Web viewThe diagram represents part of an experiment that is being used to estimate the ... in accordance with Newton's second or third law ... Newton’s Third Law

A2 Force and Momentum

323 minutes

246 marks

Q1.          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)

Q2.          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

(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)

Q3.          A golf club undergoes an inelastic collision with a golf ball and gives it an initial velocity of 60 m s–1. The ball is in contact with the club for 15 ms and the mass of the ball is4.5 × 10–2 kg.

(a)     Explain what is meant by an inelastic collision.

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(b)     Calculate

(i)      the change in momentum of the ball,

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(ii)     the average force the club exerts on the ball.

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(c)     (i)      State the value of the force exerted by the ball on the club and give its direction.

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(ii)     Explain how your answer to part (i) follows from an appropriate law of motion.

You may be awarded marks for the quality of written communication in your answer.

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(Total 9 marks)

Q4.          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.

(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)

Q5.          The graph shows how the momentum of two colliding railway trucks varies with time. TruckA 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.

(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.

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(Total 11 marks)

Q6.          (a)     State two quantities that are conserved in an elastic collision.

quantity 1: ....................................................................................................

quantity 2: ....................................................................................................(2)

(b)     A gas molecule makes an elastic collision with the walls of a gas cylinder. The molecule is travelling at 450 m s–1 at right angles towards the wall before the collision.

(i)      What is the magnitude and direction of its velocity after the collision?

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(ii)     Calculate the change in momentum of the molecule during the collision if it has a mass of 8.0 × 10–26 kg.

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(c)     Use Newton’s laws of motion to explain how the molecules of a gas exert a force on the wall of a container.

You may be awarded additional marks to those shown in brackets for the quality of written communication in your answer.

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(Total 10 marks)

Q7.          Communications satellites are usually placed in a geo-synchronous orbit.

(a)     State two features of a geo-synchronous orbit.

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(b)     The mass of the Earth 6.00 × 1024 kg and its mean radius is 6.40 × 106 m.

(i)      Show that the radius of a geo-synchronous orbit must be 4.23 × 107 m,

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(ii)     Calculate the increase in potential energy of a satellite of 750 kg when it is raised from the Earth’s surface into a geo-synchronous orbit.

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(c)     Satellites in orbits nearer the Earth than geo-synchronous satellites may be used in the future to track road vehicles.

(i)      State and explain one reason why geo-synchronous satellites would not be suitable for such a purpose.

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(ii)     Give two points you would make in arguing for or against tracking road vehicles.Explain your answers.

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(Total 12 marks)

Q8.          (a)     A spring, which hangs from a fixed support, extends by 40 mm when a mass of 0.25 kg is suspended from it.

(i)      Calculate the spring constant of the spring.

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(ii)     An additional mass of 0.44 kg is then placed on the spring and the system is set into vertical oscillation. Show that the oscillation frequency is 1.5 Hz.

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(b)     With both masses still in place, the spring is now suspended from a horizontal support rod that can be made to oscillate vertically, as shown in the diagram below, with amplitude 30 mm at several different frequencies.

The response of the masses suspended from the spring to the vertical oscillations of the support rod varies with frequency. Describe and explain, as fully as you can, the motion of the masses when the support rod oscillates at a frequency of (i) 0.2 Hz, (ii) 1.5 Hz and (iii) 10 Hz.

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

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(Total 10 marks)

Q9.          A golf club undergoes an inelastic collision with a stationary golf ball and gives it an initial velocity of 60 m s–1. The ball is in contact with the club for 15 ms and the mass of the ball is4.5 × 10–2 kg.

(a)     Explain what is meant by an inelastic collision.

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(b)     Calculate

(i)      the change in momentum of the ball,

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(ii)     the average force the club exerts on the ball.

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(Total 5 marks)

Q10.          Near the surface of a planet the gravitational field strength is uniform and for two points, 10 m apart vertically, the gravitational potential difference is 3 J kg–1. How much work must be done in raising a mass of 4 kg vertically through 5 m?

A       3 J

B       6 J

C       12 J

D       15 J(Total 1 mark)

Q11.          Which one of the following graphs correctly shows the relationship between the gravitational force, F, between two masses and their separation, r?

(Total 1 mark)

Q12.          The Earth has density ρ and radius R. The gravitational field strength at the surface is g.What is the gravitational field strength at the surface of a planet of density 2ρ and radius 2R?

A       g

B       2 g

C       4 g

D       16 g(Total 1 mark)

Q13.          Which one of the following statements is not true for a body vibrating in simple harmonic motion when damping is present?

A       The damping force is always in the opposite direction to the velocity.

B       The damping force is always in the opposite direction to the displacement.

C       The presence of damping gradually reduces the maximum potential energy of the system.

D       The presence of damping gradually reduces the maximum kinetic energy of the system.(Total 1 mark)

Q14.          The time period of a simple pendulum is doubled when the length of the pendulum is increased by 3.0 m. What is the original length of the pendulum?

A       1.0 m

B       1.5 m

C       3.0 m

D       6.0 m(Total 1 mark)

Q15.          A body moves with simple harmonic motion of amplitude 0.50 m and period 4π seconds.What is the speed of the body when the displacement of the body from the equilibrium position is 0.30 m?

A       0.10 m s–1

B       0.15 m s–1

C       0.20 m s–1

D       0.40 m s–1

(Total 1 mark)

Q16.          A particle of mass m moves horizontally at constant speed v along the arc of a circle from P1 to P2 under the action of a force. What is the work done on the particle by the force during this displacement?

A       zero

B       

C       

D       2 mv2

(Total 1 mark)

Q17.         

A model car moves in a circular path of radius 0.8 m at an angular speed of  rad s–1. What is its displacement from point P, 6 s after passing P?

A       zero

B       1.6 m

C       0.4 πm

D       1.6 πm(Total 1 mark)

Q18.          What is the value of the angular velocity of a point on the surface of the Earth?

A       1.2 × 10–5 rad s–1

B       7.3 × 10–5 rad s–1

C       2.6 × 10–1 rad s–1

D       4.6 × 102 rad s–1

(Total 1 mark)

Q19.          The rate of change of momentum of a body falling freely under gravity is equal to its

A       weight.

B       power.

C       kinetic energy.

D       potential energy.(Total 1 mark)

Q20.         

A force, F, varies with time, t, as shown by the graph and is applied to a body initially at rest on a smooth surface. What is the momentum of the body after 5.0 s?

A       zero.

B       12.5 N s.

C       25 N s.

D       50 N s.(Total 1 mark)

Q21.          A particle of mass m strikes a rigid wall perpendicularly from the left with velocity v.

If the collision is perfectly elastic, the change in momentum of the particle which occurs as a result of the collision is

A       2mv to the right.

B       2mv to the left.

C       mv to the left.

D       zero.(Total 1 mark)

Q22.          For the two physical quantities, impulse and force, which one of the following is correct?

A       Impulse is a scalar and force is a scalar.

B       Impulse is a scalar and force is a vector.

C       Impulse is a vector and force is a scalar.

D       impulse is a vector and force is a vector.(Total 1 mark)

Q23.          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)

Q24.          Water of density 1000 kg m–3 flows out of a garden hose of cross-sectional area 7.2 × 10–

4m2 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)

Q25.          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)

Q26.          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.

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

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)

Q27.          A projectile moves in a gravitational field. Which one of the following is a correct statement about the gravitational force acting on the projectile?

A       The force is in the direction of the field.

B       The force is in the opposite direction to that of the field.

C       The force is at right angles to the field.

D       The force is at an angle between 0° and 90° to the field.(Total 1 mark)

Q28.          The time period of oscillation of a simple pendulum of length l is the same as the time period of oscillation of a mass M attached to a vertical spring. The length and mass are then changed.Which row, A to D, in the table would give a simple pendulum with a time period twice that of the spring oscillations?

  new pendulum length new mass on spring

A 2l 2M

B 2l

C 2M

D

(Total 1 mark)

Q29.          Which graph, A to D, shows the variation of the kinetic energy, Ek, with displacement x for a particle performing simple harmonic motion?

(Total 1 mark)

Q30.          A body moves with simple harmonic motion of amplitude 0.90 m and period 8.9 s. What is the speed of the body when its displacement is 0.70 m?

A       0.11 m s–1

B       0.22 m s–1

C       0.40 m s–1

D       0.80 m s–1

(Total 1 mark)

Q31.          A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each revolution. What is the kinetic energy of the particle?

A       

B       

C       

D       (Total 1 mark)

Q32.          A mass on the end of a string is whirled round in a horizontal circle at increasing speed until the string breaks. The subsequent path taken by the mass is

A       a straight line along a radius of the circle.

B       a horizontal circle.

C       a parabola in a horizontal plane.

D       a parabola in a vertical plane.(Total 1 mark)

Q33.         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)

Q34.         The graph shows how the force on a glider of mass 2000 kg changes with time as it is launched from a level track using a catapult.

Assuming the glider starts at rest what is its velocity after 40 s?

A       2.5 m s–1

B       10 m s–1

C       50 m s–1

D       100 m s–1

(Total 1 mark)

Q35.         The graph shows how the force on a glider of mass 2000 kg changes with time as it is launched from a level track using a catapult.

Assuming the glider starts at rest what is its velocity after 40 s?

A       2.5 m s–1

B       10 m s–1

C       50 m s–1

D       100 m s–1

(Total 1 mark)

M1.          (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)

(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]

M2.          (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)

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(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))4

[11]

M3.          (a)     kinetic energy not conserved (1)[or velocity of approach is equal to velocity of separation]

1

(b)     (i)      (use of p = mv gives) p = 4.5 × 10–2 × 60 (1)

= 2.7 kg m s–1 (1)

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

= 180 N (1)

[or a =  = = 4000 (ms–1)

F = (ma) = 4.5 × 10–2 × 4000 = 180 N]4

(c)     (i)      180 N (1)(allow C.E. for value of F from (b) (ii)in opposite direction (to motion of the club) (1)

(ii)     body A (or club) exerts a force on body B (or ball) (1)(hence) body B (or ball) exerts an equal force on bodyA (or club) (1)correct statement of Newton’s third law (1)

max 4QWC 1

[9]

M4.          (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]

M5.          (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)     

  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]

M6.          (a)     momentum (1)

kinetic energy (1)2

(b)     (i)      450ms–1 (1)

in the opposite direction (1)

(ii)     Δp = 8.0 × 10–26 × 900 (1)

= 7.2 × 10–23Ns (1)4

(c)     force is exerted on molecule by wall (1)to change its momentum (1)molecule must exert an equal but opposite force on wall (1)in accordance with Newton's second or third law (1)

4[10]

M7.          (a)     period is 24 hours (or equal to period of Earth’s rotation) (1)

remains in fixed position relative to surface of Earth (1)

equatorial orbit (1)

same angular speed as Earth (or equatorial surface) (1)max 2

(b)     (i)        = mω2 r (1)

T =   (1)

 (1)

          (gives r = 42.3 × 103 km)

(ii)     ΔV =   (1)

= 6.67 × 10–11 × 6 × 1024 × 

= 5.31 × 107 (J kg–1) (1)

ΔEP = mΔV (= 750 × 5.31 × 107) = 3.98 ×1010 J (1)

(allow ecf for value of ΔV)6

(c)     (i)      signal would be too weak at large distance (1)

(or large aerial needed to detect/transmit signal, or any otheracceptable reason)

the signal spreads out more the further it travels (1)

(ii)     for              road pricing would reduce congestion

                            stolen vehicles can be tracked and recovered

                            uninsured/unlicensed vehicles can be apprehended

         against      road pricing would increase cost of motoring

                            possibility of state surveillance/invasion of privacy

         (1)(1) any 2 valid points (must be for both for or against)4

[12]

M8.          (a)     (i)      mg = ke (1)

k =   = 61(.3) N m–1 (1)

(ii)     T =   (1)       (= 0.667 s)

           (1)                        (= 1.5(0) Hz)4

(b)     The marking scheme for this part of the question includes an overallassessment for the Quality of Written Communication (QWC). Thereare no discrete marks for the assessment of QWC but the candidates’QWC in this answer will be one of the criteria used to assign a leveland award the marks for this part of the question.

Level Descriptoran answer will be expected to meet most of the criteria in the

level descriptor

Mark range

Good 3 –   answer supported by appropriate range of relevant points

–   good use of information or ideas about physics, going beyond those given in the question

–   argument well structured with minimal repetition or irrelevant points

–   accurate and clear expression of ideas with only minor errors of spelling, punctuation and grammar

5-6

Modest 2 –   answer partially supported by relevant points

–   good use of information or ideas about physics given in the question but limited beyond this

–   the argument shows some attempt at structure

–   the ideas are expressed with reasonable clarity but with a few errors of spelling, punctuation and grammar

3-4

Limited 1 –   valid points but not clearly linked to an argument structure

–   limited use of information or ideas about physics

–   unstructured

–   errors in spelling, punctuation and grammar or lack of fluency

1-2

0 –   incorrect, inappropriate or no response 0

examples of the sort of information or idea that might be used tosupport an argument

•        forced vibrations (at 0.2 Hz) (1)

•        amplitude fairly large (≈ 30 mm) (1)

•        in phase with driver (1)

•        resonance (at 1.5 Hz) (1)

•        amplitude very large (> 30 mm) (1)

•        oscillations may appear violent (1)

•        phase difference at 90º (1)

•        forced vibrations (at 10 Hz) (1)

•        small amplitude (1)

•        out of phase with driver or phase lag of π on driver (1)[10]

M9.          (a)     kinetic energy is not conserved (1)

(or velocity of approach equals velocity of separation)1

(b)     (i)      (use of p = mv gives) p = 4.5 × 10–2 × 60 (1)

 = 2.7kg m s–1 (1)

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

                                                      = 180 N (1)

         [or a =   = 400 (m s–1) (1)

         F = ma = 4.5 × 10–2 × 4000 = 180N (1)4

[5]

M10.          B[1]

M11.          D[1]

M12.          C[1]

M13.          B[1]

M14.          A[1]

M15.          C[1]

M16.          B[1]

M17.          B[1]

M18.          A[1]

M19.          A[1]

M20.          C[1]

M21.          B[1]

M22.          D[1]

M23.          D[1]

M24.          B[1]

M25.          C[1]

M26.          (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.

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, or

that 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

•        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]

M27.          A[1]

M28.          B[1]

M29.          A[1]

M30.          C[1]

M31.          D[1]

M32.          D[1]

M33.          D[1]

M34.          C[1]

M35.          C[1]

E1.          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.

E2.          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.

E3.          Candidates seemed quite confident in dealing with this sort of problem and there were many excellent answers. However, a significant proportion of candidates still have problems with the units for momentum and this was a commonly applied penalty. Newton’s Third Law was, for the

most part, well understood, and candidates seem more confident applying this law than they are applying the First and Second Laws.

E4.          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 usev2 = 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.

E5.          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.

E6.          This question proved to be quite discriminating. The calculation in part (b) caused some confusion and a significant proportion of candidates calculated the actual momentum of the gas molecule and not the change in momentum. Those who did calculate the change often had problems with the sign and consequently calculated the change in momentum to be zero. This was in part due to thinking that as this was an elastic collision, momentum was conserved and hence there should be no change. An incorrect unit for momentum was a common mistake.

The more able candidates had few problems with part (c) and were obviously very familiar with the derivation of the gas laws from kinetic theory. They made use of this to produce very sophisticated answers. Less able candidates were less confident in their answers and found it difficult to apply Newton’s laws appropriately. There was the usual confusion with Newton’s third

law, with a significant proportion of candidates thinking it referred to the forces acting on the molecule and not the action of the molecule on the wall producing a reaction from the wall.

E23.          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.

E24.          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.

E25.          This question, on factual knowledge of the impulse – momentum relationship, was an easy starter with a facility of 85%.

E26.          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’.

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.

E27.          The candidates in 2010 found this question to be slightly easier than their predecessors, with the facility advancing from 55% to 59%. One in four candidates demonstrated their confusion with magnetic fields by opting for distractor C, where the force was perpendicular to the field.

E28.          Knowledge of both the simple pendulum and the mass spring system was required in this question. Application of the time period formulae should have led students to conclude that, since under the new conditions Tpendulum = 2Tspring, the new pendulum length should be four times greater than the new mass on the spring. 62% of the candidates made this correct choice, but (like the preceding question) this question did not discriminate well between the candidates.

E29.          In this question, 64% of the responses were for the correct shape of Ek against x curve. By selecting distractor B, 25% of the candidates showed that they understood the trend of the Ek cbehaviour but not the exact form of it.

E30.          This question does not appear to be easy – albeit it that a correct answer only requires care when working out the result from three values substituted into v = ± 2πf (A2 – x2). Yet, as on the previous occasion when this question was used, it turned out to have the highest facility on the

paper – it was 88% this time.

E31.          Candidates found the quantitative content of this question on circular motion more to their liking, because 63% of them chose the correct answer. Both of these questions gave statistics which were very similar to those obtained when last used.

E32.          This question was the first of a pair of questions on circular motion, both of which had appeared in previous examinations. The main failing exhibited in the responses was the fact that the ball, once it had broken away from the string, would fall under gravity. Only answer D offered the possibility of some vertical motion and it was chosen by 40% of the candidates. Distractors A and C each attracted 28% of the answers.

E33.          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.

E34.          The surprising outcome of this question was that just as many candidates chose the wrong answer as the right one. This seems to have been caused by careless working. Forgetting that the mass of the glider was 2000 kg could have led candidates to the conclusion that its velocity would be 100 m s–1 (distractor D) instead of 50 m s–1, or perhaps they forgot the factor of ½ when finding the area of the triangle under the graph line. The facility of this question was 40%.

E35.          The surprising outcome of this question was that just as many candidates chose the wrong answer as the right one. This seems to have been caused by careless working. Forgetting that the mass of the glider was 2000 kg could have led candidates to the conclusion that its velocity would be 100 m s–1 (distractor D) instead of 50 m s–1, or perhaps they forgot the factor of ½ when finding the area of the triangle under the graph line. The facility of this question was 40%.