Topic 1 Forces and Motion

UNITS OF MEASUREMENT
Measurement means comparing any physical quantity with a standard to determine
its Relationship to standard this standard is called unit all measurable quantity
expressed in
a) Some number or magnitude and
b) Some unit
For example if the distance is 200km, 200 is the number or magnitude and km
(kilometer) is the unit

Standard International Unit

M.K.S. system: meters – kilogram – second system used internationally. This system
is also called SI (system international) unit

kilometer ton
meter kilogram
centimeter gram

hour
minute
second

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Powers of prefix
Prefixes are used to give multiples and submultiples. The prefix represents a power
of ten. The standard multiples and submultiples are mostly in steps of 103

Symbol Prefix Value

n Nano 10-9
μ Micro 10-6
m Milli 10-3
C Centi 10-2
K Kilo 103
M Mega 106

Scalar and Vector Quantities

 Scalar quantities: a quantity which has magnitude (size) only and is
described by its numerical value. ( distance, mass, speed, volume, time and
temperature )

 Vector quantities: a quantity which has both magnitude and direction, and both
should be mentioned to describe it ( displacement, weight, velocity and
acceleration )

Speed

Do you know how fast you can run?

The fastest runners in the world can run 100 meters in just 10
seconds. That means they cover 10 meters each second, or 10
meters per second. This is written as 10 m/s.

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  To calculate the speed

distance moved (m)

speed (m/s) =
time taken (s)

Questions

1. The runner completes 400 m is a time of 160 seconds. What is her average
speed?

Answer

distance moved 400

average speed = =
time taken 160

So the average speed = 2.5 m/s

2. An athlete completes a 4 km mountain race in at an average speed of 3 m/s.

How long does the race take?

Answer

distance moved 4000

time taken = =
average speed 3

The distance is given in 'km' and needs to be converted to meters

4 km = 4000 m.
So the time taken = 1333 seconds.

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displacement and distance

Speed and distance are difficult ones. To explain this, imagine a simple journey
If you walk 1 km to a cinema and back again, how far have you gone?

If we ignore direction, we call this the distance travelled. Distance is a scalar.

If we include the direction , it is called the displacement. Displacement is a vector.
So in the example above shown in figure 2, the distance travelled = 2km, and the
displacement = 0 km.

Speed is a scalar quantity and velocity is a vector quantity.

Acceleration
Is rate of change in velocity

When we speed up - increasing our velocity - we are accelerating , When we slow

down we are decelerating

 u = the initial, or starting velocity. (m/s)

 v = the final velocity (m/s)
 Δv = the change in velocity (m/s)
 t = the time taken (s)
 a = acceleration (m/s2)

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 change in velocity (m/s)
acceleration (m/s2)=
time taken (s)

v-u Δv
a= or a =
t t

 + acceleration (speed up )
 - acceleration (slow down )
 Zero acceleration ( constant speed )

Questions

3. The school bus accelerates from rest to a velocity of 16 m/s in a time of 20

seconds. Calculate the acceleration of the bus.

Answer

We know that the change in velocity (v-u, or Δv) is 16 m/s

Δv 16
a= =
t 20

so the acceleration = 0.8 m/s2

4. A cheetah is widely believed to be the fastest animal on Earth. It can

accelerate at 2.5 m/s2. If the cheetah starts at a velocity of 3 m/s, how long will
it take to reach a velocity of 13 m/s?

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Answer

In this question, the change in velocity Δv = 13-3 = 10 m/s.

Δv 10
t= =
a 2.5

so the time taken = 4 s.

 There is a more complicated formula that can be used to find the final
velocity if we know the distance travelled rather than the time taken

(final speed)2 = (initial speed)2 + (2 x acceleration x distance)

v2 = u2 + (2 x a x s)

Question

5. An eagle swoops down from a cliff, accelerating from an initial speed of 2

m/s at a rate of 5 m/s2. It covers a distance of 6 m. Calculate the final velocity
of the eagle.

Answer

First, write down which values are given in the question

 initial velocity u = 2 m/s

 acceleration a = 5 m/s2
 distance s = 6 m
 final velocity v = ?

Now substitute these into the formula

v2 = u2 + (2 x a x s)

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so: v2 = 22 + (2 x 5 x 6)

v2 = 4 + 60 = 64

v = √64 = 8

So the final velocity = 8 m/s.

Motion graphs

There are two main types of graph used:

 Distance - Time graphs

 Velocity - Time graphs

1- Distance - time graphs

 Graph ( A ) gradient is speed and graph is straight line so it constant speed

opposite direction of ( D )
 Graph ( B )gradient is speed and graph is increasing ( stepper up ) so
increase in speed

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  Graph ( C ) gradient is speed and graph is decreasing (stepper down ) so
decrease in speed
 Graph ( D ) gradient is speed and graph is straight line so constant speed

 Rest or stationary when graph is horizontal as

shown on figure

Questions

1. A man is taking his dog for a walk. When he opens the front door, the dog
runs off in a straight line at a steady speed. It then stops at a lamp post for
some time.

Sketch a distance-time graph for this motion. You do not need

to estimate any values for this question.

Answer

 The distance should be labeled on the y-axis.

 The time should be labeled on the x-axis.
 The first section of the line should be a straight line sloping upwards showing
"a steady speed".
 The next section should be a flat, horizontal line showing the dog remaining
in one spot at the lamp post.

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2- A student writes a program to make a small robot move
across the desk. This graph shows the motion of the robot

 a). Describe the motion in part 'X' on the graph.

 b). Describe the motion in part 'Y' on the graph.
 c). Describe the motion in part 'Z' on the graph.
 d) Calculate the speed of the robot when it is moving at
its fastest.

Answer

a) Line 'X' shows the robot moving away from the start at constant velocity.

b) Line 'Y' shows the robot stops / is stationary (100 cm from the start).

c) Line 'Z' shows the robot moving back towards the start at constant velocity.

d) The fastest speed is where the line is the steepest, (the highest gradient), which is
line X. For this section, reading from the graph, the robot covers 100 cm in 5
seconds. As we are using cm, the speed will be given in cm/s.

distance 100 cm
speed = =
time 5s

so the speed, or velocity, is = 20 cm /s.

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2- Velocity - time graphs

 Graph ( A ) gradient is acceleration and graph is straight line so it constant

deceleration as velocity decrease
 Graph ( B )gradient is acceleration and graph is increasing ( stepper up ) so
increase in acceleration
 Graph ( C ) gradient is acceleration and graph is decreasing (stepper down )
so decrease in acceleration
 Graph ( D ) gradient is acceleration and graph is straight line so constant
acceleration

 when graph is horizontal as shown on figure zero

acceleration or constant velocity

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Calculations distance using velocity-time graphs

A velocity-time graph is often used to show motion because you can use it to find
other information. The two that we will cover here are how to find
the acceleration and the distance travelled from the graph.

 The distance travelled = the area under the line.

 The acceleration = the gradient of the line.

Area under line

Area of triangle = 0.5 x base x height

Area of rectangle = base x height

Area of trapezium = 0.5 x h x (a + b)

Acceleration

This is the gradient of the line the steeper the line, the higher the acceleration. The
'steepness' of a line is the gradient

'rise' change in y-axis

gradient = or gradient =
'run' change in x-axis

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Questions

The graph shown in figure gives data on the

movement of a city tram moving away from a station.
Which section or sections of the graph shows the tram

a) Moving with constant velocity?

b) Decelerating?

c) Accelerating at the highest rate?

Answer

a) A constant velocity will be shown as a flat, horizontal line on a velocity-time

graph.
Therefore the sections showing this are Q and S.
b) Deceleration means slowing down, and this is shown by a sloping line with a
decreasing velocity.
The section showing this is section T.
c) A high acceleration is shown by a sloping line with a high gradient (like a very
steep hill).
This is shown by section R.
(Section P is also showing acceleration, but the tram is not accelerating as much as
in section R).

Using the graph shown in figure for the velocity of a

tram, calculate:

 a) The acceleration of the tram in section R.

 b) The distance travelled between 30 and 40
seconds.
 c) The distance travelled in the first 10
seconds.

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Answer

a) The acceleration is given by:

change in velocity
acceleration =
time taken

change in y-axis
or by using acceleration = gradient =
change in x-axis

15 - 5 10
acceleration = =
10 10

So the acceleration of section R = 1 m/s2.

b) The distance travelled between 30 and 40 s is given by:

Distance = area under line = rectangle of height 15 (m/s) and length 10 (s).
Therefore the area = distance = 15 x 10 = 150 m.

c) The distance travelled in the first 10 s is given by:

Distance = area under line = area of triangle = ½ base x height,
So area = ½ x 10 x 5,
Area = distance = 25 m.

Forces

A force is basically a push or a pull on something. There are many different things
that cause forces, but they are all measured in newton’s (N)

Here are a few of the most common types of forces found in nature

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  Magnetic - between magnetic materials. This can be a push (repulsion) or a
pull (attraction)
 Gravitational - between any mass. This is always a force of attraction.
 Friction - this is a force that acts against motion.
 Electrostatic - between positive and/or negative charges. This can be
attraction or repulsion, depending on the charge.
 Friction is a general term for any force produced when an object is in contact
with a solid, liquid or gas that pushes against the motion. If a liquid or gas is
involved, we often call this force of friction the 'drag'. If it is an object like a
plane moving through the air, we often call this type of friction air
resistance.

When a force acts on something, many things can

happen
 Change the speed of a body.
 Change the direction of motion (if it is already moving).
 Change the shape of a body. (For example, squashing a rubber ball).

Drawing forces
Force is a vector so can be drawn with an arrow. The length of the arrow can be
used to show the size of the force, and obviously the direction of the arrow shows
the direction of the force.

 If two or more forces are acting in the same direction, we can add them
together to find the total.
 if they act in opposite directions, we need to subtract, as shown here by
two forces acting on a ball

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Questions

1. A block of wood is pushed across a table, from left to right. Describe the
direction of the following forces:

 a) Friction.
 b) Gravity.

Answer

a) Friction always acts in the opposite direction to the motion, so it acts from right
to left.
b) Gravity always pulls objects downwards, (towards the centre of the Earth).

2. A small van is travelling down a road. The forward force provided by the
engine is 2 kN . The air resistance acting on the van is 600 N, and the friction
between road and tiers is 300 N.

a) What is the resultant force acting on the van?

b) Describe the motion of the van at this point.

Answer

 a) The forward force is 2 kN, which equals 2000 N. Air resistance AND
friction both act to prevent movement so act backwards, and total 900 N.
Therefore there is a resultant force of 1100 N (1.1kN) acting forwards.

b) As there is a resultant forward force, the van will accelerate

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Mass

Mass is a measure of the quantity of matter in an object. In simple terms, it relates to

the number of atoms present, and how big those atoms are. The international unit
for mass is the kilogram (kg).

Weight

Weight is a measure of the pull of gravity on an object. As it is a pull, it is also a force

and is therefore measured in newton (N)

Weight = mass x gravitational field strength

W=mxg

[Newton] = [kilograms] x [N/kg]

 For many of these questions, the mass of an object will remain the
same (assuming it is the same object), but the weight can change as it depends
on the strength of gravity acting on it.
 Take the weight of 1.0kg to be 10 N (acceleration of free fall = 10 m/s2)

Questions

1. If we took the 5 kg bag of fruit on a space journey, how much would it weigh
in the following locations?

 a). The Moon. On the Moon, g = 1.6 N/kg

 b). Jupiter. On Jupiter, g = 23 N/kg
 c). Deep Space. Deep space, there is no gravity g = zero.

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Answer

 a) W= m x g =5 x 1.6 = 8.0 N
 b) W= m x g =5 x 23 = 115 N
 c).In deep space, there is no gravity! g = zero.
Therefore W = 0 N (The bag is 'weightless').

2. On the Moon mission in 1972, the astronauts picked up a rock to bring back
to Earth. The weight of the rock was 32 N on the Moon. Calculate

 a). The mass of the rock.

 b). The weight of the rock when it was returned to Earth.

Answer

a). We know w = m x g. We also know that gmoon = 1.6 N/kg. Therefore

W
m=
g 32
m=
1.6

m = 20 kg.

b) If m = 20 kg, and on the Earth g = 10 N/kg,

then W = m x g
W = 20 x 10
Therefore W = 200 N

Force and acceleration

If there are unbalanced forces acting on the object, the object will accelerate. The
acceleration depends on the size of the unbalanced/net force and the mass of the
object

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  A force can make an object accelerate. The bigger the force acting on an
object, the bigger the acceleration that it gives to the object 𝑎 ∝ F
(proportional) so, doubling the force acting on an object doubles its
acceleration.
 The mass of an object affects how easily it can be accelerated or decelerated.
The bigger the mass, the smaller the acceleration given by a particular force
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𝑎∝ (inversely) so, doubling the mass of the body will halve the
𝑚

acceleration

Force= mass x acceleration

F=mxa

[newtons] = [kilograms] x [m/s2]

Questions

A mountain bike rider and bike together have a total mass of 80 kg. If the bike
is to accelerate at 1.8 m/s, what force needs to be applied?

Answer
F= m x a
Therefore F = 80 x 1.8 = 144 N

A firework rocket of mass 200 g is set alight, and the force produced at the
start is 5 N.

 a). Calculate the initial acceleration of the rocket.

 b). the rocket burns and releases fuel, making it lose mass. At what
mass will the acceleration be 40 m/s2, assuming the force produced is
still 5 N?

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Answer

a). We know F = m x a and also that 200 g = 0.2 kg (the mass MUST be in kg, not g).

Rearranging the formula gives:

F
5 a=
a= m
0.2

a = 25 m/s2

b) If a = 40 m/s2 and F = 5 N,
then F = m x a gives:

5 F
m=
m= a
40

Therefore m = 0.125 kg (or 125 g)

The Effect of a Force acting perpendicular to the direction

of motion (CH .8 ASTRO PHYSICS)
Here, the force is neither driving nor resisting, it acts as a centripetal force which is
the inward force needed to make an object move in a circle As a result Object moves
along circular path (or part of a circular path).
 For car moving round a circular road, what provides the
centripetal force? The sideway friction exerted by the road
on the tires.
 Object can have a steady speed but a changing velocity as
object moving in a circular path (acceleration change )

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