€¦ · Web view2021. 3. 14. · A positive ion with a mass of 3.4 × 10–26 kg and a charge of 3.2 × 10–19 C is initially at rest at a point P, midway between two parallel

Colonel Frank Seely School

Exampro A-level Physics (7407/7408) 3.4.1.6 Momentum

Name:

Class:

Author:

Date:

Time: 471

Marks: 370

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Q1.The figure below shows a neutron of mass 1.7 × 10–27 kg about to collide inelastically with a stationary uranium nucleus of mass 4.0 × 10–25 kg. During the collision, the neutron will be absorbed by the uranium nucleus.

(a) Calculate the velocity of the uranium nucleus immediately after the neutron has been absorbed.

(3)

(b) Collisions between neutrons and uranium nuclei can also be elastic. State, and explain briefly, how the speed of the uranium nucleus after impact would be different in the case of an elastic collision. Do not perform any further calculations.

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(c) Using the data at the beginning of the question, calculate the kinetic energy of the neutron before it collides with the uranium nucleus.

(3)(Total 9 marks)

Q2.(a) Figure 1 shows a skier descending the ramp of a ski jump. Figure 2 shows a graph of the distance travelled along the ramp against time, from the time the descent starts until the skier leaves the end of the ramp.

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Figure 1

Figure 2

The skier of mass 80 kg (including equipment) skis down the ramp and leaves it horizontally. The skier gains 55% of the available gravitational potential energy as kinetic energy when descending the ramp.

acceleration of free fall, g = 9.8 m s–2

(i) One energy transformation which occurs as the skier skis down the ramp is from gravitational potential energy to kinetic energy of the skier. State two other energy transformations that occur as the skier skis down the ramp.

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(ii) Use Figure 2 to show that the speed at which the skier leaves the ramp is about 23 m s–1. Show your reasoning clearly.

(2)

(iii) Determine the height of the ramp.(3)

(b) Figure 1 shows the path taken by the skier after leaving the ramp.Assuming that there was no lift or drag due to the air during this jump, calculate:

(i) the time for which the skier was in flight;(2)

(ii) the horizontal distance jumped by the skier before landing.(2)

(c) On landing the skier has considerable vertical momemtum that has to be reduced to zero. The surface on which the skier lands is hard-packed snow. To reduce the force experienced by the skier, the landing surface is angled at 40° to the horizontal.

Explain briefly how angling the landing surface reduces the vertical component of the force, experienced by the skier.

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(Total 14 marks)

Q3.The diagram shows the graph of force on a car against time when the car of mass 500 kg crashes into a wall without rebounding.

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Which one of the following statements is correct?

A The area under the graph is equal to the initial momentum of the car

B Momentum is not conserved in the collision

C Kinetic energy is conserved in the collision

D The average force exerted on the car is 10 × 104 N(Total 1 mark)

Q4.(a) (i) State the principle of conservation of momentum.

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(ii) Explain briefly how an elastic collision is different from an inelastic collision.

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(b) Describe and explain what happens when a moving particle collides elastically with a stationary particle of equal mass.

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(c) The diagram below shows an astronaut undertaking a space-walk. The astronaut is tethered by a rope to a spacecraft of mass 4.0 × 104 kg. The spacecraft is moving at constant velocity.

The astronaut and spacesuit have a total mass of 130 kg. The change in velocity of the astronaut after pushing off is 1.80 m s–1.

(i) Determine the velocity change of the spacecraft.(2)

(ii) The astronaut pushes for 0.60 s in achieving this speed. Calculate the average power developed by the astronaut. Neglect the change in motion of the spacecraft.

(3)

(iii) The rope eventually becomes taut. Suggest what would happen next.

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(Total 13 marks)

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Q5.The diagrams show the variation of velocity and acceleration with time for a body undergoing simple harmonic motion.

Which one of the following is proportional to the change in momentum of the body during the time covered by the graphs?

A The area enclosed by the velocity-time graph and the time axis

B The gradient of the velocity-time graph at the point P

C The area enclosed by the acceleration-time graph and the time axis

D The gradient of the acceleration-time graph at the point Q

(Total 1 mark)

Q6.A body X, moving with a velocity v, collides elastically with a stationary body Y of equal mass.

Which one of the following correctly describes the velocities of the two bodies after the collision?

velocity of X velocity of Y

A

B

C –v 0

D 0 v

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(Total 1 mark)

Q7.A body is accelerated from rest by a constant force.

Which one of the following graphs best represents the variation of the body’s momentum p with time t?

(Total 1 mark)

Q8. (a) State the principle of conservation of momentum.

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(b) The diagram below shows a sketch drawn by an accident investigator following a head-on collision between two vehicles.

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From the skid marks and debris on the road the investigator knows that the collision took place at the point marked X. The vehicles locked together on impact and vehicle A was pushed backwards a distance of 8.4 m.

For the road conditions and vehicle masses the average frictional force between the road and the vehicles immediately after the collision was known to be 7500 N.

(i) Calculate the work done against friction in bringing the vehicles to rest.

(2)

(ii) Determine the speed of the interlocked vehicles immediately after impact.

(2)

(iii) Vehicle A was known to be moving at 12.5 m s–1 just before the impact. Calculate the speed of vehicle B just before impact.

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(3)

(iv) The drivers of both vehicles have the same mass. State and explain which driver is likely to experience the higher force during the impact.

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(Total 13 marks)

Q9.The diagram below shows the rotor-blade arrangement used in a model helicopter. Each of the blades is 0.55 m long with a uniform cross-sectional area of 3.5 × 10−4 m2 and negligible mass. An end-cap of mass 1.5 kg is attached to the end of each blade.

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(a) (i) Show that there is a force of about 7 kN acting on each end-cap when the blades rotate at 15 revolutions per second.

(3)

(ii) State the direction in which the force acts on the end-cap.

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(iii) Show that this force leads to a longitudinal stress in the blade of about 20 MPa.

(2)

(iv) Calculate the change in length of the blade as a result of its rotation.

Young modulus of the blade material = 6.0 × 1010 Pa(2)

(v) Calculate the total strain energy stored in one of the blades due to its extension.

(2)

(b) The model helicopter can be made to hover above a point on the ground by directing the air from the rotors vertically downwards at speed v.

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(i) Show that the change in momentum of the air each second is Aρv2, where A is the area swept out by the blades in one revolution and ρ is the density of air.

(2)

(ii) The model helicopter has a weight of 900 N. Calculate the speed of the air downwards when the helicopter has no vertical motion.

Density of air = 1.3 kg m−3

(3)(Total 15 marks)

Q10. (a) Starting with the relationship between impulse and the change in momentum, show clearly that the unit, N, is equivalent to kg m s–2.

(2)

(b) A rocket uses a liquid propellant in order to move.Explain how the ejection of the waste gases in one direction makes the rocket move in the opposite direction.

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(c) A rocket ejects 1.5 × 104 kg of waste gas per second. The gas is ejected with a speed of 2.4 km s–1 relative to the rocket. Show that the average thrust on the rocket

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is about 40 MN.

(2)

(Total 7 marks)

Q11. (a) Explain what is meant by the principle of conservation of momentum.

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(b) A hose pipe is used to water a garden. The supply delivers water at a rate of 0.31 kg s–1 to the nozzle which has a cross-sectional area of 7.3 × 10–5 m2.

(i) Show that water leaves the nozzle at a speed of about 4 m s–1.density of water = 1000 kg m–3

(2)

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(ii) Before it leaves the hose, the water has a speed of 0.68 m s–1. Calculate the force on the hose.

(3)

(iii) The water from the hose is sprayed onto a brick wall the base of which is firmly embedded in the ground. Explain why there is no overall effect on the rotation of the Earth.

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(Total 9 marks)

Q12.

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A positive ion with a mass of 3.4 × 10–26 kg and a charge of 3.2 × 10–19 C is initially at rest at a point P, midway between two parallel conducting plates, A and B which are separated by 40 mm. The ion is accelerated and passes through a hole Q in plate B. It enters a magnetic field of uniform flux density 0.10 T at R. After following a circular path the ion leaves the field at S. Assume that the magnetic field is uniform everywhere within the dotted rectangle and that the space within the solid rectangle is evacuated.

(a) (i) Calculate the electric field strength between the plates AB.

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(ii) Calculate the force on the ion due to the electric field.

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(iii) Show that the speed of the ion just after it has passed through the hole at Q is 3.1 × 105 m s–1.

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(b) (i) State the direction of the magnetic field.

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(ii) Explain why the ion follows a circular path in the magnetic field.

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(iii) Show that the radius of the circular path is proportional to the momentum of the ion and calculate the value of the radius.

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(c) Explain how, if at all, the trajectory would be different for an ion with a slightly greater mass but carrying the same charge.

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(Total 13 marks)

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Q13.Most cars manufactured in recent years have been fitted with air bags as a safety feature. In a collision the bag inflates automatically to protect the driver as air is released from a compressed air cylinder.

(a) Explain why the driver would be less seriously injured in a collision if the air bag inflates than he would be if unrestrained.

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(b) Why is the driver of a car fitted only with seat belts more likely to be injured than if an air bag was fitted? Ignore the different deceleration times and the difference in the materials used.

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(c) With reference to pressure, volume and temperature, discuss what happens to the air as the bag inflates.

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(Total 10 marks)

Q14.A solid iron ball of mass 890 kg is used on a demolition site. It hangs from the jib of a crane suspended by a steel rope. The distance from the point of suspension to the centre of mass of the ball is 15 m.

(a) Calculate the tension in the rope when the mass hangs vertically and stationary.

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(b) The iron ball is pulled back by a horizontal chain so that the suspension rope makes an angle of 30° with the vertical. Calculate the new tension in the suspension rope.

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(c) The ball is now released from rest and hits a brick wall just as it passes through the vertical position. It can be assumed that the ball is brought to rest by the impact with the wall in 0.2 s.

Calculate

(i) the vertical height through which the ball falls,

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(ii) the speed of the ball just before impact,

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(iii) the average force exerted by the ball on the wall.

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(Total 9 marks)

Q15.A toy locomotive of mass 0.50 kg is initially at rest on a horizontal track. The locomotive is powered by a twisted rubber band which, as it unwinds, exerts a force which varies with time as shown in the table.

time/s 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

force/N 0.20 0.18 0.15 0.12 0.10 0.08 0.05 0.02 0.00

(a) (i) On the grid below plot a graph of force against time for the rubber band power source.

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(ii) State what is given by the area between the graph and the time axis.

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(b) The rubber band is wound up and released to power the locomotive. Use your graph to show that the speed of the locomotive 8.0 s after the twisted rubber band is released is 1.6 m s–1. Ignore the effects of air resistance and energy losses due to friction.

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(c) 8.0 s after release the locomotive collides with and couples to a toy truck, initially at rest, which has a mass of 1.50 kg.

(i) Calculate the speed of the coupled locomotive and truck after collision.

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(ii) Calculate the combined kinetic energy of the locomotive and truck immediately after collision.

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(iii) Show, with the aid of a calculation, whether or not the collision is elastic.

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(Total 11 marks)

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Q16.Take the acceleration due to gravity, gE, as 10 ms−2 on the surface of the Earth.

The acceleration due to gravity on the surface of the Moon is . An object whose weight on

Earth is 5.0 N is dropped from rest above the Moon’s surface. What is its momentum after falling for 3.0s?

A 2.5 kg m s−1

B 6.2 kg m s−1

C 15 kg m s−1

D 25 kg m s−1

(Total 1 mark)

Q17.(a) (i) Give an equation showing how the principle of conservation of momentum applies to the colliding snooker balls shown in the diagram.

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(ii) State the condition under which the principle of conservation of momentum applies.

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(b) A trolley, A, of mass 0.25 kg and a second trolley, B, of mass 0.50 kg are held in contact on a smooth horizontal surface. A compressed spring inside one of the trolleys is released and they then move apart. The speed of A is 2.2 m s–1.

(i) Calculate the speed of B.

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(ii) Calculate a minimum value for the energy stored in the spring when compressed.

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(c) The rotor blades of a helicopter sweep out a cross-sectional area, A. The motion of the blades helps the helicopter to hover by giving a downward velocity, υ, to a cylinder of air, density ρ. The cylinder of air has the same cross-sectional area as that swept out by the rotor blades.

Explaining your reasoning,

(i) derive an expression for the mass of air flowing downwards per second, and

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(ii) derive an expression for the momentum given per second to this air.

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(iii) Hence show that the motion of the air results in an upward force, F, on the helicopter given by

F = ρAυ2.

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(d) A loaded helicopter has a mass of 2500 kg. The area swept out by its rotor blades is 180 m2. If the downward flow of air supports 50% of the weight of the helicopter, what speed must be given to the air by the motion of the rotor blades when the helicopter is hovering? Take the density of air to be 1.3 kg m–3.

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(Total 15 marks)

Q18.(a) State the principle of conservation of linear momentum for two colliding bodies.

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(2)

(b)

A bullet of mass 0.010 kg travelling at a speed of 200 m s–1 strikes a block of wood of mass 0.390 kg hanging at rest from a long string. The bullet enters the block and lodges in the block. Calculate

(i) the linear momentum of the bullet before it strikes the block,

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(ii) the speed with which the block first moves from rest after the bullet strikes it.

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(c) During the collision of the bullet and block, kinetic energy is converted into internal energy which results in a temperature rise.

(i) Show that the kinetic energy of the bullet before it strikes the block is 200 J.

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(ii) Show that the kinetic energy of the combined block and bullet immediately after the bullet has lodged in the block is 5.0 J.

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(iii) The material from which the bullet is made has a specific heat capacity of 250 J kg–1 K–1. Assuming that all the lost kinetic energy becomes internal energy in the bullet, calculate its temperature rise during the collision.

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(d) The bullet lodges at the centre of mass G of the block. Calculate the vertical height h through which the block rises after the collision.

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(Total 13 marks)

Q19.(a) Collisions can be described as elastic or inelastic. State what is meant by an inelastic collision.

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(b) A ball of mass 0.12 kg strikes a stationary cricket bat with a speed of 18 m s–1. The ball is in contact with the bat for 0.14 s and returns along its original path with a speed of 15 m s–1.

Calculate

(i) the momentum of the ball before the collision,

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(ii) the momentum of the ball after the collision,

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(iii) the total change of momentum of the ball,

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(iv) the average force acting on the ball during contact with the bat,

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(v) the kinetic energy lost by the ball as a result of the collision,

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(Total 7 marks)

Q20.(a) When an α particle is emitted from a nucleus of the isotope , a nucleus of thallium, Tl, is formed. Complete the equation below.

→ α + Tl(2)

(b) The α particle in part (a) is emitted with 6.1 MeV of kinetic energy.

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(i) The mass of the α particle is 4.0 u. Show that the speed of the α particle immediately after it has been emitted is 1.7 × 10–7 m s–1. Ignore relativistic effects.

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(ii) Calculate the speed of recoil of the daughter nucleus immediately after the α particle has been emitted. Assume the parent nucleus is initially at rest.

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(Total 8 marks)

Q21.(a) An egg of mass 5.8 × 10–2 kg is dropped from a height of 1.5 m onto a floor. Assuming air resistance is negligible, calculate for the egg

(i) the loss of potential energy,

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(ii) the kinetic energy just before impact,

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(iii) the speed just before impact,

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(iv) the momentum just before impact.

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(b) On hitting the floor, the egg is brought to rest in a time of 0.010 s. Calculate the magnitude of the average decelerating force on the egg.

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(c) The egg is now placed in a container that crumples on impact. Explain why this type of container makes it far less likely that the egg will break.

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(Total 11 marks)

Q22.The simplified diagram shows an experimental arrangement to investigate the collision of two trolleys.

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In the experiment, trolley A is travelling at speed v. It collides with and sticks to, the initially stationary trolley B.

(a) State the measurements you would need to take so that you could determine the speed of

(i) trolley A before the collision,

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(ii) trolleys A and B after the collision.

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(b) Explain how you would verify that momentum was conserved in this collision, indicating what other measurements would be required.

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(c) State and explain what you would do to minimise the effects of friction on the motion of the trolleys.

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(Total 7 marks)

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Q23. A girl kicks a ball along the ground at a wall 2.0 m away. The ball strikes the wall normally at a velocity of 8.0 m s–1 and rebounds in the opposite direction with an initial velocity of 6.0 m s–1. The girl, who has not moved, stops the ball a short time later.

(a) Explain why the final displacement of the ball is not 4.0 m.

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(b) Explain why the average velocity of the ball is different from its average speed.

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(c) The ball has a mass of 0.45 kg and is in contact with the wall for 0.10 s. For the period of time the ball is in contact with the wall,

(i) calculate the average acceleration of the ball.

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(ii) calculate the average force acting on the ball.

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(iii) state the direction of the average force acting on the ball.

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(Total 8 marks)

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Q24. The diagram represents part of an experiment that is being used to estimate the speed of an air gun pellet.

The pellet which is moving parallel to the track, strikes the block, embedding itself. The trolley and the block then move along the track, rising a vertical height, h.

(a) Using energy considerations explain how the speed of the trolley and block immediately after it has been struck by the pellet, may be determined from measurements of h. Assume frictional forces are negligible.

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(b) The following data is collected from the experiment

mass of trolley and block 0.50 kgmass of pellet 0.0020 kgspeed of trolley and block immediately after impact 0.40 m s–1

Calculate

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(i) the momentum of the trolley and block immediately after impact,

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(ii) the speed of the pellet just before impact.

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(c) (i) State what is meant by an inelastic collision.

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(ii) Use the data from part (b) to show that the collision between the pellet and block is inelastic.

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(Total 11 marks)

Q25. The diagram shows the velocity-time graph for a vertically bouncing ball, which is released above the ground at A and strikes the floor at B. The effects of air resistance have been neglected.

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(a) (i) What does the gradient of a velocity-time graph represent?

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(ii) Explain why the gradient of the line CD is the same as line AB.

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(iii) What does the area between the line AB and the time axis represent?

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(iv) State why the velocity at C is negative.

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(v) State why the speed at C is less than the speed at B.

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(b) The ball has a mass of 0.15 kg and is dropped from an initial height of 1.2 m. After impact the ball rebounds to a height of 0.75 m.

Calculate

(i) the speed of the ball immediately before impact,

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(ii) the speed of the ball immediately after impact,

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(iii) the change in momentum of the ball as a result of the impact,

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(iv) the magnitude of the resultant average force acting on the ball during impact if it is in contact with the floor for 0.10 s.

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(Total 13 marks)

Q26. In a football match, a player kicks a stationary football of mass 0.44 kg and gives it a speed of 32 m s–1.

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(a) (i) Calculate the change of momentum of the football.

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(ii) The contact time between the football and the footballer’s boot was 9.2 m s. Calculate the average force of impact on the football.

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(b) A video recording showed that the toe of the boot was moving on a circular arc of radius 0.62 m centred on the knee joint when the football was struck. The force of the impact slowed the boot down from a speed of 24 m s–1 to a speed of 15 m s–1.

Figure 1

(i) Calculate the deceleration of the boot along the line of the impact force when it struck the football.

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(ii) Calculate the centripetal acceleration of the boot just before impact.

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(iii) Discuss briefly the radial force on the knee joint before impact and during the impact.

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(Total 7 marks)

Q27. The graph shows how the momentum of two colliding railway trucks varies with time. Truck A has a mass of 2.0 × 104 kg and truck B has a mass of 3.0 × 104 kg. The trucks are travelling in the same direction.

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(a) Calculate the change in momentum of

(i) truck A,

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(ii) truck B.

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(b) Complete the following table.

Initialvelocity/m s–1

Finalvelocity/m s–1

Initial kineticenergy/J

Final kineticenergy/J

truck A

truck B

(4)

(c) State and explain whether the collision of the two trucks is an example of an elastic collision.

......................................................................................................................

Page 38


......................................................................................................................

......................................................................................................................

......................................................................................................................(3)

(Total 11 marks)

Q28. The following figure shows an athlete of mass 65 kg about to perform a pole vault.

The centre of mass of the athlete rises 4.2 m during the vault.

(a) Calculate the change in potential energy of the athlete between take off and reaching the highest point.

change in potential energy .................. J(2)

(b) Assuming that the centre of mass falls the same distance when falling. Calculate the vertical speed, in m s−1, of the athlete when he lands.

vertical speed .............................. m s−1

(2)

(c) Explain how the landing mat reduces the force experienced by the athlete to an acceptable level when landing.

......................................................................................................................

Page 39


......................................................................................................................

......................................................................................................................

......................................................................................................................(2)

(Total 6 marks)

Q29. The graph shows the variation in the horizontal force acting on a tennis ball with time whilst the ball is being served.

(a) (i) Use the graph to show that the magnitude of the impulse that acts on the tennis ball is about 1.3 N s.

(2)

(ii) The mass of the tennis ball is 0.057 kg. Show that the impulse in part (a)(i) gives the ball a speed of about 20 m s–1 horizontally as the ball leaves the racquet.Assume that the ball had no horizontal speed before the impulse was applied.

Page 40


(2)

(b) During flight the ball accelerates due to gravity. When it reaches the ground the vertical component of the velocity is 6.1 m s–1.Calculate the speed and the angle between the direction of travel of the ball and the horizontal as it reaches the ground.Assume that air resistance is negligible.

speed ................................................. m s–1

angle ............................................... degree(3)

(Total 7 marks)

Q30. Which row, A to D, in the table correctly shows the quantities conserved in an inelastic collision?

mass momentum kinetic energy total energy

A conserved not conserved conserved conserved

B not conserved conserved conserved not conserved

C conserved conserved conserved conserved

D conserved conserved not conserved conserved

(Total 1 mark)

Page 41


Q31. Water of density 1000 kg m–3 flows out of a garden hose of cross-sectional area 7.2 × 10–4 m2 at a rate of 2.0 × 10–4 m3 per second. How much momentum is carried by the water leaving the hose per second?

A 5.6 × 10–5 N s

B 5.6 × 10–2 N s

C 0.20 N s

D 0.72 N s(Total 1 mark)

Q32. The graph shows the variation with time, t, of the force, F, acting on a body.

What physical quantity does the area X represent?

A the displacement of the body

B the acceleration of the body

C the change in momentum of the body

D the change in kinetic energy of the body(Total 1 mark)

Page 42


Q33. Deep space probes often carry modules which may be ejected from them by an explosion. A space probe of total mass 500 kg is travelling in a straight line through free space at 160 m s–1 when it ejects a capsule of mass 150 kg explosively, releasing energy. Immediately after the explosion the probe, now of mass 350 kg, continues to travel in the original straight line but travels at 240 m s–1, as shown in the figure below.

(a) Discuss how the principles of conservation of momentum and conservation of energy apply in this instance.

The quality of your written communication will be assessed in this question.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................(6)

(b) (i) Calculate the magnitude of the velocity of the capsule immediately after the explosion and state its direction of movement.

Page 43


magnitude of velocity = ....................................... m s–1

direction of movement ............................................................(3)

(ii) Determine the total amount of energy given to the probe and capsule by the explosion.

answer = ....................................... J(4)

(Total 13 marks)

Q34. A gas molecule of mass m in a container moves with velocity v. If it makes an elastic collision at right angles to the walls of the container, what is the change in momentum of the molecule?

A zero

B mv

C mv

D 2 mv

Page 44


(Total 1 mark)

Q35. A gas molecule of mass m in a container moves with velocity v. If it makes an elastic collision at right angles to the walls of the container, what is the change in momentum of the molecule?

A zero

B mv

C mv

D 2 mv(Total 1 mark)

Page 45


M1.(a) conservation of momentum equation or statement quoted or used even with incorrect data

C1

1.4 × 107 × 1.7 × 10–27 = 401.7 × 10–27 × vC1

5.9(3) × 104 m s–1

A1(3)

(b) neutron will rebound / have velocity / momentum to the leftB1

momentum transferred to the uranium will be greaterB1

velocity of uranium will be greater

(no loss of kinetic energy argument gets the final mark only)B1

(3)

(c) KE = ½mv2 seen or usedC1

0.5 × 1.7 × 10–27 × (1.4 × 107)2

C1

1.7 × 10–13 J (1.67 – 10–13 J)A1

(3)[9]

M2.(a) (i) (kinetic) energy of air / snow (pushing it out of the way)(not internal energy)

B1

melting / internal energy of snow / iceor internal energy of air / skis (condone heat)

award 1 mark only for “heat and sound”award zero marks for sound alone even if explained

B1(2)

Page 46


(ii) clear attempt to draw tangent to graph at end of runB1

correct co-ordinates and manipulation for tangentor use of 2 points on curve ≥ t = 7s

(no unit penalty) (no significant figure penalty)B1

(2)

(iii) potential energy loss = 100 / 55 × KEB1

final KE = 0.5 × m × 232

or

equates any KE to mghC1

height = 49 m allow e.c.f. from (ii)A1

(3)

(b) (i) s = ½ at2 or s = ut + ½at2 or correct numerical equivalentC1

4.0 s or 4.04 sA1

(2)

(ii) s = vt or numerical equivalentC1

92 m – 93 m

allow e.c.f. from (a)(ii) and (b)(i)A1

(2)

(c) force is lower because F = B1

skier has vertical momentum after landingB1

Page 47


change in momentum is reducedB1

or

F = B1

skier takes longer time to reduce (vertical) momentumB1

compare with time when surface is flatB1

or

F = B1

skier moves though longer distance (vertically) to come to restB1

compare with distance to come to rest when surface is flatB1

or

F = maB1

time to come to rest (vertically) is increasedB1

acceleration is reduced

allow 1 mark for idea that “(vertical) component of normal reaction is reduced”B1

(3)[14]

M3.A[1]

M4.(a) (i) momentum before (a collision) = momentum after (a collision) or

Page 48


there is no change in momentum (when bodies collide)C1

reasonably rigorous statement including reference to total momentum etc. and lack of external force or in a closed system or isolated system

A1(2)

(ii) in an elastic collision the (total) KE remains constant or in an inelastic collision the (total) KE is reduced (not just changed)

B1(1)

(b) the incident particle stops (loses all its momentum or KE)M1

the stationary particle moves off with the same velocity (speed) as the incident particle (gains same momentum or KE as incident particle had)

M1

no other possibility would conserve both momentum and KE or original total momentum = mv = final momentum and original KE = ½mv2 = final KE

A1(3)

(c) (i) 130 × 1.8 = 4.0 × 104 v or clear appreciation that momentum = mv

C1

e.g. an incorrect momentum equation 5.9 (5.85) × 10–3 m s–1 (in opposite direction to that of the astronaut)

A1(2)

(ii) energy change / timeC1

energy change of astronaut = 211 JC1

or power = Fv

C1

Page 49


force = 390 N from F = ma (need to use v = 1.8 / 2 to obtain average power 350 W)

C1

power = total energy / time = 211 / 0.6 = 350 – 352 WA1

(3)

(iii) any two sensible comments

e.g. momentum of astronaut and spacecraft reduced to zero rope stores elastic energy momentum of astronaut becomes equal and opposite to original

astronaut and spacecraft move toward one another the astronaut and spacecraft are ‘reunited’ taut rope produces a force toward the spacecraft velocity of astronaut and of spacecraft will be reversed rope acting like a spring

not astronaut pulled along at a constant speedB2

(2)[13]

M5.C[1]

M6.D[1]

M7.B[1]

M8. (a) total momentum before a collision = total momentum aftera collision or total momentum of a system is constant

Page 50


or Σmv = 0 , where mv is the momentum

B1

no external forces acting on the system/ isolated system

B12

(b) (i) work done = F s

C1

63 000 J

A12

(ii) KE = ½ mv2 or F =ma and v2 = u2 + 2as

C1

combined speed v = 4.6 (4.58) m s–1

A12

(iii) reasonable attempt at a momentum conservationequation (2 terms before and one term after any signs)

C1

(+ or -) 3600 v + (2400 × 12.5) = (6000 × 4.58) (e.c.f)

C1

16 m s–1 (cao ignoring sign)

A13

(iv) driver A is likely to experience the greater force

B1

force = rate of change of momentum(Δmv/t) or F = ma

B1

time for deceleration on impact is (approximately)

Page 51


the same

B1

change in velocity of driver B = 11.4 m s–1 (ecf from(ii) and (iii))andchange in velocity of driver A = 17.1 m s–1

(ecf from (ii) and (iii))orΔmv or Δv of A > Δmv or Δv of B

B14

[13]

M9.(a) (i) 15 rev / s = 30π rad / s or v = 51 / 52 m s–1 [could appear in subst]Bl

F = mw2r [or mv2 / r & v = ωr]Bl

appropriate sub leading to 7.33 kN [2+sf evaluation mandatory]Bl

(ii) to centre of rotor OWTTEBl

(iii) stress = F / ABl

correct substitution from aiBl

(iv) 0.55 × 2.09 × 107 / 6 ×1010 [or ε = 3.3 × 10–4 ]Cl

= 0.192 mmAl

Page 52


(v) ½ × 7.32 × 103 × 1.92 × 10–4 [ecf]Cl

= 0.702 JAl

(b) (i) volume pushed down [per second] = Av [mass = ρ × volume]Bl

Change of momentum [per second] = mass pushed down per second × v

Bl

(ii) Upward force = 900 N OWTTE [penalise use of 900g] OR area swept out by blades = π × 0.552

Cl

900 = (0.55)2 π1.3v2

Cl

= 27 m s–1

Al[15]

M10. (a) F = or Ft = mv – mu etc.

M1

substitute units

A12

(b) conservation of momentum mentioned

B1

ejected gas has momentum or velocity in one direction

B1

rocket must have equal momentum in the opposite direction

B1

Page 53

Colonel Frank Seely School3

or force = rate of change of momentum

(B1)

ejected gas has momentum or velocity in one direction

(B1)

rocket must have equal and opposite force

(B1)

(c) equation seen (F = m/t × v but not F= ma)

B1

substitution into any sensible equation leading to3.6 × 107 (N)

B12

[7]

M11. (a) total momentum of system constant/total momentum before =

B1

total momentum afterisolated system/no external force

B12

(b) (i) clear explanation of method

B1

correct numerical working leading to 4.25 m s–1

Page 54


B12

(ii) F = 0.31 × a speed

C1

use of speed difference [4.25 – 0.68]

C1

= 1.11 N [ecf]

A13

(iii) states that two momenta/forces related to hose andwall are equal in size/appreciates reaction force

B1

transmitted by hose to Earthand in opposite direction

B12

[9]

M12.(a) (i)

(ii) F = Eq = 2.5 × 105 × 3.2 × 10–19 (1) = 8.0 × 10–14 N (1)

(iii)

(= 3.1 × 105 m s–1)(max 5)

Page 55


(b) (i) field into paper (1)

(ii) force ⊥r motion in B-field (1) ∴ directed towards centre of a circular path (1)

(iii)

for B, q constant r ∝ mυ, momentum (1)

r = = 0.33 m (1)(6)

(c) greater mass, so smaller speed at Q (1)

greater momentum justified e.g. E = (E const) (1)

∴ greater radius (1)(max 2)

[13]

M13.(a) moving driver has momentum (1) in sudden impact momentum must be lost in v. short time (1) F = Δ(mv) / Δt (or F = ma) (1) air bag increases stopping contact time (1) hence reduces force (or reduces force by decreasing the deceleration) (1)

(max 5)

(b) seat belt applies force to smaller area than air bag (1) causing greater pressure on parts of body (1)

(2)

Page 56


(c) air initially at v. high pressure occupies small volume (1) on expansion volume increases and pressure decreases (1) temperature of gas decreases as it expands (1) pressure caused by momentum change in molecular collisions (1)

(max 3)[10]

M14.(a) T = mg = 890 × 10 = 8900 (1) N (1)(accept alternative correct value using g = 9.81 N kg–1)

(2)

(b)

resolve vertically T cos 30° =mg (1)

T = = 10280 (1.03 × 104) N (1)(2)

(c) (i) vertical height fallen = l(1 – cos θ) = 15(1 – 0.866) = 2.0(1) m (1) (allow e.c.f. if h calculated wrongly)

(ii) mυ2 = mgh or reference energy (1) υ = = 6.34 m s–1 (1)

(max 1 / 3 if equations of motion used)

(iii) (1)

(allow e.c.f of υ and m as before)(5)

[9]

Page 57


M15.(a) (i) points plotted correctly (1) (1) (deduct one for each incorrect)sensible scales chosen (1)line of best fit (1)

(ii) change in momentum [or impulse] (1) (accept 0.8)max 4

(b) area under graph = 0.80±0.05 (1) (kgms–1)

υ = (1) = 1.6 ms–1

alternative:state average force = 0.10(N) (1)leading to correct derivation of 1.6ms–1 (1)

2

(c) (i) Δmυ = 0 [or statement] (1) υ = 0.40ms-1 (1)

(ii) kinetic energy = 0.16 J (1)

(iii) initial kinetic energy = 0.64 (J) (1)kinetic energy lost so inelastic (1)

5[11]

M16.A[1]

M17.(a) (i) equation showing momentum before = momentum after (1)correct use of sign (1)

(ii) no external forces (on any system of colliding bodies) (1)3

Page 58


(b) (i) (by conservation of momentum m1υ1 + m2υ2 = 0)

0.25 × 2.2 = (–)0.50υ2 (1)υ2 = (–)1.1(0)ms–1 (1)

(ii) = total k.e. = × 0.25 × 2.22 + × 0.5 × 1.12 (1)

= 0.91J (1)4

(c) (i) mass of air per second = ρAυ (1)correct justification, incl ref to time (1)

(ii) momentum per second (= Mυ = υ2 Aρ) = υ2 Aρ (1)

(iii) force = rate of change of momentum (hence given result) (1)upward force on helicopter equals (from Newton third law)downward force on air (1)

5

(d) υ2 Aρ = (for 50% support) (1)

υ2 × 180×1.3 = (1)

gives υ = 7.2ms–1 (1) (or 7.3, g taken as 10)if not 50% of weight, max 1 / 3 provided all correct otherwise (gives 10.2)

3[15]

M18.(a) momentum before collision = momentum after collision (1)provided no external force acts (1)

(2)

(b) (i) p = mυ (1)

10 × 10–3 × 200 = 2.(0) (1) kg m s–1 (N s) (1)

Page 59


(ii) total mass after collision = 0.40 kg (1)0.40υ = 2.0 gives υ = 5.(0) m s–1 (1) (allow e.c.f. from (i))

(4)

(c) (i) kinetic energy = ½mv2

(1) (= 200 J)

(ii) kinetic energy = (1) (= 5.0 J)

(iii) ΔQ = 200 – 5 = 195 (J) = mcΔθ (1)

Δθ = = 78 K (1) (allow e.c.f. for incorrect ΔQ)(5)

(d) kinetic energy lost (= potential energy gained) = mgh (1)

1.3 m (1)(2)

[13]

M19.(a) kinetic energy is not conserved (1)(1)

(b) (i) (p = mv gives) p = 0.12 × 18 = 2.2 N s (1) (2.16 N s)

Page 60


(ii) p = 0.12 × (–15) = –1.8 N s (1)

(iii) Δp = 2.2 – (–1.8) = 4.0 N s (3.96 N s) (1)(allow e.c.f. from(i) and(ii))

(iv) (1)

= 28 N (1) (28.3 N)(allow e.c.f. from (iii))

(v) (Ek = ½mv2 gives) Ek = 0.5 × 0.12 × (182 – 152) = 5.9 J (1)(6)

[7]

M20.(a)

either (1) (for both atomic mass numbers, 4 and 208) and (1) (for both atomic numbers, 2 and 81) [or (1) for Tl and incorrect α]

2

(b) (i) Ek = (½mv2) = 6.1 × 106 × 1.6 × 10-19(J) (1) substitution for m = 4.0 × 1.66 × 10-27(kg) (1)

(1) (= 1.7 × 107m s-1)

(ii) correct use of conservation of momentum mTl vrecoil = mα v (1) substitution of mTl = 208u (1) (allow C.E. for mass = 208)

vrecoil = = 3.3 × 105 m s-1 (1)

Page 61


(allow C.E. for value of v)6

[8]

M21.(a) (i) Ep = mgΔh (1) = 5.8 × 10-2 × 9.8(1) × 1.5 = 0.85 J ✓

(ii) 0.85 J (1) (allow C.E. for value of Ep from (i))

(iii) (use of Ek = ½mv2 gives) 0.85 = 0.5 × 5.8 × 10-2 × v2 (1) (allow C.E. for answer from (ii)) (v2 = 29.3) v = 5.4 m s-1 (1)

(iv) (use of p = mv gives) p = 5.8 × 10-2 × 5.4 (1) (allow C.E. for value of v from (iii)) = 0.31 N s (1)

7

(b)

(allow C.E. for value of p from (iv)) = 31 N (1)

[or a = = 540 (m s-2) (1)

F = 5.8 × 10-2 × 540 = 31 N (1)]2

(c) egg effectively stopped in a longer distance (1) hence greater time and therefore less force on egg (1) [or takes longer to stop

hence force is smaller as

[or acceleration reduced as it takes longer to stop thus force will be smaller]

[or some energy is absorbed by container less absorbed by egg]

2[11]

Page 62


M22.(a) (i) length of card [or distance travelled by trolley A] (1) time at which first light gate is obscured [or time taken to travel the distance] (1)

(ii) time at which second light gate is obscured [or distance travelled after collision and time taken] (1)

3

(b) momentum = mass × velocity (1) mass of each trolley (1) (check whether) pinitial = pfinal (1)

max 2

(c) incline the ramps (1) until component of weight balances friction (1) [or identify where the friction occurs (1) sensible method of reducing (1)]

2[7]

M23. (a) displacement is a vector (1)ball travels in opposite directions (1)

max 1

(b) velocity is rate of change of displacement average speed is rate of change of distance velocity is a vector [or speed is a scalar)velocity changes direction

any two (1) (1)2

(c) (i) a = (1)

= (–)140.m s–1 (1)

Page 63


(allow C.E. for incorrect values of Δv)

(ii) F = 0.45 × (–) 140 = (–) 63N (1)(allow C.E for value of a)

(iii) away from wall (1)at right angles to wall (1)[or back to girl (1) (1)][or opposite to direction of velocity at impact (1) (1)]

5[8]

M24. (a) kinetic energy changes to potential energy (1)potential energy calculated by measuring h (1)equate kinetic energy to potential energy to find speed (1) [or use h to find s (1) use g sinθ for a (1) use v2 = u2 + 2as (1)] [or use h to find s (1) time to travel s and calculate vav (1) v = 2vav (1)]

3

(b) (i) p(= mv) = 0.5(0) × 0.4(0) = 0.2(0) (1) N s(or kg m s–1) (1)

(ii) (use of mpvp = mtvt gives) 0.002(0) v = 0.2(0) (1)v = 100 m s–1 (1)

4

(c) (i) kinetic energy is not conserved (1)

(ii) initial kinetic energy = × 0.002 × 1002 = 10 (J) (1)

final kinetic energy = × 0.5 × 0.42 = 0.040 (J) (1)

hence change in kinetic energy (1)

(allow C.E. for value of v from (b))

Page 64

Colonel Frank Seely School4

[11]

M25. (a) (i) acceleration (1)

(ii) both represent acceleration of free fall [or same acceleration] (1)

(iii) height/distance ball is dropped from above the ground [or displacement] (1)

(iv) moving in the opposite direction (1)

(v) kinetic energy is lost in the collision [or inelastic collision] (1)

5

(b) (i) v2 = 2 × 9.81 × 1.2 (1)v = 4.9 m s–1 (1) (4.85 m s–1)

(ii) u2 = 2 × 9.81 ×0.75 (1)u = 3.8 m s–1 (1) (3.84 m s–1)

(iii) change in momentum = 0.15 × 3.84 – 0.15 × 4.85 (1) = –1.3 kg m s–1 (1) (1.25 kg m s–1)

(allow C.E. from (b) (i) and (b)(ii))

(iv) F = (1)

= 13 N (1)

(allow C.E. from (b)(iii))8

[13]

Page 65


M26. (a) (i) change of momentum (= 0.44 × 32) = 14(.1) kg m s1 (1)

(ii) (use of F = gives) F = (1)

= 1.5(3) × 103N (1)

(allow C.E. for value of Δ(mv) from (i)3

(b) (i) deceleration = = 9.8 × 102m s–2 (1)

(9.78 × 102m s–2)

(ii) (use of a = gives)

centripetal acceleration = = 9.3 × 102m s–2 (1)(9.29 × 102 m s–2)

(iii) before impact: radial pull on knee joint due to centripetalacceleration of boot (1)during impact: radial pull reduced (1)

4[7]

M27. (a) (i) (change in momentum of A) = – (1) 25 × 103 (1)kg m s–1 (or N s) (1)

(ii) (change in momentum of B) = 25 × 103 kg m s–1 (1)4

(b)

Page 66


initial vel/m s–1 final vel/m s–1 initial k.e./J final k.e./J

truck A 2.5 1.25 62500 15600

truck B 0.67 1.5 6730 33750

(1) (1) (1) (1)

4

(c) not elastic (1)because kinetic energy not conserved (1)kinetic energy is greater before the collision (or less after) (1)[or justified by correct calculation]

3[11]

M28. (a) use of E = mgh (1)

2680 J (1)2

(b) use of v2 = 2 as (1)

9.1 m s-1 (1)2

(c) increases the time taken for the athlete to come to rest/reduceddeceleration

force = mass × acceleration/mass × change in velocity/time (1)

or momentum argument

or energy argument (1)2

[6]

Page 67


M29. (a) (i) clear working:

attempts to evaluate area under line(12 – 13 squares × 0.1 N s)or ½ × 106(105) × 24 × 10–3 seen

B1

1.260 – 1.272 (N s) 1.2 – 1.3 for square counting

B1

(ii) area under graph/impulse = Δmv

B1

22.3 (22.8) (m s–1)

B14

(b) use of Pythagoras or tan–1 (6.1/22.3)

B1

or or tan–1 (6.1/20)

22.8 – 23.8(m s–1) 20.9 or 21(m s–1)

B1

14.9 – 15.5 (°) (15 – 16 (°)) 17/18(°)

B13

[7]

Page 68


M30. D[1]

M31. B[1]

M32. C[1]

M33. (a) The candidate’s writing should be legible and the spelling,punctuation and grammar should be sufficiently accurate for themeaning to be clear.

The candidate’s answer will be assessed holistically. The answer will beassigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marks

The information conveyed by the answer is clearly organised, logical andcoherent, using appropriate specialist vocabulary correctly. The form andstyle of writing is appropriate to answer the question.

The candidate states that momentum is conserved, supported by reasoningto explain why the conditions required for momentum conservation aresatisfied in this case.

Page 69


The candidate also gives a statement that total energy is conserved, givingdetailed consideration of the energy conversions which take place,described in the correct sequence, when there is an explosion on a bodythat is already moving.

Intermediate Level (Modest to adequate): 3 or 4 marks

The information conveyed by the answer may be less well organised andnot fully coherent. There is less use of specialist vocabulary, or specialistvocabulary may be used incorrectly. The form and style of writing is lessappropriate.

The candidate states that momentum is conserved, but the reasoning ismuch more limited.

and/or

There is a statement that (total) energy is conserved, with basicunderstanding that some energy is released by the explosion.

Low Level (Poor to limited): 1 or 2 marks

The information conveyed by the answer is poorly organised and may notbe relevant or coherent. There is little correct use of specialist vocabulary.The form and style of writing may be only partly appropriate.

The candidate indicates that either momentum or energy is conserved, orthat both are conserved. There are very limited attempts to explain eitherof them.

The explanation expected in a competent answer should include acoherent selection of the following points concerning the physicalprinciples involved and their consequences in this case.

Momentum

• momentum is conserved because there are no external forcesacting on the overall system (probe plus capsule) – or because it’sfree space

• they are moving in free space and are therefore so far from largemasses that gravitational forces are negligible

• during the explosion, there are equal and opposite forces actingbetween the probe and the capsule

• these are internal forces that act within the overall system

Page 70


• because momentum has to be conserved, and it is a vector, thecapsule must move along the original line of movement after theexplosion

Energy

• total energy is always conserved in any physical process becauseenergy can be neither created nor destroyed

• however, energy may be converted from one form to another

• the probe is already moving and has kinetic energy

• in the explosion, some chemical energy is converted into kineticenergy (and some energy is lost in heating the surroundings)

• the system of probe and capsule has more kinetic energy than theprobe had originally, because some kinetic energy is released bythe explosion

max 6

(b) (i) conservation of momentum gives (500 × 160)= 150 v + (350 × 240) (1)from which v = (−)26(.7) (m s−1) (1)

direction: opposite horizontal direction to larger fragment[or to the left, or backwards] (1)

3

(ii) initial Ek = ½ × 500 × 1602 (1) (= 6.40 × 106 J)

final Ek = (½ × 350 × 2402) + (½ × 150 × 26.72) (1) (= 1.01 × 107 J)

energy released by explosion = final Ek − initial Ek (1)

= 3.7 × 106 (J) (1)4

[13]

M34. D

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[1]

M35. D[1]

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E1.(a) This calculation was well done by most candidates. Nearly everyone recognised it as a situation in which the principle of conservation of momentum was important. Some of the weaker candidates made statements to this effect but then proceeded in attempts to use the conservation of energy. Another mistake made by significant numbers of candidates was to use the mass of the uranium nucleus in determining the momentum after the impact, rather than the increased mass of the nucleus, incorporating an additional neutron. Significant figure errors were common in this part.

(b) Few candidates attempted to analyse this situation in terms of conservation of momentum. Very few identified the recoil of the neutron or the reversal of the direction of its momentum. Partial credit was gained by many candidates who used arguments based on kinetic energy conservation in an elastic collision. These candidates failed to appreciate that their arguments were not sufficient to explain the behaviour of the particles.

(c) This part was well done by nearly all of the candidates. There were a few who used incorrect data for the mass or the speed of the neutron and some others who forgot to square the velocity when doing the calculation. Significant figure penalties were not uncommon. Some candidates expressed their answer in a notation similar to that shown on their calculators (i.e. 1.713). This is clearly incorrect and candidates should be advised that the use of this notation will be penalised.

E2.(a) (i) Most answers referred simply to 'heat and sound'. More detail was expected for full credit. Candidates could have explained where or why heating occurs.

(ii) Many candidates drew an appropriate tangent and used it correctly. Some simply used the data from the final linear portion of the graph. There were many instances of careless reading of graph scales.

There were some who applied equations of motion incorporating the 80 m fallen after leaving the ramp into their calculations.

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(iii) The majority of the candidates attempted to equate P.E. to K.E. but the 55% factor was frequently applied incorrectly, yielding a wide range of answers. Some used the equation for uniformly accelerated motion which was not appropriate in this case.

(b) (i) This was often well done. The most common error was to include 23 m s–1 as an initial vertical speed.

(ii) Allowing errors carried forward many completed this successfully but incorrect answers to (i) produced ski jump distances of up to 500 in!

(c) Surprisingly few took the hint in the stem that momentum change needed to be considered. The expected response was that the skier retains some vertical momentum on impact, reducing the change in momentum and hence, from Ft = ∆(mv), reducing the force. Although such an approach was considered by the more able candidates the thinking of the majority was more muddled. Many took the view that somehow the vertical component of the force was split into a vertical and horizontal component, which consequently reduced the vertical component. Others took the view that vertical momentum was transformed into horizontal momentum.

E4.(a) (i) Although there was a good proportion of clear responses many gave vague statements which gained only one of the two available marks. Many simply stated ‘momentum before a collision is equal to the momentum after’ which was given credit but begs the question whether the candidate was referring to the momentum of the system or one of the colliding bodies. For full credit this needed to be clearer and the condition that no external forces act on the system stated.

(ii) Although many gave correct responses there were those who stated that energy would be conserved. This of course is true in any collision so that the inclusion of the word ‘kinetic’ was mandatory.

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(b) There were relatively few completely correct answers to this part. Surprisingly few appreciated that in such a collision the incident particle would stop and the stationary particle would move off at the speed of the colliding body before the collision. This outcome is one of the most basic situations which candidates should have been able to recall from collision experiments with trolleys or air track vehicles. Many thought that the particles would rebound in opposite directions with equal speeds, a situation that would not conserve momentum. Those who appreciated the outcome of the collision had only to state that only this result would conserve both momentus and kinetic energy for full credit.

(c) (i) There were many correct responses to this part. A common error was to treat the situation as a collision rather than an ‘explosion’ so that working included addition of masses.

(ii) The most straightforward approach to this question was to calculate the final energy and divide this by the time. Most candidates, however, used P = Fv. The force was often determined correctly (assuming this to be constant) but most candidates then used the final velocity instead of the average velocity in the power equation, arriving at 700 W.

(iii) There were not many totally realistic outcomes discussed in this part. A common answer was that the astronaut would be pulled along once the rope became taut (a sort of hang-gliding in space). There were many aspects that could have been included in a reasoned discussion of the outcome. The most common one mark answer was that the astronaut would return to the spacecraft. Further discussion was necessary for full credit.

E8. (a) Most candidates were able to gain one mark for the definition. The most common errors were either not giving the condition of no external force or failing to refer to the total momentum in the definition. It was therefore unclear whether the definition referred to the momentum of a body or the system of bodies. There was a minority who stated the principle as ‘the momentum of a body remains constant unless a force acts on it’.

(b) (i) This was usually correct.

(ii) A significant proportion of the candidates tried using the principle of conservation of momentum in this part. Those who appreciated that they had to use KE = ½ mv2 usually completed this successfully but some used an

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incorrect mass or failed to take the square root.

(iii) There were many correct answers and many structured their responses sensibly. Most were able to make a realistic attempt equating two momenta before the collision to one momentum after. Many however failed to take account of the fact that the final momentum would be to the left whilst the original momentum of A is to the right and so the signs for these terms needed to be different, regardless of the convention they used.

(iv) There were some excellent logical arguments presented in response to this question and many gained full marks. Even weak candidates usually suggested that driver A would feel the greater force but explaining why proved more difficult. Some realised that use of F = ma or F = Δ(mv)/t was useful but many went on to discuss the vehicles rather that the drivers. The most common omission was failure to state that in the comparison the time for each driver to come to rest would be (approximately) the same.

E9.(a) (i) Candidates here (and elsewhere in the paper) must recognise that questions beginning with the words ‘Show that...’ demand a clear and well-explained exposition of the calculation or proof. Examiners are not satisfied with muddled and incomplete answers or answers that simply re-state the question. Explicitly, in this part, candidates were expected to quote their answer to more than the one significant figure in the question. They were also expected to show how they converted 15 revolutions per second into a value that could be used in their equation. Failures to convince often lead to loss of marks.

(ii) Although examiners were generous here, the common answer ‘to the centre’ is not adequate for a candidate at A2 level in a physics examination; the centre of what? Strong candidates often illustrated the answer on a labelled diagram of the helicopter itself and drew the examinerߢs attention to this amendment. Candidates should take every opportunity to make their answers clear.

(iii) Again, a ‘show that’ which often did not. Those who quote the equation p = F / A for ‘stress’ without defining their symbols must not be surprised if the examiner assumes that p stands for ‘pressure’. The safest route is to supply a word equation in cases like this.

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(iv) There were many good answers here, but solutions were frequently marred by arithmetic errors and by negligent candidates who calculated the strain and then omitted to go on to evaluate the required change in length.

(v) Again, this was well done by many, but there was a significant number of candidates who attempted to go via the routestrain energy per unit volume = ½ stress × strain.This is indeed a possible method, but these candidates too often failed to multiply in the volume of the helicopter blade.

(b) (i) Only a few candidates were able to prove this simple relationship. Those who did stated clearly the volume of air being forced downwards every second, could use this to evaluate the mass of air being forced down, and could go on to show the change in momentum of this air. Otherwise candidates simply floundered around trying unsuccessfully to ‘spot’ a relationship that would yield the result they needed.

(ii) To the examiners’ surprise very many found the evaluation of the air speed beyond them. Although many recognised that the change in momentum per unit time has to equal the weight of the helicopter, candidates still turned the weight of the aircraft back to mass, and failed to spot the squared velocity term. This was a disappointing failure by many candidates.

E10. (a) Candidates’ attempts to show the relationship between impulse and momentum and their units were variable. There was considerable confusion regarding the unit of impulse (Ns–1 being fairly commonplace). Too frequently candidates simply started with F = ma without attempting to show how this related to the impulse/change of momentum equation. It was relative common to see candidates stating that impulse is equal to the rate of change of momentum.

(b) Arguments for how the ejection of waste gas propels a rocket were variable. Most candidates had an idea that conservation of momentum was involved, but tended to be rather loose in their application of this concept. Many simply said that forces are equal and opposite, or that if there was momentum in one direction there must be

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momentum in the other direction.

(c) It was not apparent from most candidates’ answers whether or not they understood that in this example it is mass changing with time rather than velocity. Although credit was

allowed;for equation F = , F = ma; F = m or F = were not allowed.

A significant minority of candidates failed to convert km s–1 into m s–1 yet still quoted their final answer in MN.

E11. (a) For full credit, candidates needed to state that the total momentum is constant (in some appropriate way) and also to mention the absence of external forces. It was common to see answers that featured one but not the other.

(b) (i) Although many candidates carried out computations which ended in anumerical solution that was plausible, the general level of explanation (whether algebraic or descriptive) was very poor. Many solutions consisted of a jumble of numbers from which examiners could make little sense. This was a ‘show that’ question and candidates needed to be much more careful about the level of their description in order to obtain full credit.

(ii) There were many poor attempts in this part also. The ostensibly correct solutions often needed considerable work by examiners to determine the line of argument. Those who write down equations at random (the dubious F = ma was a case in point) need to be exceptionally clear in their solutions if their work is to be credible.

(iii) The explanation of why water sprayed onto a wall does not alter the Earths rotation eluded many. Common misconceptions included the idea that nothing happened because the Earth is massive and the water flow is not. Also many candidates wrote wisely about the action and reaction forces at the wall but failed to gain credit because they did not consider the whole system.

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E15.The provision of the labels on the axes in the question appeared to be helpful to candidates. The great majority of candidates chose sensible scales and plotted the points clearly and accurately. Some candidates had difficulty in drawing a line of best fit, either because they insisted that the line had to pass through each of the points, or by not possessing a straight edge. In part (a)(ii) many candidates stated that the area under the graph represented momentum rather than the change in momentum.

In part (b) less than half of the candidates were able to use the area under the graph to show that the speed of the locomotive is 1.6 ms–1.

There were many correct answers to part (c)(i), with the conservation of momentum clearly stated or quoted in the form of an equation. The calculation of the combined kinetic energy of the locomotive and the truck after collision was usually correct. In part (c)(iii) it was encouraging to see that the majority of candidates referred to kinetic energy loss to show that the collision is inelastic.

E17.In part (a) most candidates were able to write down an equation relating the momentum before collision to the momentum after. However, many candidates failed to consider a sign convention even though the directions of velocity were clearly shown in the diagram. In part (a)(ii) many candidates did not state that no external forces act if momentum is to be conserved.

In part (b)(i) most candidates were able to calculate correctly the speed of trolley B after collision. Such candidates went on to calculate the kinetic energy after collision and quoted this as the minimum energy stored in the spring corresponding to an elastic collision.

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Good candidates answered part (c) well and gained full, or near full, marks. However, this part proved to be too testing for weaker candidates, with many such candidates not attempting it. In part (c)(i) good candidates derived an expression for the volume of air flowing per second and multiplied this by the density of air to obtain an expression for the mass of air flowing per second. In part (c)(ii) the change in momentum per second is given by the mass per second multiplied by the velocity of the air. There were very few good answers to part (c)(iii). By equating the change in momentum with impulse an expression is obtained for the force exerted on the air. The force exerted on the helicopter is equal in magnitude but opposite in direction to that exerted on the air, a point missed by almost all candidates.

In part (d) it was common for candidates to forget to convert the mass of the helicopter to weight and to ignore the fact that only 50% of the weight of the helicopter is supported by the downward flow of air. The remaining 50% of the weight of the helicopter is supported due to the aerofoil section of the rotor blades.

E18.This question was very well answered with many candidates scoring maximum or nearly maximum marks. Most candidates knew in part (a) that the momentum before a collision equalled the momentum after the collision, but rather few gave the condition that no external force must act on the system. Part (b) was almost always correct, although some candidates did not add the masses of the bullet and block.

Parts (c)(i) and (c)(ii) were usually correct. In section (iii) some candidates failed to subtract the remaining kinetic energy of 5.0 J from the initial kinetic energy of 200 J, or alternatively used 5.0 J for the internal energy. Part (d) was most often correct, but a number of candidates used ʋ2 = u2 + 2as and scored no marks.

E19.A surprising number of candidates failed to answer part (a) correctly and did not seem to be aware that kinetic energy was central to any discussion on inelastic collisions.

In part (b) the calculations were generally well done although it was rare for a candidate to take into account the change of direction in part (ii). Also the unit of momentum caused problems for a significant proportion of candidates. In part (v) the calculation of the loss of kinetic energy produced more difficulties than expected. The most common error was for candidates to first subtract the velocities and then use the result to calculate the loss of energy.

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E20.Most candidates were able to complete the equation in part (a) correctly, although a few lost marks by adding an extra particle to the right hand side of the equation.

In part (b)(i) the majority of candidates were aware of the correct conversions from MeV to J and from u to kg although some candidates used the proton / neutron rest mass of 1.67 x 10-27 kg instead of 1.66 x 10-27 kg to convert u to kg. The expression for kinetic energy was generally correctly used. In part (b)(ii) most candidates gave a complete and correct calculation. The conservation of momentum was generally applied correctly although a significant number of candidates incorrectly used 210 u as the mass of the recoil nucleus. Other candidates attempted to calculate the mass of the nucleus from the combined mass of its protons, neutrons and electrons. Some candidates unnecessarily converted the correct masses to kilograms.

E21.The calculations in parts (a) and (b) were well done although the unit for momentum produced the usual problems. Explaining the crumple zone in part (c) was often well answered although some candidates’ answers did tend to lack focus. The idea that the time duration of the collision was increased and that this was important, seemed to be well understood.

E22.This question was generally well answered throughout although it was clear that a significant number of candidates were unsure how light gates are used to determine velocity. Ideas for minimising friction were many and varied but very few candidates seemed aware of the concept of a friction compensated slope.

E23. This question was answered well and the only common problem was the calculation of the acceleration of the ball. Only the best candidates appreciated that in part (c) the change in velocity was (–)14 m s –1. This was not a major handicap however as allowance for consequential errors enabled most of the remaining marks to be scored.

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E24. This question was also done well although a significant minority of candidates could not explain in part (a) how the speed of the trolley and block might be determined from energy considerations. Part (b) was done well apart from the usual confusion over the unit for momentum.

The final part of the question produced more variable responses. Many candidates were able to explain correctly what is meant by an inelastic collision but were unable to carry out the necessary calculation to show that the collision of the pellet and the block was inelastic.

E25. Part (a) discriminated very well and although most candidates scored some marks, only the best were awarded the maximum. The two sections least well done were part (iii) which asked what the area under the line AB represented and part (v) which asked why the speed at C was less than the speed at B. In answering part (v) a significant proportion of candidates did not realise that the ball rebounded from the floor at C.

Part (b) also produced a variation of marks. Most candidates completed part (i) but found the other parts of the question more difficult. There was considerable confusion over signs and initial and final velocities in part (ii). This confusion was carried over into part (iii) and most candidates made no allowance for the change in direction of momentum but simply subtracted the magnitudes of the initial and final momenta. As in the past, the unit for momentum caused problems and penalising a unit error at this point was quite common.

E26. In part (a) most candidates calculated the momentum correctly, although N s–1 was commonly given as the unit of momentum. Many correct answers were seen in part (ii), although a significant number of candidates misread the impact time as 9.2 s. The final answer was often presented with too many significant figures.

Many correct calculations were seen in part (b) (i) although some candidates attempted to use v2 = u2 + 2as with s = 0.62 m. In part (ii), most candidates calculated the centripetal acceleration correctly. In both parts, incorrect units or provision of answers with too many significant figures were not uncommon. In part (iii), few candidates realized that the radial force pulled on the knee joint although a significant number of candidates knew that the force after impact was less because the speed was less. Many candidates failed to confine their answer to the limits set by the question and discussed features not relevant to the question.

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E27. This was an unusual question and a considerable amount of work was required in parts (a) and (b). There were many opportunities to make errors. Part (a) was answered quite well but common errors were omitting the 103 factor, quoting both changes as positive and the usual unit problem that appears in questions involving momentum.

Part (b) caused real problems for a significant proportion of candidates and calculation errors were common. A significant proportion of candidates confused momentum with velocity and although they were then were able to score marks for a correct calculation of kinetic energy, arithmetic errors were common.

Part (c) provided evidence that there is a common misunderstanding of what is meant by an elastic collision. A relatively frequent response was that this was an example of an elastic collision because momentum was conserved.

E29. Only a minority of the candidates made progress with part (a) (i). The working in many responses did not convey a correct physics approach to the problem. Multiplying the peak force by half the time did not show that the aim was to determine the area, but rather that the aim was to find a number that fitted that given in the question. It would help demonstration of a correct approach to a problem, and in particular to ‘show that’ questions, if candidates were to include a subject for the formula and/or numerical substitution.

Candidates were generally more successful in part (a) (ii), almost half the candidates gaining both marks.

In part (b), candidates either coped very well or not at all with the straightforward task of finding the resultant magnitude and direction of the vector addition of horizontal and vertical velocities, both of which were given in the question. Many were unsuccessful because they could not successfully use Pythagoras’ rule or identify the appropriate relationship to find the angle.

E30. This was a straightforward test of candidates’ knowledge. It required candidates to decide whether or not mass, momentum, kinetic energy and total energy would be conserved in an inelastic collision. 85% of the candidates appreciated that everything except kinetic energy would be conserved. Incorrect responses were fairly evenly spread around the other three distractors.

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E31. This question required candidates to determine the momentum of the water flowing out of a garden hose in one second. This called for mathematical application as well as knowledge and it was therefore much more demanding. 41% of the candidates selected the correct answer, and the question was not a strong discriminator. The most popular incorrect distractor, chosen by 28%, was C (0.20). This numerical value could be found by multiplying the density of water by the flow rate, ignoring the cross-sectional area value given in the question.

E32. This question, on factual knowledge of the impulse – momentum relationship, was an easy starter with a facility of 85%.

E33. It was evident from their attempts at part (a) that during their courses many candidates had considered the application of conservation of momentum to events involving an explosion. It was less clear that they had ever considered an explosion that takes place in a moving object, or considered how conservation of energy applies in an explosion. Consequently, part (a) of the question proved to be difficult, not least because it was unfamiliar territory for so many. Part (b), which was formulaic and involved much less original thinking, brought much more success for the majority.

In part (a) only a very small proportion of the candidates were able to produce answers that were well organised, coherent, detailed and contained correct physics to merit a ‘high level’ mark of five or six. More answers fell into the ‘intermediate level’ (three or four marks) and even more into the ‘low level’ (one or two marks). A major failing in most answers was to overlook the question’s requirement to address the two conservation laws ‘in this instance’. For a high level answer, it was necessary to consider an explosion on a moving space vehicle travelling in a straight line in deep space. All of the italicised section is significant. The system has momentum before exploding (unlike a straightforward recoil example); this momentum has to be conserved because there are no external forces in deep space. Hence the probe speeds up and the capsule must be ejected along the original line of movement (although it may not be possible to tell that this is ‘backwards’ until the calculation has been done). Forces between probe and capsule during the explosion are equal and opposite, but they are internal forces for the system. When considering momentum, it was common for candidates to conclude that ‘momentum must be conserved because momentum is always conserved’.

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In the explosion, chemical energy is converted into kinetic energy; this increases the total kinetic energy of the system, which is shared between probe and capsule. Examiners saw many very weak answers that showed total confusion – such as momentum being converted into energy, mass being converted into energy, or energy not being conserved. A serious omission in many answers was that of the word ‘kinetic’ before ‘energy’, whilst many answers referred to the event as an ‘inelastic collision’. There was seldom any reference to conservation of the total energy of the system taken as a whole.

Most candidates recovered from their poor attempts at part (a) to gain all three marks for the calculation in part (b) (i). There were also many awards of full marks in part (b) (ii), where the main mistake was to calculate only the kinetic energy of the system (probe + capsule) after the explosion, and to regard this as the answer to the question. Apparently, the candidates who did this had not realised that the system had an initial kinetic energy.

E34. This question tested the change of momentum of a gas molecule making an elastic collision with the walls of its container. Misunderstanding the vector nature of momentum, and therefore of the change of momentum, was responsible for the 21% of candidates who chose distractor A.

Their reasoning is likely to have been that mv – mv = 0, rather than the correct mv – (–mv) = 2mv, which 69% of the candidates selected.

E35. This question tested the change of momentum of a gas molecule making an elastic collision with the walls of its container. Misunderstanding the vector nature of momentum, and therefore of the change of momentum, was responsible for the 21% of candidates who chose distractor A.

Their reasoning is likely to have been that mv – mv = 0, rather than the correct mv – (–mv) = 2mv, which 69% of the candidates selected.

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