IGCSE Edexcel Physics 2 hours 16 questions

Exam Questions

Momentum
Momentum / Forces & Momentum / Newton's Third Law / Momentum & Safety
Features

Easy (5 questions) /34 Scan here to return to the course

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Medium (5 questions) /43

Hard (6 questions) /38

Total Marks /115

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Easy Questions
1 (a) Which row shows the correct definition for momentum and its unit?

definition unit

A. m Ns
v = p

B. p kg m/s
v = m

C. m N/kg
p = v

D. v kg m/s2
p = m

Answer

The correct answer is B because:

Momentum = mass (kg) × velocity (m/s)

Therefore, the units are kg m/s

In symbols, momentum can be written as p = mv

p
This is the same as v = m

Hence, B is the only correct answer

(1 mark)

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(b) Which of the following correctly shows the relationship between force, change in
momentum and time?

mv − mu
A. ∆p =
( )

∆t
B. F ∆ t
= ∆p
C. F = mv − mu ∆ t
( )

∆F
D. p =
∆t
Answer

The correct answer is B because:

Force is equal to the rate of change in momentum

∆p
This means F = ∆t

Therefore, the only expression that is correct is F ∆t = ∆p

This is option B

(1 mark)

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(c) Circle the correct words in the following sentences:

Momentum is a scalar / vector quantity because it has magnitude only / magnitude

and direction

In a closed system, the total momentum before a collision is greater than / less than /
equal to the total momentum after the event

Answer

The correct sentences are:

Momentum is a vector quantity because it has magnitude and direction; [1 mark]

In a closed system, the total momentum before a collision is equal to the total
momentum after the event; [1 mark]
[Total: 2 marks]
(2 marks)

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2 (a) Some quantities are vectors, others are scalars.

Complete the table by ticking the boxes to show which quantities are vectors and which
are scalars.

One has been done for you.

Quantity Vector Scalar

distance

force

momentum ✓

speed

velocity

Answer

Tick the vector and scalar quantities:

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 Quantity Vector Scalar

distance ✓

force ✓

momentum ✓

speed ✓

velocity ✓

Three correct ticks; [1 mark]

A fourth correct tick; [1 mark]
[Total: 2 marks]
(2 marks)

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(b) A truck travels at 20 m/s.

The mass of the truck is 15 000 kg.

(i) State the equation linking momentum, mass and velocity.

(ii) Calculate the momentum of the truck.

momentum = .......................... kg m/s

Answer

(i) The relationship between momentum, mass and velocity is:

Momentum = mass × velocity OR p = mv ; [1 mark]

(ii) Calculate the momentum of the truck:

List the known quantities:

Mass of the truck, m = 15 000 kg

Velocity of the truck, v = 20 m/s

Substitute values into the momentum equation:

p = 15 000 × 20 [1 mark]
p = 300 000 kg m/s [1 mark]
[Total: 3 marks]
(3 marks)

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(c) The diagram shows the force of the road on one of the tyres of the truck.

The magnitude of this force is 53 000 N.

(i) State Newton's Third Law.

(ii) Draw an arrow on the diagram to show the other force in this Newton's Third Law
force pair, and state the value of this force.

Answer

(i) Newton's Third Law states:

When two bodies interact, the forces they exert on each other are equal and act in
the opposite direction; [1 mark]

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(ii) The Newton's Third Law force pair is:

Arrow drawn from tyre downwards; [1 mark]

Arrow drawn to the same size; [1 mark]
(The force on the road from the tyre is) 53 000 N; [1 mark]
[Total: 4 marks]
(3 marks)

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3 (a) Four students draw a Newton's Third Law force pair for a book on a table.

Which student's Newton's Third Law force pair is correct?

Answer

The correct answer is A because:

A Newton's Third Law force pair must:

Have the same size

Act in the opposite direction

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 Act on different objects

Be of the same type

The force pairs shown in diagram A are the only ones to meet all of these conditions

Students often forget that Newton's Third Law force pairs must be of the same
type:

The force pair between the book and the table is a contact force

Weight is a force that occurs due to gravity, which is a non-contact force

Therefore, do not make the mistake of labelling a force pair as "Reaction force" and
"Weight" as these are not the same type of force!
(1 mark)

(b) Complete the sentences to explain how a Newton's Third Law force pair on the foot and
the ground is responsible for the motion of walking:

When a person walks on the ground, there is a force from the ..................... which pushes
the ..................... backward.

From Newton's Third Law, there is an ..................... and ..................... force from the
..................... on the ..................... which pushes the foot forward.

Answer

One mark for each correct statement:

When a person walks on the ground, there is a force from the foot which pushes
the ground backward; [1 mark]
From Newton's Third Law, there is an equal and opposite force; [1 mark]
From the ground on the foot which pushes the foot forward; [1 mark]
[Total: 3 marks]
(3 marks)

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4 (a) Airbags are one of the many safety devices used in cars to protect the driver if there is a
crash. Airbags protect the driver by

A. increasing the force acting on the driver

B. increasing the rate of change of momentum of the crash
C. increasing the time taken for the driver to stop
D. decreasing the time taken for the driver to stop

Answer

The correct answer is C because:

Safety features are designed to reduce the impact of a force by reducing the rate of
change in momentum experienced by the driver

This can be achieved by increasing the time taken for the driver to stop

∆p
This is because impact force = the rate of change of momentum or F = ∆t

So, increasing the time of the impact will reduce the size of the impact force felt
by the driver

A, B & D are incorrect as the aim of the airbag is to reduce the force experienced during
a crash, so increasing the force, rate of change of momentum, or time taken to stop
would all increase the impact force and hence, not protect the driver
(1 mark)

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(b) A driver has a mass of 67 kg and drives at a velocity of 15 m/s.

(i) State the formula linking momentum, mass and velocity.

(ii) Calculate the momentum of the driver.

Answer

(i) State the equation linking momentum, mass and velocity:

Momentum = mass × velocity OR p = mv [1 mark]

(ii) Calculate the momentum of the driver:

List the known quantities:

Mass, m = 67 kg

Velocity, v = 15 m/s

Substitute the known values into the equation:

p = 67 × 15 [1 mark]
p = 1005 kg m/s [1 mark]
[Total: 3 marks]
(3 marks)

(c) The car is involved in a crash. The driver is wearing a seatbelt and comes to rest in 0.36 s.

(i) State the equation linking force, change in momentum and time.

(ii) Show that the force on the driver is about 2800 N.

Answer

(i) State the equation linking force, change in momentum and time:

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 change in momentum mv − mu ∆p
= = ∆ t [1 mark]
( )

Force = OR F ,F
time t

(ii) Calculate the force on the driver:

List the known quantities:

Initial momentum of the driver = 1005 kg m/s

Time taken = 0.36 s

Calculate the force using the rate of change of momentum:

change in momentum
Force =
time

The final momentum mv is 0, as the driver comes to a stop, therefore change in

momentum = 1005 kg m/s

1005
Force = [1 mark]
0 . 36

Force = 2792 = 2800 N [1 mark]

[Total: 3 marks]
(3 marks)

(d) A passenger with the same mass as the driver is also in the car at the time of the crash.

They were not wearing a seatbelt and experienced an impact force that was 5 times
greater than the force experienced by the driver.

Calculate the time taken by the passenger to come to a stop.

time taken = .............................................................. s

Answer

Calculate the time taken by the passenger to come to a stop:

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List the known quantities:

Change in momentum = 1005 kg m/s

Force on driver = 2800 N

Calculate the time taken:

change in momentum change in momentum

Force =
time
⇒ time =
force

Force on the passenger = 2800 × 5 = 14 000 N [1 mark]

1005
Time taken = [1 mark]
14 000

Time taken = 0.07 s [1 mark]

[Total: 3 marks]
(3 marks)

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5 (a) The principle of conservation of momentum applies when two objects collide.

State the principle of conservation of momentum.

Answer

The principle of conservation of momentum states:

The total momentum before a collision is equal to the total momentum after the
collision; [1 mark]
In an isolated / closed system OR when no external forces act; [1 mark]
[Total: 2 marks]
(2 marks)

(b) A student is investigating conservation of momentum using an air track and two gliders,
A and B, which both have magnets attached, as shown in the diagram.

Initially, glider A has a momentum of 0.50 kg m/s and moves towards glider B, which is at
rest. The gliders collide and the magnets cause them to move in opposite directions.

Both gliders are identical and each has a mass of 0.250 kg.

(i) State and explain whether the poles of the magnets are opposite or alike.

(ii) State the total momentum of glider A and glider B after the collision.

(iii) Calculate the velocity of glider A before the collision.

velocity of glider A = ............................................... m/s

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Answer

(i) The poles of the magnets are:

Alike / the same / North-North / South-South; [1 mark]

(Because there is) repulsion (due to like magnetic poles); [1 mark]

(ii) The total momentum of glider A and glider B after the collision is:

0.50 kg m/s [1 mark]

(iii) Calculate the velocity of glider A before the collision:

List the known quantities:

Mass of glider A, m = 0.250 kg

Momentum of glider A, p = 0.5 kg m/s

Calculate velocity using the equation for momentum:

Momentum = mass × velocity or p = mv

p
p = mv ⇒ v = m

0.5
v = 0 . 250 [1 mark]

Velocity of glider A = 2 m/s [1 mark]

[Total: 5 marks]
(5 marks)

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(c) What is the velocity of glider B after the collision?

A. 1 m/s
B. 2 m/s
C. 3 m/s
D. 4 m/s

Answer

The correct answer is A because:

Momentum is equal to p = mv

From conservation of momentum: Momentum before = Momentum after

Momentum before: 0.5 kg m/s

Momentum after: mv + mv = 2mv

The two gliders will move with the same velocity in the opposite direction to each
other

Therefore:

2mv = 0.5

mv = 0.25

v = 0.25 ÷ 0.25 = 1 m/s

Therefore, option A is correct

(1 mark)

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Medium Questions
1 (a) The photograph shows a hammer just before it hits a nail.

The mass of the hammer is 0.50 kg.

When it hits the nail, the hammer is travelling downwards with a velocity of 3.1 m/s.

(i) State the relationship between momentum, mass and velocity.

(ii) Calculate the momentum of the hammer.

momentum = .............................................................. kg m/s

(iii) The hammer stops quickly when it hits the nail.

The momentum of the hammer reduces to zero in 0.070 s.

Calculate the amount of force that causes this to happen.

force = ........................................ N

Answer

(i) The relationship between momentum, mass and velocity is:

Momentum = mass × velocity OR p = mv; [1 mark]

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(ii) Calculate the momentum of the hammer:

List the known quantities:

Mass of the hammer, m = 0.50 kg

Velocity of the hammer, v = 3.1 m/s

Substitute values into the momentum equation:

p = 0.50 × 3.1 [1 mark]

p = 1.55 = 1.6 kg m/s [1 mark]
(iii) Calculate the amount of force that causes the momentum of the hammer to reduce
to zero in 0.070 s

List the known quantities:

Time for moment to reduce to zero, t = 0.070 s

Initial moment, p = 1.55 kg m/s

From the equation sheet:

change in momentum mv − mu
Force = F=
( )

time taken t

Substitute values into force equation:

1 . 55
F= [1 mark]
0 . 070

F = 22.143 = 22 N [1 mark]
[Total: 5 marks]

Force is sometimes referred to as the 'rate of change of momentum' based on the

equation

You will often see questions linking the two variables together!

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 (5 marks)

(b) As it enters the wood, the nail exerts a force on the wood. At the same time, the wood
exerts a force on the nail.

Explain how these two forces are related.

Answer

These two forces are related because:

The forces are equal; [1 mark]

and opposite / one is up and one is down; [1 mark]
OR

Newton's third law; [1 mark]

States that whenever two bodies interact, the forces they exert on each other are
equal and opposite; [1 mark]

[Total: 2 marks]

Avoid saying that the forces are 'balanced'. Newton's third law applies in all
situations, even when forces are not balanced.
(2 marks)

(c) Both ends of the nail exert pressure when the nail goes into the wood.

Explain why the nail exerts more pressure on the wood than it does on the hammer.

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Answer

The nail exerts more pressure on the wood than it does on the hammer because:

Any two from:

force
Pressure is equal to the force per unit area or [1 mark]
area

Pressure is inversely proportional to area; [1 mark]

The forces on the wood and hammer are equal; [1 mark]
Therefore, since a smaller area of the nail is in contact with the wood, it exerts more
pressure on the wood; [1 mark]
[Total: 2 marks]
(2 marks)

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2 (a) An ice skater throws a 0.23 kg snowball with a velocity of 13 m/s.

(i) State the equation linking momentum, mass and velocity.

(ii) Calculate the initial momentum of the snowball.

initial momentum = .......................... kg m/s

Answer

(i) The relationship between momentum, mass and velocity is:

momentum = mass × velocity OR p = mv ; [1 mark]

(ii) Calculate the initial momentum of the snowball:

List the known quantities:

Mass of the snowball, m = 0.23 kg

Velocity of the snowball, v = 13 m/s

Substitute values into the momentum equation:

p = 0.23 × 13 [1 mark]
p = 2.99 = 3.0 kg m/s [1 mark]
[Total: 3 marks]
(3 marks)

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(b) When the skater throws the snowball forwards, she slides backwards on the ice.

Explain why she moves in this direction.

Answer

The skater moves backwards after throwing the snowball forwards because:

Explanation 1: Conservation of momentum

Any three from:

Conservation of momentum; [1 mark]

(Means that) the momentum of snowball and skater; [1 mark]

(Are) equal and opposite; [1 mark]
(Because) momentum is initially zero; [1 mark]
OR

Explanation 2: Newton’s third law

Any three from:

Action and reaction / Newton's third law; [1 mark]

(Means that) the forces on skater and snowball; [1 mark]
(Are) equal and opposite; [1 mark]

(Because the magnitude of) rate of change of momentum is the same for both
forces; [1 mark]

An example of an answer that would score 3 marks:

Assuming that the skater is stationary (not moving) when she throws the snowball, the
combined momentum of the skater and the snowball before the throw is zero [1 mark]

From the principle of conservation of momentum, the total momentum of the system
(the snowball and the skater) must be conserved [1 mark] meaning that, after the throw,
the total momentum must also be zero.

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The snowball has a momentum of 3.0 kg m/s when thrown, therefore, the skater must
have an equal and opposite [1 mark] momentum of −3.0 kg m/s after the throw,
indicating that she is travelling in the opposite direction to the snowball

[Total: 3 marks]
(3 marks)

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(c) The skater wears soft knee pads that compress easily.

Explain how the pads protect her knees when she falls on the ice.

Answer

Soft knee pads are protective because:

Explanation 1: Momentum

Any three from:

Force is rate of change in momentum OR F = change in momentum ÷ time; [1 mark]

(Knee pads) increase the time (of impact); [1 mark]

(This means) the same change in momentum; [1 mark]
(Results in) reduced force (on knee); [1 mark]
OR

Explanation 2: Acceleration

Any three from:

(Knee pads) increase the distance / time (to slow down); [1 mark]
(This means) the same change in velocity / speed; [1 mark]
(Leads to) reduced acceleration; [1 mark]

(Results in) reduced force (on knee); [1 mark]

OR

Explanation 3: Pressure

Any three from:

Pressure = force ÷ area OR P = F ÷ A; [1 mark]

(Knee pads) increase the area (in contact with ground / knee); [1 mark]

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 (This means) pressure (on knee) is reduced; [1 mark]
(Results in) reduced force (on knee); [1 mark]
An example of an answer that would score 3 marks:

The change in momentum of the skater is equal to the skater's initial momentum, as
final momentum is zero. If the skater falls over, the compression of the pads on impact
with the ice means that the time of impact on the skater's knees is increased [1 mark]
Since force is proportional to the rate of change of momentum [1 mark] if the time of
impact is increased, the rate of change of momentum is reduced, so the force on her
knees is also reduced. [1 mark]
[Total: 3 marks]
(3 marks)

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3 (a) Some cars have a pedestrian airbag for safety. If a pedestrian is hit and lands on the
front of the car, the airbag inflates.

Use ideas about momentum to explain how this airbag can reduce injuries to
pedestrians.

Answer

This airbag can reduce injuries to pedestrians because:

Any four from:

The momentum is reduced; [1 mark]

By the same amount; [1 mark]

Over a longer period of time; [1 mark]

Therefore, the force is reduced; [1 mark]
Because force is the rate of change of momentum; [1 mark]
Less force means less damage / injuries; [1 mark]
[Total: 4 marks]
(4 marks)

(b) In a crash test, a car runs into a wall and stops.

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The momentum of the car before the crash is 22 500 kg m/s.

The car stops in 0.14 s.

(i) Calculate the average force on the car during the crash.

average force = .......................... N

(ii) Use ideas about momentum to explain how seat belts can reduce injuries to
passengers during a crash.

Answer

(i) Calculate the average force on the car:

List the known quantities:

Initial momentum = 22 500 kg m/s

Final momentum = 0

Change in momentum = 22 500 − 0 = 22 500 kg m/s

Time taken to stop = 0.14 s

Substitute the values into the equation:

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 Force = change in momentum ÷ time

22 500
Force = [1 mark]
0 . 14

Force = 160 000 N [1 mark]

(ii) Seat belts can reduce injuries to passengers during a crash because

Any three from:

(Seatbelts) increase the time (of impact); [1 mark]

For the same momentum change (with or without a seatbelt); [1 mark]
(This leads to a) reduced force; [1 mark]
Passenger stays on seat / is not thrown from the vehicle; [1 mark]
An answer that would score 3 marks:

The movement and slight stretching of a seatbelt means that during a crash, the time of
impact is longer. [1 mark]
The momentum change on the person is the same with or without a seatbelt. [1 mark]
Hence, the increased time of impact reduces the average force on the person during the
impact, reducing injury. [1 mark]
The seatbelt also keeps the person in their seat so that they are not thrown from the car
on impact, which further reduces the risk and severity of injury.

[Total: 5 marks]
(5 marks)

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4 (a) Cars have a number of features that make them safer in a collision.

Apart from seat belts, name two safety features that reduce the risk of serious injury in a
car crash.

Answer

Safety features in cars include:

Any two from:

Air bags; [1 mark]

Side impact beams / bars; [1 mark]
Crumple zones / collapsible bumpers; [1 mark]

Collapsible steering column / wheel; [1 mark]

Strong / laminated / safety glass in windows; [1 mark]
[Total: 2 marks]
The question already mentions seat belts, so you would not get any marks for
naming them. Also, if you refer to bumpers, you must state they are 'collapsible,' or
you will not gain the mark.
(2 marks)

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(b) Photograph A shows a person wearing a seat belt.

(i) Using ideas of momentum and force, explain how a seat belt reduces the risk of
serious injury in a car crash.

(ii) Photograph B shows a full-body harness used in a racing car.

Suggest why a full-body harness is used in a racing car, instead of an ordinary seatbelt.

Answer

(i) Explain how a seat belt reduces the risk of serious injury in a car crash

Any four from:

(Seatbelts) increase the time (of impact); [1 mark]

For the same momentum change (with or without a seatbelt); [1 mark]

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 (This leads to a) reduced rate of momentum change;

(Which results in a) reduced force; [1 mark]

The seat belt stretches (during collision); [1 mark]
(Which) increases the area over which the force acts; [1 mark]
(Hence) pressure (on the driver) is reduced; [1 mark]

An example of an answer that would score 4 marks:

Since force is equal to the rate of change of momentum, as the seatbelt stretches a small
amount during the collision [1 mark], it will increase the impact time [1 mark]
This increased time will reduce the rate of change of momentum [1 mark], therefore the
average force on the driver will be reduced [1 mark]

This also increases the area that the force is acting over and hence will reduce the
pressure on the driver. This will reduce the risk of injury.

(ii) Suggest why a full-body harness is used in a racing car:

Any one from:

There is a larger change in momentum (transferred in collision); [1 mark]

There is a larger force during impact; [1 mark]
Straps have a greater area over which force acts; [1 mark]
Larger area of straps reduces the pressure (on driver); [1 mark]
[Total: 5 marks]
(5 marks)

(c) Photograph C shows a crash-test dummy in a car. The car has crashed into a concrete
wall.

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State what happens to the momentum of the car during the crash.

Answer

State what happens to the momentum of the car during the crash:

Momentum (of car and dummy) reduces to zero; [1 mark]

OR

All momentum is absorbed by the wall/Earth; [1 mark]

[Total: 1 mark]
(1 mark)

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5 (a) A student is playing a game with some empty tins.

He throws a wet cloth of mass 0.15 kg at the tins. The wet cloth moves at a velocity of 6.0
m/s.

(i) State the equation linking momentum, mass and velocity.

(ii) Calculate the momentum of the wet cloth and give the unit.

Momentum = ................................ unit ...................

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(iii) The wet cloth sticks to tin 1.

The mass of tin 1 is 0.050 kg. The cloth and tin 1 move away together.

Calculate the velocity of the cloth and tin 1.

Velocity = .......................... m/s

Answer

(i) State the equation linking momentum, mass and velocity:

Momentum = mass × velocity

OR

p = mv [1 mark]
(ii) Calculate the momentum of the wet cloth and give the unit:

List the known quantities:

Mass, m = 0.15 kg

Velocity, v = 6.0 m/s

Substitute the known values into the equation from part (i):

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 p = 0 . 15 × 6 . 0 [1 mark]
p = 0.9 [1 mark]
State the unit:

kg m/s / N s [1 mark]

(iii) Calculate the velocity of the cloth and tin 1:

Rearrange the momentum equation to make velocity the subject:

p
v =m

Determine the combined mass of the cloth and tin 1:

Total mass = mass of cloth + mass of tin 1

Total mass = 0.15 + 0.050 = 0.2 kg

Substitute the known values into the equation:

0.9
v = 0 . 2 [1 mark]

v = 4.5 m/s [1 mark]

[Total: 6 marks]
(6 marks)

(b) The student throws a bigger wet cloth at the remaining tins. This wet cloth sticks to tins 2

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and 3 and they move away together.

The student concludes

Do you agree with this conclusion? Explain why.

Answer

Explain why you agree or disagree with the student's conclusion:

The student is incorrect; [1 mark]

Because the variables were not controlled; [1 mark]

[Total: 2 marks]

The student would be correct if the mass of the second cloth was 0.3 kg, or if the
momentum of the second cloth was 1.8 kg m/s, assuming that all the tins have the
same mass as tin 1, and that the cloths are thrown with the same velocity. If you
have explained this in your answer, you would still be awarded full marks.
(2 marks)

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Hard Questions
1 (a) A bowling ball rolls for 3 s and hits a pin.

The graph shows how the velocity of the ball changes with time.

Describe how can the graph be used to find the distance that the ball rolls before it hits
the pin.

Answer

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 Describe how the graph can be used to find the distance the ball rolls before it hits the
pin:

The area under the graph (from 0 s to 3 s) shows the distance

OR

6 × 3 = 18 m [1 mark]
[Total: 1 mark]
(1 mark)

(b) The mass of the ball is 6.4 kg.

(i) State the equation linking momentum, mass and velocity.

(ii) Calculate the momentum of the ball before it hits the pin. Give the unit.

Momentum = ............................ unit ......................

Answer

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 (i) State the equation linking momentum, mass and velocity:

Momentum = mass × velocity

OR

p = mv [1 mark]
(ii) Calculate the momentum of the ball before it hits the pin, giving the unit:

List the known quantities:

Mass, m = 6.4 kg

Velocity, v = 6 m/s (from part (a))

Substitute the known values into the equation:

p = 6 . 4 × 6 [1 mark]
p = 38.4 [1 mark]

State the unit:

kg m/s OR N s [1 mark]
[Total: 4 marks]
(4 marks)

(c) (i) Use the graph to determine the velocity of the ball after it hits the pin.

Velocity = ............................ m/s

(ii) After the collision, the ball and the pin have the same velocity. Calculate the mass of
the pin.

Mass = .................................... kg

Answer

(i) Determine the velocity of the ball after it hits the pin:

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From the graph:

(From graph) velocity of the ball = 4.8 m/s; [1 mark]

(ii) Calculate the mass of the pin:

List the known quantities:

Momentum of ball before collision, p1 = 38.4 kg m/s (from (b))

Velocity of ball / pin after collision, v = 4.8 m/s

Mass of ball, mball = 6.4 kg

Use conservation of momentum to write an expression for the momentum before:

Momentum before collision = momentum after collision; [1 mark]

p1 = p2
p1 = mballv + mpin v

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 p1 = (
m ball + mpin × v
)

Rearrange to make the mass of the pin the subject:

p1
Divide both sides by v:
v
= mball + mpin

p1
Subtract m ball from both sides:
v
− mball = mball + mpin − mball

p1
m pin = v
− mball

Substitute the known values into the equation to calculate mpin:

38. 4
m pin = 4 . 8 − 6 . 4 [1 mark]

mpin = 1.6 kg [1 mark]

[Total: 4 marks]

This is a difficult question because you need to manipulate the momentum equation.
Always show your working out clearly and logically. You may still gain marks from
your working out even if your final answer is incorrect.
(4 marks)

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2 (a) Newton's Cradle consists of a set of identical solid metal balls hanging by threads from a
frame so that they are in contact with each other.

A student initially pulls ball A to the side as shown. The student releases ball A and it
collides with ball B.

(i) State the equation linking momentum, mass and velocity.

(ii) Each ball has a mass of 100 g. At the time of collision, ball A has a velocity of 3 m/s.

Calculate the momentum of ball A at the time of impact and give the unit.

Momentum = ........................... unit ....................

(iii) After the collision, ball A stops. Ball E moves away. The other balls remain still.

The momentum of ball E as it moves away is the same as the momentum of ball A at the
time of impact.

Suggest a reason for this.

Answer

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(i) State the equation linking momentum, mass and velocity:

Momentum = mass × velocity

OR

p = mv [1 mark]
(ii) Calculate the momentum of ball A at the time of impact and give the unit:

List the known quantities:

Mass, m = 100 g

Velocity, v = 3 m/s

Convert mass into SI units:

100
1000
= 0 . 1 kg

Substitute the known values into the equation from part (i):

p = 0 . 1 × 3 [1 mark]
p = 0.3 [1 mark]
State the unit:

kg m/s or N s [1 mark]
(iii) Give a reason why the momentum of ball E as it moves away is the same as the
momentum of ball A at the time of impact:

Momentum is conserved; [1 mark]

[Total: 5 marks]
(5 marks)

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(b) The student then releases balls A and B together as shown below.

Predict what will happen to the other balls after the collision and gives a reason for your
answer.

Answer

Predict what will happen to the other balls:

Two balls at opposite ends of the cradle (balls D and E) move up / away OR Ball E
moves up / away with velocity 2v; [1 mark]
Give a reason for your answer:

Any one from:

Momentum is still conserved in the collision; [1 mark]

The total momentum of the system is constant; [1 mark]

There is twice the momentum of one ball so the (total) momentum is transferred to
two balls; [1 mark]
[Total: 2 marks]
(2 marks)

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3 (a) State the equation linking momentum, mass and velocity.

Answer

The equation linking momentum, mass and velocity is:

Momentum = mass × velocity OR p = mv [1 mark]

[Total: 1 mark]

If you're explaining this equation in symbols rather than words, make sure not to use
'm' or 'M' for momentum, which is always represented by 'p'.
(1 mark)

(b) A truck of mass 10 000 kg is moving with a velocity of 4.5 m/s.

A car of mass 1500 kg has the same momentum as the truck.

Calculate the velocity of the car.

velocity = .......................... m/s

Answer

List the known quantities:

Mass of truck, mT = 10 000 kg

Velocity of truck, vT = 4.5 m/s

Mass of car, mC = 1500 kg

Calculate momentum of the truck, pT:

pT = mTvT = (10 000) × 4.5 = 45 000 kg m/s [1 mark]

Rearrange momentum equation for the velocity, vC:

For the car, pC = mCvC

The car has the same momentum as the truck, so pC = pT

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 pC 45 000
vC = = [1 mark]
mC 1500

vC = 30 m/s [1 mark]
[Total: 3 marks]
(3 marks)

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4 (a) A boy of mass 43.2 kg runs and jumps onto a stationary skateboard.

The boy lands on the skateboard with a horizontal velocity of 4.10 m/s.

(i) State the relationship between momentum, mass and velocity.

(ii) The skateboard has a mass of 2.50 kg.

Using ideas about conservation of momentum, calculate the combined velocity of the
boy and skateboard just after the boy lands on it.

combined velocity = ............................................... m/s

Answer

(i) State the relationship between momentum, mass and velocity:

Momentum = mass × velocity OR p = mv [1 mark]

(ii) Calculate the combined velocity of the boy and skateboard just after the boy lands on
it:

List the known quantities:

Mass of boy, m1 = 43.2 kg

Mass of skateboard, m2 = 2.50 kg

Velocity of boy before impact, u1 = 4.10 m/s

Velocity of skateboard before impact, u2 = 0 m/s

Velocity of boy and skateboard after impact, v

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State the principle of conservation of momentum:

Momentum (of the boy just) before (he lands on the skateboard) is equal to the
combined momentum (of the skateboard and the boy just) after (he lands on it); [1
mark]
OR

Momentum before impact = momentum after impact; [1 mark]

OR

m 1u 1 + m 2u 2 = v m1 + m2 [1 mark]
( )

Calculate momentum before the impact:

Momentum of skateboard = (mass of skateboard) × (velocity of skateboard)

Momentum of skateboard = 2.50 × 0 = 0

Momentum of boy = (mass of boy) × (velocity of boy)

Momentum of boy = 43.2 × 4.10 = 177.12

Momentum before impact = Momentum of skateboard + Momentum of boy =

177.12 + 0

Momentum before impact = 177.12 kg m/s [1 mark]

Calculate the combined mass of the boy and the skateboard after the impact:

Combined mass = 43.2 + 2.5 = 45.7 kg

Apply the conservation of momentum:

Momentum before impact = momentum after impact

177.12 = (mass of boy and skateboard) × (velocity of boy and skateboard)

177.12 = 45.7 × (velocity of boy and skateboard)

Rearrange the equation to make velocity the subject and substitute the values:

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 177 . 12
Velocity of boy and skateboard = = 3.8757 [1 mark]
45. 7

Velocity of boy and skateboard = 3.88 m/s [1 mark]

[Total: 5 marks]
(5 marks)

(b) The boy holds a heavy ball as he stands on a stationary skateboard. The boy throws the
ball forwards while still standing on the skateboard.

Explain what happens to the boy and the skateboard.

Answer

Explain what happens to the boy and the skateboard:

The boy and the skateboard move backwards / in the opposite direction to the ball;
[1 mark]

(This is because of) conservation of momentum OR Newton's third law; [1 mark]

[Total: 2 marks]

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This question is a tricky one because of the combined effect of the boy on the
skateboard. The total momentum of the ball, the boy and the skateboard is
conserved. The boy and the skateboard are stationary to begin with so the total
momentum before throwing the ball is zero. For the total momentum of the system
to be conserved it must still be zero after the ball is thrown.
When the boy throws the ball forward the ball gains momentum in the forward
direction. This means that the boy and the skateboard must move backwards with
the same momentum as the ball has forwards.
(2 marks)

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5 (a) A lorry carries a load of hot asphalt – a runny mixture of small stones and tar.

The mass of the lorry and its load is 17 000 kg. The velocity is 13 m/s.

(i) State the equation linking momentum, mass and velocity.

(ii) Calculate the total momentum of the lorry and its load.

Momentum = ............................... kg m/s

Answer

(i) State the equation linking momentum, mass and velocity:

Momentum = mass × velocity OR p = mv [1 mark]

(ii) Calculate the total momentum of the lorry and its load:

List the known quantities:

Mass, m = 17 000 kg

Velocity, v = 13 m/s

Substitute the known values into the equation:

p = 17 000 × 13 [1 mark]
p = 221 000 kg m/s [1 mark]
[Total: 3 marks]
(3 marks)

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(b) The lorry stops suddenly and the load slides to the front, as shown below.

Force A and force B are upward forces from the road on the lorry.

(i) Use ideas about momentum to explain why the load slides to the front when the lorry
stops suddenly.

(ii) Use ideas about moments to explain why force B increases when the load slides to
the front.

Answer

(i) The load slides to the front when the lorry stops suddenly because:

The load has momentum; [1 mark]

The lorry stops in a shorter time because of the brakes OR the load takes longer to
stop (because there is no direct stopping force acting on it); [1 mark]

(ii) Force B increases when the load slides to the front because:

The centre of gravity (of the load and lorry) moves closer to the front (of the lorry);
[1 mark]

The clockwise and anticlockwise moments are equal; [1 mark]

Therefore, the force on B must increase to balance the moments OR to keep the
system in equilibrium; [1 mark]

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[Total: 5 marks]
(5 marks)

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6 (a) A stationary rocket in deep space ignites its engines, propelling exhaust gases at high
speed away from the rocket. Identify the pair of equal and opposite forces that show
how momentum is conserved in this situation.

Answer

The pair of equal and opposite forces that show how momentum is conserved in this
situation is:

The rocket exerts a force on the exhaust gases (backwards). [1 mark]

The exhaust gases exert a force equal in magnitude and opposite in direction on the
rocket (forwards). [1 mark]
[Total: 2 marks]
(2 marks)

(b) Suggest how rocket engineers could use the idea of momentum to design safety features
to protect the crew when the rocket ignites its engines.

Answer

Suggest how rocket engineers could use the idea of momentum to design safety features
to protect the crew when the rocket ignites its engines:

Any one from:

Harnesses / restraints could be used to reduce the forces experienced by

astronauts due to the rocket’s change in momentum; [1 mark]

Shock absorbers in seats that would extend the time interval over which the
momentum changes, reducing the forces exerted on the astronauts; [1 mark]
Protective padding around the astronauts that reduces impact by increasing the
time interval over which the forces act; [1 mark]
[Total: 1 mark]
The mark is awarded for suggesting a safety feature AND correctly linking it to
momentum. Any reasonable suggestion with a correct explanation would gain the
mark.
(1 mark)

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