P2 Revision Booklet

Momentum

Harder (Grade B-A*)

Q1.(a)     The picture shows two teenagers riding identical skateboards.
The skateboards are moving at the same speed and the teenagers have the same mass.

Why do the teenagers not have the same momentum?
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(1)

(b)     One of the skateboards slows down and stops. The teenager then jumps off the skateboard, causing it to recoil and move in the opposite direction.

The momentum of the teenager and skateboard is conserved.
(i)      What is meant by ‘momentum being conserved’?
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(1)

(ii)     The teenager, of mass 55 kg, jumps off the skateboard at 0.4 m/s causing the skateboard to recoil at 10 m/s.
Use the equation in the box to calculate the mass of the skateboard.
 
momentum    =    mass    ×    velocity	

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                                 Mass = ............................................................ kg
(3)

(c)     Once the skateboard starts to recoil, it soon slows down and its kinetic energy decreases.
Explain why.
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(2)
(Total 7 marks)
 Q2.          (a)     Complete the following sentence.
The momentum of a moving object has a magnitude, in kg m/s,
and a .................................................. .
(1)

(b)     A car being driven at 9.0 m/s collides with the back of a stationary lorry.
The car slows down and stops in 0.20 seconds. The total mass of the car and driver is 1200 kg.
Use the equations in the box to calculate the average force exerted by the lorry on the car during the collision.

Show clearly how you work out your answer.
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                                           Force = .................................................. N
(2)

(c)     Within 0.04 s of the car hitting the back of the lorry, the car driver's airbag inflates.
The airbag deflates when it is hit by the driver’s head.

Use the idea of momentum to explain why the airbag reduces the risk of the drive sustaining a serious head injury.
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(3)
(Total 6 marks)
 Q3.          (a)    In any collision, the total momentum of the colliding objects is usually conserved.
(i)      What is meant by the term ‘momentum is conserved’?
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(1)

(ii)     In a collision, momentum is not always conserved.
Why?
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(1)

(b)     The diagram shows a car and a van, just before and just after the car collided with the van.

 
Before collision	
 
After collision	

(i)      Use the information in the diagram and the equation in the box to calculate the change in the momentum of the car.
 
momentum   =   mass   ×   velocity	

Show clearly how you work out your answer and give the unit.
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                      Change in momentum =..................................................
(3)

(ii)     Use the idea of conservation of momentum to calculate the velocity of the van when it is pushed forward by the collision.
Show clearly how you work out your answer.
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                          Velocity =.................................................. m/s forward
(2)
(Total 7 marks)

Q4.          In an experiment at an accident research laboratory, a car driven by remote control was crashed into the back of an identical stationary car. On impact the two cars joined together and moved in a straight line.
(a)     The graph shows how the velocity of the remote-controlled car changed during the experiment.

(i)      How is the velocity of a car different from the speed of a car?
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(1)

(ii)     Use the graph to calculate the distance travelled by the remote-controlled car before the collision.
Show clearly how you work out your answer.
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Distance = ............................................... m
(2)

(iii)     Draw, on the grid below, a graph to show how the velocity of the second car changed during the experiment.

(2)

(iv)    The total momentum of the two cars was not conserved.
         What does this statement mean?
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(1)

(b)     The graph line shows how the force from a seat belt on a car driver changes during a collision.

          Scientists at the accident research laboratory want to develop a seat belt that produces a constant force throughout a collision.
          Use the idea of momentum to explain why this type of seat belt would be better for a car driver.
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(2)
(Total 8 marks)
 Q5.          The roads were very icy. An accident was recorded by a security camera.

          Car A was waiting at a road junction. Car B, travelling at 10 m/s, went into the back of car A. This reduced car B’s speed to 4 m/s and caused car A to move forward.
          The total mass of car A was 1200 kg and the total mass of car B was 1500 kg.
(i)      Write down the equation, in words, which you need to use to calculate momentum.
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(1)

(ii)      Calculate the change in momentum of car B in this accident.
          Show clearly how you work out your final answer and give the unit.
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Change in momentum = .........................................
(3)

(iii)     Use your knowledge of the conservation of momentum to calculate the speed, in m/s, of car A when it was moved forward in this accident.
          Show clearly how you work out your final answer.
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Speed = ............................................................ m/s
(3)
(Total 7 marks)
 Q6.          (a)     The diagram shows a hammer which is just about to drive a nail into a block of wood.

          The mass of the hammer is 0.75 kg and its velocity, just before it hits the nail, is 15.0 m/s downward. After hitting the nail, the hammer remains in contact with it for 0.1 s. After this time both the hammer and the nail have stopped moving.

(i)      Write down the equation, in words, which you need to use to calculate momentum.
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(1)

(ii)     What is the momentum of the hammer just before it hits the nail?
         Show how you work out your answer and give the units and direction.
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Momentum = ...................................................................
(3)

(iii)     What is the change in momentum of the hammer during the time it is in contact with the nail?
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(1)

(iv)    Write down an equation which connects change in momentum, force and time.
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(1)

(v)     Calculate the force applied by the hammer to the nail.
         Show how you work out your answer and give the unit.
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Force = ............................................................................
(3)

(b)     A magazine article states that:
“Wearing a seat belt can save your life in a car crash.”
          Use your understanding of momentum to explain how this is correct.
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(4)
(Total 13 marks)

 

Q7.          (a)     What is the principle of conservation of momentum?
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(2)

(b)     The diagram shows a simplified aircraft jet engine.

Adapted from GCSE Physics by Tom Duncan. John Murray (Publishers) Ltd.

(i)      What is the function of the turbine?
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(1)

(ii)     Explain how the engine produces a forward thrust.
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(4)

(c)     During flight, air enters the engine at 175 m/s and leaves at 475 m/s. A forward thrust of 105 kN is produced.
          Use the following equation to calculate the mass of air passing through the engine every second. (Ignore the mass of the burned fuel.)

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Mass of air = ............................................... kg
(2)
(Total 9 marks)
 Q8.          (a)     The picture shows two ice hockey players skating towards the puck. The players, travelling in opposite directions, collide, fall over and stop.

(i)      Use the following equation and the data given in the box to calculate the momentum of player number 3 before the collision. Show clearly how you work out your answer and give the unit.
momentum = mass × velocity
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Momentum of player 3 = .......................................
(3)

(ii)     What is the momentum of player 4 just before the collision?
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(1)

(iii)     The collision between the two players is not elastic. What is meant by an elastic collision?
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(1)

(b)     The pictures show what happened when someone tried to jump from a stationary rowing boat to a jetty.

Use the idea of momentum to explain why this happened.
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(2)

(c)     The diagram shows one type of padded body protector which may be worn by a horse rider.

          If the rider falls off the horse, the body protector reduces the chance of the rider being injured. Use the idea of momentum to explain why.
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(3)
(Total 10 marks)
 Q9.          The diagram below shows two balls on the bowling green. Ball A is moving with a velocity of 4 m/s, and is about to collide with ball B which is stationary. Both balls have a mass of
1.5 kg.

          After the collision both balls move to the right but the velocity of A is now 1 m/s.

(a)     (i)      Calculate the momentum of ball A just before the collision.
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Answer ...................... kg m/s
(1)

(ii)     What is the total momentum of balls A and B after the collision?
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Answer ....................... kg m/s
(1)

(iii)     Calculate the momentum of ball A just after the collision.
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Answer ....................... kg m/s
(1)

(iv)    Calculate the momentum of ball B just after the collision.
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Answer ....................... kg m/s
(1)

(v)     Calculate the velocity of ball B just after the collision.
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Answer ............................ m/s
(1)

(b)     Calculate the loss of kinetic energy in the collision.
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Answer ............................... J
(3)
(Total 8 marks)

 Q10.          (a)     How can the momentum of an object be calculated?
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(2)

(b)     In a collision momentum is always conserved. What does this mean?
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(2)

(c)     Two trolleys are placed on a frictionless runway as shown in the diagram below. Trolley A has a protruding pin, and trolley B is fitted with a piece of soft cork so that the trolleys will stick together after colliding.

          Trolley A has a mass of 2 kg, and trolley B has a mass of 1 kg. Trolley B is stationary. Trolley A strikes trolley B at a speed of 6 m/s. Both trolleys then move to the right together.

(i)      Calculate the speed at which trolleys A and B jointly move after the collision.
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(4)

(ii)     Calculate the change in kinetic energy which occurs during the collision. 
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(4)
(Total 12 marks)
 Momentum-Mark scheme

M1.           (a)     (moving in) different / opposite directions
accept one has positive momentum the other negative momentum
accept they have different velocities 
1

(b)     (i)      momentum before = momentum afteror(total) momentum stays the same
accept no momentum is lost
accept no momentum is gained
1

(ii)     2.2 
allow 1 mark for calculation of teenagers’ momentum as22 (kgm/s) and
allow 1 mark for correct statement, eg momentumbefore = momentum after
or
allow 2 marks for a numerical expression of above, eg
55 × 0.4 = m × 10 
or 0 = (55 × 0.4) + (m × (-10))
3

(c)     any two from:
•    work is done 
•    (against) friction
any reference to increasing friction negates this marking point
•    (transforming) (kinetic) energy into heat
2
[7]

M2.          (a)     direction
1

(b)     54 000
allow 1 mark for calculating and identifying momentum as 10 800
or
allow 1 mark for correct substitution into second equation
ie 
2

(c)     increases the time taken (for head) to stop
accept increases impact time
do not accept reference to slowing down time unless qualified
1

decreases rate of change in momentum
accept reduces acceleration / deceleration
accept increases the time taken to reduce momentum to zero is worth 2 marks
reduces momentum is insufficient
1

reduces the force (on the head)
1
[6]
M3.         (a)      (i)     momentum before = momentum afteror
(total) momentum stays the same
accept no momentum is lost
accept no momentum is gained
1

(ii)     an external force acts (on the colliding objects)
accept colliding objects are not isolated
1

(b)     (i)     9600
allow 1 mark for correct calculation of momentum before or after
ie 12000 or 2400
or
correct substitution using change in velocity = 8 m/s
ie 1200 × 8
2

kg m/s
this may be given in words rather than symbols
or
Ns
1

(ii)     3 or their (b)(i) ÷ 3200 correctly calculated
allow 1 mark for stating momentum before = momentum after
or
clear attempt to use conservation of momentum
2
[7]
 M4.          (a)     (i)      velocity includes direction
accept velocity is a vector
1

(ii)     64
allow 1 mark for obtaining values of 16 and 4 from the graph
or marking correct area or correct attempt to calculate an area
2

(iii)     any two from:
•        velocity zero from 0 to 4 seconds
•        increasing in 0.2 s (or very rapidly) to 8 m/s
•        decreasing to zero over the next 8 seconds
2

(iv)    momentum before does not equal momentum after
ignore reference to energy
         or total momentum changes
         or an external force was applied
1

(b)     to reduce the momentum of the driver
1
          a smaller (constant) force would be needed
do not accept reduces the impact / impulse on the driver
1
[8]
 M5.          (i)      momentum (change in) = mass × velocity (change in)
accept ... speed
1
(ii)      9000
1500 × 6 for 1 mark but not from incorrect equation
2
         kilogram metre(s) per second or kg m/s
1

(iii)     either 7.5 (m/s)
          or change in momentum of car B change in momentum of car A (1)
9000 = 1200 × v (1)
or v = 9000 ÷ 1200 (1)

          or error carried forward from part (ii)
examples
5 (m/s) if 6000 offered in (ii) (3)
12.5(m/s) if 15000 offered in (ii)
(3)
3
[7]
 M6.          (a)     (i)      momentum = mass × velocity
accept … × speed or any transposed version
1
(ii)     11.2 to 11.3
0.75 × 15 for 1 mark
2
kg m/s down(wards) or Ns down(ward)
n.b. both unit and direction required for this mark
1
(iii)     11.2 to 11.3
accept same numerical answer as part (a)(ii)
accept answer without any unit  or with the same unit as in part (a)(ii), even if incorrect, but any other unit cancels the mark
1

(iv)    force = 
 
accept transposed version
1
(v)     112 to 113 or numerical value from (a)(ii) × 10
11.25 ÷ 0.1 or (a)(ii) ÷ 0.1 for 1 mark
2
newton(s)
or N
accept Newton(s) 
do not credit ‘Ns’  or n
1

(b)     (the user will experience a) large change in momentum
do not credit just ‘… momentum changes’
1
          (but) seat belt increases the time for this to occur or
seat belt stops you hitting something which would stop you quickly
do not credit just ‘… stops you hitting the windscreen etc.’
1
(so) the force on the user is less(*)
1
(so) less chance of (serious / fatal) injury(*)
(*) depends on previous response   re momentum or continued movement
1
[13]
 M7.          (a)     Total momentum (of a system of bodies) remains constant
accept momentum before (a collision) = momentum after (a collision)
1
Provided no external force acts
1

(b)     (i)      rotate the compressor
1

(ii)     •        fuel is mixed with the air and ignited
•        causing an increase in the pressure 
or temperature or speed of the gases
accept air out faster than air in 
accept gases have momentum or
•        force backwards
•        exhaust gases have momentum
(backwards) or force (backwards)
if the answer is in terms of force then this third point must be scored before the fourth can be credited
•        engine or aircraft has (equal) momentum forwards or force forwards
4

(c)     m = 350
answer 0.35 one mark only
allow one mark if 105 000 or 475-175 or 300 have been used
2
[9]
M8.          (a)     (i)      direction indicated
accept to right or + or – or arrow drawn on diagram
1
         300
1
         kg m/s or Ns
1

(ii)     300 (kg m/s)
1

(iii)     there is no change in the total KE
or total KE is constant
1

(b)     momentum of person towards jetty = momentum of boat away from jetty
or total momentum is constant so as person goes one way boat goes the other
1 mark is for the idea of momentum conservation
1 is for direction
2

(c)     time of collision increases
do not accept momentum is conserved
1
so a smaller force is exerted
do not accept designed to absorb energy or momentum
1
          to produce the same change of momentum or impulse force
do not accept cushions fall
1
[10]
 M9.          (a)     (i)      6
for 1 mark
1

(ii)     6
for 1 mark
1

(iii)     1.5
for 1 mark
1

(iv)    4.5
for 1 mark
1

(v)     3
for 1 mark
1

(b)     initial ke = 12J;
final ke = 0.75J + 6.75J;
energy loss = 4.5J 
for 1 mark each
          (If wrong; any correct ke value gains 1 mark; maximum of 2
path through calculation clear and correct gains 1 mark)
(ignore either ball – max 1 mark)
3
[8]

 

M10.          (a)     mass and velocity/speed multiplied
for 1 mark each
2

(b)     total momentum before and after collision are the same 
for 1 mark each
2

(c)     (i)      MAUA + MBUB = (MA + MB)v
2 × 6 = (2 + 1)v
v = 4
m/s 
for 1 mark each
4

(ii)     1/2 mv2 (before) – 1/2 mv2 (after) 1/2 2.36 – 1/2 3.16 = 12
J 
for 1 mark each
4
[12]