0.01 - T3 = 0.2(/ T3 = 1.18N
T3 - 0.01 - T4 = 0.2a T4 = 0.88N
and T4 - 0.01 = 0 .6a

1 A tractor is connected to a trailer by a rigid, light bar. The tractor has mass 10 000 kg and the trailer has mass
2000 kg. The tractor and trailer are moving along a straight horizontal road. The tractor engine produces a
driving force of 400 N. The resistance on the tractor is 40 N and the resistance on the trailer can be ignored.
Find the tension in the bar.

2 A car of mass 1200 kg tows a trailer of mass 300 kg. The car and trailer travel along a straight horizontal
section of road. The engine of the car produces a driving force of 400 N. The car experiences a resistance of
150 N and the trailer experiences a resistance of 100 N.
a Find the acceleration of the car and trailer.
b Find the tension in the tow-bar.

CD
■
3 A box of weight 250 N is pulled across a smooth horizontal floor, using a horizontal rope. The tension in the
rope is 20N.
a Work out the acceleration of the box.
b What modelling assumptions have been made?
A box of weight I 00 N is connected to a second box of mass 150 N , using a connecting rod. The 150 N box
is pulled across a smooth horizontal floor, using a horizontal rope. The tension in this rope is 20 N and air
resistance can be ignored.
c Work out the tension in the connecting rod between the two boxes.
d Work out the tension in the connecting rod if the rope is attached to the 100 N box instead, but otherwise
the situation is unchanged.

4 A truck of mass 2000 kg tows a trailer of mass 800 kg. The engine of the truck produces a driving force of
300 N. A resistance of 120 N acts on the truck and a resistance of 40 N acts on the trailer. The truck and
trailer are moving along a straight horizontal road and initially the trailer is empty.
a Find the tension in the tow-bar when the trailer is empty.
A load of mass 1200 kg is then added to the trailer, which increases the resistance on the trailer to 80 N. The
forces on the truck are unchanged. The truck and trailer return along the same straight horizontal road.
b Find the tension in the tow-bar when the trailer carries this load.

~ 5 A car of mass 2000 kg pulls a caravan of mass 1200 kg along a straight horizontal road. The resistance on
the car is 20 N and the resistance on the caravan is 80 N. The maximum possible driving force from the car's
engine is 1900 N . The tow-bar will break if the tension exceeds 680 N.
a Find the maximum possible driving force before the tow-bar breaks.
b Find the maximum possible acceleration.
I
I
I

' Cambridge International AS & A Level Mathematics: Mechanics

I

$ 6 A bucket hangs from a vertical rod. Another rod is attached to the bottom of the bucket and a second bucket
hangs on the end of this rod. Each bucket is partially filled with water and they hang in equilibrium.
a Work out the tension in each rod when:
each bucket of water has mass 12 kg
ii the first bucket of water has mass 8 kg and the second has mass 16 kg.
b What assumptions have been made?

~ 7 A 15 kg mass hangs in equilibrium on a heavy chain. The chain is modelled as ten 1kg masses joined by short
rods of negligible mass. Including the connection at each end of the chain, this makes 11 short rods. Work out
the tension in each of the 11 short rods.

8 A horizontal bar of mass 1kg hangs from a pair of parallel vertical rods, of negligible mass, attached to either
end of the bar. A third vertical rod is connected to the middle of the bar and a 4 kg mass hangs from this ,
below the rod. Work out the tension in each of the rods.

9 A train consists of an engine and five carriages. The engine has mass 100 000 kg and each carriage has mass
20 000 kg. The engine produces a driving force of 350 000 N. The resistance force on the engine is 10 000 N and
the resistance on each carriage is 2000 N. The train moves in a straight line on a horizontal track. Find the
tension in the coupling between the third carriage and the fourth carriage.

10 A car pulls a caravan up a hill. The slope of the hill makes an angle 0 with the horizontal, where sin 0 = } .
0
The car has mass 1900 kg and the caravan has mass 600 kg . The driving force from the engine of the car is

■
1200 N. The resistance on the car is 20 N and that on the caravan is 80 N. Find the force in the tow-bar and
state whether it is a tension force or a thrust force .

11 A car tows a caravan down a hill. The slope of the hill makes an angle 0 with the horizontal, where sin 0 =} .
0
The car has mass 1900 kg and the caravan has mass 600 kg . The car is braking, so the driving force from the
engine of the car is negative. This braking force is 250 N (a driving force of -250 N). The resistance on the car
is 20 N and that on the caravan is 80 N. Find the force in the tow-bar and state whether it is a tension force or a
thrust force.

~ 12 A car tows a caravan down a hill. The slope of the hill makes an angle 0 with the horizontal, where
sin0 = 0.05 . The force from the car's engine is a braking force (a negative driving force). The car has mass
1800 kg and the caravan has mass 600 kg. The resistance on the car is 20 N and that on the caravan is 80 N.
The force in the tow-bar is a thrust of 50 N. Show that the force from the car's engine is -420 N.

5.3 Objects connected by strings

There are three main differences between rods and strings:
• a string can change direction (for example, by passing over a smooth peg or pulley)
• a string can be in tension or be slack (that is, have no tension)
• the force in a string can never be a thrust.
We use the term 'string' to mean any rope, chain or cable. You will always assume that
the string is light, so its weight can be ignored.
To keep things simple, you also assume that any strings are inextensible (they do not
stretch).
 Chapter 5: Connected particles

When a string passes over a ·mooth pulley , the magnitude of the tension is unchanged but the
direction can change.
Smooth pulley Forces on components of system
reaction from pulley

T
T

reaction from pulley

T
T

It is wrong to apply Newton's second law when the direction of travel changes. (Even
■
if this gives the right answer, it is mathematically wrong to 'bend' a vector round a
corner.)
When the acceleration of a system of connected objects is constant, you can use the
equations for constant acceleration to calculate velocities, distance travelled or time
taken.

DID YOU KNOW?

Archimedes invented many machines as 'amusements in geometry', some of which were used to
defend Syracuse when it was attacked in 212 BCE by the Romans. These machines used levers and
pulleys; for example, an 'iron hand ' that could lift the Roman ships into the air and swing them to
and fro until all the Roman soldiers were thrown out.
He also invented a compound pulley that brought him great fame when he used it to move a fully
laden ship with a crew of many men 'as smoothly and evenly as if she had been at sea ', by holding
the head of the pulley in his hand and pulling on the cords.
 , . . .. .- .. (:__ ,. l~;/:{~
Cambridge International AS & A Level Mathematics: Mechanics . .""·:·_:sj
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WORKED EXAMPLE 5.4

A box of mass 3 kg is placed on a table. The coefficient of friction between the box and the table is 0.5. A string
is attached to the box and passes over a smooth pulley at the edge of the table. The part of the string between
the box and the pulley is horizontal. After passing over the pulley, the string hangs vertically, with the other end
attached to a ball of mass 2 kg. The system is released from rest.
a Find the tension in the string.

The ball is initially 8 cm below the table top. The ball hits the ground after 1.2 s. The box has not reached the
pulley at this time.
b Find the height of the table.

Answer
a
The forces acting on the box are:
F
• its weight (30 N)
• the normal reaction from the contact with
the table (R)
• the tension in the string (T)
• the frictional resistance (F).
The forces acting on the ball are:
• its weight (20 N)

■
• the tension in the string (T) .
The two tensions are numerically equal
because the pulley is smooth.
The (horizontal) acceleration of the box is
numerically equal to the (vertical) acceleration
of the ball since otherwise the string would
either snap or would go slack.

Resolving vertically for the box : The box does not move vertically so there is no
R = 30 resultant force vertically.

F=µR The box is moving so friction is at its limiting

= 0.5 X 30 value.
= 15
Newton's second law for the box (horizontally, left to right):
The box and the ball are moving in different
T - F = 3a directions, so we must treat the (horizontal)
T - 15 = 3a motion of the box and the (vertical) motion of
Newton's second law for the ball (vertically downwards): the ball separately.
20 -T = 2a
Hence, 2T - 30 = 60 - 3T. Solve the equations simultaneously by
So T = 18. eliminating a.

The tension in the string is 18 N.

 Chapter 5: Connected particles

b a= I Substitute T = 18 into 20 - T = 2a.

2
The box and the ball each accelerate at lm s- .

For the ball:

u = 0 ms- \ a = Im s- 2 , t = 1.2 s

s = ut + -1 ar-1 Use an equation for constant acceleration.

2
= _!_ X IX 1.2 2
2
= 0.72
The ball falls 0.72 m. So the height of the table is 72 + 8 = 80 cm .

You can apply Newton's second law to any part of the connected system in which all objects are
moving with the same acceleration and in the same direction.

WORKED EXAMPLE 5.5

A crate of mass 500 kg rests on a slope and is attached to a rope that passes over
a smooth pulley. The slope is inclined at 20° to the horizontal. The coefficient of

■
friction between the slope and the crate is 0.1.
What happens to the crate when a force of 2000 N is applied to the vertical part of the rope? ..:::;..___._....::;.;;._
Answer
The pu lley is smooth so the tension in the rope at the point of
contact with the crate is also 2000 N.
If the crate slides up the slope: We are not told whether or not the crate moves
R up the slope, is stationary or moves down the
slope.

5000N
Resolving perpendicular to the slope:
R = 5000 cos 20
= 4700N
The crate is sliding,
Friction is limiting so F = µR.
so F = O.IR
= 470N
Component of weight down slope = 5000 sin 20
= 1710N
Resultant force up slope= 2000 - 470 - 1710
=-180N
This mean s that the acceleration up the slope would be negative,
and since the crate is initially at rest we can conclude that it does
not slide up the slope.
 Cambridge International AS & A Level Mathematics: Mechanics

If the crate slides down the slope: The only difference is that the friction now
R acts up the slope.

As before, R = 5000 cos 20

= 4700N
and F = 0.1 x R = 470 N
and component of weight down slope = 1710 N
Resultant force up slope = 2000 + 470 - 1710
= 760N
This means that the acceleration up the slope would be positive,
and since the crate is initially at rest we can conclude that it does
not slide down the slope.
Therefore, the frictional force is sufficient to allow the crate to
rest on the slope without sliding.
An alternative approach is to calculate the frictional force needed
for equilibrium and compare it to the limiting value of friction .
The component of the weight down the slope is 1710 N and the

■
force up the slope is 2000 N , so the forces are in equilibrium when
the frictional force is 2000 - 1710 = 290 N down the slope.
The limiting (maximum) value of F is µR = 4 70 N .
This is greater than the frictional force needed for equilibrium,
so, in this situation, friction is not limiting and is sufficient to
prevent the crate from sliding.

MODELLING ASSUMPTIONS

In all the problems in this chapter you are making assumptions about the way that objects
are connected.

The mass of a real rod will affect the equation in Newton's second law, but if the mass is
sufficiently small in comparison to the mass of the objects, the effect in the equation is negligible.

If a string moves around a pulley, the mass of the string moving in the different directions
on either side of the pulley will be constantly changing. This would make the problem much
more complicated but, because strings tend to be much lighter than the objects they pull, you
consider the mass to be negligible.

You are assuming the string is inextensible, which means the objects on either side of the
string accelerate at the same rate in the direction of the string. Strings generally do extend
slightly under tension, but the extension is sufficiently small to ignore.

You are assuming the pulley is smooth. This means that the tension in the string on either
side is the same, and also that the string slides over the pulley. With friction, the pulley
itself may rotate, and factors such as the mass of the pulley would affect the motion. This
makes the problem much more complicated.
 Chapter 5: Connected particles

1 In each of the following diagrams, the blocks are at rest and are connected by light strings passing over
smooth pulleys. Any hanging portion of a string is vertical and any other portion is parallel to the surface.
Unless marked otherwise, the surfaces are rough and horizontal. In each case, find the magnitude of the
tension in each string and the magnitude of any frictional force.
a b
4kg 4kg

7 3kg ~ 0

A
3 kg
d
C 4kg....

2
% 8
A bucket of mass 3 kg rests on scaffolding at the top of a building. The scaffolding is 22.5 m above the ground.
The bucket is attached to a rope that passes over a smooth pulley. At the other end of the rope there is another
bucket of mass 3 kg, which initially rests on the ground. The bucket at the top of the building is filled with
6 kg of bricks and is gently released. As this bucket descends, the other bucket rises .
a Find how long it will take the descending bucket to reach the ground.

■
b What modelling assumptions have been made?

3 A light inextensible string is fixed to a point on a ceiling. A box of mass 2 kg hangs from the string. Two light
inextensible strings are attached to the box and hang vertically below the box. A particle of mass 0.4kg hangs
from the lower end of one string and a particle of mass 0.6 kg hangs from the lower end of the other string.
Work out the tension in each string.

4 A block of mass 5 kg hangs from one end of a light inextensible string of length 2 m. The string passes over a
small smooth pu Iley at the edge of a smooth horizontal table of height 70 cm. The other end of the string is
connected to a block of mass 3 kg held at rest on the table. The portion of the string between the pulley and
the 3 kg mass is horizontal and of length 1.5 m. The system is released from rest.
a How long does it take for the 5 kg mass to reach the ground?
b What is the speed of the 3 kg mass when the 5 kg mass hits the ground?
The 3 kg mass continues to slide towards the pulley.
c How long does it take, from when the system is released , for the 3 kg mass to reach the pulley?

5 A mass X, of 1 kg, hangs from one end of a light inextensible string. The string passes over a small , smooth,
fixed pulley and a mass Y, of 0.6 kg, hangs from the other end of the string. Initially X is 0.4 m below the
pulley and Y is 1.8 m below the pulley. The pulley is 3 m above the ground. The system is released from rest.
a Find how long it takes from when the system is released until Y hits the pulley.
When Y hits the pulley, the string breaks.
b Find how long it takes from when the system is released until X hits the ground.
~ 6 A block of mass 4 kg is held on a rough slope that is inclined at 30° to the horizontal. The coefficient of
friction between the slope and the block is 0.2. A light inextensible string is attached to the block and runs
parallel to the slope to pass over a small smooth pulley fixed at the top of the slope. The other end of the
string hangs vertically with a block of mass I kg attached at the other end. T he system is released from rest.
a Work out the tension in the stri ng.
After 1.2 s the block of mass 4 kg reaches the bottom of the slope. The other block has not yet reached the pulley.
b Work out a lower bound for the length of the string, giving your answer to 2 significant figures.

7 A particle of mass 0.3 kg hangs from one end of a light inextensible string. The string passes over a smooth
pulley and a particle of mass 0.5 kg hangs from the other end. A second string is tied to the particle of mass
0.5 kg, and a particle of mass 0.2 kg hangs from this string, so that the particle of mass 0.2 kg hangs vertically
below the particle of mass 0.5 kg. The system is released from rest. Find the tension in each string.

8 Two smooth pulleys are 8 m apart at the same horizontal level. A light inextensible rope 4m 4m
passes over the pulleys and a box of mass 5 kg hangs at each end of the rope. A third box
of mass m kg is attached to the midpoint of the rope and hangs between the pulleys so
that all three boxes are at the same horizontal level.
The total length of the rope is 16 m. Find the value of m.

~ 9 Two smooth pulleys are fixed at the same horizontal level. A light inextensible rope passes over the pulleys
and a box of mass m kg hangs at each end of the rope. A third box of mass km kg is attached to the midpoint
of the rope and hangs between the pulleys so that all three boxes are at the same horizontal level. The portions
of the string that are not vertical make an angle of 30° with the horizontal. The system hangs in equilibrium.
Find the value of k.

10 A mass of 2 kg is held on a rough horizontal table. The coefficient of friction between the table and the
mass is 0.05. The 2 kg mass is attached by a light inextensible string to a mass of 3 kg and by a second light
inextensible string to a mass of 4 kg. The stri ngs pass over smooth pulleys at the edges of the table. The 3 kg
mass hangs on one side of the table and the 4 kg mass hangs on the other side of the table. The system is
released from rest. Find:
a the acceleration of the masses
b the tension in each string.

0 ~ 11 A wedge has two smooth sloping faces; one face makes an angle 30° with the horizontal and the other makes
an angle 60° with the horizontal. A small smooth pulley is fixed at the apex of the wedge. A light inextensible
string passes over the pulley and lies parallel to the faces of the wedge. At each end of the string there is a
particle of mass 0.3kg. The system is released from rest.
3
a Show that the tension in the string is ( J3) N.
4 1+ 3
b Work out the resultant horizontal force on each of the particles.

0 ~ 12 A box of mass 3 kg hangs from a light inextensible string, which passes over a smooth pulley
fixed below a beam and then under a smooth cylinder of mass 4 kg that is free to move.
The other end of the string is fixed to the beam.
The system is released from rest.
a Explain why the magnitude of the acceleration of the cylinder is half the magnitude of the acceleration of
the box.
b Find the acceleration of the box, including its direction.
 Chapter 5: Connected particles

0 C, 13 A horizontal shelf of mass 1kg hangs from four strings. A book of mass 0.2 kg sits on the shelf. Four strings
are attached to the underside of the shelf and a second horizontal shelf of mass 1kg hangs from these strings.
a What modelling assumptions can be made about the strings?
b Find the tension in each of the upper set of strings.
c Find the tension in each of the lower set of strings.
The book is moved to the lower shelf.
d How does this change the tensions in the strings?

EXPLORES.1

What would happen in the situation described in question 12 of Exercise 5B if the

string passed over a second fixed pulley and under a second cylinder of mass 4 kg
before being fixed to the beam? Investigate what happens if the masses are changed.

5.4 Objects in moving lifts (elevators)

When a person travels up or down in a lift, the floor of the lift acts as a connection
between the person and the lift.
person in lift person lift

■
,J I ~
For the system (person in lift), the forces are the weights and the tension in the lift cable.
The forces on the person are their weight and the normal reaction from the floor of the lift.
The forces on the lift are the reaction from the person on the floor (which, by Newton's
third law, is equal and opposite to the normal reaction from the floor on the person), the
weight of the lift and the tension in the lift cable.
When you consider the lift and the person as a single object, the normal reaction forces cancel out.
When the lift is accelerating, the normal reaction from the person on the lift is not the
same as the weight of the person.
Suppose that the tension in the cable is T , the normal reaction is R, the weight of the lift is
W and the weight of the person is w, where these forces are all measured in newtons.
T T

W IV w WR
2
If the lift is accelerating upwards with acceleration am s- , you can apply Newton's second
law to the system:
T - W - w = (M + m)a
 Cambridge International AS & A Level Mathematics: Mechanics

where m is the mass of the person and M is the mass of the lift, both in kg.
You can also apply Newton's second law to the person and the lift separately:
R- w = ma and T - R - W = Ma
You can then calculate the acceleration of the lift, the tension in the cable or the normal
reaction between the person and the floor.
If a is negative it could be because the lift is travelling upwards but slowing down or
because it is travelling downwards and speeding up.
If a is positive it could be because the lift is travelling upwards and speeding up or because
it is travelling downwards and slowing down.

WORKED EXAMPLE 5.6

A woman of mass 50 kg is travelling in a lift of mass 450 kg. The tension in the cable pulling the lift upwards is
5250 N . Calculate the acceleration of the lift.

Answer
lift and woman lift woman

5250 N
5250N
t t R

!
■
Q
4500 N SOO N
~
4500 N R
500

For the system: Apply Newton's second law to the whole system.
5250 - 4500 - 500 = (450 + 50)a
250 = 500a
a= 0.5 ms- 2

To find the acceleration you can work either with the entire system or with the individual
components. In Worked example 5.6 we used the system and in Worked example 5.7 we
show the same idea but using the individual components. To find the reaction forces you
need to consider the individual components.

WORKED EXAMPLE 5.7

A man of mass 80 kg and a woman of mass 70 kg are travelling in a lift of mass 500 kg . The tension in the cable
pulling the lift upwards is 6890 N. Calculate the acceleration of the lift and the reaction forces between the lift
floor and each of the passengers.

Answer
lift man woman
To find the reaction forces we need to use the
6890 N individual components.

Ql SOON
i'
700N

5000N R, + R 2
 Chapter 5: Connected particles

For the lift: 6890 - 5000 - R1 - R 2 = 500a Apply Newton's second law to the individual
For the man : R1 - 800 = 80a components.
For the woman : R 2 - 700 = 70a

1890 - 800 - 700 = 650a Add the equations to eliminate R1 and R2 .

a = 0.6 m s- 2

R1 - 800 = 80 x 0.6 R1 = 848 N reaction from floor on man Substitute a = 0.6 back into the previous
R2 - 700 = 70 x 0.6 R2 = 742 N reaction from floor on woman equations to find R 1 and R2•

1 A crate of mass 20 kg is put into a lift. The lift accelerates upwards at 0.3 m s- 2 . The tension in the lift cable is
5000N.
a Find the contact force between the lift floor and the crate.
b Find the mass of the lift, giving your answer to the nearest kg.
2 A crate of mass 20 kg is put into a lift. The mass of the lift is 300 kg. Find the tension in the lift cable:
a when the lift accelerates upwards at 0.3 m s-2
b when the lift travels at constant speed
c when the lift accelerates downwards at 0.3ms-2 .
~ 3 A man of mass 80 kg stands in a lift. The mass of the lift is 400 kg . The lift starts to travel downwards with an
acceleration of 8 m s- 2 . The tension in the lift cable is kR , where R is the contact force between the man and
the lift floor. Find the value of k .
4 A crate of mass 40 kg is put into a lift. The lift accelerates upwards at 0.4 m s-2 . The mass of the lift is 460 kg.
a Find the tension in the lift cable.
b Find the contact force between the lift floor and the crate.
5 A box of mass 20 kg sits on the floor of a lift. A second box of mass 10 kg sits on top of the first box and a
third box of mass 5 kg sits on top of the second box. When the tension in the lift cable is 4620 N , the lift is
accelerating upwards at 0.5 ms- 2 .
a Work out the mass of the lift.
b Work out the reaction between the floor of the lift and the first box.

c Work out the reaction between the first box and the second box.
d Work out the reaction between the second box and the third box.

~ 6 The mass of a lift is 200 kg and the maximum tension in the lift cable is 2500 N .
a Work out the maximum upwards acceleration of the lift when it is empty.
The lift carries a load of 40 kg. The lift accelerates upwards with the maximum upwards acceleration possible.
b Work out the contact force between the lift floor and the load.
Cambridge International AS & A Level Mathematics: Mechanics - -

7 A crate of mass m kg is put into a lift. The lift accelerates upwards at am s- 2 . The mass of the lift is M kg.
a Find the tension in the lift cable.
b Find the contact force between the lift floor and the crate.

0 8 A box of mass 5 kg sits in a lift. A second box of mass 3 kg sits on the first box. The lift accelerates upwards
with acceleration 0.7 m s-2 .
a Work out the contact force between the two boxes.
The tension in the lift cable is unchanged but the boxes are swapped over, so the first box sits on the
second box.
b Show that this increases the contact force between the boxes.

~ 9 A passenger lift has mass 500 kg. The breaking tension of the cable is 12 000 N. The maximum acceleration of
the lift is 0.75 ms-2 .
a If the lift travels at its maximum acceleration, calculate the maximum mass of the passengers:
when the lift is accelerating upwards
ii when the lift is accelerating downwards.
b Taking the average mass of a person to be 75 kg, what is the maximum number of passengers that should
be allowed to travel in the lift?

~ 10 The tension in a lift cable is 11070 N. The lift is accelerating upwards at am s- 1. A man of weight 800 N stands
in the lift on a set of bathroom scales. The scales suggest that the weight of the man is 820 N. Assuming that
the scales have negligible weight, find the weight of the lift.

~ 11 Two masses, A of 2 kg and B of 3 kg, are connected by a light inextensible string that passes over a small,
smooth, fixed pulley. Initially, the masses are held stationary and are then released.
a Find the acceleration of each mass.
The pulley is fixed to the wall of a lift. The lift is initially stationary. The lift has mass 400 kg and a man of
mass 75 kg is travelling in the lift, along with the pulley system. There is nothing else in the lift. The lift starts
to move upwards, from rest, by a tension in the lift cable of 4992 N.
b Find the acceleration of the lift.
c Find the acceleration of each of A and B, as viewed by a person standing stationary outside the lift.

• Newton's third law states that for every action there is an equal and opposite reaction. This
means that in every interaction there is a pair of forces that have the same magnitude, but which
act in opposing directions.

• When a string passes over a smooth pulley, the magnitude of the tension is unchanged but the
direction can change.
• Newton's second law can be applied to a system of connected objects, either to the entire system
or to individual components of the system, provided they move with the same acceleration and
in the same direction.
 Chapter 5: Connected particles

IMi #·1#i=tJ4iiililMiMiiiii4ihii
1

1 A 10 m long slope makes an angle 30° with the horizontal. A box of mass 8 kg is pulled up the slope by a rope

that is parallel to the slope. The coefficient of friction between the box and the slope is }i. At the top of the
slope, the rope passes over a smooth pulley. A ball of mass 12 kg hangs from the other end of the rope. The ball
is initially 2 m above the ground. The system is released from rest.
a Find how long it will take for the box to travel 1.5 m up the slope. [3]

b Find the tension in the rope. [2]

2 A car is travelling along a straight horizontal road. The car has mass 1800 kg and is towing a trailer of mass 600 kg.
Resistance forces are 30 N on the car and 10 N on the trailer. Find the size and type of force in the tow-bar:
a when the driving force from the engine is 400 N [3]

b when the driving force from the engine is 20 N. [2]

3 A crate of mass 15 kg rests on a platform. The platform has mass 5 kg. It is lowered using a rope. The tension in
the rope is 10 N .
a Find the acceleration of the crate. [21

b Find the contact force between the platform and the crate. [31
4 Particles P and Q, of masses 0.6 kg and 0.4 kg, respectively, are joined by a light inextensible string that
passes over a small, smooth, fixed pulley. The particles are held at rest with the string taut and its straight parts

■
vertical. Initially, both particles are 1.45 m above the ground and 0.25 m below the pulley. The system is released
from rest.
a Find the speed of P when Q reaches the pulley. [4]

The string then breaks and P falls to the ground.

b Find the time from when the system is released to when P hits the ground. [41
~ 5 Particles A and B , each of mass 0.5 kg, are attached to the ends of a light inextensible string. Particle A is held
on a smooth slope inclined at 30° to the horizontal. The string passes over a small smooth pulley at the top of
the slope and particle B hangs vertically below the pulley. Particle A is released and moves up the slope.
a Find the acceleration of particle A up the slope. [3]

Particle B hits the ground after 1.2 s. It then stays on the ground and particle A travels further up the slope.
Particle A does not reach the pulley in the subsequent motion.
b Find the distance travelled by particle A, from when it is released to when it comes to instantaneous rest. [5)
~ 6 Particles A and B , of masses 0.5 kg and 2.5 kg, respectively, are attached to the ends of a light inextensible
string. Particle A is held on a rough horizontal surface with coefficient of friction 0.2. The string passes over
a small smooth pulley at the edge of the surface, at a distance 5 m from particle A. Particle B hangs vertically
below the pulley. Particle A is released and particle B descends 1m to reach the ground. When particle B reaches
the ground, it stays there. Find the time taken from the start until particle A comes to instantaneous rest. [8]
 I

Cambridge International AS & A Level Mathematics: Mechanics :

~ C, 7 A crate of mass 20 kg is pulled vertically upwards using a rope that passes over a first pulley, under a second
pulley and over a third pulley. At the other end of the rope there is a ball of mass 30 kg. Each pulley has mass
m kg. The first and third pulleys are fixed at the same horizontal level and are 4 m apart. The second pulley is
an equal distance from the first and third pulleys and hangs at a distance 2 m below them. It is not fixed, but it
does not move.
a What modelling assumptions need to be made? Which of these assumptions is unlikely to affect the
equilibrium of the second pulley? (3]

b Find the value of m. (5]

® s - - - - - - -- - 4m--- - - - - -
p
•

A light inextensible string of length 5.28 m has particles A and B , of masses 0.25 kg and 0.75 kg respectively,
attached to its ends. Another particle, P, of mass 0.5 kg , is attached to the mid-point of the string. Two small
smooth pulleys Pi and P2 are fixed at opposite ends of a rough horizontal table of length 4 m and height 1m. The
string passes over Pi and P2 with particle A held at rest vertically below Pi, the string taut and B hanging freely
below P2 • Particle P is in contact with the table halfway between Pi and P2 (see diagram). The coefficient of fric-
tion between P and the table is 0.4. Particle A is released and the system starts to move with constant accelera-

■ tion of magnitude am s-2 . The tension in the part AP of the string is TA N and the tension in the part PB of the
string is T 8 N.
Find TA and T8 in terms of a. (3]
ii Show, by considering the motion of P that a= 2. (3]

iii Find the speed of the particles immediately before B reaches the floor. (2]

iv Find the deceleration of P immediately after B reaches the floor. [2]

Cambridge International AS & A Level Mathematics 9709 Paper 42 Q7 June 2014

0 9 Particles A and B, of masses 0.5 kg and 2.5 kg, respectively, are attached to the ends of a light inextensible
string. Particle A is held on a rough slope. The slope is inclined at 30° to the horizontal and the coefficient of
friction between the slope and particle A is 0.3. The string passes over a small smooth pulley at the top of the
slope and particle B hangs vertically below the pulley. The length of the slope is 4 m and the length of the
string is 3 m. Particle B is 1m above the ground. Particle A is released and moves up the slope. When particle B
reaches the ground the string is cut. Show that particle A does not reach the pulley. [11]
@~ 10 Particles A and B, of masses 0.2 kg and 0.45 kg respectively, are connected by a 2.lm
light inextensible string of length 2.8 m. The string passes over a small smooth Ao
pulley at the edge of a rough horizontal surface, which is 2 m above the floor.
Particle A is held in contact with the surface at a distance of 2.1 m from the pulley 2m
and particle B hangs freely (see diagram). The coefficient of friction between A and
the surface is 0.3. Particle A is released and the system begins to move.
Find the acceleration of the particles and show that the speed of B
immediately before it hits the floor is 3.95 m s- 1, correct to 3 significant figures. (7]

ii Given that B remains on the floor, find the speed with which A reaches the pulley. [4]
Cambridge International AS & A Level Mathematics 9709 Pa_per 42 Q6 June 2010
 Chapter 5: Connected particles

@~ 11

Particles A and B , of masses 0.3kg and 0.7 kg respectively, are attached to the ends of a light inextensible
string. Particle A is held at rest on a rough horizontal table with the string passing over a smooth pulley fixed at
the edge of the table. The coefficient of friction between A and the table is 0.2. Particle B hangs vertically below
the pulley at a height of 0.5 m above the floor (see diagram).
The system is released from rest and 0.25 s later the string breaks. A does not
reach the pulley in the subsequent motion. Find
the speed of B immediately before it hits the floor, [9]
ii the total distance travelled by A. [3]
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q7 June 2015
~ 12 A slope is inclined at 30° to the horizontal. A small box A , of mass 2 kg , is held on the slope. Box A is attached
to one end of a light inextensible string. The string passes over a small smooth pulley, P 1, fixed at the top of the
slope and then passes under a small smooth pulley, P 2, fixed near ground level. The other end of the string is
attached to a small box B, of mass 2 kg , at rest on the ground. The ground is horizontal and the portion of the
string between pulley P 2 and box B is horizontal (see diagram) . The coefficient of friction between box A and
the slope is 0.2 and the coefficient of friction between box B and the ground is µ.

■
The distance from the bottom of the slope to A is 1m and the distance from pulley P 2 to B is 0.6 m. Box A is
released and the system begins to move. It takes 1s for box B to reach pulley P 2 .

a Find the speed of box B just before it hits the pulley. [2]
b Show that the tension in the string is 4.14 N, to 3 significant figures. [3]

c Find the value of µ. [3]

When box B hits the pulley, the string breaks.

d Find the speed of box A just before it reaches the bottom of the slope. [6]
In this chapter you will learn how to:
■ use differentiation to calculate velocity when displacement is given as a function of time
■ use differentiation to calculate acceleration when velocity is given as a function of time
■ use integration to find displacement when velocity is given as a function of time
■ use integration to find velocity when acceleration is given as a function of time.
 Chapter 6: General motion in a straight line

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills
Chapter 1 Calculate velocity and displacement 1 A particle travels in a straight line. It has
when acceleration is constant. initial velocity of 8 m s- 1 and a constant
acceleration of 2 m s- 2 .
a Find the speed of the particle after it has
been travelling for 3 s.
b Calculate the distance travelled in the
first 3 s.
C Calculate the time it takes for the particle
to travel 65 m.
Pure Mathematics 1 Differentiate expressions of the 2 y = Sx 3 - 60x + 2
form kx'1, where n may be a fraction
or may be negative. a Find dy and find the coordinates of the
dx
Integrate expressions of the form stationary points of y.
kx"(n -:t -1).
J
b Find y dx and S:
y dx.
Evaluate definite and indefinite
integrals.

How do objects move when acceleration is not constant?

When someone drives a car through a town centre, they will need to brake and accelerate
■
to deal with traffic conditions. You might want to know the speed of the car at any moment
or how quickly the car is speeding up or slowing down.

In Chapter 1 you learnt about displacement, velocity and acceleration and how these are
connected when acceleration is constant, however, for the driver in town, acceleration is
not likely to be constant. For example, the driver may gradually increase the braking force,
so the rate of deceleration gradually increases, as the car comes to a stop at traffic lights.

In this chapter you will learn that if acceleration can be written as a function of time, you
can use calculus (differentiation and integration) to deal with variable acceleration.

Other examples of non-constant acceleration include objects attached to springs,

objects moving in circles, pendulums, and rockets leaving the Earth's surface (where air
resistance, driving force and gravity are all changing during the motion). In many of
these situations the direction of motion is changing, however, in this chapter you will just
consider one-dimensional motion.
I

Cambridge International AS & A Level Mathematics: Mechanics

'

6.1 Velocity as the derivative of displacement with respect to time

On a displacement-time graph the velocity is represented by the gradient of the graph.
Cl REWIND

This is true whether the graph is made of straight lines or curves. Look back to
Chapter 1, Section 1.4
if you need a reminder
EXPLORE 6.1
of displacement- time
graphs.
A car travels in a straight line. The diagram shows the displacement- time graph for
the car as it slows down to approach a red traffic light.
s (m)

100

80

60

40

20

0
2 4 6 8
t (s)

Use the graph to estimate the velocity of the car:

■ a between t
b at t
c
= 2s
between t
= ls and t = 3s

= 4s and t = 9s
d at t = 6s.

f._ REWIND
change in displacement
Average velocity is found by . . . If these changes are small; for
change m time You saw in Pure
example, if the change in displacement is 8s during a time 8t, then the average velocity over Mathematics 1,
Chapter 7, that the
. 11 . . 8s
t h 1s sma time 1s & ' derivative of y with
respect to x gives the
gradient of the graph
As 8t gets smaller, ~s approaches the limit ds, which is the derivative of displacements
ut dt of y against x at a
with respect to time t. We call this the instantaneous velocity or simply the velocity of the point.
object. The velocity is represented by the gradient of a displacement-time graph, whether
or not it is a straight line. If you know the displacement as a function of time, you can
WEB LINK
differentiate with respect to time to find the velocity at any instant.
You may want to
• KEY POINT 6.1 have a go at the
resource Walk-sorting
Velocity is the rate of change of displacement and is the derivative of displacement with respect at the Introducing
to time: Calculus station on
ds the Underground
v=-
dt Mathematics website.
 Chapter 6: General motion in a straight line

WORKED EXAMPLE 6.1

A particle moves in a straight line so that its displacement, s m, at time t s is given by s = t 3 - l 4t. Find an
expression for its velocity at time t.
Answer
s = t3 - 14t Differentiate this with respect to time to get velocity.

)> = -
ds
dt
= (3t 2 - 14)ms- 1 Remember to give units.

WORKED EXAMPLE 6.2

A ball moves in a straight line so that its displacement, s m, at time t sis given by s = 2t 3 - 10t 2 . Find its speed
when t = 2.
Answer

s = 2t 3 - IOt ~ Differentiate this with respect to time to get velocity.

ds
V =-
dt

■
= 6t 2 - 20t

When t = 2, v = 24 - 40 = - 16 Substitute t = 2.

So speed = l 6ms- 1• Speed = Ivelocity I

Remember to give units.

WORKED EXAMPLE 6.3

A particle moves forwards and backwards along a straight line so that its
displacement, s metres from the initial position, at time t seconds is given by
s = 2t 3 - 12t 2 + 18t. Find the distance that it travels in the first 5 s.

Answer
s = 2t 3 - 12t 2 + 18t You can calculate the displacement
when t = 5, but this is not necessarily
D isplacement when t = 5 1s
the same as the distance travelled.
250- 300 + 90 = 40m .
Although the particle moves in a
straight line, it is moving backwards
as well as forwards along the line.

Displacement measures how far an

object is from the start; distance
measures how far it has travelled to
get to that position.
 ) "I ,, ' ,. ,'i __ _
'I t , ,,, : •

Cambridge International AS & A Level Mathematics: Mechanics . · .·. ,~,

·. . ' .- ;,·,,:.' . ·\,. ~

ds Differentiate s with respect to t to

v=-
dt Sometimes an object
get an expression for v.
2 travels back the way
= 6t - 24t + 18
it came. It is useful
Stationary when v =0 Set v = 0 to find any times when to consider when it is
2 the particle is stationary. momentarily at rest
6t - 24t + 18 = 0
(the times when v = 0)
6(t- 1)(t - 3) = 0 These are the times when the as these are the times
t = 1 or t = 3 particle momentarily stops, before when it could change
possibly changing direction. direction.

Displacement when t = 1 is Set t = l and t = 3 in

2- 12 + 18 = 8rn. s = 2t 3 - l2t 2 + l 8t to find the
displacement at each stationary
Displacement when t = 3 is point.
54 - 108 + 54 = 0 m.

Distance travelled = 8 + 8 + 40 Particle travels from s = 0 to s = 8,

= 56m then from s = 8 back to s = 0, and
finally from s = 0 to s = 40.
t=O t= I

■ t=3

DID YOU KNOW?

In 1684 Edmund Halley asked Isaac Newton what orbit would be followed by a body under an
inverse square force. Newton replied immediately that it would be an ellipse and that he could prove
this using his new methods (essentially calculus methods applied to situations from mechanics) .

Halley then encouraged Newton to write up his work and in 1687 Newton published his
Philosophiae Natura/is Principia Mathematica (which is usually called the Principia). In the
Principia Newton analysed the motion of bodies, including orbits, projectiles, pendulums and
objects in free-fall near the surface of the Earth.

Throughout this
C, 1 A particle moves along the x-axis so that its x-coordinate at time t sis given chapter you may be
bys m, where s = 20t - 4. Show that the particle is moving with constant able to check numerical
velocity and find the value of this velocity. derivatives on your
calculator. If you have
2 A particle moves in a straight line so that its displacement, s m from the start
an equation solver
position, at time t s is given by s = 2t 4 + 3t 2 + 10t. Find its velocity when t = 2. on your calculator,
4D 3 A tennis ball travels vertically upwards in a straight line. The displacement of the you may also find
this helpful. However,
ball, measured from the initial position in metres, is modelled as s = - 5t 2 + 20t,
you will need to show
where t is the time from the start, in seconds.
full w_ork!ng in the j
a What modelling assumptions have been made? exammat10n. __J
 Chapter 6: General motion in a straight line

Find the speed of the ball:

b when t =0
c when t = 2.
4 A child on a fairground ride moves in a straight line. The position of the child,
measured from the start, at time t sis given by s = 0.5t 4 - t 2 for O < t < 2.
Find the speed of the child:
a when t =0
b when t =I
c when t = 2.

5 The position of a particle as it moves along a line is modelled as s = 3 + 4t - t 2,

where s is the displacement, in metres, from a fixed point O and t is the time, in
seconds, from the start.
a Show that the particle started 3 m from O.
b Find how far the particle is from O when it is instantaneously at rest.

6 A tennis ball is projected vertically upwards. The vertical displacement of

the ball, in metres, from the point of projection at time t seconds is given by
s=-5t 2 +8t.
a Find the time when the ball returns to its starting point.

7
b Find the displacement when the ball is momentarily stationary.

A small stone is dropped into a lake. The stone descends vertically so that
■
t s after entering the water it is s m below the surface of the water, where
s = 4t 2 - .f513. The stone lands at the bottom of the lake with speed 13 m s- 1•
a Show that the stone takes 2 s to reach the bottom of the lake.
b Work out the depth of the lake at the point where the stone lands.

~ 8 In a drag race, two cars, A and B, line up side by side at the start point.
When the starting flag is waved, both cars are driven as fast as possible in a
straight line. The first car to cross the finish line is the winning car. At time t s
from when the cars start to move, the distance travelled by car A is given by
s = 4t + t 2 . Car A takes 16s to reach the finish line.
a Work out the distance from the start to the finish.
b Find the speed of car A when it crosses the finish line.
At time t s from when the cars start to move, the distance travelled by car B is
given by s = 1.2t 2.
c Find the speed of car B when it crosses the finish line.
d When the winning car crosses the finish line, how far behind it is the other car?

~ 9 A burglar moves along a straight corridor from one door to the next door.
At time t s his distance, s m, from the door of the first room is given by
s = 1.8t 2 - 0.3t 4 . He starts and finishes with speed v = 0 m s- 1• Find the distance
between the two doors.
 Cambridge International AS & A Level Mathematics: Mechanics

0 ~ 10 At time t s after jumping from a plane, the distance fallen by a parachutist is

modelled as s m , where

5t 2 : 0 ,S:[ ,S: 4
s = A✓t + Bt : 4 < t ,s; 25

f
Ct + 30 : 25 < t ,s; 50

A, B and C are constants.

a Explain why A + 2B = 40.
The parachute is opened at t = 4 and the speed of the parachutist is immediately
reduced by x m s- 1.
b Show that 0.25A + B = 40 - x.

At t = 25 the speed of the parachute becomes constant.

c Write down two equations that connect A, Band C.
d Find the value of x.

~ 11 A particle moves forwards and backwards along a straight line. The

displacement of the particle, s m, from its initial position O is given by
s = -0 .6t 4 + 2.4t 3 - 3.6t 2 + 2.4t for O < t < 4, where t sis the time for which the
particle has been travelling.
a Show that the particle starts moving along the line in the positive direction.

■
The particle comes to instantaneous rest at point A, returns to pass through 0
and continues to point B , where the distance OB is the same as the distance OA.
b Find the speed of the particle when it is at B.
(Note: you will need an equation solver for this question. You will not be allowed
an equation solver in the examination.)

0 12 A robot moves along a straight line for 10s. The displacement of the robot, sm ,
from its initial position is given by s = -0.0It 4 + 0.2t 3 - l.32t 2 + 3.2t, where t is
measured in seconds and O < t < 10.
a Show that the robot is stationary when t = 2, t = 5 and t = 8.
b Find the distance that the robot travels in the 10 s.

6.2 Acceleration as the derivative of velocity with respect to time

. .
You learnt m Chapter 1 that average accelerat10n = change in.velocity
.
change m time
. If these changes
a REWIND

Look back to
are small, for example, the change in velocity is 8v during a time bt, then the average Chapter I, Section 1.5
if you need a reminder
. sma11 time
. overt h 1s
acce Ierat10n . . &
1s s: gets sma11 er, &
bv . A s ut bv approac h est h e 1·1m1t
. dt'
dv of velocity- time graphs
and acceleration.
which is the derivative of velocity with respect to time. We call this the instantaneous
acceleration or just the acceleration of the object. This means that the acceleration is
represented by the gradient of a velocity-time graph, whether or not the graph is made up
of straight lines. If you know the velocity as a function of time, you can differentiate with
respect to time to find the acceleration at any instant.
 Chapter 6: General motion in a straight line

Acceleration is a vector quantity, like velocity. Although the magnitude of the velocity is
called speed, there is no term for the magnitude of the acceleration. In this chapter you will
consider only examples of one-dimensional motion along a line. One direction along the
line will be the positive direction and the other will be the negative direction .

• KEY POINT 6.2

Acceleration is the rate of change of velocity and is the derivative of velocity with respect to time.
Acceleration is the second derivative of displacement with respect to time.
dv d 2s
a=dt =dt2

EXPLOREG.2

Positive acceleration means that the velocity is increasing. In each of the following
velocity- time graphs the acceleration is positive. Give a possible description of the The sign of the
motion in each case. What other shapes could a velocity- time graph have for there to velocity determines the
be positive acceleration? direction of travel.

A C If the velocity is
v (m s- 1) v(ms- 1) positive then when it
increases the object
speeds up, but if the

•
velocity is negative
then when it increases
0 I (s) t (s) 0 t (s) t (s)
it becomes less negative
and the object slows
down but in the
negative direction.
Negative acceleration (deceleration) means that the velocity is decreasing. In each lf the velocity is
of the following velocity-time graphs the acceleration is negative. Give a possible positive then when it
description of the motion in each case. What other shapes could a velocity- time decreases the object
graph have for there to be negative acceleration? slows down, but if the
velocity is negative
A B C
v (ms- 1) v(ms- 1) v(ms- 1) then when it decreases
it becomes more
negative and the object
speeds up but in the
negative direction .
I (s) 0 I (s) 0

I.

11
f.,
>
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 6.4

A particle moves in a straight line so that its velocity, v m s- 1, at time t s is given by

V = t2 - 4t.
Find an expression for its acceleration at time t.

Answer
V = t 2 - 4t Differentiate this with respect to time to get
acceleration.
dv
a=-
dt
= (2t- 4) m s- 2 Remember to give units.

WORKED EXAMPLE 6.5

A car moves in a straight line so that its velocity, v m s- 1, at time t s is given by v = 5t 2 - t 3 , for O < t < 4. Find its
acceleration when t = 2.

Answer

■
V = 5t 2 - t3 Differentiate this with respect to time to get
acceleration.
dv
Cl = -
dt
= lOt - 3t 2
When t = 2: Substitute t = 2.
a= 20 - 12
=8
So acceleration = 8 m s- 2. Remember to give units.

WORKED EXAMPLE 6.6

A particle moves in a straight line so that its displacement, sm, at time ts (0 ,s; t ,s; 10) is given by
s =..!..t 3 2
3 -6t +15t .
a Sketch the shape of the velocity- time graph for the particle.

b Hence, find the maximum speed of the particle.

 Chapter 6: General motion in a straight line

Answer

a s = _!_ t 3 -
6t 2 + I St First differentiate to find v as a function oft.
3
ds
V =-
dt
=t 2 - 12t + l5
The velocity- time graph is a parabola.

15 You could use a graphic calculator or a graph-

drawing package to check the shape of the
graph.
-t---- - - - - - - - - - - - - - - - ~ t (s)
0 10

b Speed is the magnitude of v, so to fi nd the maximum From the graph you can see that the min imum
speed we need to find the maximum positive va lue value occurs at the turning point and the
and the minimum negative value of v. maximum value occurs when t = 0.

v =t 2 - l 2t+l5 Differentiate again to find the acceleration.

a =-
dv
dt
= 2t - 12
■
dv
- = 0 when 2t - 12 = 0 so t = 6. This is the time when the velocity-time graph
dt
has its turning point.
Alternatively, complete the square for t he quadratic to get

V = (t -6)2 - 2J

When t = 6: This is v at the minimum turning point on the

v = 36-72+15 graph. The minimum value of vis - 21 (and the
= -21 speed at this point is 21 m s- 1).

When t = 0: These are the values of v at the end points of

v = 0 - 0+ 15 the graph. (We only really need to work out the
value when t = 0 because we can see from the
= 15 graph this is where the velocity is greatest.)
When t = 10:
The maximum value of v is 15 (and the speed at
V = 100 - [ 20 + J 5
this time is 15 m s- 1).
=-5

Hence, the maximum speed= 2lms- 1• Speed = Ivelocity I

 Cambridge International AS & A Level Mathematics: Mechanics

1 A particle moves in a straight line so that its velocity, v m s- 1, at time t s (t > 0) is given by v = 5t 2 - 20.
Find the time when the particle is stationary.

2 A particle moves in a straight line so that its velocity, v m s- 1, at time t s is given by v = t 3 + 3t 2 - 8t + 1. Find
its acceleration when t = 2.

8 3 A tennis ball travels vertically upwards in a straight line. The velocity of the ball in the upwards direction,
vms- 1, at time ts is given by v = 20 - lOt.
a Find the acceleration of the ball.
b Interpret your result.

4 A car moves in a straight line. The velocity of the car, v m s- 1, at time t sis given by v = St+ 0.5t 2 for 0 < t < 2.
Find the acceleration of the car:
a when t =0
b when t =1
c when t = 2.
0 5 A boy runs in a straight line. The velocity of the boy, vms- 1, at time ts is given by:

2 + 3t O~ t ~ I

■
6 - t2 1 ~ t ~ 1.2
v= for 0 ~ t ~ T
5-~t·
30 . 12~t
. ~T

The boy has velocity 0 ms- 1 at time T.

a Work out the value of le.
b Show that T is just under 14 s.
c Work out the average acceleration of the boy over the period 0- T.
d Show that there is no time at which the acceleration of the boy is the same as his average acceleration.

6 The motion of a cat, moving along a straight line, is modelled as s = 6t - t 2 for small values oft, where sis
measured in metres and t in seconds.
a Find an expression for the velocity of the cat as a function of time.
b Describe the motion of the cat.
c When does the cat come to momentary rest?
d Find the acceleration of the cat.

7 A particle moves forwards and backwards along a straight line. The velocity of the particle , vms- 1, at time ts
is given by v = -t 3 + 75t for 0 < t < 10. Find the maximum speed of the particle.
 Chapter 6: General motion in a straight line

8 A robot moves along a straight line for 3 s. The displacement of the robot, s m, from its initial position
is given by s = At 3 + Bt 2 + Ct for constants A, Band C, where t is measured in seconds and O < t < 3.
The robot starts with velocity 2 m s- 1. It travels 6 m before coming to rest at time t = 3. Work out the
values of A, B and C.

G 9 A particle moves along a straight line. The displacement of the particle, s m, at time t s is given by
s = 4t 2 - 0.25t 3 for O < t < 12.
a Find the maximum speed of the particle.
b Work out the difference between the time when the speed is greatest and the time when the speed is least.

10 A ball moves in a straight line. The velocity of the ball, v m s- 1, at time t sis given by v = 5 + 4t - t 2 for
0 < t < 5.
a Write down the initial velocity of the ball.
b Work out the maximum velocity of the ball.
c Find the average acceleration between the start and the end of the motion.

11 A train is travelling in a straight line. Alice is sitting on the train and is using her mobile phone, which is being
tracked. The position of the phone, measured from when the tracking began, is given by s = t 2 - 2t 3 + 75t,
where s is measured in km and t is measured in hours . The phone is tracked for 2 hours, so O < t < 2. Initially
the train speeds up but then it slows down again.
a After how long is the train travelling at its fastest speed?

■
b Find the maximum velocity.
c Find how far the train travels before it starts to slow down.

6.3 Displacement as the integral of velocity with respect to time

You can differentiate displacement (as a function of time) to find velocity. Reversing this
means that if you integrate velocity (with respect to time) you will get a function for the
displacement.

Displacement measures how far the object is from an origin. In this chapter the origin will You know that
be at the initial position for the motion, unless a question states otherwise. This means integrating a function
that the displacement, s, will usually be O when t = 0 (and the constant of integration will gives the area under
its graph and that the
usually be 0, unless the velocity function is made up of different pieces).
area under a velocity-
time graph gives the
displacement. So,
integrating velocity
V = ds
dt so s = f V dt
with respect to time
will give displacement.
I

Cambridge International AS & A Level Mathematics: Mechanics

This is the velocity- time graph for a particle:

v (ms- 1)

You can interpret the motion as follows:

• The particle starts at A with positive velocity.
• Velocity is initially decreasing so the particle is slowing down. However, the velocity is
positive so displacement is increasing and the particle is moving away from the starting
point.
• When the graph crosses the horizontal axis (at B) the particle has velocity 0 m s- 1 and is
momentarily at rest.
• The particle's velocity continues to decrease and is now negative, so it is now travelling in
the opposite direction (back towards where it started from).
• The velocity decreases to a minimum (at C) and then starts to increase, but remains
negative. This means that the particle continues to travel in the negative direction

■
(towards the start point) but is slowing down.
• When the graph again crosses the horizontal axis (at D), the particle is once more
momentarily at rest before it changes direction. The velocity continues to increase, but
now in a positive direction.

Suppose that the equation for the velocity is:

V = t2 - l 3t + 22 = (t - 2)(t - 11)
then the graph crosses the horizontal axis ( v = 0) when t =2 and when t = 11.
The particle starts with velocity 22 m s- 1• For 0 ,;;; t ,;;; 2 the particle moves in the positive
direction, for 2 ,;;; t ,;;; 11 it moves in the negative direction, and for t ;:a, 11 it moves in the
positive direction again.
To find the displacement from the starting point, you integrate the equation for v.

f f 1
s = V dt = (t 2 - 13t + 22) dt = t 3 - } t 2 + 22t ½
(the constant of integration will be 0 because s = 0 when t = 0).
The following table shows the displacement for some values oft .

.,t($)t: 0 1 2 3 4 5 6 7 8 9 10 11 12

i~ t 0 15 .8 20.7 16.5 5.3 - 10.8 -30.0 - 50.2 - 69.3 -85.5 - 96.7 - 101 -96.0
 Chapter 6: General motion in a straight line

You can see that the displacement increases as the particle travels away from s=O t=2
the start for the first 2 seconds. It then decreases as the particle returns the way
t = ll ( [ _ _ - - - - - -
A:c
s•=~O- -
_!) v=O
s = 20.7
it came and passes through the start point (when displacement is zero). The
v= 0 D
displacement continues to decrease until the particle is a distance of 101 from s = - 101
the start, but in the negative direction. The direction then changes again as the
particle travels back towards the start.

Displacement and distance travelled are not necessarily the same thing. The displacement
Unless you are told
at time t is the position of the object from the origin at that time, but the distance travelled
otherwise, in work
at time t is the sum of the distances travelled in the positive and negative directions up
involving calculus
to that time. For example, when t = 3 the displacement of the particle is 16.5. However, the displacement
the object has travelled 20.7 in the positive direction and 20.7 - 16.5 = 4.2 in the negative will be measured
direction, so the total distance travelled is 20. 7 + 4.2 = 24.9. from the initial
position, so s will be
The distance travelled is the area between the curve and the horizontal axis, but in this
0 when t = 0.
example some of the curve is below the horizontal axis. To find the total distance travelled
we need to find the times when the graph crosses the axis and then integrate the parts
above and below the axis separately to find the distances travelled in the positive and
negative directions. The total distance travelled is the sum of these distances.

WORKED EXAMPLE 6.7

A particle moves in a straight line so that its velocity, v m s- 1, at time t sis given by v = t3 - 4t. Find an expression

■
for its displacement from the initial position at time t.

Answer
V = t3 - 4t Integrate this with respect to time to get
displacement.

s = f (t 3 - 4t) dt

= .!_ !4 - 2! 2 +C
4
Displacement is measured from the initial
But s = 0 when t = 0, so c = 0. position, so s = 0 when t = 0.

s = .!_ t 4 - 2t 2 m
4
Alternatively:

s= t( t 3 - 4t) dt
Here we are using t as a variable within the
integration and also as a constant in the limit.

=[ ¾ 1
4
-2t2 I
= _!_ r• - 2t 2 m
4
 - . .

Cambridge International AS & A Level Mathematics: Mechanics · · ·

. -, . = " l '-

WORKED EXAMPLE 6 .8

A particle moves in a straight line so that its velocity, v m s- 1, at time t s is given by v = t 3 - St 2 .

a Find the displacement of the particle after 5 s.

b Sketch the velocity- time graph for O ~ t ~ 10.

c Work out the distance that the particle travels in the first 10 s.

d Find the time when the particle passes through the start position.

Answer

a v = t3 - St 2 Integrate this with respect to time to get

displacement.

In the first 5 seconds the displacement is

The limits are t =0 and t = S.
s = s: (t 3
- St 2
) dt
5
=[ 41 t4 - 35 t3 ] 0
= - 52.l m Displacement can be positive or negative.

■ b The particle is at instantaneous rest when t 3 - St2 = 0.

Bring out common factor t 2 .
t = 0 corresponds to the start of the motion.
t = 0 or t = S
Hence, it is at instantaneous rest when t = S.

t = 0 is a repeated root so the graph has a

stationary point at t = 0.
vis negative for small values of t.
The graph is part of a cubic curve.

0 10 t (s)
 Chapter 6: General motion in a straight line

c The distance travelled is the distance travelled

from t = 0 to t = 5, added to the distance The particle is momentarily at rest when t = 5.
travelled from t = 5 to t = JO.

f(
0
t
3
- St
2
) dt = [ ¾t 4 - %t 3 I From t = 0 to t = 5 the particle moves
52.08 m (in the negative direction).

= -52.08
10

f(
5
t
3
- St" ) dt = [ ¾t 4 _ %r3 l
10
From t = 5 to t = 10 the particle moves
885.42 m (in the positive direction).

= 885.42

Total distance travelled = 52.08 + 885.42

= 938111

Alternative method :

s = f (t 3 - 5t 2 ) dt

= _!_ t4 _ l {3 +C s = 0 when t = 0, so c = 0.
4 3
= _!_ r4 _ l r3

■
4 3

When t = 5, s = - 52.08. Particle moves 52.08 m in the negative

direction, then turns and travels through the
When t = 10, s = 833.33 . start position to finish 833.33 m from the start
in the positive direction.

Total distance travelled = 52.08 + 52.08 + 833.33 Particle travels 52.08 twice, but this rounds up
= 938m to 104.17 when exact values are used.

d Displacement = 0 when ¾t %13 = O.

4
- Sets = 0.

Bring out common factor t 3 •

121 t 3 (3t - 20) =0

t = 0 or -20
3
Particle passes through start position after 6f seconds.
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 6.9

A circus performer rides a unicycle along a stretched tightrope. The tightrope is modelled as a straight horizontal
line and the velocity, v ms- 1, is modelled as:

V = I for O:s; t ,s; 2

V = 2t - 3 for 2 :s; t ,s; 4

v =7-Jt for 4 :s; t ,s; T

where t is the time, in seconds, from when the performer starts to cycle along the tightrope.

The performer stops at time T s, having just reached the other end of the tightrope.

a Find the value of T.

b Calculate how far the performer cycles.

Answer
a 7 - ../T=0 v =0 at time T.

Performer stops after 49 s.

T = 49

b For 0 < t < 2: s = t + c1 Integrate v with respect to t.

■
s is measured from the start of the tightrope.
s =0 when t = 0, so c1 = 0.

Hence, s = t.

So when t = 2s, the distance cycled is s = 2 m .

For 2 < t < 4: s = t2 - 3t + c2 Integrate 2t - 3 with respect to t.

s = 2 when t = 2, so c2 = 4.

Hence, s = t2 - 3t + 4.

So when t = 4s, the total distance cycled is s = 8 m.

2 I-
For 4 < t < 49: s = 7t - - t .) + c3 Integrate 7 - t 0·5 with respect to t.
3

s =8 when t = 4, so c3 = -14 t-
2
Hence, s = 7t - - tl. 5 -14 f
3

So when t = 49 s the total distance cycled is s = 99 ½m.

 Chapter 6: General motion in a straight line

Alternatively, we ca n integrate each of the expressions fo r

v with li mits , and add the dista nces obtained.

J1dt = [t ]~
0
=2
4

f (2t - 3) dt = [ t2 - 3t J:
2
=6
49

f
4
(7 - t
05
) dt = [ 7t -¾ t
15
1
49

= 91t
The total distance travelled= 2 + 6 + 91 ½
= 99fm

WORKED EXAMPLE 6,10

A particle moves in a straight line so that its velocity, v m s- 1, at time t seconds after it starts to move is given by
v = u + at, where u and a are constants. Find the displacement after t s. ■
Answer
V = U + at Integrate this with respect to time to get displacement.

s = f
(u + at) dt u and a are constants.

= ut + .!. at 2 + c
2
But s = 0 when t = 0, so c = 0.

I 1 You should recognise this as one of the constant

Hence, s = ut +
2 ea- . acceleration formulae from Chapter 1.
 Cambridge International AS & A Level Mathematics: Mechanics

1 A particle moves in a straight line so that its velocity, v m s- 1, at time t s (t > 0) is given by v = 5t 2 - 20. Find
the displacement of the particle when it is stationary.

2 A particle moves in a straight line so that its velocity, v m s- 1, at time t sis given by v = t 3 + 3t 2 - St+ 1. Find
its displacement when t = 3.

3 A tennis ball is hit vertically upwards. The upward velocity, v m s- 1, of the ball at time t s is given by
v = 20 - lOt. Find the upward displacement of the ball, from the initial position, when v = 0.

4 A speed skater moves in a straight line with velocity v m s- 1 at time t s, given by v = 2t + t 2, for O ,e:;; t ,e:;; 2. Find
the displacement of the skater:
a when t =0
b when t =1
c when t = 2.
5 A small stone is dropped into a well. It falls down the well, from rest, with no resistance for 2 s. It then hits the
surface of the water and continues to fall vertically through the water until it reaches the bottom of the well.
In the water the downwards velocity of the stone, v m s- 1, is given by v = 20 - t, where t is the time, in seconds,
measured from when the stone hits the surface of the water. The stone takes 2.5 sin total to reach the bottom
of the well.

■
a Calculate the depth of the well.
b If air resistance is taken into account, would you expect the depth of the well to be greater or smaller than
your answer from part a?

C, 6 A ball bearing is fired vertically upwards in a straight line through a tub of butter. The upward velocity of the
ball bearing is given by v = 13 - lOt - 3t 2 cms- 1, where t is the time from when it was fired upwards.

a Find the time when the ball bearing comes momentarily to rest.

b Find how far the ball bearing has travelled upwards at this time.
The ball bearing then falls downwards through the hole it has made in the butter. The downward velocity of
the ball bearing is given by v = lOT cms-1, where T is the time from when it was momentarily at rest.
c Find the time that the ball bearing takes (from when it was momentarily at rest) to fall to its original
position.
d What assumptions have been made in the model used in part c, and how could the model be improved?

7 A particle moves on a straight line. The velocity of the particle, v m s- 1, at time ts is given by v = -t 3 + 9t m s- 1
for O < t < 5.
a Find the displacement of the particle, from its original position, when t = 5.
b Work out the distance that the particle travels from t = 0 to t = 5.

8 A car moves in a straight line with velocity v m s- 1 at time t s, given by v = 16 - 0.5tl. 5 + t for O ,e:;; t ,e:;; 25.
a Find the displacement of the car, from its original position, when t = 25.
b Work out the distance that the car travels from t = 0 to t = 25. (Note: you will need an equation solver for
this question. In the examination you will only be asked to solve the equations that can be done using
algebraic methods.)
 Chapter 6: General motion in a straight line

0 9 A ball rolls forwards and backwards in a long straight tube. The velocity, v m s- 1, at time t s after measurement
starts is given by:
for 0 ~ t ~ 25
V = A- t for t? 25
for some constant A.
a Show that A = 36.
b Find the distance from the start when the ball changes direction.

10 A particle moves in a straight line, starting from rest. At time t s after the start, the velocity, v m s- 1, of the
particle is given by:
V = 2t for O ~ t ~ 4
V = 9 - (t - 5) 2 for 4 ~ t ~ 10
a Find the maximum speed of the particle.

b Work out the distance that the particle moves in the time interval O~ t ~ 10.

~ 11 Two cars travel towards one another on the two sides of a long straight road. They start 300 m apart and each
car stops when its speed is Om s- 1• The time, in seconds, after the first car starts to move is t . The velocity,
v1 m s- 1, of the first car is given by v1 = 7.5t - 0.5t 2 • The second car starts 2 s after the first, at t = 2. The
velocity, v2 m s- 1, of the second car is given by v2 = (t - 2) 2 - 16.
a Write down the initial speed of each car.

■
b How far does the first car travel?

c How far does the second car travel?

d For what value oft are the cars alongside one another? (Note: you will need an equation solver for this
question. You will not be allowed an equation solver in the examination.)

12 A car travels in a straight line, starting and finishing at rest. At time t s after the start, the velocity of the car is
modelled as:
V =t for 0 ~ t~2

v = kt - 0.055(t - 2) 2 for 2 ~ t ~ T
a Find the value of k.

b Show that there is no change in the acceleration of the car at t = 2.

c Find the maximum velocity of the car during its journey.
d Find the time, T s, at which the car stops.

e Work out the distance that the car travels from the start to the end of the journey.
 I m ay
ible, so
Cambridge International AS & A Level Mathematics: Mechanics •lved.

city and
0

6.4 Velocity as the integral of acceleration with respect to ti r,------'

In Section 6.2 you saw that you can differentiate velocity as a function of tim (
acceleration. Reversing this means that if you integrate acceleration with res~
you will get a function for the velocity.
.--- - - - -

a= -d v so v =
dt
f a dt

------------------------------1:h respect to ti
The object is not necessarily at rest when t = 0, so you need to include a constar
integration. If you know the velocity at some time (which may bet= 0 or some o
can use this to find the constant of integration. An alternative strategy is to find
in the velocity by integrating (between, say, t = 0 and a general time t) and addin 2 m s- 1_
the velocity at the beginning of the interval. Both of these approaches are demo
Worked example 6.11.

MODELLING ASSUMPTIONS

8.
When acceleration is not constant, is it reasonable to assume the displacer
velocity and acceleration follow a neat formula?

Forces may vary according to the speed or position of an object, or accor y

t

■
time. When used with Newton's second law, this creates an equation that
one or more of these variables with acceleration. In many cases this sort o
can be solved by integration, so the displacement, velocity and acceleratio
have a neat formula . However, the integration of some functions is not po 0
sometimes you need to make approximations to allow the problem to be s
1
In this chapter you have only looked at the formulae for displacement, vel~
acceleration, not at the situations they come from. 111e

WORKED EXAMPLE 6.11

A particle moves in a straight line so that its acceleration, am s-2 , at time

velocity 2 ms- 1• Find an expression for its velocity at time t.

Answer

a= 8 - 4t Integrate this w f
,to

v = I (8 - 41) dt
ther time) you
the change
= 8t - 2t 2 + C gthis to
nstratedin
v = 2 when t = 0, so c = 2. Initial velocity

So v = 8t - 2t 2 + 2.

nent,

ling to
·elates
 Cambridge International AS & A Level Mathematics: Mechanics
'

EXPLORE 6.3 WEB LINK

You may want to have

A particle moves in a straight line so that its acceleration at time t seconds is am s-2, a go at the resource
where a is constant. The initial velocity of the particle is ums- 1• Thinking constantly
at the Calculus meets
Use calculus to find the velocity of the particle as a function oft.
functions station on
the Underground
Mathematics website.

1 A particle moves in a straight line so that its acceleration, am s- 2 , at time t s is given by a = 2t 4 + 3t 2 - 12t.
The particle starts from rest. Find its speed when t = 2.

2 A particle moves in a straight line so that its acceleration at time t s is given by a = Wt - 4 m s-2 . The initial
velocity of the particle is 15 m s- 1• Find the minimum velocity of the particle in the subsequent motion.

3 A body moves in a straight line so that its acceleration, am s- 2 , at time t sis given by a = 2t 3 + 6t 2 - 18t. The
body starts with velocity 5 m s- 1•
a Find the velocity when t = 3.
b Find the displacement when t = 3.
c When t = 3, is the body travelling towards its original position or away from it?

■ 0 4 A bird moves in a straight line from point A to point B and back to point A .
The bird has speed 5 m s- 1 when it starts and moves with acceleration, given by a =
1
~ (9t 2 - 32t - 10). At
point B the bird has velocity 0 ms- 1.
a Show that the bird takes 2s to travel from A to B.
b Find the distance from A to B.

c Show that the bird returns to A after about 4.25 s.

d Find the speed of the bird when it returns to A.

~ 5 A block slides down a sloped surface with a varying coefficient of friction. The acceleration, am s-2, of the
block (measured down the surface) is given by a = 0.0l(lOt - 3), where t is the time in seconds. At t = 0 the
block is at rest at the top of the surface. The block reaches the bottom of the surface with speed 2.24 m s- 1.
How far does the block travel down the sloped surface?

6 A ball moves with acceleration given by a = 0.01( 4t + 3t 0·5 ), where t is the time, in seconds. At t = 1 the ball is
moving with velocity 0.5 m s- 1• Find the displacement of the ball between t = 0 and t = 4.

0 7 A robot moves in a straight line with acceleration ams-2 at time ts, given by a= 40(t-1) 3 . The minimum
velocity of the robot in the subsequent motion is 0 m s- 1 (the velocity is never negative).
a Show that the robot is stationary (v = 0) when t = l.
b Find the displacement of the robot, measured from the initial position, when the robot is stationary.

8 A particle starts at the origin and moves along the x-axis. The acceleration of the particle in the direction of
the positive x-axis is a = 6t - c for some constant c. The particle is initially stationary and it is stationary
again when it is at the point with x-coordinate = -4. Find the value of c.
 Chapter 6: General motion in a straight line

9 A goods train starts from rest at point A and moves along a straight track. The train moves with acceleration
= 0.1t 2 ( 6 - t) for 0 ~ t < 6. It then moves at constant velocity for 6 ~ t < 156
am s- 2 at time t s, given by a
before decelerating uniformly to stop at point B at t = 165. Calculate the distance from A to B.

10 Two cars are travelling towards one another on the two sides of a long straight road. Each car stops when its
speed is 0 m s- 1. The time, in seconds, after the first car starts to move is t. The acceleration, a 1 m s-2 , of the
first car is given by a 1 = 5 - 2t. The maximum velocity that the first car achieves is 26.0lms- 1.
a Work out the initial velocity of the first car.

b Find the time when the first car stops moving.

The velocity, v2 ms- 1, of the second car is given by v2 = t2 - 16.
c How far does the second car travel?
Initially the cars are 200 m apart.
d Show that the cars stop before they meet.

~ 11 An ice hockey player hits the puck so that it moves across the ice in a horizontal straight line with acceleration
ams-2 at time ts, where a = -0.03t 2 . The initial speed of the puck, along the direction of motion, is 40ms- 1.
a Find the distance that the puck travels in the first 2 seconds (between t = 0 and t = 2).

b Find the speed of the puck after 2 seconds .

When t = 2 the puck is stopped by an opposing player. This player then hits the puck back the way it came,
giving it an initial speed of 30 m s- 1• The acceleration of the puck, in its direction of travel, is still given by
a = -0.03t 2 . The puck returns to its original starting point.

~ «D
c Find, to 3 significant figures , how long it takes for the puck to return to its original starting point.
12 A girl bowls a ball along a straight and horizontal skittle alley. The forces acting on the ball are its weight, the
normal contact force , friction and air resistance. The coefficient of friction between the ball and the surface of
■
the skittle alley is 0.01. The girl models the air resistance, in newtons , as m(0.9 - I.St), where m is the mass of
the ball, in kg, and t is the time, ins.
a Show that the velocity of the ball along the skittle alley, v m s- 1, is given by v = 0. 7 5t 2 - t + C for some
constant C.
The initial velocity of the ball is 8 m s- 1• The skittle alley is 7.25 m long and the ball reaches the end of the
skittle alley with velocity 2 m s- 1•
b Show that the ball takes just over 2.3 s to reach the end of the skittle alley.
(Note: you will need an equation solver for this question. You will not be allowed an equation solver in the
examination.)
c Why is the model for air resistance unreasonable?
 ds
• v= -
dt
dv d 2s
• a=-= --
dt dt 2

• s =IV dt
• =Ja dt
v

differentiate with respect to t differentiate with respect to t

displacement velocity acceleration

s V a

integrate with respect to t integrate with respect to t

• You may be able to check numerical differentiation and numerical integrals on your calculator.
If you have an equation solver on your calculator, you may also find this helpful. However, you
will need to show full working in the examination.
 Chapter 6: General motion in a straight line

liii·l·1i=tJiiiiiilMiMiiiiiMHii
1 A woman on a sledge moves in a straight line across horizontal ice. Her initial velocity is 2 m s- 1. Throughout
the journey her acceleration is given by a = - 0.0l t m s- 2, where t is the time from the start, in seconds. Find the
distance that she travels before coming to rest. [4]
2 A particle moves on a straight line, starting from rest at the point 0. It travels from Oto A with constant
acceleration 0.2 m s- 2, taking 16 s to reach A . The acceleration of the particle then changes so that the velocity,
in ms- 1, is given by v = t 0·5 - 0.8 fort> 16, where t is the time, in seconds, from the start of the motion.
a Find the acceleration of the particle immediately after passing through A. [1]
b Find the distance travelled from t = 0 to t = 36. [3]

3 A particle, P, starts from rest at a point O and travels in a horizontal straight line. For 0 < t < 20, where t is the
time ins, the velocity of P in m s- 1 is given by v = 1.2t - 0.03t 2 . When t = 20, P collides with another particle.
After the collision the direction of travel of P is reversed. For 20 < t < 30, the velocity of P in m s- 1 is given by
v = 0.3t - 9. The particle comes to rest and stops when t = 30.
a Find the speed of P immediately before the collision and immediately after the collision. [2]

b Find the total distance travelled by the particle. [2]

4 A sledge moves down a slope in a straight line. At time t s the displacement of the sledge from the start is s m,
where s = 0.4t 2 for 0 ,s; t < 10 and s = ( 7t - l~O - 20) for 10 ,s; t ,s; 50 .
a Find maximum velocity of the sledge. [3]

5
b Show that the acceleration instantaneously reduces by 1m s-2 at t = 10.
A particle travels in a tube, starting from rest. The particle does not come to rest again in the subsequent
motion. At time t s, the particle has acceleration am s-2, given by a = 9000t until it reaches a velocity of
[2]
■
28125 m s- 1, along the direction of the tube. How far does the particle travel while it is accelerating? [6)
~ 6 A particle moves in a straight line, starting at time t = 0 and continuing until it comes to rest. While it is
moving the particle has acceleration am s- 2, in the positive direction along the line. This acceleration is given by
a = 0.1 - 0.Olt, where t is the time from the start, in seconds. The particle starts with speed 4ms- 1 and finishes
with speed 0 m s- 1•
a Find the maximum speed of the particle. [4)

b Find the time when the particle comes to rest. [2)

4%, 0 7 A car is moving in a straight line. The acceleration, am s- 2, at time t s after the car starts to move is modelled as
a = A(l + 4t) for 0 ,s; t < I and a = B ( 30 - ~~) for 1 ,s; t ,s; 5, where A and Bare constants.
a Show that A = 4B . 121
At t = 5 the velocity of the car is 31.8 m s- 1•
b Show that A = 1. [2)

c Work out the distance travelled in the time interval 0 ,s; t ,s; 5. [2)
d By considering the acceleration- time graph at t = I, criticise the model. [l]
@O 8 A particle P moves on a straight line, starting from rest at a point O of the line. The time after P starts to move
is t s, and the particle moves along the line with constant acceleration_!_ m s-2 until it passes through a point A
2 4
at time t = 8. After passing through A the velocity of P is _!_ t 3 m s- 1.
2
 Cambridge International AS & A Level Mathematics: Mechanics

Find the acceleration of P immediately after it passes through A. Hence show that the acceleration of P
decreases by ~ m s- 2 as it passes through A. [2]
1
ii Find the distance moved by P from t = 0 to t = 27 . [3]

Cambridge International AS & A Level Mathematics 9709 Paper 42 Q4 June 2014

~ 9 A hockey ball is hit so that it moves in a horizontal straight line with acceleration am s- 2, along the direction of
travel; a= -0 .6t, where t is the time from when th'e ball was hit, in seconds. The initial speed of the ball is 14ms- 1•
a Find the speed of the ball when it has travelled 57.5 m. 141
b Find the distance that the ball has travelled when the ball is first momentarily stationary. 14]
c Find the value oft when the ball has travelled 40 m. 12]
@ 10 Two particles A and B start to move at the same instant from a point 0. The particles move in the same
direction along the same straight line. The acceleration of A at time t s after starting to move is am s-2, where
a = 0.05 - 0.0002t.
Find A's velocity when t = 200 and when t = 500. 141
B moves with constant acceleration for the first 200 s and has the same velocity as A wheJJ- t = 200. B moves
with constant reta rda tion from t = 200 to t = 500 and has the same velocity as A when t = 500.
ii Find the distance between A and B when t = 500 . [61

■ Cambridge International A S & A L evel Mathematics 9709 Paper 41 Q6 June 2015

@ 11 A vehicle is moving in a straight line. The velocity v m s- 1 at time t s after the vehicle starts is given by
v = A(t - 0.05t 2 ) for 0 ,s; t ,s; 15 and v =~ for t ;;;, 15, where A and Bare constants. The distance travelled by
t
the vehicle between t = 0 and t = 15 is 225 m.
Find the value of A and show that B = 3375. [5]
ii Find an expression in terms oft for the total distance travelled by the vehicle when t ?e 15. [3]
iii Find the speed of the vehicle when it has travelled a total distance of 315 m . [3]

Cambridge International AS & A Level Mathematics 9709 Paper 42 Q7 June 2010

0 ~ 12 Two walkers, P and Q, travel along a straight track ABC . Both walkers start from point A at time t = 0 s and
pass through point Bat time t = 10s. They both finish at point C. P starts from point A with speed 2 ms- 1 and
accelerates with constant acceleration 0.1 m s- 2 until reaching point B.
a Show that the distance from A to B is 25 m. [3]
b Find the speed of P on reaching point B . [2]
Q starts from point A and moves with speed v1 m s- 1, given by v1 = 0.003t 2 + 0.06t + k. When Q passes
through point B both walkers have the same speed.
c Find the value of the constant k. [3]
P moves from point B to point C with speed v2 m s- 1, given by v2 =4- 0.1 t , and comes to rest as C is
reached.
d Show that the distance from A to C is 70 m. [4]
Q moves from point B to point C with speed v3 m s- 1, given by v3 = 0.4t - 0.01 t 2 .
e Show that Q reaches point C first. [3]
 Cross-topic review exercise 2

1 Two particles, A and B , are attached to the ends of a light inextensible string, which passes over a smooth pulley.
Particle A has mass 4 kg and B has mass 6 kg. The system is released from rest and the particles move vertically.
a Find the tension in the string and the upward acceleration of particle A. [3]

b Find the magnitude of the resultant force exerted on the pulley by the string. [1]
2 A particle of mass 3 kg is at rest on a slope that is at an angle of 27° to the horizontal. SN
It is held in limiting equilibrium by a force of 5 N, which acts at an angle of 10° to the
slope, as shown. Determine in which direction the particle is on the point of
slipping and find the coefficient of friction between the particle and the slope. [5]

3 A particle P, with mass 3 kg, and a particle Q, with mass 5 kg, are attached to p
the ends of a light inextensible string. P is held at rest on a horizontal table a
and the coefficient of friction between P and the table is 0.4. The string passes
over a smooth pulley at the end of the table 0.8 m from P and Q hangs
vertically down, as shown.
l
The particles are then released from rest. Find the time until particle P hits the pulley. [5]
4 A toy train engine has mass 4 kg and pulls a carriage of mass 6 kg along a horizontal stretch of track by means of
a horizontal tow-bar. The brakes in the engine cause a deceleration of 2 m s- 2 . There is air resistance of 4 Non the
engine and of 10 N on the carriage and no other frictional forces. Find the braking force from the engine and the
force in the tow-bar, stating whether it is a tension or compression. [4] ■
5 A particle, P, starts from a point O and moves in a straight line with velocity v m s- 1, given by:

v = k for 0 < t~1

~ 24
V = 3t 2 + 2 for 1 < t ~ 5
t
where t is the time, in seconds, after leaving O.
a Find the minimum velocity for 1 ~ t ~ 5. [4]
b Find the displacement from O when P reaches the minimum velocity. [5]

6 A particle P , of mass 4 kg, is projected from a point A up a slope with speed 5 m s- 1• The slope is at an angle of
25° to the horizontal and the coefficient of friction between the slope and the particle is 0.4.
a Find the distance P travels up the slope before coming to rest. [4]

b Find the time taken for P to return to A. [5]

7 Two particles move along the same straight line. Particle P has velocity v m s- 1,
given by v = 12t + t 3 - 0.3t 5,
where t is the time in s, and is at a point O at t = 0. Particle Q has displacement s from O at time t, given by
s = 74.25 - 0.05t 6 .

a Find the displacement of P when it is moving at maximum velocity. [6]

b The particles collide at time T . Find the value of T. [2]

 Cambridge International AS & A Level Mathematics: Mechanics

8 Two particles, A and B , are attached to the ends of a light inextensible string, which passes over a smooth pulley.
Particle A has mass 8 kg and B has mass 5 kg. Both particles are held 1.2 m above the ground. The system is
released from rest and the particles move vertically.
a When particle A hits the ground, it does not bounce. Find the maximum height reached by particle B. (5]

b When particle A hits the ground, the string is cut. Find the total time from being released from rest until B hits
the ground. (3]

9 A particle, P, moves on a straight track. It has displacement sm from a point, 0, at time ts, given by
s = 0.12t + 10 - 0.0lt 3 for t > 0.
a Find the time when the particle is first stationary. [2]

b Find the total distance travelled in the first 10 s. (2]

2
Particle Q moves on a track parallel to particle P. The acceleration, am s- , of Q is given by a = 0.4 - 0.06t . Both
particles come to rest alongside each other.
c Find the displacement of Q from O after 10 s. (5]

@ 10 Two particles of masses 5 kg and 10 kg are connected by a light inextensible

string that passes over a fixed smooth pulley. The 5 kg particle is on a rough
fixed slope which is at an angle of a to the horizontal, where tan a = ¾- The IOkg

10 kg particle hangs below the pulley (see diagram). The coefficient of friction

■ between the slope and the 5 kg particle is ½. The particles are released from
rest. Find the acceleration of the particles and the tension in the string. (7]
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q5 June 2016

@ 11 A particle of mass 0.1 kg is released from rest on a rough plane inclined at 20° to the horizontal. It is given that,
5 seconds after release, the particle has a speed of 2 m s- 1•
Find the acceleration of the particle and hence show that the magnitude of the frictional force acting on the
particle is 0.302 N, correct to 3 significant figures. 13]
ii Find the coefficient of friction between the particle and the plane. [2]
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q2 November 2016

(i} 12 A particle P moves in a straight line. It starts at a point O on the line and at time t s after leaving O it has a
velocity vms- 1, where v = 6t 2 - 30t + 24 .
Find the set of values of t for which the acceleration of the particle is negative. (2]
ii Find the distance between the two positions at which P is at instantaneous rest. (4]

iii Find the two positive values oft at which P passes through 0. (3]
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q6 June 2016

@ 13 Particles P and Q are attached to opposite ends of a light inextensible string which passes over a fixed smooth
pulley. The system is released from rest with the string taut, with its straight parts vertical, and with both particles
at a height of 2 m above horizontal ground. P moves vertically downwards and does not rebound when it hits
the ground. At the instant that P hits the ground, Q is at the point X, from where it continues to move vertically
upwards without reaching the pulley. Given that P has mass 0.9 kg and that the tension in the string is 7.2 N while
P is moving, find the total distance travelled by Q from the instant it first reaches X until it returns to X. [6)
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q3 November 2011
In this chapter you will learn how to:
■ calculate the momentum of a moving body or a system of bodies
■ use the principle of conservation of momentum to solve problems involving the direct impact of
two bodies that separate after impact
■ use the principle of conservation of momentum to solve problems involving the direct impact of
two bodies that coalesce on impact.
 I •• • • •. • ,; :_• '•i • '-'.:,:tt~~;t.•)f~f.t
Cambridge International AS & A Level Mathematics: Mechanics : . . . ~. ·_, _, :::•/~:~:,:.,J) j~£l:ifi~-j

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills

Chapter 1 Calculate velocity when the 1 A car is travelling at 15 ms- 1 when the brakes
acceleration is constant. are applied. It takes 6 s for the car to come to
rest. Assume that the braking force is constant
(and hence the acceleration is constant, but
negative).
a Show that car travels 45 m under braking
before coming to rest.
b Calculate speed of the car when it has been
braking for 3 s.
C Calculate the speed of the car when it has
travelled 22.5 m under braking.
Chapter 6 Calculate velocity using 2 A car is travelling at 15 m s- 1when the brakes
calculus. are applied. It takes 6 s for the car to come
to rest. Assume that the acceleration under
b ra k"mg 1s
. given
. b y 5t(t - 6) wh ere t 1s
. t h e time
.
12
from when braking starts.
a Calculate the speed of the car when it has

■
been braking for 3 s.
b Find the speed of the car when it has
travelled 22.5 m under braking.

What is momentum?
The word momentum is used in everyday language to describe the impetus gained:

Feminism gained momentum in the early 20th century.

The fundraising campaign needs to gain momentum if it is to reach its goal.

In mechanics, momentum measures the impetus possessed by a moving object. By

considering the transfer of momentum between objects you can calculate what happens
when objects interact.

You may have pushed a supermarket trolley. Why is it easier to start the trolley moving when
it is empty than when it is full of shopping? An empty trolley has less mass than a full trolley,
so the same amount of 'push' will get an empty trolley moving much faster than a full trolley.

You explain this in mechanics by using momentum.

 Chapter 7: Momentum

7.1 Momentum

T he philosopher Rene Descartes (1 596- 1650) introduced the concept of

momentum. Descartes' built on ideas first written down by Jean Buridan (1295- 1363)
who defined the 'amount of motion' as the product of the mass of a body and its
speed. Using these ideas, Descartes form ulated his three laws of motion, which then
became the basis for Newton's laws of motion.

A body of mass m kg moving with speed v m s- 1 has momentum given by mv.

l\1omentum is a vector quantity, having the same direction as the velocity. For one-
d imensional motion along a line you only need to work out whether the momentum is
positive or negative. The units of momentum are N s.

EXPLORE 7.1

In the Systeme Internationale (SI) system of units there a re seven basic units of
measurement. These are the metre (length), kilogram (mass), second (time), ampere
(electrical current), kelvin (thermodynamic temperature), mole (amount of substance)
and candela (luminous intensity). Use the SI system of units to explain why
momentum is measured in N s. ■
WORKED EXAMPLE 7.1

Find the momentum of a body of mass 3 kg moving at 5 m s- 1•

Answer
Momentum = mv = 3 x 5 Substitute the values for m and v into the
formula fo r momentum.

= 15Ns R emember to give units .

WORKED EXAMPLE 7.2

A ball of mass 50 g hits the ground with speed 10 m s- 1 and rebounds with speed 6 m s- 1• Find the change in
momentum that occurs in the bounce.

Answer
50 g = 0.050 kg Convert the mass to kg.

Momentum before= 0.050 x 10 Calcul ate the momentum just before the
= 0.5 N s bounce.
 ' • ', • ,••• , : ',~~~.• ,' •i~~~\r:,~t)i,;~:J
Cambridge International AS & A Level Mathematics: Mechanics · ,. - ·..: ·. :/
• T O ... ••~
.\l1
-c_J• ~ l""ol, _c,/.

Momentum after = 0.050 x - 6 The direction has reversed so the sign changes.
= -0 .3Ns
So change in momentum= -0.3 - 0.5 If we use down as positive there is a loss in
= -0.8Ns momentum of 0.8 N s.

If we use up as positive there is a gain in

momentum of 0.8 N s.

IDi4iHl?i
1 Find the momentum of a body of mass 10 kg moving at 8 ms- 1•

2 Find the momentum of a car of mass 1500 kg moving at 22 m s- 1.

3 Find the momentum of a tennis ball of mass 57 g moving at 180 km h- 1.

4 A model car has mass 40 g. It slows from 2.2 m s- 1 to 0.8 m s- 1. Find the decrease in its momentum.

5 A rock of mass 4 kg is thrown upwards with an initial speed of 3 m s- 1• It is travelling at 6 m s- 1just before it
lands. Find the change in its momentum.

•
6 A girl of mass 35 kg jumps from a rock onto the beach below. Her initial vertical speed is 0 m s- 1 and she falls
2.45 m under gravity.
a Find the speed of the girl when she lands on the beach.
b Find the downward momentum of the girl just before she lands on the beach.

7 A book of mass 2 kg falls from a window ledge and drops 1.8 m to the ground. It falls freely under gravity.
a Find the speed of the book just before it hits the ground.
b Find the downward momentum of the book just before it hits the ground.

C, 8 A ball of mass 0.2 kg falls 1.25 m vertically downwards to the ground, starting from rest. It hits the ground
and rebounds. The downwards momentum of the ball changes by 1.6 N sin the bounce.
a What height does the ball reach after this bounce?
b By considering the modelling assumptions, explain why the height might be less than this.

9 A ball bearing of mass 25 g is thrown vertically upwards , and is caught on the way back down. The ball
bearing has an initial speed of 3 m s- 1 upwards and is travelling at 2 m s- 1 when it is caught. Find the change in
its momentum.

0 10 A hockey ball of mass 0.2 kg is hit so that it has an initial speed of 8 m s- 1• The ball travels in a horizontal
straight line with acceleration am s- 2 given by a = -0.5 - kt where t is the time in seconds, measured from
when the ball was hit. After 2 s the ball has travelled ~l m. It is then intercepted by a player from the other
team. This player hits the ball so that its direction of travel is reversed and its speed is now 5 m s- 1. Show that
when the ball is hit by the second player its momentum changes in magnitude by 2 N s.

11 Particle A of mass 5 kg is moving at a speed of 2 m s- 1 when it hits a stationary particle, B, of mass 2 kg . After
the impact, particle A has speed 0 m s- 1 and particle B has speed v m s- 1. The loss in momentum for particle A
equals the gain in momentum for particle B. Find the value of v.
 Chapter 7: Momentum

~ 12 A man strikes a snooker ball so that it travels horizontally across a smoker table and makes a direct hit
against the end cushion of the table. The ball rebounds from the cushion and travels to the other end of the
table where it rebounds from the cushion at that end . The ball finishes at exactly the same point at which it
started. The distance between the two cushions is 3.5 m and the initial speed of the ball is 10 m s- 1. The ball is
slowed by friction, resulting in a constant deceleration of 1m s- 2. At each rebound the direction of the ball is
reversed and the magnitude of the momentum after the rebound is 50% of the magnitude of the momentum
before. Work out the distance that the ball travels before it reaches the first cushion.

7.2 Collisions and conservation of momentum

During an impact when two bodies collide, there is a transfer of momentum between them.
Some momentum will be transferred from the first to the second and some momentum will be
transferred from the second to the first. In this chapter you will only consider one-dimensional
impacts between bodies moving in the same straight line, both before and after the impact.

Newton's cradle, shown in the diagram, is a popular toy. The first ball is released and
transfers momentum to the second , which in turn transfers momentum to the third , and
so on until the last ball swings up. The last ball then swings back down again and the
momentum is transferred back to the first ball.

In a perfect Newton's cradle, each ball comes to rest after it has hit the next one, so it looks as
if the first ball has caused the last one to move without the intermediate balls moving at all.

When a hammer is used to hit a nail, momentum is transferred from the hammer to the nail,
causing the nail to move (although resistance forces mean that the nail will not move very far).

■
Momentum is also transferred in the opposite direction, causing the hammer to rebound.

Impacts happen instantaneously, so you do not need to think about external forces , such a s
friction , when considering the impact. The change in momentum is caused by the normal
contact forces between the two objects involved.

One-dimensional instantaneous impacts happen, for example, when a snooker ball, with no
spin, hits a stationary ball and causes it to move along the same line as the original motion.

The contact forces between the two objects involved in the impact are equal and opposite,
so the momentum transferred from the first object to the second is equal and opposite to
the momentum transferred from the second object to the first.

This means that the total momentum before the impact will always be the same as the total
momentum after the impact. The total momentum is unchanged: momentum is conserved in
an impact.

MODELLING ASSUMPTIONS

In reality a snooker player would not usually want a direct, one-dimensional impact
and would probably prefer to use an oblique, two-dimensional impact, where the
motion is not all in the same straight line
'

Cambridge International AS & A Level Mathematics: Mechanics

Before After
velocity Uz Vz

0 0 0 0
mass

total momentum

Momentum is conserved in impacts. The total momentum is constant.

WORKED EXAMPLE 7.3

Two ball bearings are moving directly towards one another. The first ball bearing has mass 20 g and is moving at
3 m s- 1• The second ball bearing has mass 25 g and is moving at Im s- 1. After the collision the first ball bearing is
stationary. What is the speed of the second ball bearing after the collision?

■
Answer
before after
Draw a diagram to summarise the
3ms- 1 lms- 1 Oms- 1

- - 0-
V
information.
+--

0 0 0
0.020kg 0.025kg 0.020kg 0.025kg

Total momentum before collision Remember that momentum is a vector

= (0.020 X 3) + (0.025 X - ])
quantity.
= 0.035 N s lms- 1 f-isthesameas-lms- 1 ➔
Total momentum after coll ision
= 0 + 0.025 V
= 0.025vNs
0.035 = 0.025 V Momentum is conserved.
v = l .4ms- 1

Sometimes, instead of moving apart after an impact, the objects may coalesce. This means
that they collide and then move off together as a single object. The objects can be thought of
as having merged into a single object with a mass equal to the sum of the individual masses.
Examples of coalescence include a railway truck being pushed up to an engine and
coupling with it, a person jumping onto a moving vehicle or two ice skaters meeting up and
holding hands to continue as one.
The opposite of coalescence is called an explosion. This would happen, for example, when
the engine and truck become decoupled, when the person jumps off the moving vehicle or
when the ice skaters stop holding hands and drift apart.
 Chapter 7: Momentum

WORKED EXAMPLE 7.4

A girl is sitting on a sledge. The girl and the sledge have a combined m ass of 50 kg. When the girl and the sledge
are moving at 2 m s- 1 her sister standing in fro nt of the sledge throws a snowball at the sledge. The snowball has
mass 0.2 kg and is travel li ng at 10.55 m s- 1 when it hits the sledge, head on. The snowbal l, the girl and the sledge
continue together. Assuming that the total momentum is unchanged, find the new speed of the sledge.

Answer
before after

-z
2ms- 1 10.55 m s- 1
+-

•
-z •
V
Draw a diagram to summarise the
information.

50kg 0. 2 kg 50.2 kg

Total momentum before= (50 x 2) + 0.2 x (- 10.55) The snowball is thrown in the opposite
= 97.89N s d irection to the motion of the sledge.
Total momentum after= 50.2v N s The total mass of the girl, sledge and snowball
is 50 + 0.2 = 50.2 kg.

97.89 = 50 .2v Momentum is conserved.

V = 1.95 111 S- I

WORKED EXAMPLE 7.5 ■

A block of mass 200 g moving at 4111 s- 1 makes a direct collision with a larger block of mass 500 g moving at Im s- 1.
On impact the blocks coalesce.

a Find the speed of the blocks after the collision if the blocks were initially moving towards one another.

b Find the speed of the blocks after the collision if the blocks were initially moving in the same direction.

Answer
a before a fter
Draw a diagram to summarise the
4ms- 1 lms- 1 V
-+ +- -+ information.

I
□
D I I
0.200kg 0.500 kg 0.700kg

Total momentum before= (0.200 x 4) + (0 .500 x - 1) Momentum is a vector quantity.

= 0.3 Ns
Total momentum after= 0.700vNs

0.3 = 0.700v Momentum is conserved.

v = lms-1
7
= 0.429ms- 1(to3s.f.)
 Cambridge International AS & A Level Mathematics: Mechanics

b before after

-D -
4ms- 1 I ms- I

-I
V
The faster block must catch up with the slower
block.

□
I I

0.200kg 0.500kg 0.700kg

Total momentum before= (0.200 x 4) + (0.500 x 1)

=l.3Ns
Total momentum after = 0 .700v N s

1.3 = 0.700v Momentum is conserved.

v=llms- 1
7
= l.86ms- 1 (to3 s.f.)

MODELLING ASSUMPTIONS

By modelling objects as particles, you are ignoring the possibility of an oblique

contact. This can happen if objects collide so that the contact is not in the line of
motion and instead they bounce off each other at different angles. You will assume
this is not the case.

■ There is also a possibility that objects travelling along a surface wobble slightly or if
the objects are a different size or shape some of the momentum may cause one of the
objects to lift off the surface. The effect of this is normally quite small, but can be
significant in games where precision is required.

1 Chris and his son are skating on an ice rink. Chris skates in a straight line at a speed of 3 m s- 1 towards his
son, who is stationary on the ice. When they meet Chris lifts his son up and they continue together at a speed
of 2 m s- 1, still travelling in the same straight line. Chris has mass 80 kg. Find the mass of his son.

2 A ball of mass 0.04 kg is moving at a speed of 3 m s- 1 when it hits a stationary ball of mass 0.06 kg. After the
impact the first ball is stationary. Find the speed of the second ball.

3 A box of mass 25 kg slides down a slope until it has reached a speed of 5 m s- 1 It then travels at 5 m s- 1
horizontally across a smooth floor until it bumps into a stationary crate. Immediately after the impact
the box reverses its direction and travels at 0.5 m s- 1. The crate starts to travel at 2.75 m s- 1 . Find the mass
of the crate.

~ 4 Two snooker balls are travelling towards one another in a straight line when they make a direct impact. Before
the impact the first ball had speed 12 m s- 1 and the second ball had speed 8 m s- 1. After the impact both balls
have reversed their direction and each has speed 10 m s- 1• It is claimed that the balls are not both real snooker
balls because they have different masses. Find the ratio of the masses of the balls.
 Chapter 7: Momentum

5 Particles A, Band C, of masses 0.01 kg, 0.06 kg and 0.12 kg respectively, A (O.Ol kg) B (0.06 kg) C (0.12 kg)
are at rest in a straight line on a smooth horizontal surface, with B
between A and C. A is given an initial velocity of 4ms- 1 towards B.
After this impact A rebounds with velocity 2 m s- 1 and B goes on to
----0 ------0 ------0 ---
---+ 4m s- 1
hit C. After the second impact B comes to rest. Find the speed of C after
the second impact.

6 Three balls, A, Band C, of masses 4kg, 1kg and 2kg, respectively, are at rest in a straight line on a smooth
horizontal surface, with B between A and C. A is given an initial velocity of 5 m s- 1 towards B. When A hits B
they coalesce and continue as a single object, D, until they collide with C. After this collision Chas velocity
5 ms- 1. Work out the final velocity of D.

4D 7 Jayne is performing in a show on ice. She is pushed onto the ice while sitting on a chair. The chair slides
across the ice and Jayne then stands up and moves away from the chair. Jayne has speed 4 m s- 1 when she is
sitting on the chair and speed 5 m s- 1 when she moves away from the chair. Jayne has mass 60 kg and the chair
has mass 6 kg.
a Find the velocity of the chair as Jayne moves away from it.

b What modelling assumptions have you made?

8 A bean bag of mass 100 g is thrown at 5 m s- 1 at a stationary target. The bean bag sticks to the target and they
move off together at 0.1 m s- 1• Find the mass of the target.

9 Mariam is moving on a sledge at 2 m s- 1• The combined mass of Mariam and the sledge is 40 kg. Sarah, who
has mass 60 kg, runs up behind the sledge and jumps onto it. The sledge continues in the same straight line

■
with speed 2.3ms- 1•

a Find Sarah's speed just before she lands on the sledge.

b What assumption have you made regarding Sarah's velocity?

0 10 A simplified model of the launch of a space shuttle is as follows. The shuttle is attached to two rocket
boosters each of which contains a fuel tank. The launch is vertical and in a straight line. The initial
total mass is I million kg. The mass of the shuttle is 60 000 kg, the mass of each rocket booster is
20 000 kg and the mass of the fuel in each rocket booster is 450 000 kg. The rocket boosters accelerate the
shuttle (and themselves) to a speed ofl 500 m s- 1• At this time all the fuel in the rocket boosters has been
used up and the rocket boosters are detached. The rocket boosters have speed 0 m s- 1 immediately after
they are detached.
a Use conservation of momentum to show that the speed of the shuttle immediately after the rocket boosters
are detached is 2500 m s- 1•

Suppose that instead just the first rocket booster is used initially to accelerate the shuttle (with both rocket
boosters) to a speed of 500 m s- 1• At this time all the fuel in the first rocket booster has been used up and it is
detached (with speed 0 m s- 1). The second rocket booster is still full of fuel.
b Show that the speed of the shuttle (with the remaining rocket booster) is 518.9 m s- 1 immediately after the
first rocket booster is detached.
The second rocket booster is then used to accelerate the shuttle (and itself). When all the fuel in the second
rocket booster has been used up it is detached (with speed 0 m s- 1). After the second rocket booster has been
detached, the speed of the shuttle is 2500 m s- 1•
c Find the speed of the shuttle (and rocket booster) just before the second rocket booster was detached .
 . . ' '

Cambridge International AS & A Level Mathematics: Mechanics

.
- - ~--· . ~ ..., . ...

0 11 A car is towing a caravan at v m s- 1 in a straight line along a horizontal road. The mass of the car is m kg
and the mass of the caravan is km kg. The caravan becomes detached from the car. Immediately after the
separation the car has speed av m s- 1 and the caravan has speed 0.5v m s- 1 in the same direction as the car.
a Show that if a = 1.3 then k = 0.6.
b Find an expression for k in terms of a general value of a.

~ 12 A particle of mass 0.3 kg is travelling at speed 2 m s- 1 when it collides with a particle of mass 0.5 kg travelling
at speed 1m s- 1. After the impact the first particle has speed v m s- 1 and the second particle has speed
(v+l)ms- 1•
a By considering the directions in which the particles could be moving before and after the impact, find the
possible values for the speed of the first particle after the impact.
You are given that vis the smallest of these possible speeds.
b State whether the particles were travelling in the same direction or in opposite directions before the impact.

EXPLORE 7.2

Five small balls are placed in a line on a smooth table with 1m between each ball
and the next. The fifth ball is at the edge of the table. The first ball has mass 50 g, the
second has mass 40 g, the third has mass 30 g, the fourth has mass 20 g and the fifth
has mass 10 g. Initially the balls are all stationary. The first ball is then fired to hit the
second ball with speed 1m s- 1• In this collision, half the momentum of the first ball is

■ transferred to the second ball. It is claimed that in every impact, half the momentum
of the faster ball is transferred to the slower ball. Investigate what happens and how
long it takes until the fifth ball falls from the table.

• A body of mass m kg moving with speed v m s- 1 has momentum given by mv.

• Momentum is conserved in impacts. The total momentum is constant.
 Chapter 7: Momentum

1 Particle A moves across a smooth horizontal surface in a straight line. Particle A has mass 4 kg and speed 3 m s- 1•
Particle B, which has mass 6kg, is at rest on the surface. Particle A collides with particle B. After the collision A
is at rest and B moves away from A with speed um s- 1. Find the value of u. [3]
2 Two particles, A and B, have masses of 3 kg and 2 kg, respectively. They are moving along a straight horizontal
line towards each other. Each particle is moving with a speed of 4 m s- 1 when they collide.

The particles coalesce to form a single particle. Find the speed of the combined particle. [3]
A (3 kg) B (2 kg)

---0 ----------0 ----

4ms- 1 4ms- 1
3 A train consists of a locomotive, of 40 000 kg, pulling four coaches, each of mass 50 000 kg. The train is travelling
at 5 m s- 1 along a straight horizontal line when the coupling between the second and third coaches breaks.
The locomotive and the first two coaches continue at 7 m s- 1• The rear two coaches then decelerate at a constant
rate to come to rest after travelling 100 m. Work out how long it takes from when the coupling breaks to when
the rear two coaches come to rest. [4]
Particle A has mass 4 kg. It moves with speed 3 m s- 1 in a straight line on a smooth horizontal surface. Particle

- --
4
B has mass 6 kg and is at rest on the surface. Particle A collides with particle B. After v 4v
the collision, A and B move away from each other with speeds v m s- 1 and 4v m s- 1, as
shown in the diagram.

5
Find the value of v.
Two balls are travelling towards one another along the x-axis. The first ball has mass 2 kg and is travelling
at 3 m s- 1 in the positive x-direction. The second ball has mass 5 kg is travelling at 1 m s- 1 in the negative
x-direction. The balls collide and after the collision the balls are travelling at the same speed but in opposite
directions. Work out the speed of the balls after the collision. [41
[4]

•
6 Two particles, A and B, are moving in a straight line on a smooth horizontal surface. A has mass m kg and is
moving with velocity 5 m s- 1, B has mass 0.2 kg and is moving with velocity 2 m s- 1•
a Find, in terms of m, an expression for the total momentum of A and B. [11

Particle A collides with particle Band they coalesce to form a single particle, C . Particle Chas velocity 3 m s- 1•
b Find the value of m. [3]

7 Two particles, A and B, have masses of 3 kg and 2 kg, respectively. They are moving A (3 kg) B (2 kg)
along a straight horizontal line towards each other. Each particle is moving with a
speed of 4 m s- 1 when they collide.
·-0 ---------0 --·
--- +---
1
4 ms- I 4ms-
After the collision, particle A moves in the same direction as before the collision but
with speed 0.4 m s- 1• Find the speed of B after the collision. [4]
0 8 Ball X has mass 0.03 kg. It falls vertically from rest from a window that is 30 m above the ground. Ball Y has
mass 0.01 kg. At the same time that ball X starts to fall , ball Y is projected vertically upwards from ground level
directly towards ball X. The initial speed of ball Y is 20 m s- 1 vertically upwards.
a Find the downward momentum of each ball just before they meet. [3]

The balls coalesce and the combined object falls to the ground.
b Show that the combined object reaches the ground 2.68 seconds after ball X started to fall. [51
 ' 4 ~ • '

Cambridge International AS & A Level Mathematics: Mechanics

' . ' . '

0 9 Three balls, A, B and C, are in that order in a straight line on a smooth horizontal surface. A has mass 0 .4 kg
and is moving at 4 m s- 1 towards B. B has mass m kg and is stationary. Chas mass 0.25 kg and is moving at 0.8 m s- 1
away from B. A hits Band then B hits C. There are no further impacts. A and C now each have a speed of
1m s- 1 and are both moving in directions away from B. Find the range of possible values of m. (8]

$ 10 A ball of mass 0.6 kg is dropped from a height of 1.8 m onto a solid floor. Each time the ball bounces on the
floor it loses 10% of its speed.
a Work out how much momentum was absorbed by the floor in the first bounce. (3]

b Show that the ball first fails to reach a height of 1m after the third bounce. (4]

c What modelling assumptions have you made? 111

11 Ball X has mass 30 g and is moving at 0.51 ms- 1. The direction in which X is travelling is taken as the positive
direction. Ball Y has mass 60 g and is stationary. Ball X collides with ball Y and, after the impact, ball X moves at
0.01 m s- 1 in the positive direction. Ball Y then hits a wall and rebounds with half the speed with which it hit the wall.
a Work out how much momentum was absorbed by the wall. (5]

After rebounding from the wall, ball Y goes on to hit ball X .

b Explain why ball X must be travelling in the negative direction after being hit by ball Y. (21

After this impact ball X has speed 0.15 m s- 1•

■
c Find the final velocity of ball Y. (3]

0 12 Balls X, Y and Z lie at rest on a smooth horizontal surface, with Y between X and Z. Balls X and Z each
have mass 2 kg and ball Y has mass 1kg . Ball X is given a velocity of l m s- 1 towards ball Y. Balls X and Y
collide. After this collision the speed of ball Y is three times the speed of ball X . Ball Y goes on to collide
with ball Z. After this collision the speed of ball Y is the same as the speed of ball X, and the speed of
ball Z is twice the speed of ball Y. F inally ball Y collides with ball X again. After this collision the
speed of ball Y is twice the speed of ball X, and the speed of ball Z is four times the speed of ball Y.
Show that the balls are now all travelling in the same direction and that no further collisions occur. (10]
 .
~---.-..: - ·- ---\.
- - -I'

In this chapter you will learn how to:

■ calculate the work done by a force in moving a body

■ calculate the kinetic energy and gravitational potential energy of a body.
 Cambridge International AS & A Level Mathematics: Mechanics

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills

Chapter 3 Resolve forces. 1 A block of mass 4 kg is at rest on

a slope that is inclined at 30° to
the horizontal. A force parallel to
the slope prevents the block from
moving.
a Find the component of the
block's weight down the slope.
b Find the normal reaction that
the slope exerts on the block.

Chapter 4 Calculate frictional resistance. 2 A block of mass 4 kg is sliding

down a slope. The coefficient of
friction between the slope and the
block is ~ ../3.
1
The normal reaction that the slope
exerts on the body is 20../3 N.
Find the fr ictional force.

■
Chapter 2 Use Newton's second law. 3 A block of mass 4 kg is sliding
down a slope. The component of
the weight down the slope is 20 N
and the frictional force up the
slope is 2 N.
a Find the resultant force on the
block.
b Find the acceleration of the
block down the slope.

Chapter 1 Use the equations of constant 4 A block is initially at rest on a

acceleration. slope. It slides down the slope with
constant acceleration 4.5 m s-2
down the slope.
How far does the block slide in the
first 0.5 s?

How are work and energy used in mechanics?

The terms work and energy are used in everyday life, but what do these terms mean when
we use them in mechanics and how are they connected?

In everyday life, a student who has been studying hard for 2 hours would say that they have
been doing work, as would a gardener who has been working in a garden or an athlete who
has been training. Each of these people has spent time doing an activity and has used energy
 Chapter 8: Work and energy

in the process. This energy comes from the food that the people have eaten. The gardener and
the athlete have used energy to create movement, and the student has used energy to create
'brainpower'.

The phrase 'put more energy into it' is used to mean put more effort into a task, or apply
more force. Energy comes in many forms and can be changed from one form to another.
A person who eats a meal takes in chemical energy, which might then be converted into
movement or, if it is a cold day, used to warm the person up.

In this chapter we will show that when a force moves a body, it does work and causes a
change in the kinetic energy of the body. In Chapter 9 we will further investigate this
relationship between work and energy.

8.1 Work done by a force

In mechanics the word wort means something more than just making an effort. It has a
very specific meaning that refers to how energy changes when a force moves an object.

Mechanical work is done by a force when that force causes an object to move. For mechanical
work to happen, we need a force that causes motion and we need motion to occur.

A weightlifter does work in lifting a weight because a force acts (the tension in
the arm of the weightlifter) to cause motion (the weight is raised vertically).

However, no mechanical work is done when the weightlifter holds the weight stationary
above their head, because there is no motion (although clearly it requires a lot of effort to

■
stop the weight from falling) .

We start by considering the work done by a force acting in the direction of motion; for
example, a horizontal force pushing a box across a horizontal floor.

If the force doubles then the work done by the force doubles. The work done would also
double if the force was unchanged but the object moved twice as far.

The line ot action of a force has the same direction as the force and includes the point of
application of the force .

FAST FORWARD

When a force of magnitude F N moves a body a start end Later in this section,
distance d m, along the line of action of the force, the , we will consider what
work done by the fo rce is:

W=Fd
___
F_ __. □
0 it means for work done
to be negative.

Note that the distance moved has been represented by d here. When the motion is in
(I DID YOU KNOW?

a straight line and in a constant direction, the distance moved will be the same as the James Prescott Joule
displacement, s , and then the work done by the force is given by W = Fs . (1818-1889) studied
the nature of heat
Work is a scalar quantity; it can be positive or negative, but otherwise has no direction. and discovered
its relationship to
The work done by a force has units Nm , but it is more usual to use joules ( J) to measure mechanical work . This
work done. One joule is the amount of work done by a force of 1 newton in moving an led to the development
object a distance of 1m, along the line of action of the force. of the first law of
thermodynamics.
lJ=INm
I

Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 8.1

A boy uses a constant force of 250 N to push a box 4 m across a floor. Find the work done by the force.

Answer

Work done = Fd = 250 x 4 Substitute the values for F and d into the
formula for work done.

= 10001 Remember to give units.

A girl holds a mass of 20 kg above her head. Find the work done by the girl.

Answer

The mass does not move, so no work is done. This shows how mechanical work differs from
the everyday use of the word work.

Work done= OJ Work is done in raising the mass but no

mechanical work is done in holding it steady,

■
despite how it might feel!

A ball of mass 0.05 kg falls a distance of 1.5 m. Find the work done by the weight.

Answer

Weight = 0.05 x 10 The ball falls because of gravity, so the force

= 0.5N that is causing the ball to fall is its weight.
Work done = 0.5 x 1.5 This is the work done by the weight.
= 0.751

The work done by the weight of an object is usually referred to as the work done by gravity.
direction
If other forces act and the direction of motion is upwards, we describe this as the work done of pulling
against gravity. motion force

Sometimes the direction of motion can be in a different direction to the line of action of
the force that is causing the motion. This can occur when there are other forces acting.
For example, when a force at an angle to the horizontal pushes or pulls a box across a
horizontal floor, or when a horse-drawn barge is pulled along a narrow canal using a rope
force
from the bank of the canal, the motion is restricted by contact forces.

When the direction of a force is different from the direction of motion, we can calculate the
work done by the force by resolving it into components along the direction of motion and
perpendicular to the direction of motion.
 Chapter 8: Work and energy

start end start end

r:0
'

P: d
I
•
I I '
•
I Fcos0

d
I
•
I

The box in the diagram is on the floor and there is no motion in the perpendicular direction,
so t he per pendicular component does no work. The work done by the force is given by the
component of the fo rce in the direction of motion multiplied by the distance moved.

When a force of magnitude F N moves a body a distance

d m, at an angle 0 to the direction of the force, the work
done by the force is:
W = Fdcos 0
LF
start end Look back to
Chapter 3, Section
3.1 , if you need a
reminder about
resolving forces
into perpendicular
d
components.

You can either think of this as the component of the force in the direction of motion
multiplied by the distance moved, W = (F cos 0) x d, as in the following left-hand diagram;
or as the force multiplied by the component of the displacement in the direction of the

■
force , W = F x (d cos 0), as in the following right-hand diagram.

d d

WORKED EXAMPLE 8.4

A small truck is pulled 5 m along a railway track by a force of 100 N at an angle 60° to the track. Find the work
done by the force.

Answer

Work done= Fd cos 0 = 100 x 5 x cos 60 We can think of this as 50 N x 5 m or as

= 250] lO0N x 2.5m.

If there is a force that opposes the direction of motion, then the work done by this force will
be negative and we say that work is done against the force. T his happens, for example, when
a load is being raised vertically and work is being done against its weight (or against gravity),
as in the next Worked example.
 WORKED EXAMPLE 8.5
'

A ball of mass 0.05 kg is raised through a distance of 1.5 m. Find the work done against gravity.

Answer

Weight = 0.05 x 10 = 0.5 N There will be other forces acting to raise the
ball, but we are only asked about the work
done against gravity; that is, due to the weight.

Work done by the weight= 0.5 x -1.5 = - 0.75 1 This is negative because the weight opposes the
motion.

Work done against the weight= - (work done by the weight) We can say that we have negative work done by
Work done against weight= 0.751 the weight or that we have positive work done
Work done against gravity= 0.751 against the weight.

When several forces act on a body we can add the work done by each force to get the total
work done by all the forces . Remember that work done may be positive or negative. So to
find the total work done by the forces , we add the work done by the forces with components
in the direction of motion (forces that help to move the body) and subtract the work done
against forces with components in the direction opposite to the direction of motion (forces

■
that try to prevent the body from moving). This is illustrated in Worked example 8.6.

WORKED EXAMPLE 8.6

A box of mass 5 kg is pushed up a slope inclined at 30° to R

the horizontal by a force of 30 N at an angle 10° to the slope.
The frictional force acting on the box is 2 N. The box moves
a distance 3 m up the slope.

a Find the work done against friction.

b Find the work done against gravity.
c Find the work done by the push force.
d Find the work done by the normal reaction.
e Find the total work done on the box by all four forces. weight= S x 10 = SON

Answer

a Work done against fr iction = 2 x 3 = 6 J Friction acts along the direction of motion, but
opposing the motion.
b The component of the weight down the slope is

50sin30 = 25N

Work done against gravity = 25 x 3 = 75 J

SON
 Chapter 8: Work and energy

c T he co mponent of the push force up the slope is

The angle between the push force and the slope
30cos l0 = 29.54N is 10°.

Work do ne by pus h force = 29.54 x 3 = 88.6 J

d T here is no moveme nt in th e per pen dic ul a r

directio n, so no work is done by the no rm a l
reaction .

e Total work done = work do ne by push fo rce This is the total work done by all four forces in
- work done aga inst gravity moving the box.
- work done against friction
= 88.6 - 75 - 6 = 7.6 1

Sometimes a question may mention 'non-gravitational resistance'. This means all the components of
forces, such as friction and air resistance, that act against the motion. It does not mean any component
of the weight that would oppose the motion of a body travelling uphill or rising vertically.

1 A crate is pushed 2 m across a smooth horizontal floor by a horizontal force of 30 N. Find the work done by
the force . ■
2 A box is pulled 5 m across a smooth horizontal floor by a rope with tension 20 N . Find the work done by the
tension:
a when the rope is horizontal

b when the rope is at 40° above the horizontal.

3 A ball of mass 0.04 kg is thrown vertically upwards. It rises 2 m and then falls 2 m. Ignoring air resistance, find
the work done by gravity:
a when the ball rises 2 m

b when the ball falls 2 m

c when the ball rises 2 m and then falls 2 m.

4 A skier of mass 60 kg starts from rest at the top of a slope of vertical height 10 m. She descends the slope and
ascends the other side to come to rest at a point that is 4m vertically lower than where she started. Find the total
work done by gravity (i.e. the work done by gravity in descending minus the work done against gravity while
ascending).

CD 5 A horse-drawn barge is pulled 20 m forwards using a rope at a small angle to the direction of motion. The
barge touches against the edge of the canal. The total resistance to the motion is 100 N.
a What causes the resistance?

b Determine the work done against the resistance.

 Cambridge International AS & A Level Mathematics: Mechanics . · -

0 CD 6 A horse-drawn barge is pulled 40 m forwards using a rope at an angle 0 to the direction of motion. The
tension in the rope is 150 N . The barge is kept moving in a straight line by a contact force with the edge of the
canal. Resistance forces also act on the barge.
a Determine the work done by the tension:

when 0 = 10°
ii when 0 = 20°.
b Show that when 0 = 20° the tension would need to increase to 157.2 N to do the same work as in part ai.
Consider the barge being pulled with a tension of 150 N with 0 = 10°, or being pulled with a tension of 157.2 N
with 0 = 20°.
c Explain why the frictional resistance will be greater in the second of these situations.

7 A box is pulled 2 m across a horizontal floor, using a rope with tension 10 N at 30° to the horizontal. The
frictional resistance is 3 N . Find:
a the work done against friction
b the work done by the tension
c the work done by the weight
d the work done by the normal contact force
e the total work done by all four forces.

■
8 A crate of mass 25 kg slides 4 m down a slope that is inclined at 15° to the horizontal. Non-gravitational
resistance is 5 N . Find:
a the work done against non-gravitational resistance
b the work done against gravity
c the work done by the normal contact force
d the total work done by all these forces.

9 A crate of mass 25 kg is pushed 4 m up a slope that is inclined at 15° to the horizontal by a force of 100 N
parallel to the slope. Non-gravitational resistance is 5 N . Find:
a the work done by the force of 100 N
b the work done against non-gravitational resistance
c the work done against gravity
d the work done against the normal contact force
e the total work done by all these forces.

10 A tile of mass 0.5 kg slides 2 m down a roof, which is inclined at 60° to the vertical. The frictional force is 1.5 N.
There are no other external forces. Find:
a the work done by gravity
b the work done against friction
c the work done by the normal reaction force
d the total work done by all three forces .
 Chapter 8: Work and energy

(I' 11 A box of weight 20 N is pulled 12 m across a horizontal floor, using a rope with tension 25 N at 10° to the
horizontal. The frictional resistance is 1N. Find the total work done by all three forces.

(I' 12 A sack of mass 5 kg slides 2 m down a ramp. The ramp is inclined at 15° to the horizontal. The coefficient of
friction between the sack and the ramp is 0.25. Find the total work done.

8.2 Kinetic energy

Energy can exist in many forms: heat, light, nuclear energy, chemical energy (from food or
fuel), stored (potential) energy (such as the energy stored in a compressed spring) and so on.
Energy can be transferred from one form to another and can be used to create motion.

Energy is a scalar quantity; it can be positive or negative but otherwise has no direction.
In Mechanics we are only interested in mechanical energy. Mechanical energy can be
kinetic or potential.

Kinetic energy is the energy that a body possesses because of its motion.

A body of mass m kg moving with speed v m s- 1 has kinetic energy (KE) given by:

1
KE = - mv 2
2

■
Kinetic energy could be measured in (kg)( m s- 1)2, but this is the same as
(kgms-2)(m) = N m= J.

All forms of energy are measured in joules (J) .

WORKED EXAMPLE 8.7

Find the kinetic energy of a body of mass 3 kg moving at 5 m s- 1.

Answer

KE = -1 mv 1- = -I x 3 x 5-
?
Substitute the values for m and v into the
2 2 formula for the kinetic energy.
= 37 .51
Remember to give units.

WORKED EXAMPLE 8.8

A ball of mass 50 g hits the ground with speed 10 m s- 1 and rebounds with speed 6 m s- 1• Find the loss in kinetic
energy that occurs in the bounce.

Answer

50 g = 0.050 kg Convert the mass to kg.

Change in KE= 11nal KE initial KE
 Cambridge International AS & A Level Mathematics: Mechanics

KE before = 1 x 0.050 x 10 2 = 2.5 J Calculate the KE just before the bounce.

2
Initial KE = _!_ mu 2
2
1 Calculate the KE just after the bounce.
KE after = x 0.050 x 62 = 0.9 J
2
Final KE = !mv2
2
KE is scalar, so the change in the direction of
the velocity does not matter.
So loss in KE = 1.6 J Remember to give units.

J) KEY POINT 8.4

A common error is to use the difference between the velocities or speeds in the calculation, like this:

½x 0.050 x ( 6 + 10 ) = 6.4 J
2
or ½x 0.050 x ( 6 - 10 )2 = 0.4 J
But these are both wrong. It must be the difference of the squares of the speeds:
½ 0.050
X X (6 2 -10 2 ) = 1.6 J

■ 1

2
Find the kinetic energy of an object of mass 10 kg moving at 8 m s- 1.

Find the kinetic energy of a car of mass 1500 kg moving at 22 m s- 1.

3 Find the kinetic energy of a tennis ball of mass 57 g moving at 180 km h- 1•

4 A rock of mass 4kg is thrown upwards with an initial speed of 3 m s- 1. It is travelling at 6 m s- 1 just before it
lands. Find the increase in its kinetic energy.

5 A book of mass 2 kg falls from a window ledge and drops 10.8 m to the ground. It falls freely under gravity.
a Find the speed of the book just before it hits the ground.

b Find the kinetic energy of the book just before it hits the ground.

6 A model train of mass 1kg is moving at 3 m s- 1• It accelerates uniformly for 5 s, travelling in a straight line and
covering a distance of 40 m while accelerating.
a Find the speed of the train at the end of the 5 s.

b Find the increase in kinetic energy from the start to the end of the 5 s.

7 A ball bearing with mass 0.03 kg is projected vertically upwards. It loses 0. 735 J of kinetic energy before
coming to instantaneous rest. Find the initial speed of the ball bearing.

8 A box of mass 30 kg slides from the top of a smooth slope to the bottom of the slope. The slope is inclined at
30° to the horizontal. The box starts from rest. At the bottom of the slope the box has gained 37 5 J of kinetic
energy. Find the length of the slope.

C, 9 A boy of mass 64 kg runs at a constant speed along a straight track. He takes 16 s to run 100 m.
a Work out his kinetic energy.

b What difference would it make if the track was curved?

 Chapter 8: Work and energy
I

4D 10 At its launch a rocket has mass 2 million kg. It accelerates from rest to 75 000 m s- 1•
a Work out the increase in the kinetic energy.
b Why will the calculated value be too big?

11 Ball A, of mass 2 kg, is moving in a straight line at 5 m s- 1• Ball B, of mass 4 kg, is moving in the same straight
line at 2 m s- 1• Ball Bis travelling directly towards ball A. The balls hit each other and after the impact each
ball has reversed its direction of travel. The kinetic energy lost in the impact is 12.5 J.
a Show that the speed of ball A after the impact is ~ m s- 1.
1
b Find the speed of ball B after the impact.

0 12 Two balls, A and B, of equal mass, are travelling towards one another with velocities uA and -us, respectively.
The balls collide and their velocities after the impact are - vA and vs, respectively. The kinetic energy after
the impact is the same as the kinetic energy before the impact (i.e. a 'perfectly elastic collision'). Explain why
vA = us and vs = uA.

~ 13 Balls X, Y and Z lie at rest on a smooth horizontal surface, with Y between X and Z. Balls X and Z each
have mass 2 kg and ball Y has mass 1kg. Ball X is given a velocity of 1m s- 1 towards ball Y. Balls X and Y
collide. After this collision the speed of ball X is 0.4 m s- 1 in its original direction.
a Work out the loss in kinetic energy in this impact.
Ball Y goes on to collide with ball Z . After this collision the speed of ball Y is 0.4ms- 1 in the direction
towards ball X.
b Work out the loss in kinetic energy in this impact.
Finally, ball Y collides with ball X again. After this collision the speed of ball Y is twice the speed of ball X

■
and the speed of ball Z is four times the speed of ball Y, with the balls all travelling in the same direction.
c Work out the loss in kinetic energy in this impact.

EXPLORE 8.1

Investigate what happens in Exercise 8B, question 12, if the perfectly elastic collision
takes place between two balls that have different masses.

8.3 Gravitational potential energy

The other type of mechanical energy is potential energy. Potential energy is the energy that
a body possesses because of its position. It can be thought of as stored energy. Gravitational potential
energy (GPE) is
(ra t•1tional tlOtential energy is the energy that could be released if the body falls under
sometimes just called
gravity. 'potential energy',

aI KEY POINT 8.5

A body of mass m kg at height h m has potential energy (PE) given by:

although there
are other types of
potential energy (e.g.
elastic potential energy,
which is the energy
PE = mgh = IOmh stored in a stretched or
compressed spring).

The amount of potential energy that a body possesses depends on its height. The height is
measured from some 'base' level where PE = 0 . We can choose any level as the base, but
must measure all heights from the same level.

Potential energy is measured in joules (J).

 WORKED EXAMPLE 8.9

Find the increase in potential energy in raising a sack of mass 1kg through a height of 3 m.

Answer

Increase in PE= mgh =I x 10 x 3 Substitute the values form , g and h into the formula for PE .

= 30J Remember to give units.

When a body of mass m kg is raised through a height h m, the work done against gravity is
mgh J and the increase in gravitational potential energy is mgh J. Potential energy increases
when work is done against gravity, so objects at a higher level have more potential energy
than those that are lower.

When the same body descends through a vertical distance h m, the work done by gravity
is mgh J and the decrease in gravitational potential energy is mgh J . Potential energy
decreases when work is done by gravity.

What matters is the vertical height difference between the top and the bottom, even if the
body descends by sliding down a slope.

■
MODELLING ASSUMPTIONS

As always, we are assuming objects are particles. This means that when we consider
the kinetic energy of an object, we assume the entire object is moving at the same
speed. This is often not the case. For example, the wheels of a car are rotating, so
the point at the top of the wheel is moving more quickly than the car, but the point
at the bottom is moving more slowly. We will consider this difference as negligible.
Experimenting with a ball rolling down a slope will show that the speed at the
bottom of the slope is not as high as expected.

1 Find the increase in the potential energy of an object of mass 5 kg when it rises through 2 m.

2 Find the change in the potential energy of a body of mass 10 kg when it falls through a height of 6 m, stating
whether this is an increase or a decrease.

3 Find the increase in the potential energy of a tennis ball of mass 57 g when it rises through a height of 70 cm .

4 A box of mass 25 kg falls 2 m vertically downwards. Find:

a the loss in potential energy
b the work done by gravity
c the increase in kinetic energy.

5 A tile of mass 1.2 kg slides 3 m down a roof that makes an angle of 35° to the horizontal. Find the decrease in
potential energy.
 Chapter 8: Work and energy

6 A person of mass 70 kg climbs three flights of stairs to reach the third floor of a building. Each of the flights
of stairs consists ofl5 stairs, each of depth 18 cm. Find the increase in the potential energy of the person when
they climb from the ground floor to the third floor.

7 A crate is pulled up a smooth slope using a rope that is parallel to the slope. The slope is inclined at an angle 0
to the horizontal, where sin0 = 0.28, and the tension in the rope is 50 N . The work done by the tension is 75 J
and the increase in the potential energy of the crate is 168 J. Find the mass of the crate.

8 A ramp rises 10 cm for every 80 cm along the sloping surface. A box of mass 50 kg slides down the ramp, starting
from rest at the top of the ramp. The coefficient of friction between the ramp and the box is 0.03 and no other
resistance forces act.
a Draw a diagram to show the forces acting on the box.
The box is travelling at 2 m s- 1 when it reaches the bottom of the ramp.
b Find the length of the ramp.

c Find the loss in the potential energy of the box.

CD 9 A boy of mass 60 kg slides down a slope that makes an angle of 45° to the horizontal. The coefficient of
friction between the boy and the surface of the slope is 0.2 . The boy starts from rest at the top of the slope and
finishes at the end with speed v m s- 1•
a Show that the acceleration of the boy is 5.66ms-2 down the slope.
b Work out the length of the slope in terms of v.

■
c Find an expression for the loss in the boy's potential energy when he slides down the slope.
d What modelling assumptions have you made and what effect would each of these have on your answer to
part c?

~ 10 The recommended slope for wheelchair ramps is 1: 12. This means that the ramp rises 1cm vertically for every
12 cm along the slope.
A person in their wheelchair, with total mass 90 kg, descend along a 1: 12 wheelchair ramp that has slope
length Sm.
They start the descent with speed 2 m s- 1 and finish with speed 4 m s- 1.
Work out the change in total mechanical energy (kinetic energy + potential energy) after descending
the ramp.

11 A ball of mass 0.02 kg is projected vertically upwards through oil. The ball has initial speed~ m s- 1• The oil
12
exerts a resistance of0.lt 0·5 N , where t sis the time from when the ball was projected .
Work out the increase in the potential energy of the ball from the start to when it comes to instantaneous rest.
(Note: you will need an equation solver for this question. You will not be allowed an equation solver in the
examination.)

0 12 A particle of mass m kg is projected up a slope. The particle has initial speed v m s- 1 up the slope. The slope is
inclined at an angle 0 to the horizontal and the coefficient of friction between the particle and the slope is µ.
2
· Ie comes to rest its
Sb ow t h at wh en t h e partlc · potentia
· I energy h as mcrease
· d b y -mv
-- tan-0-
2(µ + tan 0)
I§
Cambridge International AS & A Level Mathematics: Mechanics

• The work done, in joules, by a force of magnitude F N in moving a body a distance d m in the
direction of the force is:

W=Fd

• The work done, in joules, by a force of magnitude F Nin moving a body a distance d m at an
angle 0 to the direction of the force is:

W = Fdcos 0

• The kinetic energy, in joules, of a body of mass m kg moving with speed v m s- 1 is:

1
KE= - mv 2
2
• The gravitational potential energy, in joules, of a body of mass m kg at height h m above the
base level is:
GPE = mgh
I

Chapter 8: Work and energy

@ 1 A block is pulled for a distance of 50 m along a horizontal floor, by a rope that is inclined at an angle of a to
0

the floor. The tension in the rope is 180 N and the work done by the tension is 8200 J. Find the value of a. 13)
Cambridge International AS & A Level Mathematics 9709 Paper 43 QI June 201 I
2 A ball of mass 30 g is thrown vertically upwards with an initial speed of 4 m s- 1• Air resistance can be ignored.
The ball reaches a maximum height of 80cm. Find:
a the decrease in kinetic energy (21
b the increase in potential energy. 121
1
3 A car of mass 1600 kg is driven 200 m along a straight horizontal road. The car starts with a speed of 3 m s-
and finishes with a speed of 20 m s- 1• A constant resistance of 40 N acts.
a Find the acceleration of the car. 121
b Find the work done by the driving force. (21
0 4) 4 A box of mass 20 kg is pushed 3 m up a slope inclined at an angle a to the horizontal. The work done by the
push force is 900 J and non-gravitational resistance (friction and air resistance) is 40 N .
a Work out the push force . 111
b Show that the acceleration of the box up the slope, am s-2, is given by a = 13 - 10 sin a. 131
c What assumptions have you made? Ill
@0 5 A and Bare two points 50 metres apart on a straight path inclined at an angle 0 to the horizontal, where sin 0 = 0.05,

■
with A above the level of B . A block of mass 16 kg is pulled down the path from A to B. The block starts from rest at
A and reaches B with a speed of 10 m s- 1• The work done by the pulling force acting on the block is 1150 J.
Find the work done against the resistance to motion. 13)
The block is now pulled up the path from B to A. The work done by the pulling force and the work done against
the resistance to motion are the same as in the case of the downward motion.
ii Show that the speed of the block when it reaches A is the same as its speed when it started at B . (2)

Cambridge International AS & A Level Mathematics 9709 Paper 42 Q2 June 2013

4) 6 A basketball of mass 0.625 kg is thrown from a height of 2 m with a speed of 6 m s- 1• It passes through the hoop
at a height of 3 m with speed 4 m s- 1•
a Find the change in the kinetic energy of the ball, stating whether this is an increase or a decrease. (31
b Find the change in the potential energy of the ball, stating whether this is an increase or a decrease. (3)
c What difference does changing the ball's angle of projection make? 111
@~ 7 A lorry of mass 16 000 kg moves on a straight hill inclined at angle a to the horizontal. The length of the hill is
0

500m .
While the lorry moves from the bottom to the top of the hill at constant speed, the resisting force acting
on the lorry is 800 N and the work done by the driving force is 2800 kJ. Find the value of a. 14)
ii On the return journey the speed of the lorry is 20 m s- 1 at the top of the hill. While the lorry travels down
the hill, the work done by the driving force is 2400 kJ and the work done against the resistance to motion
is 800 kJ. Find the speed of the lorry at the bottom of the hill. 14)
Cambridge International AS & A Level Mathematics 9709 Paper 43 Q5 June 2012
 Cambridge International AS & A Level Mathematics: Mechanics

8 A box of mass 20 kg moves across a horizontal floor. The coefficient of friction between the box and the floor is
0.2, and friction is the only resistance force. The box has initial speed 3 m s- 1 and moves until it comes to rest.
a Find the retardation (negative acceleration) of the box. [3]
b Find the distance that the box travels. [3]
c Find the work done against friction. [3]
9 Sam and his skateboard have a combined mass of 100 kg. He accelerates from 0 m s- 1 to 20 m s- 1 while descending
a hill. The hill is modelled as a slope at an angle of sin- 1 0.2 to the horizontal. The non-gravitational resistance is
200 N. The bottom of the hill is 10 m below the top of the hill. Find:
a the increase in the kinetic energy of Sam and his skateboard [3]
b the decrease in the potential energy of Sam and his skateboard [2]
c the distance that the skateboard travels [3]
d the work done against resistance. [2]
10 Kiera climbs up a ladder to sit at the top of a slide 2 m above the ground. Her potential energy increases by 1280 J.
a Find K iera's weight. [1]
Kiera then slides down the slide, starting from rest. The slide is modelled as a slope at an angle 0 to the
horizontal. The resistance force is a constant 20 N. The work done against resistance by Kiera when she is
sliding is 80 J.
b Find the length of the slide. [2]

■ c
d
Find the value of 0.
Find Kiera's speed when she reaches the bottom of the slide.
[3]
[4]
0 11 A ramp is inclined at an angle sin- 1 0.1 to the horizontal. A box of mass 40 kg is projected up the ramp with
initial speed 5 m s- 1. The coefficient of friction between the ramp and the box is 0.05, and no other resistance
forces act.
a Find the acceleration of the box, stating its direction. [4]
The box comes to rest when it reaches the top of the ramp.
b Find the length of the ramp. [3]
c Find the gain in the potential energy of the box. [3]
The total mechanical energy is the sum of the kinetic energy and the potential energy.
d Show that the overall loss in the mechanical energy of the box is 166 J. [3]
~ $ 1
12 Jack has mass 70 kg. He works as a 'human cannon ball'. Jack is projected with speed 12 m s- at an angle of
45° above the horizontal. He lands on a trampoline when the angle between his flight and the horizontal is 50°.
Model Jack as a particle with no air resistance.
a Explain why the horizontal component of Jack's velocity is constant. [2]
b Find Jack's speed when he hits the trampoline. [4]
c Find the kinetic energy gained during the flight. [3]
The gain in Jack's kinetic energy equals the loss in his gravitational potential energy.
d Find the difference in height between the mouth of the cannon and the trampoline. [3]
By changing the angle of projection, Jack can change the angle between his flight and the horizontal when he
lands. Suppose that Jack lands on the trampoline at an angle a to the horizontal.
e What could happen if a is very small? [1]
f What could happen if a is close to 90°? [1]
In this chapter you will learn how to:
■ use the work-energy principle
■ understand when mechanical energy is conserved
■ calculate the power of a moving body
■ use power to calculate the maximum speed of a moving body.
 Cambridge International AS & A Level Mathematics: Mechanics

PREREQUISITE KNOWLEDGE

Where it comes from What you should be able to do Check your skills
Chapter 8 Calculate kinetic energy. 1 A box of mass 5 kg is pushed up a slope.
The box has initial speed 2 m s- 1 and final
speed 3ms- 1• Find the increase in the
kinetic energy of the box.
Chapter 8 Calculate the work done by a 2 A box of mass 5 kg is pushed 5 m up a slope
force in moving a body. inclined at 30° to the horizontal by a force
of 30 N parallel to the slope. The frictional
force acting on the box is 3 N.
a Find the work done by the push force.

b Find the work done against friction.

C Find the work done against gravity.

-
How is power used in mechanics?
We talk about a 'powerful argument' to mean a persuasive argument, or a 'power lifter' as
someone who lifts great weights. Political activists talk about giving 'power to the people'
when they mean giving rights to a group of people or acting on the wishes of the majority.

■
In everyday use the word power means something like strength, but in mechanics the word
strength relates to the force needed to break something (such as the breaking strength of a
cable). Power in mechanics is a way of measuring the rate at which a machine generates motion.

In this chapter you will learn how energy can be converted from one form to another
and how work can increase or decrease the mechanical energy (kinetic and gravitational
potential energy) of a body. You will also learn how the relationship between the power
generated by the engine of a vehicle and the work that is done by the driving force lets us
find the maximum speed that can be achieved by the vehicle.

9.1 The work-energy principle

When a force moves a body, it does work and causes a change in the kinetic energy of the body.
For motion in a straight line with constant acceleration we know that
v2 = u2 + 2as
Using Newton's second law we can replace a by F to give
m
F
v2 = u2 + 2-s
m

Multiplying by ½m and rearranging gives

If I vl < Iuj, then
1
-mv 2 1
--mu 2
= Fs 1 2 1 2 .
2 2 -mv --mu 1s
2 2
You know from Chapter 8 that ½mv -½ mu
2 2
is the increase in kinetic energy. When the negative and there is
a decrease in kinetic
motion is in a straight line and in a constant direction, the distance moved will be the same energy.
as the displacement, s, and so the work done by the force is given by W = Fs.
 Chapter 9: The work-energy principle and power

This means that the previous equation can be expressed as:

increase in kinetic energy = work done by force
This relationship between work done and kinetic energy is not restricted to motion in a
straight line, nor to motion with constant acceleration. We do not need to know the exact
path taken to get from the start to the finish. This means that we can easily deal with non- The path of the body
linear motion or situations where we know what happens at the start and at the finish but can be any curve, or
not the exact path taken in between . even unknown . For
example, the work-
This result tells us how the forces acting cause the kinetic energy to increase or decrease. energy principle applies
The total work done by all the forces acting (driving force, weight, non-gravitational to a child on a slide,
resistance etc.) equals the increase in kinetic energy. helter-skelter or roller
coaster; a person skiing
in a zigzag path or up
and down hills; or the
The work-energy principl states that for any motion: motion of a particle
moving in a circle.
increase in kinetic energy = total work done by all forces
½mv ½mu = L Fs
2
-
2 Application of the
work-energy principle
where the 'total work done' is the sum of the work done by forces (including weight) with a is the only method that
component in the direction of motion (forces that speed up the motion) minus the work done against
forces with a component in the direction opposing the motion (forces that slow down the motion).
can be used when the
path is not a straight
line.
j
-
The total work done will include work done by any force that is not perpendicular to the

■
direction of motion. This includes any driving force , push or pull, tension or compression,
weight, air resistance, friction etc.

The work- energy principle applies whatever the path taken during the motion.

WORKED EXAMPLE 9.1

A boy uses a constant force of 250 N to push a box, of mass 20 kg, a distance 4 m in a curved path across a
horizontal floor. The box starts from rest. Find the final speed of the box:
a when the floor is smooth

b when the coefficient of friction between the floor and the box is 0. I 2.

Answer
a Work done= 250 x 4 The only force that does work is the push
= 10001 force.

Using the work- energy pri nciple: The path is not a straight line, so you
I 1 1 1 need to use the work- energy principle.
- 111 1r - - mu - = work done
2 2
u=O The box starts from rest.

-1 mv-1 - 0 = 10 00
2
 ·.,·;. i .• -~ ,;f/;}.~-~~:~;~ ;~"}\:
Cambridge International AS & A Level Mathematics: Mechanics ,. r ~-.,.J~~~fj;~~~!(·:11:t··~\-_,,
~-~1l.Jt,,,
,1..:...,'-';~· ...... ~ ..... T:.~

v2 = 1000 + 10 1
- m = 10
= 100 2
V = IQ
1
So the final speed of the box is I Oms

b Work done by push force = IO00J Work is done by the push force and work
is done against friction.
Friction force= 0.12 x 200
= 24N

so work done against friction = 24 x 4

= 96]
Total work done = WD by push force - WO against friction 'WD ' = work done
= 1000 - 96 = 904 J
u =0
l
- mv 2 - 0 = 904
2

v2 = 904 + JO
= 90.4

■
. V = 9.51
So final speed is 9.Slms- 1•

When the motion involves a change in the height of the body, work will be done by or
against the weight of.the body.
The total work done can be written as the sum of the work done by the weight and the
work done by the other forces.
When the height of a body increases, the work done against the weight (or against gravity) is
the same as the increase in gravitational potential energy; when the height decreases, the work
done by the weight (or by gravity) is the same as the decrease in gravitational potential energy.
We have:

increase in kinetic energy = total work done

and:
total work done = work done by the weight + total work done by the other forces ! ) TIP
= decrease in gravitational potential energy + total work done by other forces Kinetic and potential
energy are types of
This gives an alternative form for the work- energy principle: mechanical energy.
Other forms of energy
increase in kinetic energy+ increase in gravitational potential energy= total work done by forces
(heat, light, sound,
(where 'forces' here excludes the weight of the body). chemical, electrical,
nuclear etc.) are non-
The sum of the kinetic energy and the gravitational potential energy is the total mechanical mechanical.
energy.
 Chapter 9: The work-energy principle and power

We can write the work-energy principle as

increase in mechanical energy = total work done by forces that act to speed the body up
- total work done by forces that act to slow the body down
(in both cases 'forces' excludes the weight of the body).

WORKED EXAMPLE 9.2

A ball of mass 0.05 kg is thrown vertically upwards with an initial speed of um s- 1• It rises through a distance
of 1.5 m and then falls through 2.5 m before hitting the floor. It hits the floor with speed v m s- 1• Throughout the
motion air resistance of 0.0 IN acts on the ball. Calculate the initial speed, um s- 1, and the final speed , v m s- 1•

Answer
Using the work- ene rgy princ iple:
State that you are using the work-energy
increase in mechanical energy = work done principle.

increase in KE+ increase in GPE = 0 - WD against resistance

To find u we cons ider the motion from the start to the top:

increase in KE = 0 - J_ x 0.05 x u 2
2

■
= - 0.025u 2 J The KE decreases by 0.025u 2 J.

increase in GPE = 0.05 x 10 x 1. 5

= 0.75 ]

so increase in mechanical energy= 0.75 - 0.025u2 J

Work done against resistance = 0.01 x 1.5 = 0.0 15 J

0.75 - 0.025u 2 = - 0.0 15 Increase in mechanical energy = - WO

2
u = 30.6 against resistance, so it is negative (it is a
decrease).
u = 5.53
Initial speed = 5.53ms- 1

To find v we consider the motion fro m the top to the floo r: Use the work- energy principle again for the
increase in KE = _!_ x 0.05 x F2 - 0 = 0.025v 2 J second part of the motion.
2
increase in GPE = 0.05 x 10 x - 2.5
= - 1.251
So increase in mechanica l energy fro m the top to the flo or

= (0.0251•2 - 1.25)]

Work done against resistance = 0.0 1 x 2.5 Air resistance is constant throughout the
= 0.0251 motion.
 Cambridge International AS & A Level Mathematics: Mechanics

0 .025v 2 - 1.25 = - 0.025 Increase in mechanical energy = -WD

2
0.025v = 1.225 against resistance.
v=7
Final speed= 7 ms- 1
Alternatively, we could consider the entire motion together.
From the start to the end:
increase in KE = 0.025( v2 - u2 )
= 0 .025v 2 - 0.765

increase in GPE = 0.05 x 10 x (1.5 - 2.5)

= -0.5J
So increase in mechanical energy = (0.025v 2 - 1.265) J

Work done against resistance = 0.01 x (1.5 + 2.5) Note that the resistance acts for a total
= 0.04J distance of 4 m of travel, although the
displacement is only 1m downwards.
0.025v 2 - 1.265 = -0.04
0.025v 2 = 1.225
v=7ms- 1

■ WORKED EXAMPLE 9.3

A woman snowboards down a hill of varying gradient. The mass of the woman and her snowboard is 64 kg. She
starts from rest at the top of the hill and accelerates under gravity. Throughout the descent the woman does no
work to accelerate or decelerate the snowboard, the average frictional force is 1.5 N and all other resistance forces
are negligible. The snowboarder reaches the bottom of the hill with a speed of 30 m s- 1, having travelled a distance
of 500 m in a zigzag route down the hill. Find the height of the hill, h metres.

Answer
Using the work- energy principle:
increase in mechanical energy = work done The motion is non-linear so the work- energy
principle must be used.
increase in KE + increase in PE = 0 - WD against friction

Initial speed = 0 m s- 1 The snowboarder starts from rest.

Final speed = 30 m s- 1

So increase in kinetic energy = ½x 64 x 30 2

= 28800J
 Chapter 9: The work-energy principle and power

Initial gravitational potential energy= 64 x 10 x h Take the bottom of the hill as the zero level
Final gravitational potential energy = 0 for potential energy.
So increase in potential energy = - 640h J

The work done against friction = 1.5 x 500 (Average) force x distance
= 7501

Hence, 28800 - 640/, = - 750 Substitute the values into the work- energy
/z = 46.2 equation.

The height of the hill is 46.2 m.

1 A box of mass 25 kg is pulled 5 m across a smooth floor by a rope with tension 22 N. The rope is horizontal.
There is a frictional force with average value 12 N. The box starts from rest. Find:
a the work done against friction
b the work done by the tension
c the total work done by all the forces

■
d the final speed of the box.

2 For the situation described in question 1, find the final speed when the rope is inclined at 40° above the horizontal.

3 A crate of mass 50 kg slides down a smooth slope. At the top of the slope the crate has speed 0 m s- 1 and at the
bottom of the slope it has speed 4 m s- 1•
Find:
a the increase in kinetic energy
b the decrease in potential energy
c the vertical height through which the crate has descended.

4 A boy sledges down a hill. The boy and his sledge have a combined mass of 85 kg. He starts from rest and
descends through a vertical height of 3 m. Friction and air resistance are negligible.
a Find the work done by gravity.
b Use the work-energy principle to find the boy's speed at the end of the descent.
The boy descends the hill again, starting from rest, but this time he is joined on the sledge by his little brother, of
mass 35kg.
c Find their speed at the end of the descent.

5 A girl of mass 50 kg travels down a water slide. She starts at the top with a speed of 2 m s- 1 and descends
through a vertical height of 5 m .
a Assuming that there is no resistance, find her speed when she reaches the bottom of the slide.
b The girl's actual final speed is 8 m s- 1 because there is resistance of average value 40 N . Find the length of
the water slide.
 Cambridge International AS & A Level Mathematics: Mechanics

~ 6 A child of mass 45 kg travels down a water chute. The child has speed 1m s- 1 at the top of the chute and
speed 5 m s- 1 at the bottom of the chute. The length of the water chute is 20 m and the height through which it
descends is 4 m. Work out the average resistance force that acts.

7 A boy sits on a sledge at the top of an icy hill. He gently sets the sledge in motion. When he reaches the
bottom of the hill he is moving at 10 m s- 1• Assuming that friction is negligible, find the height of the hill.

G) 8 A girl of mass 50 kg sits on a sledge at the top of a grassy hill. She gently sets the sledge in motion. When she
reaches the bottom of the hill she is moving at 9.9 ms- 1 . The hill is 5 m high and the sledge slides 100 m down
the hill.
a Work out the resistance force .
b Comment on your answer.

9 A child of mass 40 kg slides down a playground slide. The child starts from rest at the top of the slide, 2 m
above the ground. At the bottom of the slide its slope levels off.
a Find the child's loss of gravitational potential energy.
There is a constant resistance of 112 N throughout.
b Find the distance the child has travelled when she comes to rest.
The slide is inclined at an angle of 30° to the horizontal.
c Find the distance the child travels on the level part of the slide.

~ 10 A car of mass 1600 kg travels 200 m along a level road. The average driving force is 2000 N and the average

■
resistance is 800 N. The driver claims that the speed throughout the journey was less than 30 m s- 1• What can
you say about the initial speed of the car?

0 11 A roller-coaster car has mass 100 kg. It carries two passengers, each of mass between 50 kg and 80 kg. The
car becomes detached from the drive chain and continues to travel along the ride with no drive force and no
braking force. The car comes to instantaneous rest at the highest point of the ride and then descends under
gravity to reach the lowest point of the ride. The highest point is 12 m vertically above the lowest point. The
car travels 100 m along the track while descending through 12 m. When the car passes through the lowest
point it has speed 15ms-1.
a Show that the average frictional force is less than 20 N.
b Ifno other non-gravitational resistances act, show that the average frictional force must be at least 15N.

12 A ball, of mass 1kg, moves in an arc of a vertical circle of radius 1m by rotating on the end of a light rod. Air
resistance can be ignored. Initially the rod hangs vertically. The ball is then given an initial horizontal speed
of v ms- 1. It travels in a circular arc through an angle 0.
a Find the gain in the gravitational potential energy of the ball in rising to 0 = 120°.
b Show that the speed of the ball at this position is ✓ v2 - 30 m s- 1.
c In the first case to be considered, v = 8. Find the speed of the ball when 0=120°.
d In the second case to be considered, the ball comes to rest when 0=120°. What was its initial speed, v?
e In the third case to be considered, v = 3.5. What is the value of 0 when the ball comes to instantaneous
rest?
f In the final case to be considered, the ball is just able to make a complete circle (so its speed at the top of
the circular path is 0 m s-1). What was its initial speed, v?
 Chapter 9: The work-energy principle and power

EXPLORE 9. 1

In the situation described in Exercise 9A, question 12, suppose that the ball is
rotating on a string instead of a rod and that the string breaks when 0 =120°. The
ball moves freely under gravity from that point onwards. This means that once the
string has broken, the horizontal component of the velocity is constant and the
vertical component is subject to a constant acceleration of 10 m s-2 downwards. Use a
spreadsheet to investigate where the ball passes through the original vertical circle for
different values of the initial speed, v.
You might also investigate the effect of changing the angle at which the string breaks.

9.2 Conservation of energy in a system of conservative forces

A cow ·enative force is any force for which the work done by that force in moving a particle
between two points is independent of the path taken.
Weight is an example of a conservative force because the work done by the weight depends
only on the change in the vertical height between the initial and final positions, and not on
the shape of the path taken. Friction and driving force are not conservative forces because
the work done depends on the length of the particular path travelled.
When work is done by a conservative force it changes stored potential energy into kinetic
energy, with no loss of mechanical energy. All this energy can be recovered again as
potential energy by reversing the effect.
In a closed system of conservative forces all energy transfers will be between potential and
kinetic energy. You have already seen examples of this in Exercise 9A, questions 5 and 7. In
question 7, a boy sat on a sledge at the top of an icy hill. He gently set the sledge in motion.
■
When he reached the bottom of the hill he was moving at 10 m s- 1• The initial mechanical
energy was all gravitational potential energy, which was then transferred into kinetic energy
as the sledge descended the hill. There were no resistance forces so all of the gravitational
potential energy was converted into kinetic energy. Taking the bottom of the hill as the zero
level for potential energy, the initial potential energy was !Omh J, where m kg is the mass of
the boy and his sledge and h m is the height of the hill. The initial kinetic energy was OJ.
The final potential energy was OJ and the final kinetic energy was 50m J. As all the potential
energy was converted into kinetic energy, this means that the height of the hill is 5 m.
GPE= l Omh
KE=O

GPE=O
KE= 50111

A consequence of the work-energy principle is that for a closed system of conservative forces the
total mechanical energy, KE + GPE, is constant:
initial KE + initial GPE = KE at any point + GPE at that point = final KE+ fina l GPE
Alternatively, we can think of this as:
loss in GPE = gain in KE (or gain in GPE = loss in KE)
We call this conservation of mechanical energy.
 Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 9.4

A box of mass m kg is initially at rest. It slides down a smooth slope that is inclined at 30° to the horizontal.
Find the speed of the box after sliding a distance of 3 m .

Answer
We can ignore friction and air resistance, There is no mention of resistances so this is a
so GPE + KE is constant and closed system of conservative forces.

increase in KE = loss of GPE.

1
m ( v2 - u2 ) = mgh
2
1
- v2 = 10h Cancel m and set u = 0, g = 10.
2
The speed is independent of the mass of
the box.
After sliding 3 m down the slope:
h = 3 sin 30° = 1.5
The box is 1.5 m lower than at the start.

V = J3o
= 5.48

■ The speed of the box is 5.48 m s- 1•

WORKED EXAMPLE 9.5

A ball of mass 0.05 kg is thrown vertically upwards from a height of 1.5 m above the ground. The ball rises through
a height of 2 m to reach its maximum height at 3.5 m above the ground. Use the conservation of mechanical energy
to find the initial speed of the ball.

Answer
Let the initial speed be um s- 1•

We can ignore friction and air resistance, so There is no mention of resistances so this is a
GPE + KE is constant. closed system of conservative forces .

Measuring heights from ground level:

initial GPE= 0.05 x 10 x 1.5 = 0.75 J

initial KE = _!_ x 0.05 x u2 J

2
Final GPE = 0.05 x 10 x 3.5 = l.75J
Final KE= OJ
Hence, 0.75 + 0 .025u 2 = 1.75
0.025u 2 = 1
u = 6.32m s- 1
 Chapter 9: The work-energy principle and power

Alternatively, we could measure heights from the point

where the ball was thrown :

initial GPE + initial KE = final GPE + final KE

0 + 0.025u 2 = 0.05 x JO x 2 + 0
u = 6.32 m s- 1

1 A parcel of mass 3 kg slides 3.5 m down a smooth slope inclined at 20° to the horizontal. When it reaches the
bottom of the slope it has speed 8 m s- 1. Find the speed of the parcel at the top of the slope.

2 A waiter drops a plate and it falls 1.43 m to the floor, where it smashes. Find the speed of the plate when it hits
the floor.

3 A tennis ball of mass 57 g is hit to give it an initial speed of 180 km 11- 1 . It rises through a height of Im.
Ignoring air resistance, find:
a the increase in the gravitational potential energy of the ball
b the horizontal speed of the ball at the top of its flight.

4 A box slides down a smooth ramp. The height of the ramp is 20 cm and the length of the ramp is 2.5 m. The

■
box starts from rest. What is the speed of the box when it reaches the bottom of the ramp?

5 A ball is launched up a smooth slope that makes an angle 30° to the horizontal. The ball travels a distance
2.5 m up the slope before coming to instantaneous rest. Find the launch speed of the ball.

6 In a pinball machine, ball bearings are fired up a slope inclined at 10° to the horizontal. After travelling 1.2 m
the ball bearings reach a curved barrier. Find the maximum initial speed of a ball bearing if it comes to rest
before getting to the curved barrier.

7 A boy sits on a sledge at the top of an icy hill. He gently sets the sledge in motion. When he reaches the bottom
of the hill he is moving at v ms- 1• Assuming that friction is negligible, find an expression for the height of the
hill.

C, 8 A diver jumps from a 10 m tall board into a swimming pool.

a The diver has an initial velocity of ums- 1 upwards. Find his speed when he hits the water.
b What modelling assumptions have been made?

0 ~ C, 9 A football is kicked from ground level with speed 15 m s- 1 and rises to a height of 1.45 m. Assume that air
resistance is negligible.
a Find the speed of the ball when it is 1m above the ground.
At the top of its flight the ball is travelling horizontally.
b Explain why the horizontal component of the velocity is constant throughout the motion .
c Show that the ball was kicked at an angle of 21.0° with the horizontal.
 Cambridge International AS & A Level Mathematics: Mechanics

0 10 A crate of mass M kg sits at the bottom of a smooth slope that is inclined at an angle 0 to the horizontal.
A light inextensible rope is attached to the crate and passes over a smooth pulley at the top of the slope.
The part of the rope between the crate and the pulley is parallel to the slope. The other end of the rope hangs
vertically and at the other end there is a ball of mass m kg. The system is released from rest and the ball
reaches the ground with speed v m s- 1 after descending a distance of h m.
a Find expressions for:
the decrease in potential energy for the ball
ii the increase in kinetic energy for the ball
111the increase in mechanical energy for the crate.~ ~ - - - - - - , -
8
b Use the work-energy principle to show that v = 20h ( m - M sin )-
m+M
a, 11 A piece of sculpture includes a vertical metal circle with radius 2.45 m. A particle of mass 0.2 kg sits at point A on
top of the sculpture at the top of the circle (on the outside of the circle). The particle is gently displaced and slides
down the circle until it reaches point B, which is level with the centre of the circle. It then falls a further 3.6 m
vertically to hit the ground at point C.
a Use the work-energy principle to find:
the speed of the particle when it reaches point B
ii the speed of the particle when it reaches point C.
b What modelling assumptions have you made?
c How do your answers change if the mass of the particle is doubled?

■ ~ 12 A boy is performing tricks on his skateboard. He skates inside a vertical circle and
accelerates until he is moving just fast enough to reach the top of the circle with
speed 2 m s- 1, using just gravity.
We can model the boy and his skateboard as a particle positioned at his centre of
mass, moving in a circle of radius 0.4 m.
a Find the boy's speed at the bottom of the circle.
b Find the angle between the upward vertical and the radius from the centre of
the circle to the boy when his speed is JIT5 ms- 1 .

9.3 Conservation of energy in a system with non-conservative forces

A non-conservative force is any force for which the work done by that force in moving a
particle between two points is different for different paths taken. Driving force , friction DID YOU KNOW?
and air resistance are examples of non-conservative forces .
Leibniz thought of
When work is done by a non-conservative force it converts energy into movement, such as a
energy as a 'living
driving force converting chemical energy from fuel into kinetic energy. The total energy is
force' and believed that
conserved, but mechanical energy increases.
the total living force of
When work is done against a non-conservative force it converts movement into other forms a body was constant.
of energy, such as heat energy when friction acts. This energy is 'lost' from the mechanical To account for slowing
system. The original situation cannot be recovered by reversing the effect because some due to friction, Leibniz
mechanical energy has been dissipated into non-mechanical energy. The total energy is said that heat consisted
conserved, but some mechanical energy is lost. of the random motion
of the constituent
For example, when a box slides across a rough floor and comes to rest because of friction, parts of matter.
the kinetic energy is being converted into heat energy (and also some sound energy).
 Chapter 9: The work-energy principle and power

WORKED EXAMPLE 9.6

A ball of m ass 50 g falls from rest through a height of 80 cm. It hits the ground and rebounds to a height of 30 cm. Find
the mechanical energy lost in the motion from the start at height 80 cm to the end at height 30 cm above the ground.

Answer
T he initia l and fin al KE are both 0.
T he loss o f GPE is 0.050 x 10 x (0.80 - 0. 30) = 0.25 J Convert d ata to kg and m.

Energy lost = 0.25 J This will be dissipated as heat and sound.

The system in Worked example 9.6 is not a closed system of conservative forces because the
reaction from the ground is a non-conservative force. We cannot recover the original position
by reversing the bounce.

WORKED EXAMPLE 9.7

A crate of mass 50 kg slides across a rough horizontal floor. The crate has an initial speed of 3 m s- 1 and is brought
to rest by friction. The distance travelled by the crate is 4 m. Find the coefficient of friction between the floor and
the crate.

■
Answer

T he motion is horizont al so there is no change in GPE.

The loss of KE is ½x 50 x ( 32
- 0 2 ) = 225 J

Energy lost = 225 J

This will be dissipated as heat and sound .

So work done against frict ion = 225 J

U se the work- energy principle.

We now need to fin d t he fri ctio na l fo rce.

R

weight, 500 N

R = SOON
Resolve vertically.

and F = µR Crate is moving so friction is limiting.

= 500µ
so work done by fr iction = -(-500µ ) x 4 = 2000 µ
Work done = force x distance

Hence, 2000µ = 225

µ = 0. 11 25
I

Cambridge International AS & A Level Mathematics: Mechanics

WORKED EXAMPLE 9.8

A parcel of mass 3 kg slides 3.5 m down a rough slope inclined at 20° to the horizontal. The coefficient of friction
between the parcel and the slope is 0.5. When it reaches the bottom of the slope the parcel has speed 8 m s- 1• Use
the work- energy principle to find the speed of the parcel at the top of the slope.

Answer
R

~ 30N

R = 30cos20 = 28.2 N Resolve perpendicular to slope.

F = 0.5 x 28 .2 = 14.1N Parcel is moving so friction is limiting.

F = µR = 0.5 x R

Increase in KE+ increase in GPE = work done The only force that does work is friction and
work will be done against friction .

Let the speed at the top of the slope be um s- 1•

Increase in KE=½ x 3 x (8 2 - u2 )

■
= (96-1.5u 2 )J

Increase in GPE = 3 x 10 x (-3.5sin20)

Parcel slides 3.5 m down slope so vertical
= -35.9J
drop = 3.5 sin 20 = 1.20 m. The GPE decreases
by35.9J.
Increase in KE+ increase in GPE
= (96 - l.5u 2 ) - 35.9 = (60.l- l.5u 2 )J

The work done against friction= 14.1 x 3.5

= 49.3 J
2
Hence, (60.l - l.5u ) = 0 - 49.3 Use the work- energy principle.

l.5u 2 = 109.4 WD by forces that speed up the parcel= 0 J.

u = 8.54ms- 1 WD against forces that slow down the

parcel= 49.3J.
 Chapter 9: The work-energy principle and power

WORKED EXAMPLE 9.9

A car of mass 1500 kg (including the driver) is travelling at 64 km h- 1 along a level road when the driver sees a
ball roll out onto the road and in front of the car. The driver takes 2 s to react and then applies the brakes, using
the maximum braking force. The car comes to rest, just missing the ball, after travelling total distance of 80 m
from when the driver first saw the ball. Assuming that the wheels lock as soon as the brakes are applied (so the car
slides) and that air resistance can be ignored, find the coefficient of friction between the tyres and the road.

Answer

Convert speed to ms- 1•

Distance travelled at 17.78 m s- 1 = 2 x 17.78 = 35.56 m Distance travelled while the driver is reacting
to seeing the ball.

Distance travelled under braking= 80 - 35.56 = 44.44 m

Increase in kinetic energy= 0 - 0.5 x 1500 x 17.78 2

= -237037 J
So work done against friction = 237 037 J The car is slowed by the friction between the
tyres and the road.

. . l .- 237 037 Work done against friction =

Average fnct1ona 1orce = - - -
44.44 average frictional force x distance
= 5333 .33 N
R = mg
= 1500 X 10
= 15000N
F =µR
5333 .33
µ = 15000
= 0.356

4J 1 A helter-skelter ride at a fairground consists of a spiral-shaped slide that people slide down on mats. The top
of the slide is 7 m higher than the bottom. The average frictional resistance is 50 N. A boy of mass 60 kg slides
down the helter-skelter, starting from rest. At the bottom of the ride the boy has speed 10 m s- 1.
a Find the length of the slide.
b Work out the amount of mechanical energy that has been lost.

c What form of energy has most of this loss of mechanical energy been changed into?

2 A box of mass 10 kg slides 3 m down a rough slope inclined at 30° to the horizontal. At the top of the slope the
speed of the box is 3.25 m s- 1 and at the bottom of the slope the speed of the box is 5 m s- 1. Find the coefficient
of friction between the box and the slope.
 Cambridge International AS & A Level Mathematics: Mechanics

3 A sack of mass 12 kg slides down a ramp, starting from rest at a height of 2m above the ground. The sack
reaches the ground with speed 6 m s- 1• Work out the amount of mechanical energy that has been dissipated.

4 A skydiver of mass 80 kg falls 1000 m from rest and then opens his parachute for the remaining 2000 m of his
fall. Air resistance is negligible until the parachute opens. The skydiver is travelling at 5 m s- 1 just before he
hits the ground. Find the average resistance force when the skydiver is falling with the parachute open.

~ 5 A tile of mass 1kg slides 3 m down a roof inclined at 20° to the horizontal. The tile then falls 5 m under gravity
to hit the ground with speed 8 ms- 1. Find the frictional force between the tile and the roof.

6 A model racing car of mass 50 g is released from rest at the top of a downward-sloping track. It travels along
the track under the action of gravity. The resistance to the motion of the car is 0.05 N. The car comes to a stop
on a horizontal piece of track that is 2 m lower than the top of the track. Find the distance that the car has
travelled.

7 A diver jumps from a 10 m board above a swimming pool. The diver has an initial velocity of um s- 1 upwards.
The horizontal component of the diver's path is negligible. A constant resistance of 0.5 N kg- 1 acts on the diver.
Find an expression for the height of the diver above the pool at the highest point of her dive.

C, 8 A golf ball of mass 45.9 g is hit from a tee with speed 50 ms- 1• The ball lands in a pond that is 5 m lower than
the tee. When the ball lands in the pond it has travelled along a curved path of length 160 m. The resistance
acting on the ball has magnitude 0.3 N.
a Find the speed of the ball just before it hits the water.
The water immediately absorbs 8 J of energy from the ball. The ball then sinks vertically downwards to reach

■
the bottom of the pond. The resistance acting on the ball has magnitude 3 N and the ball just comes to rest as
it reaches the bottom of the pond.
b Find the depth of the pond.

0 9 A golf ball of mass 45.9 g is hit from a tee with speed 180 kmh- 1• The ball rises to a height of 20 m, having travelled
along a curved path oflength 61.87 5 m. At the highest point of its path the ball is travelling at 144 km h- 1•
a Find the magnitude of the average resistance force acting on the golf ball.
The ball travels a further 105.8 m along a curved path to land on the green. The green is 4 m lower than the
tee. The average resistance remains unchanged.
b Find the speed of the ball just before it lands on the green.
The ball is travelling vertically when it lands on the green, where it is immediately brought to rest.
c Show that the energy absorbed by the green is 30.3 J.

~ 10 Two particles, A and B, are connected by a light inextensible string. Particle A has mass 2 kg and particle B
has mass 5 kg. The string passes over a pulley and hangs vertically with particle A and particle B on each
side of the pulley. The pulley, however, is not smooth and 10 J of energy is dissipated for each rotation of the
pulley. The system is released from rest, and the particles reach a speed of 0.2 m s- 1 after each moving 1.6 m.
a Work out how many rotations the pulley has made.
b If the string passes over the pulley without slipping, work out the radius of the pulley.

11 A woman of weight 54 N skis from point X to point Y. The distance from point X to point Y is 16.2 m.
Point Y is 3m lower than point X. At point X she has speed lms- 1 and at point Y she has speed 7 ms- 1.
a Use the work-energy principle to work out the average resistance force that acts on the woman.
b Give an expression for the average resistance force if, instead, her speed at point Y is v ms- 1.
 Chapter 9: The work-energy principle and power

~ - 12 A piece of sculpture includes a vertical metal circle with radius 2.45 m. A particle of mass 0.2 kg sits at point
A on top of the sculpture at the top of the circle (on the outside of the circle). The particle is gently displaced
and slides down the circle until it reaches point B, which is level with the centre of the circle. It then falls a
further 3.6 m vertically to hit the ground at point C. When the particle reaches point C it has speed 10 m s- 1•
Air resistance can be ignored.
a Work out how much mechanical energy has been lost by the particle in travelling from A to C.
b Show that the average frictional force between the surface and the particle is 0.546 N.
c It is claimed that the coefficient of friction between the surface and the particle is 0.273. Explain how this
value has been calculated and why it is too small.

9.4 Power
Energy can be put into a system by an engine converting fuel (chemical energy) into a
driving force. The work done by the engine is given by:
work done = force x displacement
= Fs
Power is the rate of doing work, so the average power generated by an engine is given by:
average power = work d~ne by the engine
time taken
Fs

a REWIND

■
It always takes the same amount of work to make a car speed up from Oto 100 kmh- 1, but a
more powerful engine will get the car to 100 km h- 1 more quickly than a less powerful one. Recall, from
Chapter 6, that when
Over a very small time interval, bt, the driving force Fis constant and the rate of doing
we differentiate
work is given by: displacement with
8s respect to time we get
power= F 8t velocity, and when we
differentiate distance
As bt gets smaller, 8s approaches the limit ds and we get: with respect to time
8t dt
we get speed.
ds
power = F - = Fv
dt

The rate at which an engine works is called the power of the engine.
Power = rate of doing work = Fv
where F, the driving force, is constant.

Power is measured in J s- 1 or watts (W). We often use units of 1000 watts (kW).

Power is a scalar quantity. Strictly speaking, this involves a product of vectors, but as we
are usually only concerned with motion in one direction along a line, we can say that
power = force x speed
 Cambridge International AS & A Level Mathematics: Mechanics

MODELLING ASSUMPTIONS

When an engine converts energy from fuel into other forms of energy, some energy is
lost in the form of heat and sound. We will ignore this and assume that the stated power
of an engine is the measure of the rate of energy conversion to mechanical energy by the
engine and that no energy is lost. However, energy losses do need to be considered by
manufacturers of machines so they can try to minimise energy losses and improve efficiency.

WORKED EXAMPLE 9.10

A car of mass 1500 kg is being driven along a level road. It accelerates from Okmh- 1 to 100 kmh- 1 in 10s. Air
resistance and friction may be ignored. Find the average power generated by the engine.

Answer
lOOkmh- 1 = 27.8ms~1
Conver t speed to ms- 1•

Work done = 0.5 x 1500 x ( 27.8 2 - 0 2 ) = 578 7041 Increase in KE = WD by engine

Average power = 578 704 + 10
Average power = WD 7 time taken
= 57870.4 W
= 57.9 kW (to 3 significant figures)

■ For a given power, the driving force generated will be greater at lower speeds and smaller
at higher speeds.
It is important to be
For example, a car pulling off from stationary under maximum power has a low speed and so has
very careful to use
a greater driving force than when it is moving more quickly. This means that the resultant force is the resultant force (or
greater and so, using Newton's second law, the acceleration is greater and the car speeds up quickly. net force) in Newton's
second law but only
As the speed increases, the driving force (under maximum power) decreases and so the resultant
the driving force in the
force decreases and, hence, the acceleration decreases. If the maximum power is maintained,
power equation. You
the acceleration will eventually become Oand the driving force is balanced by the resistances. may find it helpful to
At this point, the car is moving at its maximum speed and this cannot be increased further. denote the driving force
We can use the maximum power output of an engine to find the maximum speed that it can by D rather than F .
generate. At the maximum speed the acceleration is Om s- 2 and, hence, the resultant force is ON.

WORKED EXAMPLE 9.11

A car of mass 1500 kg has an engine that has a maximum power output of 200 kW. The resistance to motion is typically
5000 N. Find the maximum speed that the car can achieve on a level road (ignoring legal speed restrictions).

Answer

$
R
A diagram is helpful.
resistance driving force
Resistance = 5000 N
weight
 Chapter 9: The work-energy principle and power

Resultant force = driving force - 5000

At the maximum speed, acceleration= 0.
Use Newton's second law.
Resultant force = 0
Driving force = 5000 N

Power = driving force x speed Note: we have not used the mass of the car,
= 5000v so the maximum speed is the same for any
200 000 engine that has this maximum power and this
\I=--- = 40rns- l resistance to motion .
5000

WORKED EXAMPLE 9.12

A car of mass 1500kg has an engine that has a maximum power output of200kW. The resistance to motion is
typically 5000 N . Find the instantaneous acceleration of the car when the engine is working at its maximum power
and the car is travelling at 20 m s- 1•

Answer
R
A diagram is helpful.
resistance driv ing force
Resistance = 5000 N
weight

Power = driving force x speed

Driving force = 200 000 + 20
= 10 OOON
Resultant force = driving force - resistance
= 10000 - 5000
= 5000N
Resultant force = mass x acceleration Use Newton's second law.
Instantaneous acceleration = 5000 + 1500
= 3.33 m s- 2

WORKED EXAMPLE 9.13

A car of mass 1500 kg has an engine that has a maximum power output of 200 kW. The resistance to motion is
typically 5000 N. Find the instantaneous acceleration of the car when the engine is working at its maximum power
and the car is travelling at 20 ms- 1 up a hill that is inclined at sin- 1 0.28 to the horizontal.

Answer

D A diagram is helpful.
Resistance = 5000 N
Component of weight down slope
= 15000 sin 0 = 4200N
sin 0= 0.28
 Cambridge International AS & A Level Mathematics: Mechanics

Power = driving force D x speed

D = 200000-:- 20
= 10000N
Resultant force (up slope) = 10 000 - 5000 - 4200
= SOON
Resultant force = mass x acceleration Use Newton's second law.
Instantaneous acceleration = 800 -:- 1500
= 0.533ms- 2

1 A car of mass 2000 kg travels in a straight line on a horizontal road. The car accelerates from 10 m s- 1 to
20 m s- 1 in 8 s. Assume that resistance can be ignored.
a Use the work-energy principle to find the work done by the driving force .
b Find the average power generated by the engine.

2 The engine of a 5 tonne truck has a power output of 400 kW. The truck is travelling in a straight line on a
horizontal road. The resistance to motion is 20 000 N . Find the maximum speed the truck could achieve.

The maximum power of a boat engine is 140 kW. The boat is subject to a resistance force of 10 000 N. Find

■
3
the maximum speed the boat can achieve when travelling in a straight line.

4 A model train of mass 200 g is moving in a straight line on a level track. The train accelerates from 2 m s- 1 to
8 m s- 1 in 3 s. Find the average power generated by the engine.

5 The maximum power ofa boat engine is 120kW. The maximum speed the boat can achieve is lOms- 1. Find
the resistance force acting on the boat when it is travelling at its maximum speed and the engine is working at
maximum power.

6 A car of mass 1800 kg is being driven, at a constant 15 ms-1, against a resistance of 2000 N, up a hill inclined
at 10° to the horizontal. Find the rate at which the engine is working.

7 A car of mass 1600 kg is being driven up a hill inclined at 10° to the horizontal. The car has an initial speed
of 10 m s- 1 and a final speed of 12 m s- 1 after 60 s. Air resistance and friction may be ignored. Find the average
power generated by the engine.

8 A car of mass 1200 kg accelerates up a hill against a resistance of 263 N. At a certain point on the hill the road
is inclined at 8° to the horizontal. The engine is working at 75kW and the car is travelling at 25ms- 1. Find the
acceleration of the car at this instant.

$ 9 A small van of mass 1600 kg accelerates from rest in a straight line along a horizontal road. The resistance
from friction and air resistance is 2400 N throughout the motion. The engine works at a constant rate of
48kW.
a Write down an expression for the driving force when the van is travelling at v ms- 1.
b Write down an expression for the acceleration of the van when it is travelling at v m s- 1.
c Explain why the power cannot be constant.
 Chapter 9: The work- energy principle and power

4D 10 A powerboat of mass 500 kg travels in a straight line at its maximum velocity against a resistance of 15 N.
The engine of the powerboat has a maximum power output of 750 W.
a Find the maximum velocity.
In different weather conditions the same powerboat has a maximum velocity of only 25 m s- 1
b State what has changed in the model and give the new value of this quantity.

~ 11 A van of mass m kg moves up a hill that is inclined at 2° to the horizontal. The engine works at a constant rate
of 30 kW and the resistance (from friction and air resistance) is a constant 400 N. When the van is travelling at
20 ms- 1 it has acceleration 0.1 ms-2 • Find the value of m .

0 ~ 12 Car A, of mass 1250 kg, is travelling along a straight horizontal road at speed 10 m s- 1• The engine works at a
constant rate of25 kW and the resistance is a constant 500 N. After 5 s the speed of the car has increased to v ms- 1.
a Use the work-energy principle to find the amount of energy that is dissipated and , hence, find the distance
travelled in the 5 s.
b Find an expression for the acceleration at time 5 s as a function of v and show that the acceleration is not
constant.
Car B travels along the same road , starting with speed 10 m s- 1 and accelerating at a constant rate for 5 s. After
5 s the two cars have the same speed and also have the same acceleration as one another.
c Show that v must satisfy the equation v2 - 8v = 100 and, hence, find the speed of the cars at the end of the 5s.

• The work-energy principle states that for any motion:

increase in kinetic energy = total work done by all forces
1
2 mv
2
- 1 mu-,
2
"°'
=.£. Fs

where the 'total work done' is the sum of the work done by forces with a component in the direction of motion
(forces that speed up the motion) minus the work done against forces with a component in the direction opposing
the motion (forces that slow down the motion).
• The work-energy principle can also be written as:
increase in mechanical energy= total work done by forces that help to speed the body up
- total work done by forces that help to slow the body down
(In both cases 'forces' excludes the weight of the body.)
• Kinetic energy and potential energy are types of mechanical energy. Other forms of energy (heat, light, sound,
chemical , electrical , nuclear etc.) are non-mechanical energy.
• A consequence of the work- energy principle is that for a system of conservative forces the total mechanical energy
is constant. We call this conservation of mechanical energy.
• Power is measured in Watts. The power of an engine, in watts, is the rate at which that engine can work:
power = work done -'- time taken
• The power of an engine is also the product of the driving force of the engine and the velocity in the direction of
the driving force:
power= Fv , where Fis the driving force of the engine.
 Cambridge International AS & A Level Mathematics: Mechanics

Fifri·i#i=MiiiiU@iifiiUMhii
@ 1 A car of mass 1250 kg is travelling along a straight horizontal road with its engine working at a constant rate of
PW. The resistance to the car's motion is constant and equal to RN. When the speed of the car is 19 ms- 1 its
acceleration is 0.6 m s-2 , and when the speed of the car is 30 m s- 1 its acceleration is 0.16 ms-2 . Find the values
of P and R . (6)

Cambridge International AS & A Level Mathematics 9709 Paper 43 Q2 June 2011

~ 2 Particle X, of mass 2 kg, and particle Y, of mass m kg, are attached to the ends of a light inextensible string of length
4.8 m. The string passes over a fixed small smooth pulley and hangs vertically either side of the pulley. Particle X is
held at ground level, 3 m below the pulley. Particle X is released and rises while particle Y descends to the ground.
a Find an expression, in terms of m, for the tension in the string while both particles are moving. (2)

b Use the work- energy principle to find how close particle X gets to the pulley in the subsequent motion. (21
3 A van of mass 1500 kg starts from rest. It is driven in a straight line up a slope inclined at angle a to the
horizontal, where sin a = _!__ _The driving force of the engine is 2000 N and the non-gravitational resistances
10
total 350 N throughout the motion. The speed of the van is v ms- 1 when it has travelled x m from the start. Use
the work- energy principle to find v in terms of x. [6)
9 4 A car of mass 1000 kg travels in a straight line up a slope inclined at angle a to the horizontal, where
sin a = 0.05. The non-gravitational resistances are 200 N throughout the motion.
a When the power produced by the engine is 50 kW, the car is accelerating at 1.2 m s- 2 . Find the speed of the
car at this instant. [4]

■ b What would happen to the speed if the mass of the car increased?
c What would happen to the speed if the power produced by the engine decreased?
(1)
[1]
5 A truck of mass 3000 kg starts from rest. It is driven in a straight line up a slope inclined at angle a to the
horizontal, where sin a = 0.08. The driving force of the engine is 7000 N and the non-gravitational resistances
total 4000 N throughout the motion. The speed of the truck is v ms- 1 when it has travelled x m from the start.
Find, to 3 significant figures, the value of k for which x = kv 2 . (6)
6 A car of mass 1200 kg is driven along a straight horizontal road against a resistance of 5000 N. The engine has
a maximum power output of 100 kW.
a Find the maximum speed the car can reach. [2)
1
b Find the power being used when the car is travelling at a speed of 15 ms- and accelerating at -2 ms-2. (4)
7 A box of mass 2 kg is pulled up a rough slope by a rope. The rope passes over a smooth pulley and is attached,
at the other end, to a block of mass 4 kg with that end of the rope hanging vertica11y. The slope is inclined at 20°
to the horizontal and the coefficient of friction between the slope and the box is 0.3. The system is released from
rest. Use the work-energy principle to find the speed of the box when it has moved lm up the slope. (6)
8 A car of mass 1600 kg travels down a straight road inclined at an angle 0 to the horizontal. The power produced
by the engine is 24 kW and the non-gravitational resistance is a constant 1600 N.
a Find the driving force when the car is travelling at a constant 20 ms- 1• (2)

b Find the value of sin 0. (4)

 Chapter 9: The work-energy principle and power

@ 9 A lorry of mass 14 000 kg moves along a road starting from rest at a point 0. It reaches a point A, and then
continues to a point B which it reaches with a speed of 24ms- 1. The part OA of the road is straight and
horizontal and has length 400 m. The part AB of the road is straight and is inclined downwards at an angle of
0° to the horizontal and has length 300 m.
For the motion from Oto B, find the gain in kinetic energy of the lorry and express its loss in potential
energy in terms of 0. [3]
The resistance to the motion of the lorry is 4800 N and the work done by the driving force of the lorry from
0 to B is 5000 kJ.
ii Find the value of 0. [3]
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q4 June 2015
10 A block of mass 25 kg is dragged across a rough horizontal floor, using a rope that makes an angle of 30° with
the floor. The coefficient of friction between the floor and the block is 0.25. The tension in the rope is TN and
air resistance can be ignored. After travelling a distance of 5 m, the speed of the box has increased by 2 m s- 1•
a Find the work done against friction, in terms of T. [3]

b Use the work-energy principle to find, in terms of T, the average of the initial and final speeds. [4]
@0 11 A light inextensible rope has a block A of mass 5 kg attached at one end, and a
block B of mass 16kg attached at the other end. The rope passes over a smooth B
pulley which is fixed at the top of a rough plane inclined at an angle of 30° to the
horizontal. Block A is held at rest at the bottom of the plane and block B hangs
below the pulley (see diagram). The coefficient of friction between A and the plane
is ~. Block A is released from rest and the system starts to move. When each of ■
the blocks has moved a distance of x m each has speed v m s- 1•
Write down the gain in kinetic energy of the system in terms of v. [1]
ii Find, in terms of x,
a the loss of gravitational potential energy of the system, [21

b the work done against the frictional force. [31

iii Show that 21 v2 = 220x. (21

Cambridge International AS & A Level Mathematics 9709 Paper 42 Q5 June 2014
~ 12 Particle W, of mass 3 kg, and particle X, of mass 5 kg, are attached to the ends of B
a light, inextensible string of length 4m. The string passes over a small smooth
pulley fixed at the top of a fixed triangular wedge, ABC. The angles BAC and
BCA are each 45° and the side AC is fixed to horizontal ground. The distance
from A to C is 3../2. m.
Surface AB is smooth and surface BC is rough , with coefficient of friction Js. 3{2m

Particle W is held at the bottom of the slope AB and is then gently released.
a Find the work done against friction when particle X moves a distance x m. [31

b Find the change in the total potential energy when particle X moves a distance x m. (31
c Use the work- energy principle to find the speed of the particles when particle X reaches
the ground at C. [21

d Explain why the work done by the tension does not need to be included in the work-energy calculation. (11
 Cambridge International AS & A Level Mathematics: Mechanics

lii·Mii•jltiilMi@iiiiMHii
1 A ball of mass 0.5 kg is dropped from a height of 0.45 m and bounces to a height of 0.2 m . Find the change
in momentum of the ball during the bounce. [5)
2 A box of mass 3 kg is pulled 10 m from rest up a smooth slope, which is at an angle of 8° to the horizontal.
The box is pulled by a string with constant tension, parallel to the line of greatest slope, for 6 s. Find the work
done by the tension in the string. [5]

3 Object A has mass 3 kg and is moving with velocity 10 m s- 1 towards object B, which has mass 6 kg and is
stationary. After they collide, object A bounces back with speed 2 m s- 1• Object B then collides with object C,
which has mass 17 kg and is stationary. Object C moves at 3 m s- 1 after this collision. Deduce whether or
not there will be a third collision and explain your reasoning. [5]

4 A cyclist and his cycle have a combined mass of 80 kg. He works at a rate of 600 W while cycling along a
straight horizontal road. There is a constant resistance of RN .
a Given the cyclist has a maximum velocity of 12ms- 1, find the value of R. [2)
b Find the speed of the cyclist when he is accelerating at 0.625 m s-2 • [3)

5 A particle of mass 3 kg is projected with speed 4 m s- 1 towards a stationary particle of mass 5 kg .

a The particles coalesce on impact. Find the speed at which the particles move after the collision. [2]
b The coalesced particles then move towards a particle of mass 2 kg. After the collision the coa lesced
particles remain stationary and the 2 kg particle moves with speed 3 m s- 1• Find the speed and direction of

■
motion of the 2 kg particle before the collision. [3)

6 A car of mass 1200 kg travels along a straight horizontal road, starting at a point A. The resista nce to motion
of the car is 800 N.
a The car travels from A to a point Bat a constant speed in 12s. The power of the engine is 20kW. Find the
distance AB. [3)
1
b The car travels from B to a point C with an increased power of 24kW, reaching C with a speed of 28ms-
after 37 s. Find the distance BC. [3)

7 The diagram shows a vertical cross-section, ABCD, of a fixed surface. AB and CD A 1: D

are smooth curves and BC is a rough horizontal surface. A is at a vertical I
3.2m : I
I
height 3.2 m above BC. A particle of mass 2 kg is released from A and travels : I
I

along the surface to D. ·---- -- B C

a Find the speed of the particle at B. [2)

b Given that the particle reaches C with a speed of 4ms- 1, find the work done against the resistance to
motion as the particle moves from B to C. [2)

c The particle reaches the point D. Find the maximum vertical height of D above BC. [3)

8 A car of mass 2000 kg climbs a straight hill, ABC, which makes an angle 0 with the horizontal, where sin 0 =/ .
6
For the motion from A to B , the work done by the car's engine is 256 kJ and the resistance to motion is RN.
The length of AB is 200 m. The speed of the car is 20 m s- 1 at A and 16 m s- 1 at B .
a Find the value of R . [4)

b From B to C, the work done by the engine is 388 kJ. The resistance to motion remains the same as that
between A and B. The speed of the car at C is 12 m s- 1• Find the distance BC. [3)
 Cross-topic review exercise 3

9 A particle, P, of mass 4 kg is projected with speed 9 m s- 1 along rough horizontal ground. The coefficient of
friction between P and the ground is 0.4. After 7 m it strikes a particle, Q, of mass 3 kg. The coefficient of
friction between Q and the ground is also 0.4. Q comes to rest after 2m.
a Show that the speed of P immediately before the collision is 5 m s- 1 and find the speed of Q immediately
after the collision. (5]

b Find the distance between Q and P when both have come to rest. (3)

@ 10 A lorry of mass 24 000 kg is travelling up a hill which is inclined at 3° to the horizontal. The power developed
by the lorry's engine is constant, and there is a constant resistance to motion of 3200 N.
When the speed of the lorry is 25 m s- 1, its acceleration is 0.2 m s- 2• Find the power developed by the
lorry's engine. (4)

ii Find the steady speed at which the lorry moves up the hill if the power is 500 kW and the resistance
remains 3200 N. (5)
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q3 November 2015
@ 11 A box of mass 25 kg is pulled, at a constant speed, a distance of 36 m up a rough plane inclined at an
angle of 20° to the horizontal. The box moves up a line of greatest slope against a constant frictional force
of 40 N. The force pulling the box is parallel to the line of greatest slope. Find
the work done against friction, (1)

■
ii the change in gravitational potential energy of the box, [2)

iii the work done by the pulling force. (2)

Cambridge International AS & A Level Mathematics 9709 Paper 41 Q2 June 2016

@ 12 A car of mass 1000 kg is moving along a straight horizontal road against resistances of total magnitude 300 N.
Find, in kW, the rate at which the engine of the car is working when the car has a constant speed of 40 m s- 1. (3)

ii Find the acceleration of the car when its speed is 25 m s- 1 and the engine is working at 90% of the power
found in part i. [3)

Cambridge International AS & A Level Mathematics 9709 Paper 41 Q3 June 2016

@ 13 A block of mass 25 kg is pulled along horizontal ground by a force of magnitude 50 N inclined at 10° above
the horizontal. The block starts from rest and travels a distance of 20 m. There is a constant resistance force
of magnitude 30 N opposing motion.
Find the work done by the pulling force. [2)
ii Use an energy method to find the speed of the block when it has moved a distance of 20 m. [2)

iii Find the greatest power exerted by the 50 N force. [2)

SON

~~ _1_u~L- - ~
~-~--~-~
~oJC__:-:_-:-::
__::-_:::: __:-=-_ -=--
__-::-: =-=-

After the block has travelled the 20 m, it comes to a plane inclined at 5° to the horizontal. The force of 50 N is
now inclined at an angle of 10° to the plane and pulls the block directly up the plane (see diagram).
The resistance force remains 30 N.
iv Find the time it takes for the block to come to rest from the instant when it reaches the foot of the
inclined plane. [4)
Cambridge International AS & A Level Mathematics 9709 Paper 41 Q6 November 2016
 Cambridge International AS & A Level Mathematics: Mechanics

iijifii4itiitffiil1Hiililiil
Time allowed is 1 hour 15 minutes (50 marks).

Answer all the questions.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees,
unless a different level of accuracy is specified in the question.

Where a numerical value for the acceleration due to gravity is needed, use 10 m s-2 .

The use of an electronic calculator is expected, where appropriate.

You are reminded of the need for clear presentation in your answers.
1 Four horizontal forces act at a point. The forces have magnitudes F N, 5 N, 4 N and 10 N. The F N force acts
at an angle of 90° to the 5 N force and at an angle of 120° to the 10 N force. The 4 N force acts at an angle a to
the 5 N force, as shown in the diagram. The forces are in equilibrium. Show that a = 23.8° and find the value
~F. ~
5N
4N

FN

IO N

■
2 A car has a maximum power output of 60 kW. The car is driven at its maximum power in a straight line
on a horizontal road against a constant resistance. The car travels 200 m at a constant speed of 32 m s- 1•
a Find the resistance. (2)
b Find the work done by the engine. (2)

3 Two balls are travelling directly towards one another in a straight line. The first ball has mass 1.2 kg and is
initially moving at 3ms- 1; the second ball is initially moving at Sms- 1. The balls hit each other and after the
impact each ball has reversed its direction of travel and is moving at half its original speed.
a Find the mass of the second ball. (3)

b Find the loss of kinetic energy in the impact. (2)

4 A girl roller-skates in a straight line along a horizontal track. She starts from rest and accelerates at a constant
rate for 20 s, during which time she covers a distance of 100 m. She travels at a constant speed for the next 10 s
and then slows at a constant rate for 5 s until she stops.
a Find the total distance that the girl skates. (4)
On another occasion the girl skates along the same track, starting from rest, with the same acceleration as
before. This time she accelerates for only 10 s before travelling at a constant speed.
b How long does she take to travel 100 m? (2)
 Practice exam-style paper

5 A man runs along a straight track, starting from rest. He accelerates so that his velocity, v1 m s- 1, is given by
v1 = _!_ t 2 - __!_ t 3, where t is the time, in seconds, from when he starts to run. The man runs until his acceleration
3 27
a 1 = 0 and then runs at constant speed V for 50 m. He then accelerates again, with acceleration a2 m s-2 , given
by a2 = 0.048T 2 - 0.008T 3, where T is the time, in seconds, from when he starts the second acceleration phase.
The man runs until he comes to rest and then stops.
a Find V.
b Show that the man comes to rest when T = 10. [3)
c Find the total time for which the man runs. (6)

6 Particles A and B, of masses 0.4kg and 0.1kg, respectively, are attached to the ends of a light inextensible string.
Particle A sits on a rough horizontal table and the string passes over a small smooth pulley at the edge of the table.
A
l

The system is released from rest and particle B descends 1 m in 2 s.

■
a Calculate the frictional force acting on part A. [5)

Particle Bis now on the floor. Particle A continues until it comes to rest, without having reached the pulley.
b Find the total distance travelled by particle A. [5]
7 A crate of mass 29 kg slides down a slope inclined at an angle 0 to the horizontal, where sin 0 = _!2_. For the
145
first part of the slope the coefficient of friction between the slope and the crate is } .
5
a Find the acceleration of the crate down the slope on this part of the slope. 15]
When the crate is moving at 0.3 m s- 1 down the slope, the surface of the slope changes, although the angle of
the slope is unchanged. After travelling 0.58 m on this second part of the slope the crate is moving at 0.1 m s- 1•
b Find the loss in the kinetic energy of the crate. [2]
c Find the loss in the potential energy of the crate. 121
d Find the average resistance force on the crate while it is travelling on this second part of the slope. [3]
I

Cambridge International AS & A Level Mathematics: Mechanics

Answers 7 48m

8 a 0.6ms-2
lV city a ac eleration b The sprinter can maintain a constant accelera-
tion and we are ignoring the shape of the
Prerequisite knowledge
3 sprinter's body and the different positions it
1 a x = - 3 or x = 5 b x = I or x = - -
2 takes when running, by considering the
c x = -0.907 or x = 2.57 sprinter having a single position at any point
in time.
2 a x = I and y = 2 b x = I and y = 3

Exercise lA
10 8ms- 1
1 8ms- 1
11 He can pedal because doing nothing he will
2 63m
arrive at the bend with velocity 10.8 m s-1•
3 a 6s
b The cheetah can instantly reach that speed. Exercise lC
The gazelle remains stationary. 1 a s=32m b s = 72m
c a= 2ms-2 d a= 3 ms- 2
4 8 minutes 20 seconds
e a= 4ms-2 f u=3ms- 1
5 2.94s g v = lms- 1 h s = 14m
6 a 6.3ms- 1 2 a t = 4s b t = 6s

■
b The speeds are average speeds or the runner C t = 4s
instantaneously changes speed between sections.
3 v=-lms- 1
7 a 45m
c Sms- 1
5 a v=9ms- 1
8 12.Sms- 1
b Positive acceleration means v must be larger
9 0.09s than u.
10 80s 6 50m
11 15840m 7 4.Sms- 2
12 1.014m in 0.78s 8 800m
13 a Proof b Proof

14 a Proof b Proof 10 a lOms- 1

b The deceleration is constant.
Exercise 1B
1 2 m s-2 11 0.4m past the target

2 2.5 ms- 2 12 No, the ball stops 0.4 m short of the hole.

3 1.5s 13 The car cannot brake safely in time, but can

accelerate to get past the lights before they turn
4 19ms- 1 red, so must accelerate.
5 3ms- 1 14 a Proof b Proof
6 14ms- 1
 Answers

15 Proof C s(m)

16 Proof 175

JOO
Exercise 1D 75
1 a s(m) 0
5 15 I (s)
30
d s(m)

0
10 t (s)
1.25
b 0
0.5 I t (s)

-18 .75

3 a s(m)
C s(m)
187.5 - -------

0
37 .5 - ---
t (s) 0
5 15 I (s)

■
-10
b s(m)

85
d s(m) 60

C s(m)
50: I (s)
I

-200

0 I------;:-;--:,-:;- ,t~)
2 a s(m)
-20
200

-60

d s(m)

b s(m)
68 ---------------
2.25
50

Ol--- - - - - ; ; -
0_5:;-- "t(:;;s\)
-88

-100
'

Cambridge International AS & A Level Mathematics: Mechanics

4 s = l6t - 60 so t = 3.75s C v(ms- 1)

5 a p = - 0.5, q = 10 and r = 0

b s = - 50 + lOt - -1 t 2
2
So a= - l ms-2 and u = lOms- 1 0
t(s)
-4
6 a s(m)

3000
d v(ms- 1)

0
t (s)
6
b s = 30t and s = 3000 - 20t
C They meet at t = 60 s at a distance of 1800 m 0
20 t (s)
from junction 1.
2 a v(ms- 1)
7 a s(m)

5000 20

0
40 t (s)
0
t (s)

■
b t = 140s and s = 3500m

8 s = 8 t and s = 25 + l0(t - 5) - 20
- 21
12.5ms- 2
9 a 115s b v(ms- 1)
b Rowing boats can travel at constant speed (in
reality they tend to increase speed with the
strokes and decrease speed between strokes).

10 0.02ms-2 0
5.5: t (s)
I
11 60m -5

12 50s
C v(ms- 1) d v(m s- 1)
13 1.2m
35
14 6s

15 h = gt1t2 0
2 21 t (s) 0
7 10.5 t (s)

Exercise lE
1 a v(m s- 1) b v(ms- 1)
3 160m
4 6.25m
5 96m
6 t (s)
5 t (s) 6 1260m

7 22.5s
 Answers

8 11.25 s s(m)

9 12
20
17 .5
10 a 8
b The boat accelerates at a constant rate and
moves instantaneously to constant speed at
the change between the two stages.

11 1.19 o----~------+
2.5 I (s)

12 a 62m b 23. ls

13 a He leads by 69 m. b 64.7 s 3 s(m)

14 a v(ms- 1)

4
01---..___ _ __ _ :i,..

-6
6 t (s)

b v=26 - 3t and v= - S(t-6) c 2s

15 a The graph is a triangle and area under graph 4 s(m)
20
is s = l_ tv independent of gradients of lines.

■
2
b The graph is a trapezium and area under graph is 5
0
s= ½(t + T) v, independent of gradients oflines. I (s)
2 3 t(s)

Exercise lF
1 v (m s- 1) s(m)
20
10 5 v(m s- 1) s (m)
5
4.9

0 1.4
0 t (s)
•2
I
7 t (s) 0
I 0 2.5 9.5 t (s)
I
I
2,5 9.5 t (s)
I
I I
-8 - 20 I
- 1.75
- 2.25

·:~
2
6 a v(ms- 1) s(rn)
3
3.1
2.9 I
I I
I I
I
.J ____
I 0.918 I
I 0 8 t(s)
I
9.18 t s) - 0.892
I
I
I
- 1.31
- 2.03
-v •I 4:S9

-3 1:1
2 --------- ~ i~i

0
2.5 5 t (s)
Cambridge International AS & A Level Mathematics: Mechanics

b The ball is modelled as a particle so has no 16

width (otherwise the distance the ball travels
between the cushions would be less than 6m)
and the change in velocity is instantaneous.

7 Proof

End-of-chapter review exercise 1

1 a 25m b 5.67 s
T3 =V
2 a 112.5m b 23s
ii Proof, V = 24
3 a 0.008 ms-2 b 0.075ms-2
17 3.2ms- 1 ii 6
2
4 a 6.25ms- 2 b 46.9ms- iii 8.5 iv 1m s-2
5 s(m)

2 Force and motion in one

dim sion
Prerequisite knowledge
54
o-=--.,- - - - ~ - - - 1 t = 0.4s
6 15 t (s)
2 v = 5ms- 1

Exercise 2A
1 1000N
7 a 4.17s
b When the footballer kicks the ball it
instantaneously starts moving at 4 m s- 1.
3 300kg
8 a 135m b 11.6s
4 35m
9 a Proof
5 a 0.8ms- 1
b The closest the lion gets is 0.5 m away at
b The balls are considered as particles and so
t = lOs.
the 1m distance does not need to include the
10 a Proof b 6.5s thickness of the balls.
C 320m 6 2.5N
11 55m 7 2000N
12 640m 8 60kg
13 Proof 9 33 600N
2u 2 s - 2usv - as 2
14
2(u - v) 2 10 80kg

15 0.0167ms- 2 , - 0.2ms- 2 11 15s

ii 40.5m iii 84.5m 12 80 000kg

13 15kg
 Answers

Exercise 2B 14 7.56ms- 1
1 40N
15 3.05m
2 SON
16 109m
3 8
17 5.6N
4 0.625 ms-2
Exercise 2D
5 4950N
1 1050N
6 600N
2 354N
7 25s
3 195N
8 26.25m
4 604.8N
9 a 5340kg
5 420N
b The air resistance is constant or the variations
in air resistance are assumed to be negligible. 6 200N

10 l60ms- 1 7 7.07N

11 7N 8 600N

12 136N 9 a 630N
b The girl is being modelled as a particle, so has

■
13 4000N
only one point of contact with the helicopter,
14 Reduce the driving force to 125 N. otherwise there may be contact forces where
her feet are on the helicopter as well as from
Exercise 2C her seat.
1 700N
10 4 N acting from the top pad pushing downwards
2 18kg on the parcel.

3 2s End-of-chapter review exercise 2

4 Sm 1 44m

2 203N

6 1.25 m 3 40N

7 76.7 ms- 1 4 a 26.3ms- 1

b 3.16s
8 4.5s
5 a 120s
9 150m b 50.7ms- 1

10 a 5.22s 6 a 302 SOON b 6480m

I
b The force provided by the flare remains vertical c The submarine can reach a higher speed I
(often the flare may be blown at an angle). underwater, suggesting there is not as much b:

11 0.085N
resistance when the submarine is underwater. I
7 a 13.Sms- 1 b 3.93 m
12 I0 .6ms- 1
8 Accelerate with force 75N .
13 a 3.2m b Proof
 Cambridge International AS & A Level Mathematics: Mechanics

9 a 0.35N b 6.24ms- 1 2 a = 10.6N, Fy = 3.64N upwards

F
b F = 8.83N, f'( = 3.73N right
10 68.75m
C F = 12.8N, 0 = 38.7°
11 a 140m b 84mm d Fy = 18.3 N upwards, 0 = 47.2°
12 3s e Fx = 2.33 N left, 0 = 52.1 °

13 8.27m 3 F = 6.36, 0 = 62.4°

14 a Proof b 0.0607 4 a T

15 i 5.66ms- 1 ii 0.234s

16 2s
ii P = 8ms- 1,Q = 17ms-1
17 13.5N
2gN
ii v(ms- 1)
b Tsin0 = 2.74
Tcos0 = 12.5
0
C T = 12.8, 0 = 12.4°
t (s)
5 F = 95.3N , 0 = 70.5°
6 a 30N

■
The particle enters the liquid at time 1s with
velocity -lOms- 1 and reaches the bottom of the
F
container at time 1.36 s with velocity -12 m s-1•

3 Forces in two dimensions 20N

Prerequisite knowledge b F = 29.5N, R = 14.8N

1 13m
7 a
2 AB = 6.13m, BC= 5.14m

3 LABC = 33.4°, AB = 10.4m

4
4
5
5 750gN
5
12
b 0 = 34.6°, R = 7020N
Exercise 3A 8 a
1 i a 11.5N right b 9.64 N upwards
ii a 6.88N left b 9.83 N upwards
iii a 7.52N left b 2.74N downwards
iv a 7.18N right b 19.7N downwards
SON
V a 7.46N right b 10.6N upwards
vi a 72.5 N right b 33.8 N downwards b F = 10.4N, R = 47.3N
vii a 8.47N left b 0.741 N downwards
viii a 41.0N left b 113 N upwards
 Answers

9 T = 4260N, R = 51500N b They can prevent the box from moving

horizontally if the strongest person holds the
10 F = 50 N, G = 34 N one at 10° and the next strongest holds the one
at 25°. However, to remain in equilibrium the
11 T1 = 17N, T2 = 39N
contact force would be negative, so the box
12 At 30° the box cannot remain on the ground. cannot remain on the ground.
The force is not large enough to break equilibrium
horizontally, so the box lifts off the ground first. Exercise 3C
13 Proof, F = ( 75../3 - 100)
1 e = 103.5°, </> = 116.9°

2 30 N force makes an angle of 53.1°, 40 N force

14 a = 53.1°, f3 = 67.4°
makes an angle of 36.9°.
Exercise 3B 3 T=56.2N,0 = 13.2°
1 a 123 N in the given direction
4 0 = 153.2°, F = 68.0N
b 12. 7 N in the given direction
c 5.26 N in the opposite direction 5 13.3 Nat 100°, 26.2 N at 210°
d 3.59 N in the opposite direction
6 SN at 120°, 8.66N at 150°
2 a 38.1 N anticlockwise from given direction
7 108.2°
b 9.14 N clockwise from given direction
8 a Up to 376N b Less than 220 N
c 0.212 N anti-clockwise from given direction
C 28.1°
d 3.86 N clockwise from given direction

4
F = 2.90N, 0 = 53 .5°

T = 68.4N, F = 53.2N
9

10
Proof

Proof ■
5 F = 7.76N, R = 29.0N Exercise 3D
1 a = 3.88ms-2 , R = 45.2N
6 0 = 17.5°, R = 38.2N

8
F = 4.57N, R = 18.7N
0
0 = 42.l ,R = 80.6N
2 a
-----
a= 0.3ms- 2

9 F = 26.6N, R = 76.SN
2gN
10 0 = 33.5°,R = l55N
b T = 0.603N, R = 19.9N
11 F = 20.4N, R = 23.0N
3 0 = 53.1°, R = 84N
12 a Any arrangement works. If the man holds the
rod at 40°, each child can pull with force 4 T = 58.5N, a = 0.129ms- 2
71.0 N. If the man holds the rod at 50°, each
child can pull with force 80.6N. If the man 5 e = 20.2°, a = 3.34ms-2
holds the rod at 60°, each child can pull with 6 a
force 89.7N.
SOON
b They can hold it in equilibrium provided the
man holds the one at 40°.
T
13 a They can hold it in equilibrium if the strongest b T = 5910N, a= 0.543ms-2
person holds the one at 10° and the next
strongest holds the one at 25°. 7 T = 1040N, and 0 = 75°
 Cambridge International AS & A Level Mathematics: Mechanics

8 0.433ms- 2 b The people continue to pull at these angles

once the motion starts, otherwise the answer
9 a T will only be the initial direction of motion.

9 Bearing 36.4°, a = 0.860ms-2

10 The mass moves on a bearing of 088.1°, so closest

to Bob.
200g N
11 Bearing 021°, a = 0.463 m s-2
b 543N
12 295N
10 F = 54.9N , a= 0.73lms-2
13 Proof
11 0 = 16.8°, a= 3.25ms-2
14 Akhil pulls at 10°, Ben pulls north and Khadijah
12 5.37 s
pulls at 30° to give a net force of 734 N .
13 3.02ms- 1
End-of-chapter review exercise 3
14 a 4.61m 1 F = 10.2N, 0 = 42.8°
b The ball is being modelled as particle and
2 17.7N
slides up the slope.
3 a 170N b 0.36lms-2
15 4.34s
4 0.629 m s-2 on a bearing of 358.8°
16 1.5 s

■
5 0 = 52.5°, T = 35.7N
17 5.31m
6 0.257ms-2
18 4.14ms-1
7 a 064.6° b 0.629ms-2
19 14.1 N
8 23.6kg
20 6150N
9 a 2.40s b 0.784s
Exercise 3E
10 a 0.576ms- 1
1 a = 1.94 m s- 2 at an angle of 30.9° to the right of
the positive y-direction b The force is constant and the angle remains
unchanged despite the motion starting.
2 26.6° C 5.46m
3 43.9° above the positive x-direction 11 a 6.72ms- 1 b 13.0m

4 a = l.83ms- 2 at an angle of 14.2° above the 12 a 9.80 m s- 1 b 60.3s

positive x -direction
13 Proof
5 15.8 N at an angle of 18.4° below the positive
x-direction 14 Proof

6 a 53.1° 15 Resultant = 73 N , direction 41. 1° from positive

x-direction
b 15.5 N at an angle of 75° below the negative
x-direction 16 AP = 6.5N, BP = ION
7 45° below the negative x-direction 17 35.3
8 a a = 0.619ms- 2 at an angle of 35.0° to the right
of the direction AB
 Answers

ic review exercise 1 Exercise4A

1 a 40 N horizontally to the left
1 700N
b 23.5 N horizontally to the right
2 a At t = 5s,v = 4ms- 1 andat t = 40s,v = 3ms- 1 C ON

2 a 36.2 N up the slope

4
I b 13.8 N down the slope
3 - ,-------,---
'
I
I
I
I
I
c 55.9 N up the slope
I I
d 22.1 N up the slope
5 30 40 t(s)

3 a 204N b 193N
3 a 2.39ms- 1 b 2.87ms- 1
c 218N
4 406.4N 4 a 40N b 41.5N
5 a F = 30.2 N, a = 34.2° c 36.8 N
b R = 36.4 N, 0 = 16.4° 5 1.5N
6 a 528m b Proof 6 a 1.80N b 23.6N
7 T = 87 N, F = 60 N 7 ll.4N
8 a 71.3° 8 708N
b m = 2 kg, angle 54.3°
9 0.258

■
9 a 12.9 Nin direction AB and 7.34 N
10 a 0.221
perpendicular to AB above AB
b The roller is being modelled as a particle so it
b Magnitude 14.8 Nat angle 29.6° to AB
does not roll down the slope.
above AB
11 0.400
10 a 0.5ms- 2 b 2.9°
12 µ ~ 0.275
11 a = 84.8°, F = 5.52
13 1.84
12 45s
ii 33s 14 39.4N ~ T ~ 399N

13 F = 1.90, R = 12.4 15 a 80.9° with the upward slope

ii 0.526kg b 83° with the upward slope

16 a At that value of tension there will be no

14 Tension in S 1 is 13 N, tension in S2 is 20 N.
normal contact force, so no friction.
b As the tension increases, the normal contact
force increases, thereby increasing friction, so
a smaller coefficient of friction may be enough
Prerequisite knowledge to prevent motion.
1 17m
17 34.6N ~ T ~ 198N
2 12.2m
Exercise 4B
3 3.34ms-2 1 a 140 N b 14N c 0.5ms-2

2 a 29 OOON b 7240N c O.l 73ms-2

 Cambridge International AS & A Level Mathematics: Mechanics

3 0.182ms- 2 10 9.2lms- 1

4 a 1.29 m s-2 11 18.5m

b The gardener will not move the wheelbarrow.
12 0.259
5 0.447 13 Proof
6 0.236 14 0.0956
2
7 3.73ms-
Exercise4D
8 0.516 1 48N
9 0.268 2 45.5N
10 25.9ms- 1 3 253N
11 37.4m 4 a 27.4° b 0.518
12 Pulling with the string gives an acceleration of 5 14.3°
0.544 ms-2 compared to an acceleration of
0.5 ms-2 by pushing. 6 120N

13 a 0.244 7 36.lN
b 1.19ms- 2 8 1610kg
14 0.448m

■
9 Proof, 61.7N
15 a 2.97ms-2 b 0.485 ms- 2 10 31.5°
16 a 13m 11 Proof
b The tension instantly falls to zero when the
string breaks. 12 Proof

Exercise4C End-of-chapter review exercise 4

1 12.7m, proof 1 42N

2 a 0.954s, proof 2 22.6N < P < 104N

b A ball would always roll, whereas a particle 3 8.14ms-1
would not slide.
4 1.69ms- 2
3 a 0.422 b 5.18ms- 2
4 a 82.7N b 6.l8ms-2 5 2.44ms- 1

5 Proof, 1340 N 6 R = 400N, T = ISON

6 Proof, 1.60 m s-2 7 a Proof, 3.58 N b Proof, 0.530ms-2

7 3.32ms- 1 8 It accelerates at 0.178 m s-2 towards the

8 4.32s younger boy.

9 a 0.972m 9 a Proof, 61.4 N

b The ball is being modelled as a particle, so b 60 N at 76° to the upward slope
slides rather than rolls and has no thickness,
so the size of the ball does not affect the height 10 a 92.2N b 0.393
reached up the slope. 11 a 1.83m b 3.26s
 Answers

12 a 0.517ms- 2 b 3.22ms- 1 Exercise SB

C 11.7m d 7.27 s 1 a Tension 30 N, friction 30 N
b Tensions SON, 30N, friction 20N
13 Proof
c Tension 30 N, friction 10 N
14 a Proof b Proof d Tension 20 N, friction with horizontal
15 Proof ii 2.43 m surface 20N
2 a 3s
16 Proof ii Proof
b The rope is modelled as a light inextensible
17 0.578 ~ P ~ 4.49 string and the buckets as particles.

3 30N, 4N, 6N

4 a 0.25s b 1.58 m s- 1
Prerequisite knowledge
C 1.08s
1 a 1.25m b 0.5s
5 a 1.2 s b 1.4s
2 a 200N
b Surface is horizontal, no other forces act, 6 a 10.6N
2
acceleration due to gravity is 10 m s- . b 0.66m

3 a 2N b ( 4 sin 0 - 1) N 7 4.2N, 1.2N

8 6
Exercise SA

■
1 60N 9

2 a O.lms-2 b 130N 10 a lms- 2 b 36N, 33N

3 a 0.8ms-2 11 a Proof b 0.475 N, 0.275 N

b Model box as a particle so air resistance can 12 a There are two lengths of string at the cylinder,
be ignored. so the distance moved by the cylinder in a given
c 8N d 12N time is half the distance moved by the box,
hence the speed and the magnitude of the
4 a 80N b 130N acceleration are also half those of the box.
5 a 1700N b O.Sms- 2 b 2.5 m s- 2 downwards

6 a i Upper rod 240N, lower rod 120N 13 a Strings are light, inextensible and hang
vertically.
ii Upper rod 240 N, lower rod 160 N
b 5.5N in each C 2.5 Nin each
b The masses of the rods are negligible, the
second rod is vertical, the buckets of water can d Upper unchanged, lower changed to 3 N.
be modelled as particles.
Exercise SC
7 From top: 250 N, 240 N, 230 N, 220 N, 210 N,
1 a 206N b 465kg
200N, 190N, 180N, 170N, 160N, 150N
2 a 3300N b 3200N
8 25N, 25N, 40N
C 3100N
9 70000 N
3 6 I
10 344 N tension
4 a 5200N b 416N
11 4 N thrust
5 a 405kg b 368N
12 Proof C 158N d 52.5N
I

' Cambridge International AS & A Level Mathematics: Mechanics

6 a 2.5ms- 2 b 417N 6 Gen r I otion ·n a straight line

7 a (M + m)(I0 + a)N b m(lO + a)N Prerequisite knowledge
1 a 14ms- 1 b 33m
8 a 32.lN b Proof
C 5s
9 a i 616kg ii 797kg
b 8 2 a 15x 2 - 60, (- 2, 82), (2, - 78)

10 10000N b ~ x4 - 30x 2 + 2x + c - 96
4 '
11 a 2 m s- 2 , upwards for A and downwards for B
Exercise GA
b 0.4ms- 2 upwards 1 20ms- 1
2
c A 2.4 m s- upwards, B l.6 m s-2 downwards
2 86ms- 1
End-of-chapter review exercise 5 3 a Ball is modelled as a particle, air resistance
1 a ls b 84 N can be ignored.
2 a 100 N tension b 5N tension b 20ms- 1 c 0ms- 1

3 a 9.5ms- 2 b 7.5N 4 a 0ms- 1

c 12 units per second
4 a lms- 1 b 0.9s
5 a Proof
5 a 2.5ms-2 b 2.7m
b 7m

■
6 2.5s
6 a 1.6s b 3.2m
7 a Pulleys are smooth, crate and ball are modelled
7 a Proof
as particles, rope is light and inextensible. If the
pulleys are not smooth they might 'stick' and b 12m
the tension in the rope might be different on 8 a 320m b 36ms- 1
the two sides of the pu lleys, but the second
pulley would (probably) still not move. c 39.2ms- 1 d 12.8 m

b 24./2 = 33.9 9 2.7m

8 TA = 2.5 + 0.25a, T8 = 7.5 - 0.75a 10 a s is continuous at t = 4,

ii Proof so 5 X 4 2
= A./4 + B X 4
iii l.2ms- 1 iv 6ms- 2 80 = 2A + 4B
9 Proof 40 = A + 2B
b Proof
10 6ms- 2 ii 4.84ms- 1 c A+ SB = SC + 6, A + 10B = lOC
11 2.75ms- 1 ii 0.89m d 19

12 a l.2ms- 1 b Proof 11 a Proof

C 0.0868 d 2.01 ms- 1 b 4.04 ms- 1

12 a Proof b 6.74 m

Exercise GB
1 t =2
2 l6ms-2
 Answers

3 a - lOms- 2 7 a - 43.8m b 84.3m

b This is the acceleration due to gravity. It is
negative because the upward direction is 8 a 87.5m b 271m
positive in this question. 9 a Proof
4 a 5ms-2 b 6ms-2 b 294m
c 7ms-2 10 a 9 b 57.3
5 a 11 b Proof 11 a First car Om s- 1, second car l6ms- 1
c -11
- ms -2 d Proof b 281m
75
C 42.7m d 12.3
6 a v = 6-2tms- 1
b It starts with speed 6 m s- 1 but slows down and 12 a k = I b Proof
then stops and returns back the way it came, C 6.55ms- 1 d T = 22s
speeding up all the time. e 95.3m
C t = 3s
d a = - 2 m s-2 (constant acceleration) Exercise GD
1 3.2ms- 1
7 250 ms- 1
2 14.2 ms- 1
8 A= - IB=IC = 2
9' 3'
1
3 a 18.5ms- 1 b -1.2m
9 a 21.3ms- b 5.33s
C Towards

■
10 a 5ms- 1 b 9ms- 1
4 a Proof b 5.78m
c -1 ms- 2
C Proof d 3.43ms- 1
11 a 10 minutes
5 4.98m
b 75.2kmh- 1 (or 20.9ms- 1)
c 12.5km 6 2.52m

7 a Proof
Exercise GC
b 2m
80
1 8 6
3
2 14.25m 9 1690m
3 20m 10 a 19.8ms- 1 b 7.6s
4 a Orn b 1.33m C 42.7m d Proof
20
C -m 11 a 79.9m b 39.8ms- 1
3
C 2.67s
5 a 29.9m
b Smaller 12 a Proof b Proof
6 a ls b 7cm C According to the model there is still air
C 1.18 s resistance when t = 2.3, but the ball will stop
d There is no resistance on the downward when it reaches the end of the skittle alley.
journey, so, for example, the ball bearing does End-of-chapter review exercise 6
not touch the sides of the hole it has made. 1 8.43m
Improve model by incorporating a factor to
represent friction. 2 a 0.125 ms- 2 b lllm
 Cambridge International AS & A Level Mathematics: Mechanics

3 a 12ms-1, 3ms- 1 b 175m 12 i t < 2.5 ii 27m

4 a 8ms- 1 b Proof iii t = 2.31 and t = 5.19

5 23.4km 13 0.8m
6 a 4.5ms- 1 b 40s
7 o e u
7 a Proof b Proof
Prerequisite knowledge
C 69.2m
1 a Proof b 7.5 m s- 1
d The gradient of the acceleration- time graph
c I0.6ms- 1
changes suddenly at t = 1.
2 a 7.5ms- 1 b 12.4ms-1
8 ii 71.3m

9 a 6.5ms- 1 b 63.8m Exercise 7A

C 3.06s 1 80Ns

6ms- 1, 0ms- 1 2 33000Ns

ii 583m 3 2.85Ns

11 A= 4, proof 4 0.056Ns

iii 5.4ms- 1 5 36Ns

12 Proof b 3ms- 1 6 a 7ms- 1 b 245Ns

■
a
C 2.1 d Proof 7 a 6ms- 1 b 12Ns
e Proof
8 a 0.45m
Cross-topic r vie xer ise b Ball is modelled as a particle with no size and
it is assumed there is no air resistance. Air
1 a Tension = 48 N , acceleration = 2 m s-2 resistance would slow the ball, so it would
b 96N have a smaller velocity when it reaches the
2 On the point of slipping down, µ = 0.336 ground and consequently a smaller velocity
after the bounce. If the ball has size then the
3 0.580s centre of the ball does not reach the ground so
the distance travelled is reduced and again
4 Braking force = 6 N, compression of 2 N in the this will reduce the velocity after the bounce.
tow-bar.
Reduced rebound velocity will reduce the
5 a 7.5ms- 1 b 59m height reached.

6 a 1.59m b 2.94s 9 0.125Ns

7 a 24.8m b 3s 10 Proof

8 a 2.68m b 1.99s 11 1.2ms- 1

9 a 2s b 9.12m 12 1.33m
C 14m Exercise 7B
2 1 40kg
10 a = }= 3.33ms-2, T = ~g = 66.7N

11 i a = 0Ams-2 , proof ii µ = 0.321

3 50kg
 Answers

4 9: 11 This is negative so at least one ball must be

travelling in the negative direction after
5 0.Sms- 1 impact.
6 2 m s- 1 in the same direction as C (the original Y cannot pass through X, so X must reverse
direction of travel for A) its direction of travel.
c - 0.045 m s- 1
7 a 6 m s- 1 in the opposite direction to Jayne
12 Proof
b Motion is in a straight line. Jayne just stands
and does not 'push off' with her skates . Jayne
and chair can be modelled as particles.
Prerequisite knowledge
8 4.9kg
1 a 20N b 34.6N
9 a 2.Sms- 1 b It is horizontal.
2 6 N up the slope
10 a Proof b Proof
3 a 18 N down the slope b 4.Sms- 2
C l880ms- 1
4 56.3cm
11 a Proof b 2(a - 1)

12 a 0.75ms- 1, 0.Sms- 1, 3ms- 1, 2ms- 1

Exercise SA
b Opposite. 1 60J
End-of-chapter review exercise 7 2 a IO0J b 76.6J
2

■
1
3 a -0.8J b 0.8J
2 0.8ms- 1 C OJ
3 90.9s 4 2400]

4 0.6 5 a Friction from the edge of the canal , resistance

from the water and some air resistance
5 _!_ms- 1 b 2000J
3
6 a (Sm+ 0.4)Ns b 0.1 6 a i 5910J ii 5640J
7 b Proof
c The component of the tension perpendicular
8 a X: 0.45Ns, Y: - 0.0SNs b Proof
to the direction of motion is more than double
9 1.95 < m < 2.5 in the second situation (53.8 N) compared
with the first (26.0 N) so the frictional
10 a 6.84Ns resistance will be greater.
b Height reached after second bounce 1.18 m ,
7 a 6J b l 7.3J
after third bounce 0.957 m.
C OJ d OJ
c Ball can be modelled as a particle, so ball has
no size, and there is no air resistance. e I 1.3J

11 a 0.0225 N s 8 a 20J b 2591

b Let the original direction of travel for X be C OJ d 239J
l,
the positive direction.
9 a 400J b 20J
Total momentum before impact= - 0.0072 N s 259]
C d OJ
e 121J
 Cambridge International AS & A Level Mathematics: Mechanics
I

10 a 5J b 3J Exercise SC
C OJ d 2J 1 lOOJ

11 283J 2 600 J decrease

12 l.73J 3 0.399J
4 a 500J b 5001
Exercise 8B C 500J
1 320J
5 20.6J
2 363 000 J or 363 kJ
6 5670J
3 71.3J 7 40kg
4 54J 8 a R
1
5 a 14.7ms- b 216J
6 a 13ms- 1 b 80J

SOON

8 2.5m b 2.10m C 131J

v2
9 a 1250J 9 a Proof b --m
11.32
b No difference, provided the speed is still C 37.5v 2 J

■
constant. d The boy is modelled as a particle, this means
air resistance is ignored. Air resistance would
10 a 5.63 x 10 15 J slow the boy down, so the slope would be
b Fuel will be used while the rocket is longer and the loss in GPE would be greater
accelerating, so the mass will decrease. than the values given.
10 13 The slope is modelled as a straight line.
11 a or 3.33 b or2.17ms- 1
3 6 In reality it would flatten out towards the
12 Momentum is conserved bottom, so the boy would slow down while
muA - mu 8 = -mvA + mv8 travelling horizontally, his speed at the
bottom of the descent would be greater than
souA+vA =uB+ v8
v and the loss in GPE would be greater than
Kinetic energy is conserved
the value given.
0.5mu~ + 0.5muj = 0.5mv~ + 0.5mvj
10 -60J
so u~ - v~ = vj - uj
(uA - vA)(uA + vA) = (vs - us)(vB + uB) 11 0.075J
But uA + vA = vB + uB 12 Proof
SO UA - VA = VB - UB

Add: (uA + vA) + (uA -vA) = (vs + uB) + (vs -uB) End-of-chapter review exercise 8
so 2uA = 2vB 1 24.3
uA = vB and vA = uB 2 a 0.24J b 0.24 J

13 a 0.12J b OJ 3 a 0.978ms- 2 b 321kJ

C 0.21J
4 a 300N b Proof
c The push force and resistance are constant.
 Answers

5 750] ii Proof 2 l.39ms- 1

6 a 6.25] decrease b 6.25 J increase 3 a 400J b 400J
C No difference to the numerical answers, but it C 0.80m
would affect how far the ball travels horizon-
tally, the height the ball reaches and also the 4 a 2550J b 7.75ms- 1
angle that the path of the ball makes with the c 7.75 ms- 1
vertical when the ball passes through the
hoop. 5 a 10.2ms- 1 b 25m

7 1.7 ii 30ms- 1 6 63N

8 a 2ms- 2 b 2.25m 7 5m
C 90] 8 a 0.498N

9 a 20000] b 10000] b Very small resistance force so grass is very

slippery, perhaps the grass is wet.
C 50m d 10000]

10 9 a 800J b 7.14m
a 640N b 4m
C 3.14m
C 30° d 6.12ms- 1
10 If the initial speed is um s- 1 and the final speed is
11 a 1. 50 m s-2 parallel to the slope and down the
vms- 1 then v2 - u 2 = 300. According to the
slope.
driver, v < 30 sou must be less than 24.5.
b 8.33m c 334J

■
d Proof 11 a Proof
b Proof
12 a The only force acting on Jack during the flight
is his weight, which is vertical, so there is no 12 a 15 J b Proof
horizontal resultant force and hence no C 5.83ms- 1 d 5.48 m s- 1
horizontal acceleration.
e 67.2° f 6.32 ms- 1
b 13.2ms- 1
C 1060 J Exercise 9B
d 1.51m 1 6.33ms- 1
e He could easily slide off the trampoline.
2 5.35 m s- 1
f He would bounce up to quite a height and
could bounce several times before coming to 3 a 0.571
rest. b 49.8ms- 1 = 179kmh- 1

9 The work-energy principle and

Prerequisite knowledge 6 2.04ms- 1

1 12.5 J
7 0.05 v2
2 a 150J b 15J
C 125J
b Diver modelled as a particle so no air
Exercise 9A resistance, no spin etc. End of the board
1 a 60J b l lOJ assumed to be 10 m above the water at take off,
C 50J d 2ms- 1
 Cambridge International AS & A Level Mathematics: Mechanics

but if it is a flexible board it may be less (or Exercise9D

more) than 10 m. 1 a 300kJ b 37.5kW
9 a 14.3ms- 1 2 20ms-1
b The only force acting is the weight, which is
vertically downwards, so there is no horizontal 3 14ms- 1
component to the acceleration. 4 2W
c Proof
5 12kN
10 a i IOmhJ ii 0.5mv 2 J
6 76.9kW
iii (0.5Mv 2 + IOMh sin0)J
b Proof 7 3.6kW
11 a i 7ms- 1 ii l lms- 1 8 0.889ms-2
b The surface is smooth.
48000 N
c No difference. 9 a b (3vO -l.5)m s- 2
V

12 a .J20 = 4.47ms- 1 b 75.5° c If the initial value of v is zero the initial

driving force and the initial acceleration
would both be infinite.
Exercise 9C
1 a 24m b 1200J 10 a 50ms- 1 b Resistance is now 30 N.
C Heat 11 2450

■
2 0.30 12 a Energy dissipated = 625(300 - v2 ) J,

3 24J distance travelled = 1.25(300 - v2 ) m

4 1200N 2
b ( vO - 0.4 )ms- 2, Proof

5 9.42N
c Proof, 14.8 ms- 1
6 20m
2 End-of-chapter review exercise 9
7 (10+;1 u )m
1 P=28500, R=750
8 a 22.5ms- 1 b 1.44m a 40m N 0.6(m + 10)
m
2 b
m+2 (m+2)
9 a 0.2N b 35ms- 1 = 126kmh- 1
C Proof
3 v=H
10 a 4.79 b 5.32cm
4 a 26.3ms- 1
2
61 - v b The speed would decrease
11 a 2N b N
6
c The speed would decrease
12 a 2.lJ b Proof
5 2.50
c At the start the normal contact force is 2 N
and 0.546 + 2 = 0.273. However as the particle b 39kW
moves down the surface the frictional force
7 3.03ms- 1
reduces to 0. The value 0.546 is an average and
at the start the frictional force is greater 8 a 1200N b 0.025
than this.
 Answers

9 Gain in KE = 4030k1, 8 a 750 b 250m

loss in PE = 42 000 sin 0 kl
9 a Proof, 4ms- 1 b 1.5m
ii 3.26°
10 514kW ii 31.7ms- 1
10 a 0.625(500 - T)l b (0.99T - l.56)ms- 1
11 14401 ii 30801
11 10.5v 2 1
iii 45201
ii a 135x1
b 25x1 12 12kW ii 0.132ms-2
iii Proof 13 9851 5.55ms- 1
ii
12 a 12.5x 1 b 10-Jlxl = 14.lxl iii 273 W iv 54.4s
C 0.906ms- 1
d Work done by tension in pulling X up slope - tyle paper
AB is cancelled by work done against tension 1 Proof, F = 6.61 N
when Y slides down BC.
2 a 1875N
b 375k1
C iew exercise 3
1 2.5 N s upwards 3 a 0.72kg
b 10.81
2 58.51
4 a 225m
3 Yes, there will be a third collision, because object

■
B is moving faster than object A and will catch it b 25s
up. 5 a V=4ms- 1
4 a 50 b 6ms- 1 b Proof
C 28.5s
5 a 1.5ms- 1
b 3 m s- 1 in the opposite direction from the 6 a 0.75N
coalesced particle b 1.27m

6 a 300m b 991m 7 a 0.775ms-2

b 1.161
7 a 8ms- 1 b 481
C 19.71
C 0.8m
d 36.0N
 A position (in a gravitational field) . The potential energy is the
Acceleration: rate of change of velocity product of the weight and the height. It is a scalar quantity
measured in joules (J)
Angle of friction: the angle between the normal contact
force and the total contact force when friction is limiting Gravity: attraction between two objects as a result of their
masses, usually thought of as a force acting on an object
C towards the Earth
Coalesce: when two bodies join together in an impact and
continue as one object; the opposite of an explosion I
Coefficient of friction: the ratio between the frictional force Impact: a collision or other interaction between two
and the normal contact force when friction is limiting bodies
Components: the parts of a force acting parallel to given Instantaneous acceleration: the acceleration at an instant,
axes, usually two perpendicular axes which is the gradient at a point on a velocity-time graph;
usually just referred to as acceleration
Compression: force in a rod or other connecting object,
but not a string, which provides a force in the direction of Instantaneous velocity: the velocity at an instant, which
the rod towards the object it is connected to is the gradient at a point on a displacement- time graph;
usually just referred to as velocity
Connected objects: objects that are attached together with
forces acting between them K
Conservative force: a force for which the work done in Kinetic energy (or linear kinetic energy): the energy that a
moving an object between two points is independent of

■
body possesses because of its motion; calculated as half
the path taken the product of the mass and the square of the speed; a
Conserved: unchanged, as in 'total momentum is scalar quantity, measured in joules (J)
conserved in an impact'
Contact force: the combined effect of two objects L
touching, comprising two components: the normal Light: having no, or negligible, mass
contact force and friction Limiting equilibrium: when friction is at its maximum
possible value but there is no net force on the object
D
Line of action: the direction in which a force acts
Displacement: distance relative to a fixed point or origin in
a given direction Line of greatest slope: the steepest path up or down a
surface which is at an angle to the horizontal
Dissipated: mechanical energy lost by being converted into
non-mechanical energy, such as heat, sound and light
M
Distance: length of path between two points
Momentum, or linear momentum: the product of the mass
E and the velocity of an object; a vector quantity, measured in
N s; its direction is the same as the direction of the velocity.
Equilibrium: state of an object when there is no net force
acting on it
N
Explosion: when a single object splits into two or more
Negligible: small enough to be ignored for the purposes of
separate parts; the opposite of coalescence
the mathematical model
F Newton's first law: the principle that an object continues
Force: influence on an object that can alter its motion moving in the same direction at the same speed unless a
net force acts on the object
Friction: force between two surfaces acting parallel to the
contact between the surfaces, as a result of the roughness Newton's second law: the principle that the rate of change
of the surfaces in contact of momentum is proportional to force acting on an object,
which leads to the equation F = ma in the case where
G mass is constant
Gravitational potential energy (GPE) (or potential energy Newton's third law: the principle that for every action there
(PE)): the energy that a body possesses because of its is an equal and opposite reaction
 Glossary

Non-conservative force: any force for which the work done Speed: rate of moving over a distance
in moving a particle between two points is different for String: any flexible connector joining two objects; it can be
different paths taken in tension but not in thrust; it is assumed to be light and
Normal contact force: influence of one object on another inextensible (does not stretch)
through being in contact; the force acts perpendicular to
the touching surfaces T
0 Tension: force in a string, or other connecting object, which
provides a force in the direction of the string away from the
'On the point of slipping': state of an object when friction
object it is connected to
is limiting so that any increase in the force applied to the
object will cause it to move Thrust: the force provided by, for example, a rod when
under compression, acting along the rod towards an object
Origin: reference point from which displacement is
measured
V
p
Vector: quantity having a numerical value in an assigned
Power: the rate of doing work, measured in watts; the power direction , which may be negative
generated by the engine of a vehicle is the product of the Velocity: rate of change of displacement
driving force and the speed at which the vehicle is moving

R w
Reaction force: alternative name for normal contact force Work: the work done by a force that causes an object to
move along the line of action of the force is the product
Resistance: force opposing motion possibly caused by the
air or other medium through which the object moves of the magnitude of the force and the distance the object
moves in the direction of the force ; a scalar quantity,
Resolving: process of splitting forces into components in measured in joules (J)
given (usually perpendicular) directions
Work done against gravity: the work done by the weight

■
Resultant: single force equivalent to the net total of other of a body when the body is raised vertically; if the body
forces has mass m and rises through a vertical height h then the
Rod: any light rigid connector joining two objects; it can be work done against gravity is mgh ; equal to the increase in
in tension or in thrust potential energy
Rough: having friction Work done by gravity: the work done by the weight of a
body when the body falls vertically; if the body has mass m
s and falls through a vertical height h then the work done by
Scalar: quantity having a numerical value but no assigned gravity is mgh ; equal to the decrease in potential energy
direction
Work-energy principle: for any motion the increase in
Smooth: having no friction kinetic energy is equal to the work done by all forces or the
Smooth pulley: a pulley for which the magnitude of the tension increase in mechanical energy is equal to the work done by
in a string passed over it is the same on each side of the pulley all forces (excluding weight)
Cambridge International
AS & A Level Mathematics:
Mechanics
Coursebook
Jan Dangerfield and Stuart Haring
Series Editor: Julian Gilbey
Written by an experienced author team, this series supports
teachers and students following the Cambridge International AS &
A Level Mathematics syllabus (9709). The materials introduce new
mathematical skills with clear explanations as well as providing
practice questions to help students consolidate their knowledge and
progress through the syllabus. The series includes activities that
encourage discussion of mathematical concepts and provides the
opportunity for deeper investigation into how mathematics can be
applied to solve a variety of problems.
Cambridge International AS & A Level Mathematics: Mechanics
matches the corresponding unit of the syllabus. It contains materials
on topics such as velocity and acceleration, force and motion, friction,
connected particles, motion in a straight line, momentum, and work
and energy.
Features:
• Prerequisite knowledge sections to check prior learning
• Detailed explanations and worked examples
• End-of-chapter and cross-topic revision exercises
• ‘Explore’ tasks to encourage deeper discussions of key concepts

Additional Components
Pure Mathematics 1 978-1-108-40714-4
Pure Mathematics 2 & 3 978-1-108-40719-9
Probability & Statistics 1 978-1-108-40730-4
Probability & Statistics 2 978-1-108-40734-2

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