Momentum and Newton’s laws of motion — Paper 2 · A Level Physics 9702

2 A girl G is riding a bicycle at a constant velocity of 3.5 m s–1. At time t = 0, she passes a boy B sitting on a bicycle that is stationary, as illustrated in Fig. 2.1. B G 3.5 m s–1 t = 0 Fig. 2.1 At time t = 0, the boy sets off to catch up with the girl. He accelerates uniformly from time t = 0 until he reaches a speed of 5.6 m s–1 in a time of 5.0 s. He then continues at a constant speed of 5.6 m s–1. At time t = T, the boy catches up with the girl. T is measured in seconds. (a) State, in terms of T, the distance moved by the girl before the boy catches up with her. distance = … m [1] (b) For the boy, determine (i) the distance moved during his acceleration, distance = … m [2] (ii) the distance moved during the time that he is moving at constant speed. Give your answer in terms of T. distance = … m [1] Examiner’s Use (c) Use your answers in (a) and (b) to determine the time T taken for the boy to catch up with the girl. T = … s [2] (d) The boy and the bicycle have a combined mass of 67 kg. (i) Calculate the force required to cause the acceleration of the boy. force = … N [3] (ii) At a speed of 4.5 m s–1, the total resistive force acting on the boy and bicycle is 23 N. Determine the output power of the boy’s legs at this speed. power = … W [2]

11 marks

Mark scheme: 2 (a) 3.5 T B1 [1] (b) (i) distance = average speed × time (however expressed) C1 = 14 m A1 [2] (ii) distance = 5.6 × (T – 5) (or 3.5T – 14) A1 [1] (c) 3.5T = 14 + 5.6(T – 5) C1 T = 6.7 s A1 [2] (d) (i) acceleration = (5.6 / 5 =) 1.12 m s–2 C1 force = ma C1 = 75 N A1 [3] (ii) power = (force × speed =) {75 + 23} × 4.5 C1 = 440 W A1 [2] (allow 1/2 for 234 W, 0/2 for 338 W or 104 W)

This question in 9702/21 Oct/Nov 2007

3 A shopping trolley and its contents have a total mass of 42 kg. The trolley is being pushed For along a horizontal surface at a speed of 1.2 m s–1. When the trolley is released, it travels a Examiner’s distance of 1.9 m before coming to rest. Use (a) Assuming that the total force opposing the motion of the trolley is constant, (i) calculate the deceleration of the trolley, deceleration = … m s–2 [2] (ii) show that the total force opposing the motion of the trolley is 16 N. [1] (b) Using the answer in (a)(ii), calculate the power required to overcome the total force opposing the motion of the trolley at a speed of 1.2 m s–1. power = … W [2] (c) The trolley now moves down a straight slope that is inclined at an angle of 2.8° to the For horizontal, as shown in Fig. 3.1. Examiner’s Use 2.8° Fig. 3.1 The constant force that opposes the motion of the trolley is 16 N. Calculate, for the trolley moving down the slope, (i) the component down the slope of the trolley’s weight, component of weight = … N [2] (ii) the time for the trolley to travel from rest a distance of 3.5 m along the length of the slope. time = … s [4] (d) Use your answer to (c)(ii) to explain why, for safety reasons, the slope is not made any steeper. … … [1]

12 marks

Mark scheme: 3 (a) (i) v2 = 2as 1.22 = 2 × a × 1.9 M1 a = 0.38 m s–2 A1 [2] (ii) F = ma = 42 × 0.38 M1 = 16 N A0 [1] (b) power = Fv C1 = 16 × 1.2 = 19 W A1 [2] (c) (i) component = 42 × 9.8 × sin2.8 C1 = 20.1 N A1 [2] (ii) accelerating force = 20.1 – 16 = 4.1 N C1 acceleration of trolley = 4.1 / 42 = 0.098 m s–2 C1 s = ½at2 3.5 = ½ × 0.098 × t2 C1 t = 8.5 s A1 [4] GCE A/AS LEVEL – May/June 2008 9702 02 (d) either allows plenty of time to stop runaway trolley or speed of trolley increases gradually or trolley will travel faster B1 [1] (answer must be unambiguous when read in conjunction with question)

This question in 9702/21 May/June 2008

2 A ball B of mass 1.2 kg travelling at constant velocity collides head-on with a stationary ball S For of mass 3.6 kg, as shown in Fig. 2.1. Examiner’s Use v ball B ball S mass 1.2 kg mass 3.6 kg Fig. 2.1 Frictional forces are negligible. The variation with time t of the velocity v of ball B before, during and after colliding with ball S is shown in Fig. 2.2. +4 +3 v / m s–1 +2 +1 0 0 0.1 0.2 0.3 0.4 0.5 t / s –1 –2 Fig. 2.2 (a) State the significance of positive and negative values for v in Fig. 2.2. … … [1] (b) Use Fig. 2.2 to determine, for ball B during the collision with ball S, For Examiner’s (i) the change in momentum of ball B, Use change in momentum = … N s [3] (ii) the magnitude of the force acting on ball B. force = … N [3] (c) Calculate the speed of ball S after the collision. speed = … m s–1 [2] (d) Using your answer in (c) and information from Fig. 2.2, deduce quantitatively whether For the collision is elastic or inelastic. Examiner’s Use … … [2]

11 marks

Mark scheme: 2 (a) ball moving in opposite direction (after collision) … B1 [1] (b) (i) change in momentum = 1.2 (4.0 + 0.8) … C2 (correct values, 1 mark; correct sign {values added}, 1 mark ) = 5.76 N s …(allow 5.8) … A1 [3] (ii) force = ∆p / ∆t or m∆v / ∆t … C1 = 5.76 / 0.08 or 1.2 × 4.8 / 0.08 … C1 = 72 N … A1 [3] (c) 5.76 = 3.6 × V … C1 V = 1.6 m s-1 … A1 [2] (d) either speed of approach = 4.0 m s-1 and speed of separation = 2.4 m s-1 … M1 not equal and so inelastic … A1 or kinetic energy before = 9.6 J and kinetic energy after collision = 4.99 J … M1 kinetic energy after is less / not conserved so inelastic … A1 [2]

This question in 9702/21 May/June 2009

2 A sky-diver jumps from a high-altitude balloon. For Examiner’s (a) Explain briefly why the acceleration of the sky-diver Use (i) decreases with time, … … … [2] (ii) is 9.8 m s–2 at the start of the jump. … … [1] (b) The variation with time t of the vertical speed v of the sky-diver is shown in Fig. 2.1. 40 v / m s–1 30 20 10 0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 t / s Fig. 2.1 Use Fig. 2.1 to determine the magnitude of the acceleration of the sky-diver at time For t = 6.0 s. Examiner’s Use acceleration = … m s–2 [3] (c) The sky-diver and his equipment have a total mass of 90 kg. (i) Calculate, for the sky-diver and his equipment, 1. the total weight, weight = … N [1] 2. the accelerating force at time t = 6.0 s. force = … N [1] (ii) Use your answers in (i) to determine the total resistive force acting on the sky-diver at time t = 6.0 s. force = … N [1]

9 marks

Mark scheme: 2 (a) (i) (air) resistance increases with speed … M1 resultant / accelerating force decreases … A1 [2] (ii) either (air) resistance is zero or weight / gravitational force is only force … B1 [1] (b) use of gradient of a tangent … M1 acceleration = 1.9 ± 0.2 m s-2 … A2 [3] (for values > ± 0.2 but ≤ 0.4, allow 1 mark) (answer 3.3 m s-2 scores no marks) (c) (i) 1 weight = 90 × 9.8 = 880 N … A1 [1] (use of g = 10 m s-2 then deduct mark but once only in the Paper) 2 accelerating force = 90 × 1.9 = 170 N …(allow ecf) … A1 [1] (ii) resistive force = 880 – 170 = 710 N … A1 [1] (allow ecf but only if resistive force remains positive) [Total: 9]

This question in 9702/21 Oct/Nov 2009

3 (a) (i) Define force. For Examiner’s … Use … [1] (ii) State Newton’s third law of motion. … … … … [3] (b) Two spheres approach one another along a line joining their centres, as illustrated in Fig. 3.1. sphere sphere A B Fig. 3.1 When they collide, the average force acting on sphere A is FA and the average force acting on sphere B is FB. The forces act for time tA on sphere A and time tB on sphere B. (i) State the relationship between 1. FA and FB, … [1] 2. tA and tB. … [1] (ii) Use your answers in (i) to show that the change in momentum of sphere A is equal in magnitude and opposite in direction to the change in momentum of sphere B. … … [1] (c) For the spheres in (b), the variation with time of the momentum of sphere A before, For during and after the collision with sphere B is shown in Fig. 3.2. Examiner’s Use 151515 momentum to right / N s 101010 spherespheresphere AAA 55 00 timetimetime spherespheresphere BBB –-5-55 –10-10-10 –15-15-15 Fig. 3.2 The momentum of sphere B before the collision is also shown on Fig. 3.2. Complete Fig. 3.2 to show the variation with time of the momentum of sphere B during and after the collision with sphere A. [3]

10 marks

Mark scheme: 3 (a) (i) force is rate of change of momentum ………………………………………… B1 [1] (ii) force on body A is equal in magnitude to force on body B (from A) …………M1 forces are in opposite directions ……………………………………………… A1 forces are of the same kind ………………………………………………………A1 [3] (b) (i) 1 FA = – FB ……………………………………………………………………. B1 [1] 2 t A = t B ……………………………………………………………………… B1 [1] (ii) ∆p = FA t A = – FB t B ………………………………………………………….. B1 [1] (c) graph: momentum change occurs at same times for both spheres …………. B1 final momentum of sphere B is to the right …………………………………….. M1 and of magnitude 5 N s …………………………………………………………… A1 [3]

This question in 9702/22 May/June 2010

3 (a) State the relation between force and momentum. For Examiner’s … [1] Use (b) A rigid bar of mass 450 g is held horizontally by two supports A and B, as shown in Fig. 3.1. ball 45 cm A C B 50 cm 25 cm Fig. 3.1 The support A is 45 cm from the centre of gravity C of the bar and support B is 25 cm from C. A ball of mass 140 g falls vertically onto the bar such that it hits the bar at a distance of 50 cm from C, as shown in Fig. 3.1. The variation with time t of the velocity v of the ball before, during and after hitting the bar is shown in Fig. 3.2. 6 4 velocity downwards / m s–1 2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 time / s –2 –4 –6 Fig. 3.2 For the time that the ball is in contact with the bar, use Fig. 3.2 For Examiner’s (i) to determine the change in momentum of the ball, Use change = … kg m s–1 [2] (ii) to show that the force exerted by the ball on the bar is 33 N. [1] (c) For the time that the ball is in contact with the bar, use data from Fig. 3.1 and (b)(ii) to calculate the force exerted on the bar by (i) the support A, force = … N [3] (ii) the support B. force = … N [2]

9 marks

Mark scheme: 3 (a) force = rate of change of momentum (allow symbols if defined) B1 [1] (b) (i) ∆ρ = 140 × 10–3 × (5.5 + 4.0) C1 = 1.33 kg m s–1 A1 [2] (ii) force = 1.33 / 0.04 M1 = 33.3 N A0 [1] (c) (i) taking moments about B C1 (33 × 75) + (0.45 × g × 25) = FA × 20 C1 FA = 129 N A1 [3] (ii) FB = 33 + 129 + 0.45g C1 = 166 N A1 [2] GCE AS/A LEVEL – October/November 2010 9702 21

This question in 9702/21 Oct/Nov 2010

2 (a) Explain what is meant by work done. For Examiner’s … Use … [1] (b) A car is travelling along a road that has a uniform downhill gradient, as shown in Fig. 2.1. 25 m s–1 7.5° Fig. 2.1 The car has a total mass of 850 kg. The angle of the road to the horizontal is 7.5°. Calculate the component of the weight of the car down the slope. component of weight = … N [2] (c) The car in (b) is travelling at a constant speed of 25 m s–1. The driver then applies the brakes to stop the car. The constant force resisting the motion of the car is 4600 N. (i) Show that the deceleration of the car with the brakes applied is 4.1 m s–2. [2] (ii) Calculate the distance the car travels from when the brakes are applied until the car comes to rest. distance = … m [2] (iii) Calculate For Examiner’s 1. the loss of kinetic energy of the car, Use loss of kinetic energy = … J [2] 2. the work done by the resisting force of 4600 N. work done = … J [1] (iv) The quantities in (iii) part 1 and in (iii) part 2 are not equal. Explain why these two quantities are not equal. … … [1]

11 marks

Mark scheme: 2 (a) work done is the force × the distance moved / displacement in the direction of the force or work is done when a force moves in the direction of the force B1 [1] (b) component of weight = 850 × 9.81 × sin 7.5° C1 = 1090 N A1 [2] (use of incorrect trigonometric function, 0/2) (c) (i) Σ F = 4600 – 1090 = (3510) M1 deceleration = 3510 / 850 A1 = 4.1 m s–2 A0 [2] (ii) v2 = u2 + 2as 0 = 252 + 2 × – 4.1 × s C1 s = 625 / 8.2 = 76 m A1 [2] (allow full credit for calculation of time (6.05 s) & then s) (iii) 1. kinetic energy = ½ mv2 C1 = 0.5 × 850 × 252 = 2.7 × 105 J A1 [2] 2. work done = 4600 × 75.7 = 3.5 × 105 J A1 [1] (iv) difference is the loss in potential energy (owtte) B1 [1] GCE AS/A LEVEL – May/June 2011 9702 21

This question in 9702/21 May/June 2011

2 A climber is supported by a rope on a vertical wall, as shown in Fig. 2.1. For Examiner’s Use P T 18° R wall W Fig. 2.1 The weight W of the climber is 520 N. The rope, of negligible weight, is attached to the climber and to a fixed point P where it makes an angle of 18° to the vertical. The reaction force R acts at right-angles to the wall. The climber is in equilibrium. (a) State the conditions necessary for the climber to be in equilibrium. … … … [2] (b) Complete Fig. 2.2 by drawing a labelled vector triangle to represent the forces acting on the climber. W Fig. 2.2 [2] (c) Resolve forces or use your vector triangle to calculate For Examiner’s (i) the tension T in the rope, Use T = … N [2] (ii) the reaction force R. R = … N [1] (d) The climber moves up the wall and the angle the rope makes with the vertical increases. Explain why the magnitude of the tension in the rope increases. … … … [1]

8 marks

Mark scheme: 2 (a) resultant moment = zero / sum of clockwise moments = sum of anticlockwise moments B1 resultant force = 0 B1 [2] (b) shape and orientation correct and forces labelled and arrows correct M1 angles correct / labelled A1 [2] (c) (i) T cos18° = W Scale diagram: C1 T = 520 / cos18° = 547 N ± 20 N A1 [2] (ii) R = T sin18° = 169 N ± 20 N A1 [1] (d) θ is larger hence cos θ is smaller, T = W / cos θ M1 hence T is larger A0 [1]

This question in 9702/23 May/June 2011

3 A helicopter has a cable hanging from it towards the sea below, as shown in Fig. 3.1. For Examiner’s Use helicopter cable sea Fig. 3.1 A man of mass 80 kg rescues a child of mass 50.5 kg. The two are attached to the cable and are lifted from the sea to the helicopter. The lifting process consists of an initial uniform acceleration followed by a period of constant velocity and then completed by a final uniform deceleration. (a) Calculate the combined weight of the man and child. weight = … N [1] (b) Calculate the tension in the cable during (i) the initial acceleration of 0.570 m s–2, tension = … N [2] (ii) the period of constant velocity of 2.00 m s–1. tension = … N [1] (c) During the final deceleration the tension in the cable is 1240 N. Calculate this For deceleration. Examiner’s Use deceleration = … m s–2 [2] (d) (i) Calculate the time over which the man and child are 1. moving with uniform acceleration, time = … s [1] 2. moving with uniform deceleration. time = … s [1] (ii) The time over which the man and child are moving with constant velocity is 20 s. On Fig. 3.2, sketch a graph to show the variation with time of the velocity of the man and child for the complete lifting process. 2.0 velocity / m s–1 1.0 0 0 5 10 15 20 25 30 35 time / s Fig. 3.2 [2]

10 marks

Mark scheme: 3 (a) weight = m × g = 130.5 × 9.81 = 1280 N A1 [1] (b) (i) F = ma T – 1280 = 130.5 × 0.57 C1 T = 1280 + 74.4 = 1350 N A1 [2] (ii) 1280 N A1 [1] (c) 1240 – 1280 = 130.5 × a C1 a = (–) 0.31 m s–2 A1 [2] (d) (i) 1. 3.5 s A1 [1] 2. 6.5 s A1 [1] GCE AS/A LEVEL – May/June 2011 9702 23 (ii) basic shape M1 correct points A1 [2]

This question in 9702/23 May/June 2011

2 A ball is thrown vertically down towards the ground with an initial velocity of 4.23 m s–1. The For ball falls for a time of 1.51 s before hitting the ground. Air resistance is negligible. Examiner’s Use (a) (i) Show that the downwards velocity of the ball when it hits the ground is 19.0 m s–1. [2] (ii) Calculate, to three significant figures, the distance the ball falls to the ground. distance = … m [2] (b) The ball makes contact with the ground for 12.5 ms and rebounds with an upwards velocity of 18.6 m s–1. The mass of the ball is 46.5 g. (i) Calculate the average force acting on the ball on impact with the ground. magnitude of force = … N direction of force … [4] (ii) Use conservation of energy to determine the maximum height the ball reaches after it hits the ground. height = … m [2] (c) State and explain whether the collision the ball makes with the ground is elastic or inelastic. … … … [1]

11 marks

Mark scheme: 2 (a) (i) v = u + at C1 = 4.23 + 9.81 × 1.51 M1 = 19.0(4) m s–1 (Allow 2 s.f.) A0 [2] (Use of –g max 1/2. Use of g = 10 max 1/2. Allow use of 9.8. Allow 19 m s–1) (ii) either s = ut + ½ at2 (or v2 = u2 + 2as etc.) = 4.23 × 1.51 + 0.5 × 9.81 × (1.51)2 C1 = 17.6 m (or 17.5 m) A1 [2] (Use of –g here wrong physics (0/2)) (b) (i) F = ∆P / ∆t need idea of change in momentum C1 = [0.0465 × (18.6 + 19)] / 12.5 × 10–3 C1 = 140 N A1 (Use of – sign max 2/4. Ignore –ve sign in answer) Direction: upwards B1 [4] (ii) h = ½ × (18.6)2 / 9.81 C1 = 17.6 m (2 s.f. –1) A1 [2] (Use of 19 m s–1, 0/2 wrong physics) (c) either kinetic energy of the ball is not conserved on impact or speed before impact is not equal to speed after hence inelastic B1 [1]

This question in 9702/21 May/June 2012

3 (a) State Newton’s first law. For Examiner’s … Use … [1] (b) A log of mass 450 kg is pulled up a slope by a wire attached to a motor, as shown in Fig. 3.1. motor log wire 12° Fig. 3.1 The angle that the slope makes with the horizontal is 12°. The frictional force acting on the log is 650 N. The log travels with constant velocity. (i) With reference to the motion of the log, discuss whether the log is in equilibrium. … … … … [2] (ii) Calculate the tension in the wire. tension = … N [3] (iii) State and explain whether the gain in the potential energy per unit time of the log is equal to the output power of the motor. … … … … [2]

8 marks

Mark scheme: 3 (a) A body continues at rest or constant velocity unless acted on by a resultant (external) force B1 [1] (b) (i) constant velocity/zero acceleration and therefore no resultant force M1 no resultant force (and no resultant torque) hence in equilibrium A1 [2] (ii) component of weight = 450 × 9.81 × sin 12° (= 917.8) C1 tension = 650 + 450 g sin12° = (650 + 917.8) C1 = 1600 (1570) N A1 [3] GCE AS/A LEVEL – May/June 2012 9702 22 (iii) work done against frictional force or friction between log and slope M1 output power greater than the gain in PE / s A1 [2]

This question in 9702/22 May/June 2012

1 (a) Explain the differences between the quantities distance and displacement. Use … … … [2] (b) State Newton’s first law. … … … [1] (c) Two tugs pull a tanker at constant velocity in the direction XY, as represented in Fig. 1.1. tug 1 T1 X 25.0° tanker Y 15.0° T2 tug 2 Fig. 1.1 Tug 1 pulls the tanker with a force T1 at 25.0° to XY. Tug 2 pulls the tanker with a force of T2 at 15.0° to XY. The resultant force R due to the two tugs is 25.0 × 103 N in the direction XY. (i) By reference to the forces acting on the tanker, explain how the tanker may be described as being in equilibrium. … … … … [2] (ii) 1. Complete Fig. 1.2 to draw a vector triangle for the forces R, T1 and T2. [2] For Examiner’s Use R 25.0 × 103 N Fig. 1.2 2. Use your vector triangle in Fig. 1.2 to determine the magnitude of T1 and of T2. T1 = … N T2 = … N [2]

9 marks

Mark scheme: 1 (a) displacement is a vector, distance is a scalar B1 displacement is straight line between two points / distance is sum of lengths moved / example showing difference B1 [2] (either one of the definitions for the second mark) (b) a body continues at rest or at constant velocity unless acted on by a resultant (external) force B1 [1] (c) (i) sum of T1 and T2 equals frictional force B1 these two forces are in opposite directions B1 [2] (allow for 1/2 for travelling in straight line hence no rotation / no resultant torque) (ii) 1. scale vector triangle with correct orientation / vector triangle with correct orientation both with arrows B1 scale given or mathematical analysis for tensions B1 [2] 2. T1 = 10.1 × 103 (± 0.5 × 103) N A1 T2 = 16.4 × 103 (± 0.5 × 103) N A1 [2]

This question in 9702/23 May/June 2012

2 A motor drags a log of mass 452 kg up a slope by means of a cable, as shown in Fig. 2.1. For Examiner’s Use 10.0m motor start and finish cable P position of log 14.0° S Fig. 2.1 The slope is inclined at 14.0° to the horizontal. (a) Show that the component of the weight of the log acting down the slope is 1070 N. [1] (b) The log starts from rest. A constant frictional force of 525 N acts on the log. The log accelerates up the slope at 0.130 m s–2. (i) Calculate the tension in the cable. tension = … N [3] (ii) The log is initially at rest at point S. It is pulled through a distance of 10.0 m to For point P. Examiner’s Use Calculate, for the log, 1. the time taken to move from S to P, time = … s [2] 2. the magnitude of the velocity at P. velocity = … m s–1 [1] (c) The cable breaks when the log reaches point P. On Fig. 2.2, sketch the variation with time t of the velocity v of the log. The graph should show v from the start at S until the log returns to S. [4] v 0 0 t Fig. 2.2

11 marks

Mark scheme: 2 (a) weight = 452 × 9.81 component down the slope = 452 × 9.81 × sin 14° M1 = 1072.7 = 1070 N A0 [1] (b) (i) F = ma C1 T – (1070 + 525) = 452 × 0.13 C1 T = 1650 (1653.76) N any forces missing 1/3 A1 [3] (ii) 1. s = ut + ½at2 hence 10 = 0 + ½ × 0.13t2 C1 t = [(2 × 10) / 0.13]1/2 = 12.4 or 12 s A1 [2] 2. v = (0 + 2 × 0.13 × 10)1/2 = 1.61 or 1.6 m s–1 A1 [1] (c) straight line from the origin B1 line down to zero velocity in short time compared to stage 1 B1 line less steep negative gradient B1 final velocity larger than final velocity in the first part – at least 2× B1 [4]

This question in 9702/23 May/June 2012

1 (a) (i) Define acceleration. … … [1] (ii) State Newton’s first law of motion. … … [1] (b) The variation with time t of vertical speed v of a parachutist falling from an aircraft is shown in Fig. 1.1. 60 B C 50 v / m s–1 40 30 20 D 10 E A 0 0 10 20 30 t / s Fig. 1.1 (i) Calculate the distance travelled by the parachutist in the first 3.0 s of the motion. For Examiner’s Use distance = … m [2] (ii) Explain the variation of the resultant force acting on the parachutist from t = 0 (point A) to t = 15 s (point C). … … … … … [3] (iii) Describe the changes to the frictional force on the parachutist 1. at t = 15 s (point C), … … … [1] 2. between t = 15 s (point C) and t = 22 s (point E). … … … [1] (iv) The mass of the parachutist is 95 kg. For Examiner’s Calculate, for the parachutist between t = 15 s (point C) and t = 17 s (point D), Use 1. the average acceleration, acceleration = … m s–2 [2] 2. the average frictional force. frictional force = … N [3] Please turn over for Question 2.

14 marks

Mark scheme: 1 (a) (i) acceleration = change in velocity / time (taken) or acceleration = rate of change of velocity B1 [1] (ii) a body continues at constant velocity unless acted on by a resultant force B1 [1] (b) (i) distance is represented by the area under graph C1 distance = ½ × 29.5 × 3 = 44.3 m (accept 43.5 m for 29 to 45 m for 30) A1 [2] (ii) resultant force = weight – frictional force B1 frictional force increases with speed B1 at start frictional force = 0 / at end weight = frictional force B1 [3] (iii) 1. frictional force increases B1 [1] 2. frictional force (constant) and then decreases B1 [1] (iv) 1. acceleration = (v2 – v1) / t = (20 – 50) / (17 – 15) C1 = (–) 15 m s–2 A1 [2] 2. W – F = ma C1 W = 95 × 9.81 (= 932) C1 F = (95 × 15) + 932 = 2400 (2360) (2357) N A1 [3]

This question in 9702/21 Oct/Nov 2012

1 (a) The drag force D on an object of cross-sectional area A, moving with a speed v through Use a fluid of density ρ, is given by 1 2 D = CρAv 2 where C is a constant. Show that C has no unit. [2] (b) A raindrop falls vertically from rest. Assume that air resistance is negligible. (i) On Fig. 1.1, sketch a graph to show the variation with time t of the velocity v of the raindrop for the first 1.0 s of the motion. 10.0 8.0 6.0 v / m s–1 4.0 2.0 0 0 1.0 2.0 3.0 4.0 5.0 t / s Fig. 1.1 [1] (ii) Calculate the velocity of the raindrop after falling 1000 m. velocity = … m s–1 [2] (c) In practice, air resistance on raindrops is not negligible because there is a drag force. For This drag force is given by the expression in (a). Examiner’s Use (i) State an equation relating the forces acting on the raindrop when it is falling at terminal velocity. [1] (ii) The raindrop has mass 1.4 × 10–5 kg and cross-sectional area 7.1 × 10–6 m2. The density of the air is 1.2 kg m–3 and the initial velocity of the raindrop is zero. The value of C is 0.60. 1. Show that the terminal velocity of the raindrop is about 7 m s–1. [2] 2. The raindrop reaches terminal velocity after falling approximately 10 m. On Fig. 1.1, sketch the variation with time t of velocity v for the raindrop. The sketch should include the first 5 s of the motion. [2]

10 marks

Mark scheme: 1 (a) units for D identified as kg m s–2 M1 all other units shown: units for A: m2 units for v2: m2 s–2 units for ρ: kg m–3 kg m s −2 C = with cancelling / simplification to give C no units A1 [2] kg m − 3 m 2 m 2 s − 2 (b) (i) straight line from (0,0) to (1,9.8) ± half a square B1 [1] (ii) ½ mv2 = mgh or using v2 = 2 as C1 v = (2 × 9.81 × 1000)1/2 = 140 m s–1 A1 [2] (c) (i) weight = drag (D) ( + upthrust) B1 [1] Allow mg or W for weight and D or expression for D for drag (ii) 1. mg = 1.4 ×10–5 × 9.81 C1 1.4 × 10–5 × 9.81 = 0.5 × 0.6 × 1.2 × 7.1 × 10–6 × v2 M1 v = 7.33 m s–1 A0 [2] 2. line from (0,0) correct curvature to a horizontal line at velocity of 7 m s–1 M1 line reaches 7 m s–1 between 1.5 s and 3.5 s A1 [2]

This question in 9702/22 Oct/Nov 2012

2 (a) State Newton’s second law. For Examiner’s … Use … [1] (b) A ball of mass 65 g hits a wall with a velocity of 5.2 m s–1 perpendicular to the wall. The ball rebounds perpendicularly from the wall with a speed of 3.7 m s–1. The contact time of the ball with the wall is 7.5 ms. Calculate, for the ball hitting the wall, (i) the change in momentum, change in momentum = … N s [2] (ii) the magnitude of the average force. force = … N [1] (c) (i) For the collision in (b) between the ball and the wall, state how the following apply: 1. Newton’s third law, … … … … [2] 2. the law of conservation of momentum. … … [1] (ii) State, with a reason, whether the collision is elastic or inelastic. … … [1]

8 marks

Mark scheme: 2 (a) (resultant) force = rate of change of momentum / allow proportional to or change in momentum / time (taken) B1 [1] (b) (i) ∆p = (–) 65 × 10–3 (5.2 + 3.7) C1 = (–) 0.58 N s A1 [2] (ii) F = 0.58 / 7.5 × 10–3 = 77(.3) N A1 [1] (c) (i) 1. force on the wall from the ball is equal to the force on ball from the wall M1 but in the opposite direction A1 [2] (statement of Newton’s third law can score one mark) 2. momentum change of ball is equal and opposite to momentum change of the wall / change of momentum of ball and wall is zero B1 [1] (ii) kinetic energy (of ball and wall) is reduced / not conserved so inelastic B1 [1] (Allow relative speed of approach does not equal relative speed of separation.) GCE AS/A LEVEL – October/November 2012 9702 22

This question in 9702/22 Oct/Nov 2012

6 (a) State Hooke’s law. For Examiner’s … Use … [1] (b) A spring is attached to a support and hangs vertically, as shown in Fig. 6.1. An object M of mass 0.41 kg is attached to the lower end of the spring. The spring extends until M is at rest at R. spring M R S Fig. 6.1 The spring constant of the spring is 25 N m–1. Show that the extension of the spring is about 0.16 m. [2] (c) The object M in Fig. 6.1 is pulled down a further 0.060 m to S and is then released. For M, just as it is released, (i) state the forces acting on M, … [1] (ii) calculate the acceleration of M. acceleration = … m s–2 [3] (d) Describe and explain the energy changes from the time the object M in Fig. 6.1 is For released to the time it first returns to R. Examiner’s Use … … … … [2] Please turn over for Question 7.

9 marks

Mark scheme: 6 (a) extension is proportional to force / load B1 [1] (b) F = mg C1 x = (mg / k ) = 0.41 × 9.81 / 25 = (4.02 / 25) M1 x = 0.16 m A0 [2] GCE AS/A LEVEL – October/November 2012 9702 22 (c) (i) weight and (reaction) force from spring (which is equal to tension in spring) B1 [1] (ii) F – weight or 0.06 × 25 = ma C1 F = 0.2209 × 25 = 5.52 (N) or 0.22 × 25 = 5.5 a = (5.52 – 0.41× 9.81) / 0.41 or 1.5 / 0.41 and (5.5 – 4.02) C1 a = 3.7 (3.66) m s–2 gives 3.6 m s–2 A1 [3] (d) elastic potential energy / strain energy to kinetic energy and gravitational potential energy B1 stretching / extension reduces and velocity increases / height increases B1 [2] 3 3 4 1

This question in 9702/22 Oct/Nov 2012

2 Two planks of wood AB and BC are inclined at an angle of 15° to the horizontal. The two For wooden planks are joined at point B, as shown in Fig. 2.1. Examiner’s Use M C A 0.26 m 0.26 m 15° B 15° Fig. 2.1 A small block of metal M is released from rest at point A. It slides down the slope to B and up the opposite side to C. Points A and C are 0.26 m above B. Assume frictional forces are negligible. (a) (i) Describe and explain the acceleration of M as it travels from A to B and from B to C. … … … … [3] (ii) Calculate the time taken for M to travel from A to B. time = … s [3] (iii) Calculate the speed of M at B. speed = … m s–1 [2] (b) The plank BC is adjusted so that the angle it makes with the horizontal is 30°. M is released from rest at point A and slides down the slope to B. It then slides a distance along the plank from B towards C. Use the law of conservation of energy to calculate this distance. Explain your working. distance = … m [2]

10 marks

Mark scheme: 2 (a) (i) accelerations (A to B and B to C) are same magnitude B1 accelerations (A to B and B to C) are opposite directions or both accelerations are toward B B1 (A to B and B to C) the component of the weight down the slope provides the acceleration B1 [3] (ii) acceleration = g sin15 ° C1 s = 0 + ½ at2 s = 0.26 / sin 15 ° = 1.0 C1 2 1 . 0 × 2 t = t = 0.89 s A1 [3] 9 . 8 × sin15 ° (iii) v = 0 + g sin15t or v2 = 0 + 2g sin15 × 1.0 C1 v = 2.26 m s–1 A1 [2] (using loss of GPE = gain KE can score full marks) (b) loss of GPE at A = gain in GPE at C or loss of KE at B = gain in GPE at C B1 h1 = h2 = 0.26 m or ½ mv2 = mgh h2 = 0.5 × (2.26)2 / 9.81 = 0.26 m x = 0.26 / sin 30° = 0.52 m A1 [2]

This question in 9702/23 Oct/Nov 2012

2 (a) Distinguish between mass and weight. For Examiner’s mass: … Use … weight: … … [2] (b) An object O of mass 4.9 kg is suspended by a rope A that is fixed at point P. The object is pulled to one side and held in equilibrium by a second rope B, as shown in Fig. 2.1. P rope A Ƨ O rope B Fig. 2.1 Rope A is at an angle θ to the horizontal and rope B is horizontal. The tension in rope A is 69 N and the tension in rope B is T. (i) On Fig. 2.1, draw arrows to represent the directions of all the forces acting on object O. [2] (ii) Calculate For Examiner’s 1. the angle θ, Use θ = … ° [3] 2. the tension T. T = … N [2]

9 marks

Mark scheme: 2 (a) mass is the property of a body resisting changes in motion / quantity of matter in a body / measure of inertia to changes in motion B1 weight is the force due to the gravitational field/force due to gravity or gravitational force B1 [2] Allow 1/2 for ‘mass is scalar weight is vector’ (b) (i) arrow vertically down through O B1 tension forces in correct direction on rope B1 [2] (ii) 1. weight = mg = 4.9 × 9.81 (= 48.07) C1 69 sin θ = mg C1 θ = 44.(1)° scale drawing allow ± 2° A1 [3] use of cos or tan 1/3 only 2. T = 69 cos θ C1 = 49.6 / 50 N scale drawing 50 ±2 (2/2) 50 ±4 (1/2) A1 [2] correct answers obtained using scale diagram or triangle of forces will score full marks cos in 1. then sin in 2. (2/2)

This question in 9702/21 May/June 2013

2 (a) Define force. For Examiner’s … [1] Use (b) A resultant force F acts on an object of mass 2.4 kg. The variation with time t of F is shown in Fig. 2.1. 10.0 8.0 F / N 6.0 4.0 2.0 0 0 1.0 2.0 3.0 4.0 t / s Fig. 2.1 The object starts from rest. (i) On Fig. 2.2, show quantitatively the variation with t of the acceleration a of the For object. Include appropriate values on the y-axis. Examiner’s Use a / m s–2 0 0 1.0 2.0 3.0 4.0 t / s Fig. 2.2 [4] (ii) On Fig. 2.3, show quantitatively the variation with t of the momentum p of the object. Include appropriate values on the y-axis. p / N s 0 0 1.0 2.0 3.0 4.0 t / s Fig. 2.3 [5]

10 marks

Mark scheme: 2 (a) force = rate of change of momentum A1 [1] (b) (i) horizontal line on graph from t = 0 to t about 2.0 s ± ½ square, a > 0 M1 horizontal line at 3.5 on graph from 0 to 2 s A1 vertical line at t = 2.0 s to a = 0 or sharp step without a line B1 horizontal line from t = 2 s to t = 4 s with a = 0 B1 [4] (ii) straight line and positive gradient M1 starting at (0,0) A1 finishing at (2,16.8) A1 horizontal line from 16.8 M1 from 2.0 to 4.0 A1 [5]

This question in 9702/22 May/June 2013

3 (a) (i) State the principle of conservation of momentum. For Examiner’s … Use … … [2] (ii) State the difference between an elastic and an inelastic collision. … [1] (b) An object A of mass 4.2 kg and horizontal velocity 3.6 m s–1 moves towards object B as shown in Fig. 3.1. A B 3.6 m s–1 1.2 m s–1 4.2 kg 1.5 kg before collision Fig. 3.1 Object B of mass 1.5 kg is moving with a horizontal velocity of 1.2 m s–1 towards object A. The objects collide and then both move to the right, as shown in Fig. 3.2. A B 3.0 m s–1 v 4.2 kg 1.5 kg after collision Fig. 3.2 Object A has velocity v and object B has velocity 3.0 m s–1. (i) Calculate the velocity v of object A after the collision. velocity = … m s–1 [3] (ii) Determine whether the collision is elastic or inelastic. [3]

9 marks

Mark scheme: 3 (a) (i) the total momentum of a system (of interacting bodies) remains constant M1 provided there are no resultant external forces / isolated system A1 [2] (ii) elastic: total kinetic energy is conserved, inelastic: loss of kinetic energy B1 [1] [allow elastic: relative speed of approach equals relative speed of separation] GCE AS/A LEVEL – May/June 2013 9702 23 (b) (i) initial mom: 4.2 × 3.6 – 1.2 × 1.5 (= 15.12 – 1.8 = 13.3) C1 final mom: 4.2 × v + 1.5 × 3 C1 v = (13.3 – 4.5) / 4.2 = 2.1 m s–1 A1 [3] (ii) initial kinetic energy = ½ mA(vA)2 + ½ mB(vB)2 = 27.21 + 1.08 = 28(.28) M1 final kinetic energy = 9.26 + 6.75 = 16 M1 initial KE is not the same as final KE hence inelastic A1 [3] provided final KE less than initial KE [allow in terms of relative speeds of approach and separation]

This question in 9702/23 May/June 2013

3 (a) State what is meant by work done. For Examiner’s … Use … [1] (b) A trolley of mass 400 g is moving at a constant velocity of 2.5 m s–1 to the right as shown in Fig. 3.1. 2.5 m s–1 trolley 400 g Fig. 3.1 Show that the kinetic energy of the trolley is 1.3 J. [2] (c) The trolley in (b) moves to point P as shown in Fig. 3.2. 2.5 m s–1 trolley F 400 g P Q x Fig. 3.2 At point P the speed of the trolley is 2.5 m s–1. A variable force F acts to the left on the trolley as it moves between points P and Q. The variation of F with displacement x from P is shown in Fig. 3.3. 20 F / N 10 0 0 x P Q Fig. 3.3 The trolley comes to rest at point Q. For Examiner’s (i) Calculate the distance PQ. Use distance PQ = … m [3] (ii) On Fig. 3.4, sketch the variation with x of velocity v for the trolley moving between P and Q. 2.5 v / m s–1 0 P Q x Fig. 3.4 [2]

8 marks

Mark scheme: 3 (a) (work =) force × distance moved / displacement in the direction of the force OR when a force moves in the direction of the force work is done B1 [1] (b) kinetic energy = ½ mv2 C1 kinetic energy = ½ 0.4 (2.5)2 = 1.25 / 1.3 J A1 [2] (c) (i) area under graph is work done / work done = ½ Fx C1 1.25 = (14 x) / 2 C1 x = 0.18 (0.179) m [allow x = 0.19 m using kinetic energy = 1.3 J] A1 [3] (ii) smooth curve from v = 2.5 at x = 0 to v = 0 at Q M1 curve with increasing gradient A1 [2] GCE AS/A LEVEL – October/November 2013 9702 22

This question in 9702/22 Oct/Nov 2013

3 (a) State Newton’s first law of motion. … … [1] (b) A box slides down a slope, as shown in Fig. 3.1. v box 20° horizontal Fig. 3.1 The angle of the slope to the horizontal is 20°. The box has a mass of 65 kg. The total resistive force R acting on the box is constant as it slides down the slope. (i) State the names and directions of the other two forces acting on the box. 1. … 2. … [2] (ii) The variation with time t of the velocity v of the box as it moves down the slope is shown in Fig. 3.2. 8.0 6.0 v / m s–1 4.0 2.0 0 0 1.0 2.0 t / s Fig. 3.2 1. Use data from Fig. 3.2 to show that the acceleration of the box is 2.6 m s–2. [2] 2. Calculate the resultant force on the box. resultant force = … N [1] 3. Determine the resistive force R on the box. R = … N [3]

9 marks

Mark scheme: 3 (a) a body / mass / object continues (at rest or) at constant / uniform velocity unless acted on by a resultant force B1 [1] (b) (i) weight vertically down B1 normal / reaction / contact (force) perpendicular / normal to the slope B1 [2] (ii) 1. acceleration = gradient or (v – u) / t or ∆v / t C1 = (6.0 – 0.8) / (2.0 – 0.0) = 2.6 m s–2 M1 [2] 2. F = ma = 65 × 2.6 = 169 N (allow to 2 or 3 s.f.) A1 [1] 3. weight component seen: mg sinθ (218 N) C1 218 – R = 169 C1 R = 49 N (require 2 s.f.) A1 [3] GCE AS/A LEVEL – May/June 2014 9702 22

This question in 9702/22 May/June 2014

3 (a) Force is a vector quantity. State three other vector quantities. 1. … 2. … 3. … [2] (b) Three coplanar forces X, Y and Z act on an object, as shown in Fig. 3.1. Y object θ X Z Fig. 3.1 The force Z is vertical and X is horizontal. The force Y is at an angle θ to the horizontal. The force Z is kept constant at 70 N. In an experiment, the magnitude of force X is varied. The magnitude and direction of force Y are adjusted so that the object remains in equilibrium. Fig. 3.2 shows the variation of the magnitude of force Y with the magnitude of force X. 130 Y / N 110 90 70 50 0 20 40 60 80 100 120 X / N Fig. 3.2 (i) Use Fig. 3.2 to estimate the magnitude of Y for X = 0. Y = … N [1] (ii) State and explain the value of θ for X = 0. … … … [2] (iii) The magnitude of X is increased to 160 N. Use resolution of forces to calculate the value of 1. angle θ, θ = … ° [2] 2. the magnitude of force Y. Y = … N [2] (c) The angle θ decreases as X increases. Explain why the object cannot be in equilibrium for θ = 0. … … … [1]

10 marks

Mark scheme: 3 (a) displacement / velocity / acceleration / momentum / etc. three correct (none wrong) 2, two correct (none or one wrong) 1 A2 [2] (b) (i) Y = 70 N [allow 71 N as +½ small square on graph] A1 [1] (ii) θ = 90° M1 (for equilibrium) the direction of Y must be opposite to Z or using Y sin θ = Z, hence sin θ = 70 / 70 = 1, θ = 90° A1 [2] (iii) 1. Y cos θ = 160 and Y sin θ = 70 C1 tan θ = 70 / 160 hence θ = 23.6° (24°) A1 [2] 2. Y = 160 / cos 23.6° or 70 / sin 23.6° C1 = 174.6 or 175 or 170 N A1 [2] or: 1602 + 702 = Y2 (C1) Y = 174.6 or 175 or 170 N (A1) (c) (equilibrium not possible as) there is no vertical component from Y to balance Z B1 [1]

This question in 9702/23 Oct/Nov 2014

3 Two balls X and Y are supported by long strings, as shown in Fig. 3.1. X Y 4.5 m s–1 2.8 m s–1 Fig. 3.1 The balls are each pulled back and pushed towards each other. When the balls collide at the position shown in Fig. 3.1, the strings are vertical. The balls rebound in opposite directions. Fig. 3.2 shows data for X and Y during this collision. ball mass velocity just before velocity just after collision / m s–1 collision / m s–1 X 50 g +4.5 –1.8 Y M –2.8 +1.4 Fig. 3.2 The positive direction is horizontal and to the right. (a) Use the conservation of linear momentum to determine the mass M of Y. M = … g [3] (b) State and explain whether the collision is elastic. … … … [1] (c) Use Newton’s second and third laws to explain why the magnitude of the change in momentum of each ball is the same. … … … … … [3]

7 marks

Mark scheme: 3 (a) 4.5 × 50 – 2.8 × M ( = ...) C1 (...) = –1.8 × 50 + 1.4 × M C1 (M = ) 75 g A1 [3] (b) total initial kinetic energy/KE not equal to the total final kinetic energy/KE or relative speed of approach is not equal to relative speed of separation so not elastic or is inelastic B1 [1] (c) force on X is equal and opposite to force on Y (Newton III) M1 force equals/is proportional to rate of change of momentum (Newton II) M1 time of collision same for both balls hence change in momentum is the same A1 [3]

This question in 9702/21 May/June 2015

4 (a) A gas molecule has a mass of 6.64 × 10−27 kg and a speed of 1250 m s−1. The molecule collides normally with a flat surface and rebounds with the same speed, as shown in Fig. 4.1. flat surface flat surface molecule molecule before collision after collision Fig. 4.1 Calculate the change in momentum of the molecule. change in momentum = … N s [2] (b) (i) Use the kinetic model to explain the pressure exerted by gases. … … … … … … [3] (ii) Explain the effect of an increase in density, at constant temperature, on the pressure of a gas. … … [1]

6 marks

Mark scheme: 4 (a) (p =) mv C1 ∆p (= – 6.64 × 10–27 × 1250 – 6.64 × 10–27 × 1250) = 1.66 × 10–23 N s A1 [2] (b) (i) molecule collides with wall/container and there is a change in momentum B1 change in momentum / time is force or ∆p = Ft B1 many/all/sum of molecular collisions over surface/area of container produces pressure B1 [3] (ii) more collisions per unit time so greater pressure B1 [1]

This question in 9702/22 May/June 2015

3 (a) Define power. … … [1] (b) Fig. 3.1 shows a car travelling at a speed of 22 m s–1 on a horizontal road. speed 22 m s–1 1200 N resistive force horizontal road Fig. 3.1 The car has a mass of 1500 kg. A resistive force of 1200 N acts on the car. Calculate (i) the force F required from the car to produce an acceleration of 0.82 m s–2, F = … N [3] (ii) the power required to produce this acceleration. power = … W [2] (c) The resistive force on the car is proportional to v 2, where v is the speed of the car. Suggest why the car has a maximum speed. … … [1]

7 marks

Mark scheme: 3 (a) (power =) work done / time (taken) or rate of work done A1 [1] (b) (i) F – R = ma C1 F = 1500 × 0.82 + 1200 C1 = 2400 (2430) N A1 [3] (ii) P = Fv C1 = (2430 × 22) = 53 000 (53 500) W A1 [2] (c) (there is maximum power from car and) resistive force = force produced by car hence no acceleration or suggestion in terms of power produced by car and power wasted to overcome resistive force B1 [1]

This question in 9702/23 May/June 2015

4 Fig. 4.1 shows a metal cylinder of height 4.5 cm and base area 24 cm2. metal cylinder 4.5 cm base area 24 cm2 Fig. 4.1 The density of the metal is 7900 kg m–3. (a) Show that the mass of the cylinder is 0.85 kg. [2] (b) The cylinder is placed on a plank, as shown in Fig. 4.2. plank cylinder 40° horizontal Fig. 4.2 The plank is at an angle of 40° to the horizontal. Calculate the pressure on the plank due to the cylinder. pressure = … Pa [3] (c) The cylinder then slides down the plank with a constant acceleration of 3.8 m s–2. A constant frictional force f acts on the cylinder. Calculate the frictional force f. f = … N [3]

8 marks

Mark scheme: 4 (a) density = mass / volume C1 mass = 7900 × 4.5 × 24 × 10–6 = 0.85 (0.853) kg M1 [2] (b) pressure = force / area C1 force = W cos 40° C1 pressure = (0.85 × 9.81 cos 40°) / 24 × 10–4 = 2.7 (2.66) × 103 Pa A1 [3] (c) F = ma C1 W sin 40° – f = ma C1 0.85 × 9.81 × sin 40° – f = 0.85 × 3.8 f (= 5.36 – 3.23) = 2.1 N [5.38 – 3.242 if 0.8532 kg is used for the mass] A1 [3]

This question in 9702/21 Oct/Nov 2015

2 A ball of mass 0.030 kg moves along a curved track, as shown in Fig. 2.1. ball mass 0.030 kg speed 1.3 m s–1 A wall 0.31 m B Fig. 2.1 The speed of the ball is 1.3 m s–1 when it is at point A at a height of 0.31 m. The ball moves down the track and collides with a vertical wall at point B. The ball then rebounds back up the track. It may be assumed that frictional forces are negligible. (a) Calculate the change in gravitational potential energy of the ball in moving from point A to point B. change in gravitational potential energy = … J [2] (b) Show that the ball hits the wall at B with a speed of 2.8 m s–1. [3] (c) The change in momentum of the ball due to the collision with the wall is 0.096 kg m s–1. The ball is in contact with the wall for a time of 20 ms. Determine, for the ball colliding with the wall, (i) the speed immediately after the collision, speed = … m s–1 [2] (ii) the magnitude of the average force on the ball. force = … N [2] (d) State and explain whether the collision is elastic or inelastic. … … [1] (e) In practice, frictional effects are significant so that the actual increase in kinetic energy of the ball in moving from A to B is 76 mJ. The length of the track between A and B is 0.60 m. Use your answer in (a) to determine the average frictional force acting on the ball as it moves from A to B. frictional force = … N [2] [Total: 12]

12 marks

Mark scheme: 2 (a) ∆E = mg∆h C1 = 0.030 × 9.81 × (–)0.31 = (–)0.091 J A1 [2] (b) E = ½mv 2 C1 (initial) E = ½ × 0.030 × 1.32 (= 0.0254) C1 0.5 × 0.030 × v 2 = (0.5 × 0.030 × 1.32) + (0.030 × 9.81 × 0.31) so v = 2.8 m s–1 or 0.5 × 0.030 × v 2 = (0.0254) + (0.091) so v = 2.8 m s–1 A1 [3] (c) (i) 0.096 = 0.030 (v + 2.8) C1 v = 0.40 m s–1 A1 [2] (ii) F = ∆p / (∆)t or F = ma = 0.096 / 20 × 10–3 or 0.030 (0.40 + 2.8) / 20 × 10–3 C1 = 4.8 N A1 [2] (d) kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic or (relative) speed of approach (of ball and wall) not equal to/greater than (relative) speed of separation, so inelastic. B1 [1] (e) force = work done / distance moved = (0.091 – 0.076) / 0.60 C1 = 0.025 N A1 [2]

This question in 9702/22 Oct/Nov 2016

3 (a) Define velocity. … … [1] (b) A car travels in a straight line up a slope, as shown in Fig. 3.1. ms–1 9.0 car mass 850 kg slope Fig. 3.1 The car has mass 850 kg and travels with a constant speed of 9.0 m s–1. The car’s engine exerts a force on the car of 2.0 kN up the slope. A resistive force FD, due to friction and air resistance, opposes the motion of the car. The variation of FD with the speed v of the car is shown in Fig. 3.2. 0.70 FD / kN 0.60 0.50 0.40 0.30 7 8 9 10 11 12 13 14 15 16 v / m s–1 Fig. 3.2 (i) State and explain whether the car is in equilibrium as it moves up the slope. … … … [2] (ii) Consider the forces that act along the slope. Use data from Fig. 3.2 to determine the component of the weight of the car that acts down the slope. component of weight = … N [2] (iii) Show that the power output of the car is 1.8 × 104 W. [2] (iv) The car now travels along horizontal ground. The output power of the car is maintained at 1.8 × 104 W. The variation of the resistive force FD acting on the car is given in Fig. 3.2. Calculate the acceleration of the car when its speed is 15 m s–1. acceleration = … m s–2 [3] [Total: 10]

10 marks

Mark scheme: 3(a) change of displacement / time (taken) B1 3(b)(i) constant velocity, so resultant force is zero M1 (so car is) in (dynamic) equilibrium A1 3(b)(ii) FD = 0.40 (kN) or 0.40 × 103 (N) C1 component of weight = 2.0 × 103 – 0.40 × 103 = 1.6 × 103 N A1 3(b)(iii) P = Fv C1 = 2.0 ×103 × 9.0 = 1.8 × 104 W A1 3(b)(iv) (driving) force = 1.8 × 104 / 15 (= 1.2 × 103) C1 FD = 0.66 (kN) or 0.66 × 103 (N) C1 acceleration = (1.2 × 103 – 0.66 × 103) / 850 = 0.64 (0.635) m s–2 A1

This question in 9702/22 Feb/March 2017

4 (a) State Newton’s first law of motion. … … [1] (b) An object A of mass 100 g is moving in a straight line with a velocity of 0.60 m s–1 to the right. An object B of mass 200 g is moving in the same straight line as object A with a velocity of 0.80 m s–1 to the left, as shown in Fig. 4.1. A B 0.60 m s–1 0.80 m s–1 100 g 200 g Fig. 4.1 Objects A and B collide. Object A then moves with a velocity of 0.40 m s–1 to the left. (i) Calculate the magnitude of the velocity of B after the collision. magnitude of velocity = … m s–1 [2] (ii) The collision between A and B is inelastic. Explain how the collision is inelastic and still obeys the law of conservation of energy. … … … [1] [Total: 4]

4 marks

Mark scheme: 4(a) a body/mass/object continues (at rest or) at constant/uniform velocity unless acted on by a resultant force B1 4(b)(i) initial momentum = final momentum m1u1 + m2u2 = m1v1 + m2v2 C1 0.60 × 100 − 0.80 × 200 = −0.40 × 100 + v × 200 v = (−) 0.3(0) m s–1 A1 4(b)(ii) kinetic energy is not conserved/is lost (but) total energy is conserved/constant or some of the (initial) kinetic energy is transformed into other forms of energy B1

This question in 9702/22 May/June 2017

1 (a) Two forces, with magnitudes 5.0 N and 12 N, act from the same point on an object. Calculate the magnitude of the resultant force R for the forces acting (i) in opposite directions, R = … N [1] (ii) at right angles to each other. R = … N [1] (b) An object X rests on a smooth horizontal surface. Two horizontal forces act on X as shown in Fig. 1.1. 18 N 115° X 55 N Fig. 1.1 (not to scale) A force of 55 N is applied to the right. A force of 18 N is applied at an angle of 115° to the direction of the 55 N force. (i) Use the resolution of forces or a scale diagram to show that the magnitude of the resultant force acting on X is 65 N. [2] (ii) Determine the angle between the resultant force and the 55 N force. angle = … ° [2] (c) A third force of 80 N is now applied to X in the opposite direction to the resultant force in (b). The mass of X is 2.7 kg. Calculate the magnitude of the acceleration of X. acceleration = … m s–2 [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) R = 7(.0) N B1 1(a)(ii) R = 13 N B1 1(b)(i) forces resolved: 18 sin 65° (vertical) and 55 + 18 cos 65° (horizontal) or scale drawing: correct triangle drawn for forces B1 F = [(18 sin 65°)2 + (55 + 18 cos 65°)2]1/2 = 65 (64.7) N or scale drawing: scale given, length of resultant given correctly, ± 1 N A1 1(b)(ii) angle = tan–1 [18 sin 65° / (55 + 18 cos 65°)] = tan–1 (16.3 / 62.6) or scale drawing: correct angle measured/direction correct on diagram below the 55 N force C1 angle = 15 (14.6)° (below the 55 N force) or scale drawing: angle = 15° ± 1° A1 1(c) (resultant) force = mass × acceleration C1 80 − 65 = 2.7a C1 a = 5.6 m s–2 [5.7 if 64.7 N used from (i)] A1

This question in 9702/23 May/June 2017

2 (a) State Newton’s second law of motion. … … [1] (b) A constant resultant force F acts on an object A. The variation with time t of the velocity v for the motion of A is shown in Fig. 2.1. 9.0 v / m s–1 8.0 7.0 6.0 5.0 4.0 0 1.0 2.0 3.0 4.0 t / s Fig. 2.1 The mass of A is 840 g. Calculate, for the time t = 0 to t = 4.0 s, (i) the change in momentum of A, change in momentum = … kg m s–1 [2] (ii) the force F. F = … N [1] (c) The force F is removed at t = 4.0 s. Object A continues at constant velocity before colliding with an object B, as illustrated in Fig. 2.2. A B 840 g 730 g at rest Fig. 2.2 Object B is initially at rest. The mass of B is 730 g. The objects A and B join together and have a velocity of 4.7 m s–1. (i) By calculation, show that the changes in momentum of A and of B during the collision are equal and opposite. [2] (ii) Explain how the answers obtained in (i) support Newton’s third law. … … … … [2] (iii) By reference to the speeds of A and B, explain whether the collision is elastic. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) (resultant) force is proportional/equal to the rate of change of momentum B1 2(b)(i) change in momentum = m(v2 − v1) = 0.84 × (8.8 − 4.2) C1 = 3.9 (3.86) kg m s–1 A1 2(b)(ii) F = (3.9 / 4.0) = 0.97 (0.965) N A1 2(c)(i) change in momentum for A: 0.84 × (4.7 − 8.8) = −3.4 (3.44) change in momentum for B: 0.73 × (4.7 − 0) = 3.4 (3.43) M1 change in momentum for B is equal and opposite to A A1 2(c)(ii) change in momentum equal (for A and B) M1 force is change in momentum / time and time (of collision) is the same hence force on A and B equal and opposite as for Newton’s third law A1 2(c)(iii) inelastic as relative speed of approach not equal to relative speed of separation B1

This question in 9702/23 May/June 2017

1 (a) The drag force FD acting on a sphere moving through a fluid is given by the expression FD = Kρv 2 where K is a constant, ρ is the density of the fluid and v is the speed of the sphere. Determine the SI base units of K. base units … [3] (b) A ball of weight 1.5 N falls vertically from rest in air. The drag force F D acting on the ball is given by the expression in (a). The ball reaches a constant (terminal) speed of 33 m s–1. Assume that the upthrust acting on the ball is negligible and that the density of the air is uniform. For the instant when the ball is travelling at a speed of 25 m s–1, determine (i) the drag force FD on the ball, FD = … N [2] (ii) the acceleration of the ball. acceleration = … m s–2 [2] (c) Describe the acceleration of the ball in (b) as its speed changes from zero to 33 m s–1. … … … … [3] [Total: 10]

10 marks

Mark scheme: 1(a) C1 units of ρ: kg m–3 and units of v: m s–1 C1 units of K: kg m s–2 / [kg m–3 (m s–1)2] = m2 A1 1(b)(i) Kρ = 1.5 / 332 C1 = 1.38 × 10–3 FD = 1.38 × 10–3 × 252 or FD / 1.5 = 252 / 332 FD = 0.86 N A1 1(b)(ii) a = (1.5 – 0.86) / (1.5 / 9.81) or a = 9.81 – [0.86 / (1.5 / 9.81)] C1 a = 4.2 m s–2 A1 1(c) initial acceleration is g/9.81 (m s–2)/acceleration of free fall B1 acceleration decreases B1 final acceleration is zero B1

This question in 9702/21 Oct/Nov 2017

3 A spring is attached at one end to a fixed point and hangs vertically with a cube attached to the other end. The cube is initially held so that the spring has zero extension, as shown in Fig. 3.1. spring with zero extension cube weight 4.0 N 5.1 cm 5.1 cm water 7.0 cm density 1000 kg m–3 Fig. 3.1 Fig. 3.2 The cube has weight 4.0 N and sides of length 5.1 cm. The cube is released and sinks into water as the spring extends. The cube reaches equilibrium with its base at a depth of 7.0 cm below the water surface, as shown in Fig. 3.2. The density of the water is 1000 kg m–3. (a) Calculate the difference in the pressure exerted by the water on the bottom face and on the top face of the cube. difference in pressure = … Pa [2] (b) Use your answer in (a) to show that the upthrust on the cube is 1.3 N. [2] (c) Calculate the force exerted on the spring by the cube when it is in equilibrium in the water. force = … N [1] (d) The spring obeys Hooke’s law and has a spring constant of 30 N m–1. Determine the initial height above the water surface of the base of the cube before it was released. height above surface = … cm [3] (e) The cube in the water is released from the spring. (i) Determine the initial acceleration of the cube. acceleration = … m s–2 [2] (ii) Describe and explain the variation, if any, of the acceleration of the cube as it sinks in the water. … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a) C1 ∆p = 1000 × 9.81 × (7.0 × 10–2 – 1.9 × 10–2) or 686 – 186 = 500 Pa A1 3(b) F = pA or (∆)F = ∆p × A C1 upthrust = 500 × (5.1 × 10–2)2 = 1.3 N or upthrust = (686 – 186) × (5.1 ×10–2)2 = 1.3 N or upthrust = 1000 × 9.81 × 5.1 ×10–2 × (5.1 × 10–2)2 = 1.3 N A1 3(c) force = 4.0 – 1.3 = 2.7 N A1 Question Answer Marks 3(d) extension/x/e = 2.7 / 30 C1 = 0.09 (m) or 9 (cm) C1 height above surface = 9 – 7 = 2 cm A1 3(e)(i) mass = 4.0 / 9.81 C1 acceleration = 2.7 / (4.0 / 9.81) = 6.6 m s–2 A1 3(e)(ii) viscous force increases (and then becomes constant) M1 (weight and upthrust constant so) acceleration decreases (to zero) A1

This question in 9702/22 Oct/Nov 2017

5 (a) Define the coulomb. … [1] (b) Two vertical metal plates in a vacuum have a separation of 4.0 cm. A potential difference of 2.0 × 102 V is applied between the plates. Fig. 5.1 shows a side view of this arrangement. 4.0 cm smoke particle weight 3.9 × 10–15 N charge –8.0 × 10–19 C metal plate metal plate +2.0 × 102 V s Fig. 5.1 A smoke particle is in the uniform electric field between the plates. The particle has weight 3.9 × 10–15 N and charge –8.0 × 10–19 C. (i) Show that the electric force acting on the particle is 4.0 × 10–15 N. [2] (ii) On Fig. 5.1, draw labelled arrows to show the directions of the two forces acting on the smoke particle. [1] (iii) The resultant force acting on the particle is F. Determine 1. the magnitude of F, magnitude = … N 2. the angle of F to the horizontal. angle = … ° [3] (c) The electric field in (b) is switched on at time t = 0 when the particle is at a horizontal displacement s = 2.0 cm from the left-hand plate. At time t = 0 the horizontal velocity of the particle is zero. The particle is then moved by the electric field until it hits a plate at time t = T. On Fig. 5.2, sketch the variation with time t of the horizontal displacement s of the particle from the left-hand plate. 4.0 s / cm 2.0 0 0 T t Fig. 5.2 [2] [Total: 9]

9 marks

Mark scheme: 5(a) (coulomb is) ampere second B1 5(b)(i) E = V / d or E = F / Q C1 F = VQ / d F = (2.0 × 102 × 8.0 × 10–19) / 4.0 × 10–2 = 4.0 × 10–15 N A1 5(b)(ii) arrow pointing to the left labelled ‘electric force’ and arrow pointing downwards labelled ‘weight’ B1 5(b)(iii) 1. resultant force = √ [(3.9 × 10–15)2 + (4.0 × 10–15)2] C1 = 5.6 × 10–15 N A1 2. angle = tan–1 (3.9 × 10–15 / 4.0 × 10–15) = 44° A1 5(c) downward sloping line from (0, 2.0) M1 magnitude of gradient of line increases with time and line ends at (T, 0) A1

This question in 9702/22 Oct/Nov 2017

2 A liquid of density ρ fills a container to a depth h, as shown in Fig. 2.1. container liquid h base area A Fig. 2.1 The base of the container has area A. (a) Derive, from the definitions of pressure and density, the equation p = ρgh where p is the pressure exerted by the liquid on the base of the container and g is the acceleration of free fall. [3] (b) A small solid sphere falls with constant velocity through the liquid. (i) State 1. the names of the three forces acting on the sphere, … … 2. a word equation that relates the magnitudes of these forces. … [2] (ii) State and explain the changes in energy that occur as the sphere falls. … … … … [2] (c) The liquid in the container is liquid L. Liquid M is now added to the container. The two liquids do not mix. The total depth of the liquids is 0.17 m. Fig. 2.2 shows how the pressure p inside the liquids varies with height x above the base of the container. 9.25 p / 104 Pa liquid L 9.20 9.15 liquid M 9.10 0 0.05 0.10 0.15 0.20 x / m Fig. 2.2 Use Fig. 2.2 to (i) state the value of atmospheric pressure, atmospheric pressure = … Pa [1] (ii) determine the density of liquid M. density = … kg m–3 [2] [Total: 10]

10 marks

Mark scheme: 2(a) B1 p = F / A or p = W / A B1 p = [ρAhg] / A or p = [ρVg] / [V / h] (so) p = ρgh A1 Question Answer Marks 2(b)(i) 1. weight/gravitational (force) upthrust (force)/buoyancy (force) drag/viscous/frictional (force)/fluid resistance/resistance B1 2. weight = upthrust + viscous (force) B1 2(b)(ii) • decrease in (gravitational) potential energy (of sphere) due to decrease in height (since Ep = mgh) • increase in thermal energy due to work done against viscous force/drag • loss/change of (total) Ep equal to gain/change in thermal energy Any 2 points. B2 2(c)(i) atmospheric pressure = 9.1(0) × 104 Pa A1 2(c)(ii) (∆)p = ρg(∆)h (9.15 – 9.10) × 104 = ρ × 9.81 × (0.17 – 0.10) C1 ρ = 730 (728) kg m–3 A1

This question in 9702/23 Oct/Nov 2017

3 (a) State the principle of conservation of momentum. … … … [2] (b) Ball A moves with speed v along a horizontal frictionless surface towards a stationary ball B, as shown in Fig. 3.1. 6.0 m s–1 4.0 kg A v initial path θ A B of ball A 30° 4.0 kg 12 kg 12 kg B 3.5 m s–1 after collision before collision Fig. 3.1 Fig. 3.2 (not to scale) Ball A has mass 4.0 kg and ball B has mass 12 kg. The balls collide and then move apart as shown in Fig. 3.2. Ball A has velocity 6.0 m s–1 at an angle of θ to the direction of its initial path. Ball B has velocity 3.5 m s–1 at an angle of 30° to the direction of the initial path of ball A. (i) By considering the components of momentum at right-angles to the direction of the initial path of ball A, calculate θ. θ = … ° [3] (ii) Use your answer in (i) to show that the initial speed v of ball A is 12 m s–1. Explain your working. [2] (iii) By calculation of kinetic energies, state and explain whether the collision is elastic or inelastic. … … [3] [Total: 10]

10 marks

Mark scheme: 3(a) sum/total momentum (of system of bodies) is constant or sum/total momentum before = sum/total momentum after M1 for an isolated system/no (resultant) external force A1 3(b)(i) p = mv C1 (4.0 × 6.0 × sin θ) – (12 × 3.5 × sin 30°) = 0 or (mAvA × sinθ) – (mBvB × sin 30°) = 0 M1 θ = 61° A1 Question Answer Marks 3(b)(ii) shows the horizontal momentum component of ball A or of ball B as (4.0 × 6.0 × cos θ) or (12 × 3.5 × cos 30°) C1 (4.0 × 6.0 × cos 61°) + (12 × 3.5 × cos 30°) = 4.0v so v = 12 (m s–1) A1 3(b)(iii) initial EK (= ½ × 4.0 × 122) = 290 (288) (J) M1 final EK (= ½ × 4.0 × 6.02 + ½ × 12 × 3.52) = 150 (145.5) (J) M1 (initial EK > final EK) so inelastic [both M1 marks required to award this mark] A1

This question in 9702/23 Oct/Nov 2017

1 (a) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar. quantity vector or scalar acceleration speed power Fig. 1.1 [2] (b) A ball is projected with a horizontal velocity of 1.1 m s–1 from point A at the edge of a table, as shown in Fig. 1.2. table ball 1.1 m s–1 A path of ball B horizontal ground 0.43 m Fig. 1.2 The ball lands on horizontal ground at point B which is a distance of 0.43 m from the base of the table. Air resistance is negligible. (i) Calculate the time taken for the ball to fall from A to B. time = … s [1] (ii) Use your answer in (b)(i) to determine the height of the table. height = … m [2] (iii) The ball leaves the table at time t = 0. For the motion of the ball between A and B, sketch graphs on Fig. 1.3 to show the variation with time t of 1. the acceleration a of the ball, 2. the vertical component sv of the displacement of the ball from A. Numerical values are not required. a sv 0 0 0 t 0 t Fig. 1.3 [2] (c) A ball of greater mass is projected from the table with the same velocity as the ball in (b). Air resistance is still negligible. State and explain the effect, if any, of the increased mass on the time taken for the ball to fall to the ground. … … [1] [Total: 8]

8 marks

Mark scheme: 1(a) acceleration: vector speed: scalar power: scalar All three correct scores 2 marks. Only two correct scores 1 mark. B2 1(b)(i) time = 0.43 / 1.1 = 0.39 s A1 1(b)(ii) s = ut + ½at 2 = ½ × 9.81 × 0.392 C1 = 0.75 m A1 1(b)(iii) 1 horizontal line at a non-zero value of a. B1 2 curved line from origin with increasing gradient. B1 1(c) acceleration (of free fall) is unchanged / not dependent on mass and so no effect (on time taken). A1

This question in 9702/22 Feb/March 2018

2 (a) State Newton’s first law of motion. … … [1] (b) A block of weight 15 N hangs by a wire from a remotely controlled aircraft, as shown in Fig. 2.1. aircraft wire block weight 15 N Fig. 2.1 The aircraft is used to move the block only in a vertical direction. The force on the block due to air resistance is negligible. The variation with time t of the vertical velocity v of the block is shown in Fig. 2.2. The velocity is taken to be positive in the upward direction. 4.0 3.0 v / m s–1 2.0 1.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 t / s –1.0 –2.0 –3.0 –4.0 –5.0 –6.0 –7.0 Fig. 2.2 (i) Determine, for the block, 1. the displacement from time t = 0 to t = 3.0 s, magnitude of displacement = … m direction of displacement … [3] 2. the change in gravitational potential energy from time t = 0 to t = 3.0 s. change in gravitational potential energy = … J [2] (ii) Calculate the magnitude of the acceleration of the block at time t = 2.0 s. acceleration = … m s–2 [2] (iii) Use your answer in (b)(ii) to show that the tension T in the wire at time t = 2.0 s is 20 N. [2] (iv) The wire has a cross-sectional area of 2.8 × 10–5 m2 and is made from metal of Young modulus 1.7 × 1011 Pa. The wire obeys Hooke’s law. Calculate the strain of the wire at time t = 2.0 s. strain = … [3] (v) At some time after t = 3.0 s the tension in the wire has a constant value of 15 N. State and explain whether it is possible to deduce that the block is moving vertically after t = 3.0 s. … … … … [2] [Total: 15]

15 marks

Mark scheme: 2(a) a body continues at (rest or) constant velocity unless acted upon by a resultant force B1 2(b)(i)1. from 0–2 s, distance = ½ × 2 × 6.8 (= 6.8 m) and from 2–3 s, distance = ½ × 1 × 3.4 (= 1.7 m) C1 magnitude of displacement = 5.1 m A1 direction of displacement is down(wards) B1 2(b)(i)2. (∆E) = mg∆h or (E) = mgh or (E) = Wh C1 (∆)E = 15 × 5.1 = (–) 77 J A1 2(b)(ii) a = (v – u) / t or a = gradient or a = dv / dt C1 a = 3.4 m s–2 A1 2(b)(iii) T – W = ma or T – mg = ma C1 T = 15 + (15 / 9.81) × 3.4 = 20 N or 20.2 N A1 2(b)(iv) E = F / Aε or E = σ / ε and σ = F / A C1 ε = 20 / (2.8 × 10–5 × 1.7 × 1011) C1 = 4.2 × 10–6 A1 2(b)(v) block is in equilibrium/has no resultant force B1 block could be stationary (or have constant velocity/speed) (so no, not possible to deduce) B1

This question in 9702/21 May/June 2018

3 (a) State what is meant by the mass of a body. … … [1] (b) Two blocks travel directly towards each other along a horizontal, frictionless surface. The blocks collide, as illustrated in Fig. 3.1. 0.40 m s–1 0.25 m s–1 0.20 m s–1 v block A block B mass mass mass mass 3M M 3M M before after Fig. 3.1 Block A has mass 3M and block B has mass M. Before the collision, block A moves to the right with speed 0.40 m s–1 and block B moves to the left with speed 0.25 m s–1. After the collision, block A moves to the right with speed 0.20 m s–1 and block B moves to the right with speed v. (i) Use Newton’s third law to explain why, during the collision, the change in momentum of block A is equal and opposite to the change in momentum of block B. … … … … [2] (ii) Determine speed v. v = … m s–1 [3] (iii) Calculate, for the blocks, 1. the relative speed of approach, relative speed of approach = … m s–1 2. the relative speed of separation. relative speed of separation = … m s–1 [2] (iv) Use your answers in (b)(iii) to state and explain whether the collision is elastic or inelastic. … … [1] [Total: 9]

9 marks

Mark scheme: 3(a) mass is the property (of a body/object) resisting changes in motion or mass is the quantity of matter (in a body) B1 3(b)(i) force on A (by B) equal and opposite to force on B (by A) or both A and B exert equal and opposite forces on each other B1 force is rate of change of momentum and time (of contact) is same B1 3(b)(ii) p = mv or 3M × 0.40 or M × 0.25 or 3M × 0.2 or Mv C1 (3M × 0.40) – (M × 0.25) = (3M × 0.2) + Mv C1 v = (3 × 0.40) – 0.25 – (3 × 0.2) = 0.35 m s–1 A1 3(b)(iii) 1. relative speed of approach = 0.40 + 0.25 = 0.65 m s–1 A1 2. relative speed of separation = 0.35 – 0.20 = 0.15 m s–1 A1 3(b)(iv) (relative) speed of separation not equal to/less than (relative) speed of approach or answers (to (b)(iii) are) not equal and so inelastic collision B1

This question in 9702/21 May/June 2018

1 (a) Define force. … [1] (b) State the SI base units of force. … [1] (c) The force F between two point charges is given by Q1Q2 F = 4πr 2ε where Q1 and Q2 are the charges, r is the distance between the charges, ε is a constant that depends on the medium between the charges. Use the above expression to determine the base units of ε. base units … [2] [Total: 4]

4 marks

Mark scheme: 1(a) rate of change of momentum B1 1(b) kg m s–2 A1 1(c) units for Q: A s and for r: m C1 units for ε = (A s × A s) / (kg m s–2 × m2) = A2 kg–1 m–3 s4 A1

This question in 9702/22 May/June 2018

3 A child on a sledge slides down a steep hill and then travels in a straight line up an ice-covered slope, as illustrated in Fig. 3.1. ice-covered slope child and sledge total mass 70 kg B 18 m s–1 A Fig. 3.1 (not to scale) The sledge passes point A with speed 18 m s–1 at time t = 0 and then comes to rest at point B. The child applies a brake to the sledge at point B. The brake does not keep the sledge stationary and it immediately slides back down the slope towards A. The variation with time t of the velocity v of the sledge from t = 0 to t = 24 s is shown in Fig. 3.2. 20 v / m s–1 10 0 0 4 8 12 16 20 24 t / s –10 Fig. 3.2 (a) State the time taken for the sledge to travel from A to B. time = … s [1] (b) Determine the displacement of the sledge up the slope from point A at time t = 24 s. displacement = … m [3] (c) Show that the acceleration of the sledge as it moves from B back towards A is 0.50 m s–2. [2] (d) The child and sledge have a total mass of 70 kg. The component of the total weight of the child and sledge that acts down the slope is 80 N. Determine (i) the frictional force on the sledge as it moves from B towards A, frictional force = … N [2] (ii) the angle θ of the slope to the horizontal. θ = … ° [2] (e) The child on the sledge blows a whistle between t = 4.0 s and t = 8.0 s. The whistle emits sound of frequency 900 Hz. The speed of the sound in the air is 340 m s–1. A man standing at point A hears the sound. Use Fig. 3.2 to (i) determine the initial frequency of the sound heard by the man, initial frequency = … Hz [2] (ii) describe and explain qualitatively the variation, if any, in the frequency of the sound heard by the man. … … [1] [Total: 13]

13 marks

Mark scheme: 3(a) time = 12 s A1 3(b) distance (up slope) = ½ × 12 × 18 (= 108) C1 distance (down slope) = ½ × 12 × 6 (= 36) C1 displacement from A = 108 – 36 = 72 m A1 3(c) v = u + at or a = gradient or a = ∆v / (∆)t C1 a = 6 / 12 = 0.50 (m s–2) (other points from the line may be used) A1 or v2 = u2 + 2as and u = 0 or v2 = 2as (C1) a = 6.02 / (2 × 36) = 0.50 (m s–2) (A1) or s = ut + ½at2 and u = 0 or s = ½at2 (C1) a = 2 × 36 / 122 = 0.50 (m s–2) (A1) or s = vt – ½at2 (C1) a = 2 × (6 × 12 – 36) / 122 = 0.50 (m s–2) (A1) Question Answer Marks 3(d)(i) F = 70 × 0.50 (= 35) C1 frictional force = 80 – 35 = 45 N A1 3(d)(ii) sin θ = 80 / (70 × 9.81) C1 θ = 6.7° A1 3(e)(i) f0 = (900 × 340) / (340 + 12) C1 = 870 Hz A1 3(e)(ii) speed/velocity (of sledge) decreases and (so) frequency increases B1

This question in 9702/22 May/June 2018

3 A ball is thrown vertically upwards towards a ceiling and then rebounds, as illustrated in Fig. 3.1. ceiling ball leaving speed 3.8 m s–1 ceiling ball thrown speed 9.6 m s–1 upwards Fig. 3.1 The ball is thrown with speed 9.6 m s–1 and takes a time of 0.37 s to reach the ceiling. The ball is then in contact with the ceiling for a further time of 0.085 s until leaving it with a speed of 3.8 m s–1. The mass of the ball is 0.056 kg. Assume that air resistance is negligible. (a) Show that the ball reaches the ceiling with a speed of 6.0 m s–1. [1] (b) Calculate the height of the ceiling above the point from which the ball was thrown. height = … m [2] (c) Calculate (i) the increase in gravitational potential energy of the ball for its movement from its initial position to the ceiling, increase in gravitational potential energy = … J [2] (ii) the decrease in kinetic energy of the ball while it is in contact with the ceiling. decrease in kinetic energy = … J [2] (d) State how Newton’s third law applies to the collision between the ball and the ceiling. … … … … [2] (e) Calculate the change in momentum of the ball during the collision. change in momentum = … N s [2] (f) Determine the magnitude of the average force exerted by the ceiling on the ball during the collision. average force = … N [2] [Total: 13]

13 marks

Mark scheme: 3(a) v = u + at v = 9.6 – (9.81 × 0.37) = 6.0 m s–1 A1 3(b) s = ½ × (9.6 + 6.0) × 0.37 or 6.02 = 9.62 – (2 × 9.81 × s) or s = (9.6 × 0.37) – (½ × 9.81 × 0.372) or s = (6.0 × 0.37) + (½ × 9.81 × 0.372) C1 s = 2.9 m A1 3(c)(i) (∆)E = mg(∆)h C1 ∆E = 0.056 × 9.81 × 2.9 = 1.6 J A1 3(c)(ii) E = ½mv 2 C1 ∆E = ½ × 0.056 × (6.02 – 3.82) = 0.60 J A1 3(d) force on ball (by ceiling) equal to force on ceiling (by ball) M1 and opposite (in direction) A1 3(e) (p =) mv or 0.056 × 6.0 or 0.056 × 3.8 C1 change in momentum = 0.056 × (6.0 + 3.8) = 0.55 N s A1 Question Answer Mark 3(f) resultant force = 0.55 / 0.085 (= 6.47 N) C1 force by ceiling = 6.47 – (0.056 × 9.81) = 5.9 N A1

This question in 9702/23 May/June 2018

2 A wooden block moves along a horizontal frictionless surface, as shown in Fig. 2.1. 45 m s –1 2.0 m s –1 block steel ball mass 85 g mass 4.0 g horizontal surface Fig. 2.1 The block has mass 85 g and moves to the left with a velocity of 2.0 m s –1. A steel ball of mass 4.0 g is fired to the right. The steel ball, moving horizontally with a speed of 45 m s –1, collides with the block and remains embedded in it. After the collision the block and steel ball both have speed v. (a) Calculate v. v = … m s –1 [2] (b) (i) For the block and ball, state 1. the relative speed of approach before collision, relative speed of approach = … m s–1 2. the relative speed of separation after collision. relative speed of separation = … m s–1 [1] (ii) Use your answers in (i) to state and explain whether the collision is elastic or inelastic. … … [1] (c) Use Newton’s third law to explain the relationship between the rate of change of momentum of the ball and the rate of change of momentum of the block during the collision. … … … … [2]

6 marks

Mark scheme: 2(a) C1 (4.0 × 45) – (2.0 × 85) = 89 v v = 0.11 m s–1 A1 2(b)(i) 1. speed of approach = 47 m s–1 and 2. speed of separation = 0 A1 2(b)(ii) speed of separation less than/not equal to speed of approach and so inelastic collision A1 2(c) force is equal to rate of change of momentum B1 force on ball (by block) equal and opposite to force on block (by ball) so rates of change of momentum are equal and opposite B1 or force on ball (by block) equal and opposite to force on block (by ball) (B1) force is equal to rate of change of momentum so rates of change of momentum are equal and opposite (B1)

This question in 9702/21 Oct/Nov 2018

1 A golfer strikes a ball so that it leaves horizontal ground with a velocity of 6.0 m s–1 at an angle θ to the horizontal, as illustrated in Fig. 1.1. vY 6.0 m s–1 4.8 m s–1 ball θ ground vX Fig. 1.1 (not to scale) The magnitude of the initial vertical component vY of the velocity is 4.8 m s–1. Assume that air resistance is negligible. (a) Show that the magnitude of the initial horizontal component vX of the velocity is 3.6 m s–1. [1] (b) The ball leaves the ground at time t = 0 and reaches its maximum height at t = 0.49 s. On Fig. 1.2, sketch separate lines to show the variation with time t, until the ball returns to the ground, of (i) the vertical component vY of the velocity (label this line Y), [2] (ii) the horizontal component vX of the velocity (label this line X). [2] 5.0 4.0 velocity / m s–1 3.0 2.0 1.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 t / s –1.0 –2.0 –3.0 –4.0 –5.0 Fig. 1.2 (c) Calculate the maximum height reached by the ball. maximum height = … m [2] (d) For the movement of the ball from the ground to its maximum height, determine the ratio kinetic energy at maximum height . change in gravitational potential energy ratio = … [4] (e) In practice, significant air resistance acts on the ball. Explain why the actual time taken for the ball to reach maximum height is less than the time calculated when air resistance is assumed to be negligible. … … … [1] [Total: 12]

12 marks

Mark scheme: 1(a) or 6.0 sinθ = 4.8 (so θ = 53.1°) and vx = 6.0 cos 53.1° = 3.6 (m s–1) A1 1(b)(i) straight line from (0, 4.8) to (0.49, 0) M1 straight line continues with same slope to (0.98, –4.8) (labelled Y) A1 1(b)(ii) a horizontal line M1 from (0, 3.6) to (0.98, 3.6) (labelled X) A1 1(c) s = ut + ½at2 = (4.8 × 0.49) + (½ × –9.81 × 0.492) or s = ½(u + v)t or area under graph = ½ × (4.8 + 0) × 0.49 or s = vt – ½at2 = ½ × 9.81 × 0.492 or v2 = u2 + 2as s = 4.82 / (2 × 9.81) C1 s = 1.2 m A1 Question Answer Marks 1(d) (∆)E = mg(∆)h C1 E = ½mv2 C1 ratio = (½ × m × 3.62) / (m × 9.81 × 1.2) or ratio = [(½ × m × 6.02) – (m × 9.81 × 1.2)] / (m × 9.81 × 1.2) or ratio = (½ × m × 3.62) / (½ × m × 4.82) C1 ratio = 0.56 A1 1(e) (force due to) air resistance acts in opposite direction to the velocity or (with air resistance, average) resultant force is larger (than weight) B1

This question in 9702/22 Oct/Nov 2018

3 (a) State the principle of conservation of momentum. … … … [2] (b) The propulsion system of a toy car consists of a propeller attached to an electric motor, as illustrated in Fig. 3.1. propeller moving air 0.045 m speed 1.8 m s–1 electric motor of car body of car 0.045 m ground Fig. 3.1 The car is on horizontal ground and is initially held at rest by its brakes. When the motor is switched on, it rotates the propeller so that air is propelled horizontally to the left. The density of the air is 1.3 kg m–3. Assume that the air moves with a speed of 1.8 m s–1 in a uniform cylinder of radius 0.045 m. Also assume that the air to the right of the propeller is stationary. (i) Show that, in a time interval of 2.0 s, the mass of air propelled to the left is 0.030 kg. [2] (ii) Calculate 1. the increase in the momentum of the mass of air in (b)(i), increase in momentum = … N s 2. the force exerted on this mass of air by the propeller. force = … N [3] (iii) Explain how Newton’s third law applies to the movement of the air by the propeller. … … … [2] (iv) The total mass of the car is 0.20 kg. The brakes of the car are released and the car begins to move with an initial acceleration of 0.075 m s–2. Determine the initial frictional force acting on the car. frictional force = … N [2] [Total: 11]

11 marks

Mark scheme: 3(a) sum/total momentum (of a system of bodies) is constant or sum/total momentum before = sum/total momentum after M1 for an isolated system or no (resultant) external force A1 3(b)(i) m = ρV C1 = 1.3 × π × 0.0452 × 1.8 × 2.0 = 0.030 (kg) A1 3(b)(ii) 1. (∆)p = (∆)mv C1 = 0.030 × 1.8 = 0.054 N s A1 2. F = 0.054 / 2.0 or 0.030 × 1.8 / 2.0 = 0.027 N A1 3(b)(iii) force on air (by propeller) equal to force on propeller (by air) M1 and opposite (in direction) A1 3(b)(iv) resultant force = 0.20 × 0.075 (= 0.015 N) frictional force = 0.027 – 0.015 C1 = 0.012 N A1

This question in 9702/22 Oct/Nov 2018

3 (a) State Newton’s second law of motion. … … [1] (b) A toy rocket consists of a container of water and compressed air, as shown in Fig. 3.1. container compressed air water density 1000 kg m–3 nozzle radius 7.5 mm Fig. 3.1 Water is pushed vertically downwards through a nozzle by the compressed air. The rocket moves vertically upwards. The nozzle has a circular cross-section of radius 7.5 mm. The density of the water is 1000 kg m–3. Assume that the water leaving the nozzle has the shape of a cylinder of radius 7.5 mm and has a constant speed of 13 m s–1 relative to the rocket. (i) Show that the mass of water leaving the nozzle in the first 0.20 s after the rocket launch is 0.46 kg. [2] (ii) Calculate 1. the change in the momentum of the mass of water in (b)(i) due to leaving the nozzle, change in momentum = … N s 2. the force exerted on this mass of water by the rocket. force = … N [3] (iii) State and explain how Newton’s third law applies to the movement of the rocket by the water. … … … [2] (iv) The container has a mass of 0.40 kg. The initial mass of water before the rocket is launched is 0.70 kg. The mass of the compressed air in the rocket is negligible. Assume that the resistive force on the rocket due to its motion is negligible. For the rocket at a time of 0.20 s after launching, 1. show that its total mass is 0.64 kg, 2. calculate its acceleration. acceleration = … m s–2 [3] [Total: 11]

11 marks

Mark scheme: 3(a) (resultant) force proportional/equal to rate of change of momentum B1 3(b)(i) ρ = m / V C1 V = π × (7.5 × 10–3)2 × 13 × 0.2 (= 4.59 × 10–4 m3) m = π × (7.5 × 10–3)2 × 13 × 0.2 × 1000 = 0.46 kg A1 3(b)(ii) 1. (∆)p = (∆m)v C1 (∆)p = 0.46 × 13 = 6.0 N s A1 2. F = 6.0 / 0.20 = 30 N A1 3(b)(iii) force on water (by rocket/nozzle) equal to force on rocket/nozzle (by water) M1 in the opposite direction A1 3(b)(iv) 1. mass = 0.40 + 0.70 – 0.46 = 0.64 kg A1 2. acceleration = [30 – (0.64 × 9.81)] / 0.64 or 30 / 0.64 – 9.81 C1 = 37 m s–2 A1

This question in 9702/23 Oct/Nov 2018

2 (a) Define: (i) displacement … … [1] (ii) acceleration. … … [1] (b) A man wearing a wingsuit glides through the air with a constant velocity of 47 m s–1 at an angle of 24° to the horizontal. The path of the man is shown in Fig. 2.1. 47 m s–1 A man in wingsuit glide path total mass 85 kg h 24° B horizontal Fig. 2.1 (not to scale) The total mass of the man and the wingsuit is 85 kg. The man takes a time of 2.8 minutes to glide from point A to point B. (i) With reference to the motion of the man, state and explain whether he is in equilibrium. … … … … [2] (ii) Show that the difference in height h between points A and B is 3200 m. [1] (iii) For the movement of the man from A to B, determine: 1. the decrease in gravitational potential energy decrease in gravitational potential energy = … J [2] 2. the magnitude of the force on the man due to air resistance. force = … N [2] (iv) The pressure of the still air at A is 63 kPa and at B is 92 kPa. Assume the density of the air is constant between A and B. Determine the density of the air between A and B. density = … kg m–3 [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) distance in a specified direction (from a point) B1 2(a)(ii) change in velocity / time (taken) B1 2(b)(i) constant velocity so no resultant force B1 no resultant force so in equilibrium B1 2(b)(ii) (difference in height =) 47 × 2.8 × 60 × sin24° = 3200 m A1 Question Answer Marks 2(b)(iii) 1 (∆)E = mg(∆)h = 85 × 9.81 × 3200 C1 = 2.7 × 106 J A1 2 In terms of energy: work done = 2.7 × 106 J force = 2.7 × 106 / (47 × 2.8 × 60) C1 = 340 N A1 In terms of forces: component of weight along path = force due to air resistance force = 85 × 9.81 × sin24° (C1) = 340 N (A1) 2(b)(iv) (∆)p = ρg(∆)h (92 – 63) × 103 = ρ × 9.81 × 3200 C1 ρ = 0.92 kg m–3 A1

This question in 9702/22 Feb/March 2019

2 A block X slides along a horizontal frictionless surface towards a stationary block Y, as illustrated in Fig. 2.1. momentum 0.40 kg m s–1 X Y surface Fig. 2.1 There are no resistive forces acting on block X as it moves towards block Y. At time t = 0, block X has momentum 0.40 kg m s−1. A short time later, the blocks collide and then separate. The variation with time t of the momentum of block Y is shown in Fig. 2.2. 0.60 0.50 block Y momentum / kg m s–1 0.40 0.30 0.20 0.10 0 0 20 40 60 80 100 120 140 160 t / ms – 0.10 – 0.20 – 0.30 – 0.40 – 0.50 – 0.60 Fig. 2.2 (a) Define linear momentum. … [1] (b) Use Fig. 2.2 to: (i) determine the time interval over which the blocks are in contact with each other time interval = … ms [1] (ii) describe, without calculation, the magnitude of the acceleration of block Y from: 1. time t = 80 ms to t = 100 ms … 2. time t = 100 ms to t = 120 ms. … [2] (c) Use Fig. 2.2 to determine the magnitude of the force exerted by block X on block Y. force = … N [2] (d) On Fig. 2.2, sketch the variation of the momentum of block X with time t from t = 0 to t = 160 ms. [3] [Total: 9]

9 marks

Mark scheme: 2(a) B1 2(b)(i) time = 40 ms A1 2(b)(ii) 1. (the magnitude of the acceleration is) constant B1 2 . (the magnitude of the acceleration is) zero B1 2(c) F = ∆p / (∆)t or F = gradient C1 e.g. F = 0.50 / 40 × 10–3 = 13 N A1 2(d) horizontal line from (0, 0.40) to (60, 0.40) B1 straight line from (60, 0.40) to (100, –0.10) B1 horizontal line from (100, –0.10) to (160, –0.10) B1

This question in 9702/21 May/June 2019

2 (a) State Newton’s second law of motion. … … [1] (b) A car of mass 850 kg tows a trailer in a straight line along a horizontal road, as shown in Fig. 2.1. car trailer tow-bar mass 850 kg horizontal road Fig. 2.1 The car and the trailer are connected by a horizontal tow-bar. The variation with time t of the velocity v of the car for a part of its journey is shown in Fig. 2.2. 15 14 v / m s–1 13 12 11 10 9 8 0 5 10 15 20 25 t / s Fig. 2.2 (i) Calculate the distance travelled by the car from time t = 0 to t = 10 s. distance = … m [2] (ii) At time t = 10 s, the resistive force acting on the car due to air resistance and friction is 510 N. The tension in the tow-bar is 440 N. For the car at time t = 10 s: 1. use Fig. 2.2 to calculate the acceleration acceleration = … m s−2 [2] 2. use your answer to calculate the resultant force acting on the car resultant force = … N [1] 3. show that a horizontal force of 1300 N is exerted on the car by its engine [1] 4. determine the useful output power of the engine. output power = … W [2] (c) A short time later, the car in (b) is travelling at a constant speed and the tension in the tow-bar is 480 N. The tow-bar is a solid metal rod that obeys Hooke’s law. Some data for the tow-bar are listed below. Young modulus of metal = 2.2 × 1011 Pa original length of tow-bar = 0.48 m cross-sectional area of tow-bar = 3.0 × 10−4 m2 Determine the extension of the tow-bar. extension = … m [3] (d) The driver of the car in (b) sees a pedestrian standing directly ahead in the distance. The driver operates the horn of the car from time t = 15 s to t = 17 s. The frequency of the sound heard by the pedestrian is 480 Hz. The speed of the sound in the air is 340 m s−1. Use Fig. 2.2 to calculate the frequency of the sound emitted by the horn. frequency = … Hz [2] [Total: 14]

14 marks

Mark scheme: 2(a) (resultant) force proportional/equal to/is rate of change of momentum B1 2(b)(i) distance = area under graph or s = ½ (u + v) t = ½ × (9 + 13) × 10 or s = ut + ½at 2 = (9 × 10) + (½ × 0.40 × 102) or s = vt – ½at 2 = (13 × 10) – (½ × 0.40 × 102) or v 2 = u 2 + 2as 132 = 92 + (2 × 0.40 × s) C1 distance = 110 m A1 Question Answer Marks 2(b)(ii) 1. a = gradient or a = (v – u) / t or a = ∆v / (∆)t e.g. a = (14 – 9) / 12.5 or (13 – 9) / 10 C1 a = 0.40 m s–2 A1 2. resultant force = 850 × 0.40 = 340 N A1 3. (F =) 510 + 440 + 340 = 1300 (N) A1 4. P = Fv C1 = 1300 × 13 = 1.7 × 104 W A1 2(c) E = σ / ε C1 E = (F / A) / (∆L / L) or E = FL / A∆L C1 ∆L = (480 × 0.48) / (3.0 × 10–4 × 2.2 × 1011) = 3.5 × 10–6 m A1 2(d) fo = fs v / (v – vs) 480 = fs × 340 / (340 – 14) C1 fs = 460 Hz A1

This question in 9702/22 May/June 2019

3 A cylindrical disc of mass 0.24 kg has a circular cross-sectional area A, as shown in Fig. 3.1. cross-sectional force X area A 8.9 N constant 30° speed 0.60 m s–1 disc, disc mass 0.24 kg ground Fig. 3.1 Fig. 3.2 The disc is on horizontal ground, as shown in Fig. 3.2. A force X of magnitude 8.9 N acts on the disc in a direction of 30° to the horizontal. The disc moves at a constant speed of 0.60 m s−1 along the ground. (a) Determine the rate of doing work on the disc by the force X. rate of doing work = … W [2] (b) The force X and the weight of the disc exert a combined pressure on the ground of 3500 Pa. Calculate the cross-sectional area A of the disc. A = … m2 [3] (c) Newton’s third law describes how forces exist in pairs. One such pair of forces is the weight of the disc and another force Y. State: (i) the direction of force Y … [1] (ii) the name of the body on which force Y acts. … [1] [Total: 7]

7 marks

Mark scheme: 3(a) P = Fv C1 P = 8.9 cos 30° × 0.60 = 4.6 W A1 3(b) p = F / A C1 F = 8.9 sin 30° + (0.24 × 9.81) ( = 6.80 N) C1 A = 6.80 / 3500 = 1.9 × 10–3 m2 A1 3(c)(i) upwards/up B1 3(c)(ii) the Earth/planet B1

This question in 9702/23 May/June 2019

4 Two vertical metal plates in a vacuum are separated by a distance of 0.12 m. Fig. 4.1 shows a side view of this arrangement. 0.080 m sand X particle 2.0 m 0 V + 900 V path of particle metal plate metal plate Y 0.12 m Fig. 4.1 (not to scale) Each plate has a length of 2.0 m. The potential difference between the plates is 900 V. The electric field between the plates is uniform. A negatively charged sand particle is released from rest at point X, which is a horizontal distance of 0.080 m from the top of the positively charged plate. The particle then travels in a straight line and collides with the positively charged plate at its lowest point Y, as illustrated in Fig. 4.1. (a) Describe the pattern of the field lines (lines of force) between the plates. … … … [2] (b) State the names of the two forces acting on the particle as it moves from X to Y. … [1] (c) By considering the vertical motion of the sand particle, show that the time taken for the particle to move from X to Y is 0.64 s. [2] (d) Calculate the horizontal component of the acceleration of the particle. horizontal component of acceleration = … m s−2 [2] (e) (i) Calculate the magnitude of the electric field strength. electric field strength = … N C−1 [2] (ii) The sand particle has mass m and charge q. Use your answers in (d) and (e)(i) to q determine the ratio m. ratio = … C kg−1 [2] q(f) Another particle has a smaller magnitude of the ratio than the sand particle. This particle is m also released from point X. For the movement of this particle, state the effect, if any, of the decreased magnitude of the ratio on: (i) the vertical component of the acceleration … [1] (ii) the horizontal component of the acceleration. … [1] [Total: 13]

13 marks

Mark scheme: 4(a) straight (horizontal) lines and from the +0.90 kV plate/to the 0 V plate B1 (lines are) equally spaced B1 4(b) weight/gravitational force and electric force B1 4(c) s = ½ at 2 or s = ut + ½at 2 and u = 0 C1 2.0 = ½ × 9.81 × t 2 so t = 0.64 s A1 4(d) 0.080 = ½ × a × 0.642 C1 a = 0.39 m s–2 A1 4(e)(i) E = (∆)V / (∆)d C1 E = 0.90 × 103 / 0.12 = 7.5 × 103 N C–1 A1 4(e)(ii) ma = Eq or F = ma and F = Eq C1 q / m = 0.39 / 7.5 × 103 = 5.2 × 10–5 C kg–1 A1 4(f)(i) no effect B1 4(f)(ii) decreases/smaller B1

This question in 9702/23 May/June 2019

3 A small remote-controlled model aircraft has two propellers, each of diameter 16 cm. Fig. 3.1 is a side view of the aircraft when hovering. body of 16 cm 16 cm aircraft propeller propeller air air speed speed 7.6 m s–1 7.6 m s–1 Fig. 3.1 Air is propelled vertically downwards by each propeller so that the aircraft hovers at a fixed position. The density of the air is 1.2 kg m–3. Assume that the air from each propeller moves with a constant speed of 7.6 m s–1 in a uniform cylinder of diameter 16 cm. Also assume that the air above each propeller is stationary. (a) Show that, in a time interval of 3.0 s, the mass of air propelled downwards by one propeller is 0.55 kg. [3] (b) Calculate: (i) the increase in momentum of the mass of air in (a) increase in momentum = … N s [1] (ii) the downward force exerted on this mass of air by the propeller. force = … N [1] (c) State: (i) the upward force acting on one propeller force = … N [1] (ii) the name of the law that explains the relationship between the force in (b)(ii) and the force in (c)(i). … [1] (d) Determine the mass of the aircraft. mass = … kg [1] (e) In order for the aircraft to hover at a very high altitude (height), the propellers must propel the air downwards with a greater speed than when the aircraft hovers at a low altitude. Suggest the reason for this. … … [1] (f) When the aircraft is hovering at a high altitude, an electric fault causes the propellers to stop rotating. The aircraft falls vertically downwards. When the aircraft reaches a constant speed of 22 m s–1, it emits sound of frequency 3.0 kHz from an alarm. The speed of the sound in the air is 340 m s–1. Determine the frequency of the sound heard by a person standing vertically below the falling aircraft. frequency = … Hz [2] [Total: 11]

11 marks

Mark scheme: 3(a) C1 V = π × (0.16 / 2)2 × 7.6 × 3.0 (= 0.458 m3) C1 m = π × (0.16 / 2)2 × 7.6 × 3.0 × 1.2 = 0.55 kg A1 3(b)(i) ∆p = 0.55 × 7.6 = 4.2 N s A1 3(b)(ii) F = 4.2 / 3.0 or 0.55 × 7.6 / 3.0 = 1.4 N A1 3(c)(i) F = 1.4 N A1 3(c)(ii) Newton’s third law (of motion) B1 3(d) 2 × 1.4 = m × 9.81 m = 0.29 kg A1 3(e) the density of air is less at high altitude B1 3(f) fo = fsv / (v – vs) = 3000 × 340 / (340 – 22) C1 = 3200 Hz A1

This question in 9702/21 Oct/Nov 2019

3 (a) State Newton’s third law of motion. … … … [2] (b) A block X of mass mX slides in a straight line along a horizontal frictionless surface, as shown in Fig. 3.1. speed 5v speed v mass mX mass mY X Y X Y Fig. 3.1 Fig. 3.2 The block X, moving with speed 5v, collides head-on with a stationary block Y of mass mY. The two blocks stick together and then move with common speed v, as shown in Fig. 3.2. mY (i) Use conservation of momentum to show that the ratio is equal to 4. mx [2] (ii) Calculate the ratio total kinetic energy of X and Y after collision . total kinetic energy of X and Y before collision ratio = … [3] (iii) State the value of the ratio in (ii) for a perfectly elastic collision. ratio = … [1] (c) The variation with time t of the momentum of block X in (b) is shown in Fig. 3.3. momentum 0 0 10 20 30 40 50 60 t / ms Fig. 3.3 Block X makes contact with block Y at time t = 20 ms. (i) Describe, qualitatively, the magnitude and direction of the resultant force, if any, acting on block X in the time interval: 1. t = 0 to t = 20 ms … 2. t = 20 ms to t = 40 ms. … … [3] (ii) On Fig. 3.3, sketch the variation of the momentum of block Y with time t from t = 0 to t = 60 ms. [3] [Total: 14]

14 marks

Mark scheme: 3(a) force on body A (by body B) is equal (in magnitude) to force on body B (by body A) B1 force on body A (by body B) is opposite (in direction) to force on body B (by body A) B1 3(b)(i) mX × 5v or (mX + mY) × v C1 mX × 5v = (mX + mY) × v (so) mY / mX = 4 A1 3(b)(ii) (E =) ½mv2 C1 ratio = [½ × (mX + mY) × v2] / [½ × mX × (5v)2] C1 ratio = 0.2 A1 3(b)(iii) ratio = 1 A1 3(c)(i) 1. (magnitude of resultant force is) zero B1 2. (magnitude of resultant force is) constant B1 (direction of resultant force is) opposite to the momentum B1 3(c)(ii) horizontal line from (0 ms, 0 squares) ending at (20 ms, 0 squares) B1 straight line from (20 ms, 0 squares) ending at (40 ms, 4.0 squares [= 4.0 cm vertically]) B1 horizontal line from (40 ms, 4.0 squares) ending at (60 ms, 4.0 squares) B1

This question in 9702/22 Oct/Nov 2019

4 (a) A sphere in a liquid accelerates vertically downwards from rest. For the viscous force acting on the moving sphere, state: (i) the direction … [1] (ii) the variation, if any, in the magnitude. … [1] (b) A man of weight 750 N stands a distance of 3.6 m from end D of a horizontal uniform beam AD, as shown in Fig. 4.1. FB FC A B C D 2.0 m 2.0 m 380 N 750 N 3.6 m 9.0 m Fig. 4.1 (not to scale) The beam has a weight of 380 N and a length of 9.0 m. The beam is supported by a vertical force FB at pivot B and a vertical force FC at pivot C. Pivot B is a distance of 2.0 m from end A and pivot C is a distance of 2.0 m from end D. The beam is in equilibrium. (i) State the principle of moments. … … … [2] (ii) By using moments about pivot C, calculate FB. FB = … N [2] (iii) The man walks towards end D. The beam is about to tip when FB becomes zero. Determine the minimum distance x from end D that the man can stand without tipping the beam. x = … m [2] [Total: 8]

8 marks

Mark scheme: 4(a)(i) (vertically) upwards/up B1 4(a)(ii) increases (with time/velocity/depth) B1 4(b)(i) for a body in (rotational) equilibrium B1 sum/total of clockwise moments about a point = sum/total of anticlockwise moments about the (same) point B1 4(b)(ii) (FB × 5.0) or (380 × 2.5) or (750 × 1.6) C1 (FB × 5.0) = (380 × 2.5) + (750 × 1.6) FB = 430 N A1 4(b)(iii) taking moments about C: (380 × 2.5) = 750 × (2.0 – x) C1 (2.0 – x) = 1.3 x = 0.7 m A1 or moments may be taken about other points, e.g. about D: (380 × 4.5) + (750 × x) = 1130 × 2.0 (C1) x = 0.7 m (A1)

This question in 9702/22 Oct/Nov 2019

2 (a) State Newton’s second law of motion. … … [1] (b) A delivery company suggests using a remote-controlled aircraft to drop a parcel into the garden of a customer. When the aircraft is vertically above point P on the ground, it releases the parcel with a velocity that is horizontal and of magnitude 5.4 m s–1. The path of the parcel is shown in Fig. 2.1. 5.4 m s–1 X parcel path of parcel h P Q horizontal ground d Fig. 2.1 (not to scale) The parcel takes a time of 0.81 s after its release to reach point Q on the horizontal ground. Assume air resistance is negligible. (i) On Fig. 2.1, draw an arrow from point X to show the direction of the acceleration of the parcel when it is at that point. [1] (ii) Determine the height h of the parcel above the ground when it is released. h = … m [2] (iii) Calculate the horizontal distance d between points P and Q. d = … m [1] (c) Another parcel is accidentally released from rest by a different aircraft when it is hovering at a great height above the ground. Air resistance is now significant. (i) On Fig. 2.2, draw arrows to show the directions of the forces acting on the parcel as it falls vertically downwards. Label each arrow with the name of the force. parcel velocity Fig. 2.2 [2] (ii) By considering the forces acting on the parcel, state and explain the variation, if any, of the acceleration of the parcel as it moves downwards before it reaches constant (terminal) speed. … … … … … … [3] (iii) Describe the energy conversion that occurs when the parcel is falling through the air at constant (terminal) speed. … … [1] [Total: 11]

11 marks

Mark scheme: 2(a) (resultant) force proportional to rate of change of momentum B1 2(b)(i) arrow drawn vertically downwards from point X B1 2(b)(ii) s = ut + ½at2 h = ½ × 9.81 × 0.812 C1 = 3.2 m A1 2(b)(iii) d = 5.4 × 0.81 = 4.4 m A1 2(c)(i) downward pointing arrow labelled weight B1 upward pointing arrow labelled air resistance B1 2(c)(ii) air resistance increases B1 weight constant or resultant force decreases B1 (so) acceleration decreases B1 2(c)(iii) gravitational potential energy to thermal/internal energy B1

This question in 9702/21 May/June 2020

1 (a) Define velocity. … … [1] (b) The drag force FD acting on a car moving with speed v along a straight horizontal road is given by FD = v 2Ak where k is a constant and A is the cross-sectional area of the car. Determine the SI base units of k. SI base units … [2] (c) The value of k, in SI base units, for the car in (b) is 0.24. The cross-sectional area A of the car is 5.1 m2. The car is travelling with a constant speed along a straight road and the output power of the engine is 4.8 × 104 W. Assume that the output power of the engine is equal to the rate at which the drag force FD is doing work against the car. Determine the speed of the car. speed = … m s–1 [3] [Total: 6]

6 marks

Mark scheme: 1(a) (velocity =) change in displacement / time (taken) B1 1(b) units of F: kg m s–2 C1 units of k: kg m s–2 / [m2 × (m s–1)2] = kg m–3 A1 1(c) P = Fv C1 4.8 × 104 = 0.24 × 5.1 × v 3 C1 v = 34 m s–1 A1

This question in 9702/22 May/June 2020

2 (a) State Newton’s first law of motion. … … [1] (b) A skier is pulled in a straight line along horizontal ground by a wire attached to a kite, as shown in Fig. 2.1. kite wire skier mass 89 kg 28° horizontal ground Fig. 2.1 (not to scale) The mass of the skier is 89 kg. The wire is at an angle of 28° to the horizontal. The variation with time t of the velocity v of the skier is shown in Fig. 2.2. 5.0 4.0 v / m s–1 3.0 2.0 1.0 0 0 1.0 2.0 3.0 4.0 5.0 t / s Fig. 2.2 (i) Use Fig. 2.2 to determine the distance moved by the skier from time t = 0 to t = 5.0 s. distance = … m [2] (ii) Use Fig. 2.2 to show that the acceleration a of the skier is 0.80 m s–2 at time t = 2.0 s. [2] (iii) The tension in the wire at time t = 2.0 s is 240 N. Calculate: 1. the horizontal component of the tension force acting on the skier horizontal component of force = … N [1] 2. the total resistive force R acting on the skier in the horizontal direction. R = … N [2] (iv) The skier is now lifted upwards by a gust of wind. For a few seconds the skier moves horizontally through the air with the wire at an angle of 45° to the horizontal, as shown in Fig. 2.3. 45° horizontal Fig. 2.3 (not to scale) By considering the vertical components of the forces acting on the skier, determine the new tension in the wire when the skier is moving horizontally through the air. tension = … N [2] [Total: 10]

10 marks

Mark scheme: 2(a) a body continues at (rest or) constant velocity unless acted upon by a resultant force B1 2(b)(i) distance = [½ × (2.0 + 4.4) × 3.0] + [4.4 × 2.0] C1 = 9.6 + 8.8 = 18 m A1 2(b)(ii) a = (v – u) / t or gradient or Δv / (Δ)t C1 e.g. a = (4.4 – 2.0) / 3.0 = 0.80 m s–2 A1 2(b)(iii) 1. force = 240 cos 28° or 240 sin 62° = 210 N A1 2. resultant force = 89 × 0.80 (= 71.2 N) C1 R = 210 – 71 = 140 N A1 2(b)(iv) T sin 45° = mg C1 T = (89 × 9.81) / sin 45° = 1200 N A1

This question in 9702/23 May/June 2020

2 A small block is lifted vertically upwards by a toy aircraft, as illustrated in Fig. 2.1. aircraft string velocity block Fig. 2.1 As the block is moving upwards, the string breaks at time t = 0. The block initially continues moving upwards and then falls and hits the ground at time t = 0.90 s. The variation with time t of the velocity v of the block is shown in Fig. 2.2. 1.96 v / m s–1 0 0 0.20 0.900.90 t / s –6.86 Fig. 2.2 Air resistance is negligible. (a) State the feature of the graph in Fig. 2.2 that shows the block has a constant acceleration. … [1] (b) Use Fig. 2.2 to determine the height of the block above the ground when the string breaks at time t = 0. height = … m [3] (c) The block has a weight of 0.86 N. Calculate the difference in gravitational potential energy of the block between time t = 0 and time t = 0.90 s. difference in gravitational potential energy = … J [2] (d) On Fig. 2.3, sketch a line to show the variation of the distance moved by the block with time t from t = 0 to t = 0.20 s. Numerical values of distance are not required. distance moved 0 0 0.20 t / s Fig. 2.3 [2] (e) A block of greater mass is now released from the same height with the same upward velocity. Air resistance is still negligible. State and explain the effect, if any, of the increased mass on the speed with which the block hits the ground. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) constant gradient B1 2(b) (displacement until 0.20 s =) ½ × 1.96 × 0.20 (= 0.196 m) or (displacement after 0.20 s =) ½ × 6.86 × 0.70 (= 2.401 m) C1 height = 2.401 – 0.196 C1 = 2.2 m (alternative methods are possible using equations of uniformly accelerated motion) A1 2(c) (Δ)E = mg(Δ)h or W(Δ)h C1 (Δ)E = 0.86 × 2.2 = 1.9 J A1 2(d) curved line from the origin M1 gradient of curved line decreases and is zero at t = 0.20 s only A1 2(e) acceleration (of free fall) is unchanged/is not dependent on mass and (so) no effect B1

This question in 9702/21 Oct/Nov 2020

3 (a) Define force. … … [1] (b) A ball falls vertically downwards towards a horizontal floor and then rebounds along its original path, as illustrated in Fig. 3.1. ball reaching ball leaving speed the floor speed the floor 3.8 m s–1 1.7 m s–1 Fig. 3.1 The ball reaches the floor with speed 3.8 m s–1. The ball is then in contact with the floor for a time of 0.081 s before leaving it with speed 1.7 m s–1. The mass of the ball is 0.062 kg. (i) Calculate the loss of kinetic energy of the ball during the collision. loss of kinetic energy = … J [2] (ii) Determine the magnitude of the change in momentum of the ball during the collision. change in momentum = … N s [2] (iii) Show that the magnitude of the average resultant force acting on the ball during the collision is 4.2 N. [1] (iv) Use the information in (iii) to calculate the magnitude of: 1. the average force of the floor on the ball during the collision average force = … N 2. the average force of the ball on the floor during the collision. average force = … N [2] [Total: 8]

8 marks

Mark scheme: 3(a) (force =) rate of change of momentum B1 3(b)(i) E = ½mv2 or ½ × 0.062 × 3.82 or ½ × 0.062 × 1.72 C1 loss of KE = ½ × 0.062 × (3.82 – 1.72) = 0.36 J A1 3(b)(ii) p = mv or 0.062 × 3.8 or 0.062 × 1.7 C1 change in momentum = 0.062 × (1.7 + 3.8) = 0.34 N s A1 3(b)(iii) (average resultant force =) 0.34 / 0.081 = 4.2 (N) or (average resultant force =) 0.062 × (1.7 + 3.8) / 0.081 = 4.2 (N) A1 3(b)(iv) 1. average force = 4.2 + (0.062 × 9.81) = 4.8 N A1 2. average force = 4.8 N A1

This question in 9702/21 Oct/Nov 2020

2 (a) A cylinder is suspended from the end of a string. The cylinder is stationary in water with the axis of the cylinder vertical, as shown in Fig. 2.1. cylinder string weight 0.84 N water h density 1.0 × 103 kg m–3 0.031 m Fig. 2.1 (not to scale) The cylinder has weight 0.84 N, height h and a circular cross-section of diameter 0.031 m. The density of the water is 1.0 × 103 kg m−3. The difference between the pressures on the top and bottom faces of the cylinder is 520 Pa. (i) Calculate the height h of the cylinder. h = … m [2] (ii) Show that the upthrust acting on the cylinder is 0.39 N. [2] (iii) Calculate the tension T in the string. T = … N [1] (b) The string is now used to move the cylinder in (a) vertically upwards through the water. The variation with time t of the velocity v of the cylinder is shown in Fig. 2.2. 12.5 10.0 v / cm s–1 7.5 5.0 2.5 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 t / s Fig. 2.2 (i) Use Fig. 2.2 to determine the acceleration of the cylinder at time t = 2.0 s. acceleration = … m s−2 [2] (ii) The top face of the cylinder is at a depth of 0.32 m below the surface of the water at time t = 0. Use Fig. 2.2 to determine the depth of the top face below the surface of the water at time t = 4.0 s. depth = … m [2] (c) The cylinder in (b) is released from the string at time t = 4.0 s. The cylinder falls, from rest, vertically downwards through the water. Assume that the upthrust acting on the cylinder remains constant as it falls. (i) State the name of the force that acts on the cylinder when it is moving and does not act on the cylinder when it is stationary. … [1] (ii) State and explain the variation, if any, of the acceleration of the cylinder as it falls downwards through the water. … … … [2] [Total: 12]

12 marks

Mark scheme: 2(a)(i) (Δ)p = ρg(Δ)h 520 = 1000 × 9.81 × h C1 h = 0.053 m A1 2(a)(ii) (upthrust =) (Δ)p × A = (Δ)p × π(d / 2)2 or (Δ)p × πr 2 C1 = 520 × π(0.031 / 2)2 = 0.39 (N) A1 2(a)(iii) T = 0.84 – 0.39 = 0.45 N A1 Question Answer Marks 2(b)(i) a = (v – u) / t or (Δ)v / (∆)t or gradient C1 = e.g. 8.0 × 10–2 / 2.0 = 4.0 × 10–2 m s–2 A1 2(b)(ii) distance = (½ × 2.5 × 0.10) + (½ × 1.5 × 0.10) or (½ × 4.0 × 0.10) (= 0.20 (m)) C1 depth = 0.32 – 0.20 = 0.12 m A1 2(c)(i) viscous (force) B1 2(c)(ii) viscous force increases (with speed/time/depth) B1 (so) acceleration decreases B1

This question in 9702/22 Oct/Nov 2020

4 A rigid plank is used to make a ramp between two different horizontal levels of ground, as shown in Fig. 4.1. 45 N 1.10 m B 0.30 m D C 1.50 m rope 38° T 96 N A Fig. 4.1 (not to scale) Point A at one end of the plank rests on the lower level of the ground. A force acts on, and is perpendicular to, the plank at point B. The plank is held in equilibrium by a rope that connects point D on the plank to the ground. The plank has a weight that may be considered to act from its centre of gravity C. The rope is perpendicular to the plank and has tension T. The plank is at an angle of 38° to the vertical. The forces and the distances along the plank of points A, B, C and D are shown in Fig. 4.1. (a) Show that the component of the weight that is perpendicular to the plank is 59 N. [1] (b) By taking moments about end A of the plank, calculate the tension T. T = … N [3] [Total: 4]

4 marks

Mark scheme: 4(a) (component =) 96 sin 38° = 59 (N) or 96 cos 52° = 59 (N) 4(b) (45 × 2.9) or (T × 1.8) or (59 × 1.5) C1 (45 × 2.9) = (T × 1.8) + (59 × 1.5) C1 T = 23 N A1

This question in 9702/22 Oct/Nov 2020

1 (a) Complete Table 1.1 by stating whether each of the quantities is a vector or a scalar. Table 1.1 quantity vector or scalar acceleration power work [2] (b) The variation with time t of the velocity v of an object is shown in Fig. 1.1. 1.50 1.25 1.00 v / m s–1 0.75 0.50 0.25 0 0 2.0 4.0 6.0 8.0 10.0 12.0 t / s Fig. 1.1 (i) Determine the acceleration of the object from time t = 0 to time t = 4.0 s. acceleration = … m s−2 [2] (ii) Determine the distance moved by the object from time t = 0 to time t = 4.0 s. distance = … m [2] (c) (i) Define force. … … [1] (ii) The motion represented in Fig. 1.1 is caused by a resultant force F acting on the object. On Fig. 1.2, sketch the variation of F with time t from t = 0 to t = 12.0 s. Numerical values of F are not required. F 0 00 2.02.0 4.04.0 6.06.0 8.08.0 10.010.0 12.012.0 tt // ss Fig. 1.2 [3] [Total: 10]

10 marks

Mark scheme: 1(a) acceleration: vector work: scalar power: scalar Three correct scores 2 marks. Two correct scores 1 mark. B2 1(b)(i) a = (v – u) / t or a = gradient or a = Δv / (Δ)t e.g. a = (1.40 – 0.70) / 4.0 C1 = 0.18 m s–2 A1 1(b)(ii) distance = 0.5 × (0.70 + 1.40) × 4.0 or (0.70 × 4.0) + (0.5 × 0.70 × 4.0) C1 = 4.2 m A1 1(c)(i) (force equal to) rate of change of momentum B1 1(c)(ii) horizontal line starting from t = 0 and ending at t = 4.0 s at a positive value of F B1 horizontal line starting from t = 4.0 s and ending at t = 8.0 s at F = 0 B1 horizontal line starting from t = 8.0 s and ending at t = 12.0 s at a negative value of F and the magnitude of F is larger than from t = 0 to 4.0 s B1

This question in 9702/22 Feb/March 2021

2 (a) State what is meant by work done. … … [1] (b) A beach ball is released from a balcony at the top of a tall building. The ball falls vertically from rest and reaches a constant (terminal) velocity. The gravitational potential energy of the ball decreases by 60 J as it falls from the balcony to the ground. The ball hits the ground with speed 16 m s−1 and kinetic energy 23 J. (i) Show that the mass of the ball is 0.18 kg. [2] (ii) Calculate the height of the balcony above the ground. height = … m [2] (iii) Determine the average resistive force acting on the ball as it falls from the balcony to the ground. average resistive force = … N [2] (c) State and explain the variation, if any, in the magnitude of the acceleration of the ball in (b) during the time interval when the ball is moving downwards before it reaches constant (terminal) velocity. … … … … … … [3] [Total: 10]

10 marks

Mark scheme: 2(a) force × displacement in the direction of the force B1 2(b)(i) E = ½mv 2 C1 (m =) 23 × 2 / 162 = 0.18 (kg) A1 2(b)(ii) (Δ)E = mg(Δ)h 60 = 0.18 × 9.81 × h C1 h = 34 m A1 2(b)(iii) (work done =) 60 – 23 = 37 (J) C1 average resistive force = 37 / 34 = 1.1 N A1 2(c) air resistance (acting on ball) increases B1 resultant force (on ball) decreases or weight constant and air resistance increases B1 acceleration decreases B1

This question in 9702/22 Feb/March 2021

2 A person uses a trolley to move suitcases at an airport. The total mass of the trolley and suitcases is 72 kg. (a) The person pushes the trolley and suitcases along a horizontal surface with a constant speed of 1.4 m s–1 and then releases the trolley. The released trolley moves in a straight line and comes to rest. Assume that a constant total resistive force of 18 N opposes the motion of the trolley and suitcases. (i) Calculate the power required to overcome the total resistive force on the trolley and suitcases when they move with a constant speed of 1.4 m s–1. power = … W [2] (ii) Calculate the time taken for the trolley to come to rest after it is released. time = … s [3] (b) At another place in the airport, the trolley and suitcases are on a slope, as shown in Fig. 2.1. trolley and suitcases 18 N F, 54 N X slope 9.5 m Y Fig. 2.1 (not to scale) The person releases the trolley from rest at point X. The trolley moves down the slope in a straight line towards point Y. The distance along the slope between points X and Y is 9.5 m. The component F of the weight of the trolley and suitcases that acts along the slope is 54 N. Assume that a constant total resistive force of 18 N opposes the motion of the trolley and suitcases. (i) Calculate the speed of the trolley at point Y. speed = … m s–1 [3] (ii) Calculate the work done by F for the movement of the trolley from X to Y. work done = … J [1] (iii) The trolley is released at point X at time t = 0. On Fig. 2.2, sketch a graph to show the variation with time t of the work done by F for the movement of the trolley from X to Y. Numerical values of the work done and t are not required. work done 0 0 t Fig. 2.2 [2] (c) The angle of the slope in (b) is constant. The frictional forces acting on the wheels of the moving trolley are also constant. Explain why, in practice, it is incorrect to assume that the total resistive force opposing the motion of the trolley and suitcases is constant as the trolley moves between X and Y. … … [1] [Total: 12]

12 marks

Mark scheme: 2(a)(i) P = Fv C1 = 18 × 1.4 = 25 W A1 2(a)(ii) a = F / m C1 a = 18 / 72 = 0.25 (m s–2) t = 1.4 / 0.25 C1 = 5.6 s A1 2(b)(i) a = (54 – 18) / 72 or 36 / 72 (= 0.50 m s–2) C1 v 2 = 2 × 0.50 × 9.5 C1 v = 3.1 m s–1 A1 2(b)(ii) W = 54 × 9.5 = 510 J A1 2(b)(iii) curved line from the origin M1 gradient of line increases A1 2(c) (force due to) air resistance increases/changes/not constant or air resistance increases with speed B1

This question in 9702/21 May/June 2021

3 A pendulum consists of a solid sphere suspended by a string from a fixed point P, as shown in Fig. 3.1. P θ string 0.93 m Y sphere h X momentum 0.72 N s Fig. 3.1 (not to scale) The sphere swings from side to side. At one instant the sphere is at its lowest position X, where it has kinetic energy 0.86 J and momentum 0.72 N s in a horizontal direction. A short time later the sphere is at position Y, where it is momentarily stationary at a maximum vertical height h above position X. The string has a fixed length and negligible weight. Air resistance is also negligible. (a) On Fig. 3.1, draw a solid line to represent the displacement of the centre of the sphere at position Y from position X. [1] (b) Show that the mass of the sphere is 0.30 kg. [3] (c) Calculate height h. h = … m [2] (d) The distance between point P and the centre of the sphere is 0.93 m. When the sphere is at position Y, the string is at an angle θ to the vertical. Show that θ is 47°. [1] (e) For the sphere at position Y, calculate the moment of its weight about point P. moment = … N m [2] (f) State and explain whether the sphere is in equilibrium when it is stationary at position Y. … … [1] [Total: 10]

10 marks

Mark scheme: 3(a) solid straight line drawn between centre of sphere at X and at Y B1 3(b) p = mv or 0.72 = mv C1 E = ½mv 2 or 0.86 = ½mv 2 C1 (m =) 0.722 / (2 × 0.86) = 0.30 (kg) or v = 2EK / p v = (0.86 × 2) / 0.72 = 2.4 (to 2 s.f.) m = 0.72 / 2.4 = 0.30 (kg) A1 3(c) (Δ)E = mg(Δ)h C1 h = 0.86 / (0.30 × 9.81) = 0.29 m A1 3(d) cosθ = (0.93 – 0.29) / 0.93 so θ = 47° A1 3(e) moment = (0.30 × 9.81) × 0.93 × (sin 47° or cos 43°) or moment = (0.30 × 9.81) × [0.932 – (0.93 – 0.29)2]0.5 C1 = 2.0 N m A1 Question Answer Mark 3(f) there is a resultant force (acting on sphere) or there is a resultant moment (about P acting on pendulum) (so) not in equilibrium B1

This question in 9702/21 May/June 2021

2 A ball is thrown vertically downwards to the ground, as illustrated in Fig. 2.1. ball speed u path of ball 1.5 m speed 8.7 m s–1 ground Fig. 2.1 The ball is thrown with speed u from a height of 1.5 m. The ball then hits the ground with speed 8.7 m s–1. Assume that air resistance is negligible. (a) Calculate speed u. u = … m s–1 [2] (b) State how Newton’s third law applies to the collision between the ball and the ground. … … … … [2] (c) The ball is in contact with the ground for a time of 0.091 s. The ball rebounds vertically and leaves the ground with speed 5.4 m s–1. The mass of the ball is 0.059 kg. (i) Calculate the magnitude of the change in momentum of the ball during the collision. change in momentum = … N s [2] (ii) Determine the magnitude of the average resultant force that acts on the ball during the collision. average resultant force = … N [1] (iii) Use your answer in (c)(ii) to calculate the magnitude of the average force exerted by the ground on the ball during the collision. average force = … N [2] (d) The ball was thrown downwards at time t = 0 and hits the ground at time t = T. On Fig. 2.2, sketch a graph to show the variation of the speed of the ball with time t from t = 0 to t = T. Numerical values are not required. speed 0 0 T t Fig. 2.2 [1] (e) In practice, air resistance is not negligible. State and explain the variation, if any, with time t of the gradient of the graph in (d) when air resistance is not negligible. … … … … [2] [Total: 12]

12 marks

Mark scheme: 2(a) v2 = u2 + 2as u2 = 8.72 – (2 × 9.81 × 1.5) C1 u = 6.8 m s–1 A1 2(b) (magnitude of) force on ball (by ground) equal to force on ground (by ball) B1 (direction of) force on ball (by ground) opposite to force on ground (by ball) B1 2(c)(i) (p = ) 0.059 × 8.7 or 0.059 × 5.4 C1 change in momentum = 0.059 (8.7 + 5.4) = 0.83 N s A1 2(c)(ii) resultant force = 0.83 / 0.091 or 0.059 [(8.7 + 5.4) / 0.091] = 9.1 N A1 2(c)(iii) (W =) 0.059 × 9.81 C1 (W =) 0.58 (N) force = 9.1 + 0.58 = 9.7 N A1 2(d) straight line with a positive gradient and starting from a non-zero value of speed at t = 0 and ending when t = T B1 2(e) air resistance increases B1 resultant force/acceleration decreases so gradient (of curve) decreases B1

This question in 9702/22 May/June 2021

2 (a) Define acceleration. … … [1] (b) A stone falls vertically from the top of a cliff. Fig. 2.1 shows the variation with time t of the velocity v of the stone. 40 v / m s–1 30 20 10 0 0 5 10 15 20 25 30 t / s Fig. 2.1 (i) Explain, with reference to forces acting on the stone, the shape of the curve in Fig. 2.1. … … … … … [3] (ii) Use Fig. 2.1 to determine the speed of the stone when the resultant force on it is zero. speed = … m s–1 [1] (iii) Use Fig. 2.1 to calculate the approximate height through which the stone falls between t = 0 and t = 30 s. height = … m [3] (iv) On Fig. 2.2, sketch the variation with t of the acceleration a of the stone between t = 0 and t = 30 s. 20 a / m s–2 15 10 5 0 0 5 10 15 20 25 30 t / s Fig. 2.2 [3] [Total: 11]

11 marks

Mark scheme: 2(a) change in velocity / time (taken) B1 2(b)(i) air resistance increases (with speed/with time) B1 resultant force decreases (as speed increases/with time) so acceleration decreases (as speed increases/with time) B1 when air resistance equals the weight the speed/velocity/v becomes constant B1 2(b)(ii) speed = 36 m s–1 A1 2(b)(iii) height given by area under the curve C1 height = 950 m Round to two significant figures and award 2 marks for a value in the range 920–980 m and 1 mark for a value in the range 900–910 m or 990–1000 m. A2 2(b)(iv) line starting at (0, 9.8) B1 curve with negative gradient between t = 0 and t = 20 s B1 line showing zero acceleration between t = 20 s and t = 30 s B1 Question Answer Marks

This question in 9702/23 May/June 2021

2 (a) Define momentum. … … [1] (b) Two balls X and Y, of equal diameter but different masses 0.24 kg and 0.12 kg respectively, slide towards each other on a frictionless horizontal surface, as shown in Fig. 2.1. mass 0.24 kg mass 0.12 kg X Y 2.3 m s–1 2.3 m s–1 frictionless surface Fig. 2.1 Both balls have initial speed 2.3 m s–1 before they collide with each other. Fig. 2.2 shows the variation with time t of the force FY exerted on ball Y by ball X during the collision. 400 FY / N 200 0 0 1 2 3 4 5 t / ms –200 – 400 Fig. 2.2 (i) Calculate the kinetic energy of ball X before the collision. kinetic energy = … J [3] (ii) The area enclosed by the lines and the time axis in Fig. 2.2 represents the change in momentum of ball Y during the collision. Determine the magnitude of the change in momentum of ball Y. change in momentum = … N s [2] (iii) Calculate the magnitude of the velocity of ball Y after the collision. velocity = … m s–1 [2] (c) On Fig. 2.3, sketch the variation with time t of the force FX exerted on ball X by ball Y during the collision in (b). 400 FX / N 200 0 0 1 2 3 4 5 t / ms –200 – 400 Fig. 2.3 [3] [Total: 11]

11 marks

Mark scheme: 2(a) mass × velocity B1 2(b)(i) kinetic energy = ½mv2 C1 = ½ × 0.24 × 2.32 C1 = 0.63 J A1 2(b)(ii) change in momentum = ½ × 240 × 5.0 × 10–3 C1 = 0.60 N s A1 2(b)(iii) (change in velocity of Y) = 0.60 / 0.12 ( = 5.0 m s–1) C1 final velocity of Y = 5.0 – 2.3 = 2.7 m s–1 A1 or (final momentum of Y) = 0.60 – 0.12 × 2.3 ( = 0.324 N s) (C1) final velocity of Y = 0.324 / 0.12 = 2.7 m s–1 (A1) 2(c) sloping straight line from (0, 0) to t = 3.0 ms and another straight line continuous with the first from t = 3.0 ms to (5.0, 0) B1 lines showing maximum force of magnitude 240 N B1 lines wholly in the negative F region of the graph B1

This question in 9702/21 Oct/Nov 2021

4 An α-particle moves in a straight line through a vacuum with a constant speed of 4.1 × 106 m s–1. The α-particle enters a uniform electric field at point A, as shown in Fig. 4.1. uniform electric field α-particle, A B speed 4.1 × 106 m s–1 Fig. 4.1 The α-particle continues to move in the same straight line until it is brought to rest at point B by the electric field. The deceleration of the α-particle by the electric field is 2.7 × 1014 m s–2. (a) State the direction of the electric field. … [1] (b) Calculate the distance AB. distance = … m [2] (c) Calculate the electric field strength. electric field strength = … V m–1 [3] (d) The α-particle is at point A at time t = 0. On Fig. 4.2, sketch the variation with time t of the momentum of the α-particle as it travels from point A to point B. Numerical values are not required. momentum 00 t Fig. 4.2 [1] (e) State the name of the quantity that is represented by the gradient of the graph in (d). … [1] (f) A β– particle now enters the electric field along the same initial path as the α-particle and with the same initial speed of 4.1 × 106 m s–1. (i) Calculate the kinetic energy, in J, of the β– particle at point A. kinetic energy = … J [3] (ii) State and explain the differences between the electric force on the β– particle in the electric field and the electric force on the α-particle in the electric field. … … … … … [3] (iii) The β– particle is produced by the decay of a nucleus. State the name of another lepton that is produced at the same time as the β– particle. … [1] [Total: 15]

15 marks

Mark scheme: 4(a) to the left/from the right/from B to A/opposite (direction) to (α-particle) velocity B1 4(b) v2 = u2 + 2as s = (4.1 × 106)2 / (2 × 2.7 × 1014) C1 = 0.031 m A1 4(c) E = F / Q or E = ma / Q C1 = (4 × 1.66 × 10–27 × 2.7 × 1014) / (2 × 1.60 × 10–19) C1 = 5.6 × 106 V m–1 A1 4(d) straight line with negative gradient that intercepts both the momentum and t axes B1 4(e) force (on α-particle) B1 4(f)(i) E = ½mv2 C1 = ½ × 9.11 × 10–31 × (4.1 × 106)2 C1 = 7.7 × 10–18 J A1 4(f)(ii) particles have opposite charges B1 (so) forces (on charges) are opposite (directions) B1 β– has less/half the charge so less/half the force B1 4(f)(iii) (electron) antineutrino B1

This question in 9702/23 Oct/Nov 2021

1 A sphere of radius 2.1 mm falls with terminal (constant) velocity through a liquid, as shown in Fig. 1.1. sphere, radius 2.1 mm weight 7.2 × 10–4 N constant velocity downwards liquid, density ρ Fig. 1.1 Three forces act on the moving sphere. The weight of the sphere is 7.2 × 10–4 N and the upthrust acting on it is 4.8 × 10–4 N. The viscous force FV acting on the sphere is given by FV = krv where r is the radius of the sphere, v is its velocity and k is a constant. The value of k in SI units is 17. (a) Determine the SI base units of k. SI base units … [2] (b) Use the value of the upthrust acting on the sphere to calculate the density ρ of the liquid. ρ = … kg m–3 [3] (c) (i) On the sphere in Fig. 1.1, draw three arrows to show the directions of the weight W, the upthrust U and the viscous force FV. Label these arrows W, U and FV respectively. [1] (ii) Determine the magnitude of the terminal (constant) velocity of the sphere. velocity = … m s–1 [2] [Total: 8]

8 marks

Mark scheme: 1(a) Fv: kg m s–2 k: kg m s–2 / m × m s–1 = kg m–1 s–1 A1 1(b) F = ρgV V = 4 / 3 × π × (2.1 × 10–3)3 (=3.88 × 10–8 m3) C1 ρ = 4.8 × 10–4 / 9.81 × V C1 = 1300 kg m–3 A1 1(c)(i) W downwards, U upwards, Fv upwards B1 1(c)(ii) FV = 7.2 × 10–4 – 4.8 × 10–4 = 2.4 × 10–4 (N) C1 velocity = 2.4 × 10–4 / (17 × 2.1 × 10–3) = 6.7 × 10–3 m s–1 A1

This question in 9702/22 Feb/March 2022

3 A jet of water hits a vertical wall at right angles, as shown in Fig. 3.1. horizontal jet of water, vertical density 1.0 × 103 kg m–3 wall water runs velocity cross-sectional area down the wall 5.0 m s–1 1.5 × 10–4 m2 Fig. 3.1 (not to scale) The water hits the vertical wall with a velocity of 5.0 m s–1 in a horizontal direction. The cross-sectional area of the jet is 1.5 × 10–4 m2. The density of the water is 1.0 × 103 kg m–3. The water runs down the wall after hitting it. (a) Show that, over a time of 1.6 s, the mass of water hitting the wall is 1.2 kg. [2] (b) Calculate: (i) the decrease in the horizontal momentum of the mass of water in (a) due to hitting the wall decrease in momentum = … N s [1] (ii) the magnitude of the horizontal force exerted on the water by the wall. force = … N [1] (c) State and explain the magnitude of the horizontal force exerted on the wall by the water. … … [1] (d) Calculate the pressure exerted on the wall by the water. pressure = … Pa [2] [Total: 7]

7 marks

Mark scheme: 3(a) (m =) ρV C1 = 1.0 × 103 × 1.5 × 10–4 × 5.0 × 1.6 = 1.2 (kg) A1 3(b)(i) (∆)p = 1.2 × 5.0 = 6.0 N s A1 3(b)(ii) F = 6.0 / 1.6 or 1.2 × 5.0/1.6 = 3.8 N A1 3(c) Newton’s third law applies (so) 3.8 N. B1 3(d) p = F / A = 3.8 / 1.5 × 10–4 C1 = 2.5 × 104 Pa A1

This question in 9702/22 Feb/March 2022

1 (a) Define velocity. … … [1] (b) A rock of mass 7.5 kg is projected vertically upwards from the surface of a planet. The rock leaves the surface of the planet with a speed of 4.0 m s–1 at time t = 0. The variation with time t of the velocity v of the rock is shown in Fig. 1.1. 5 4 v / m s–1 3 2 1 0 0 1 2 3 4 t / s –1 –2 –3 Fig. 1.1 Assume that the planet does not have an atmosphere and that the viscous force acting on the rock is always zero. (i) Determine the height of the rock above the surface of the planet at time t = 4.0 s. height = … m [3] (ii) Determine the change in the momentum of the rock from time t = 0 to time t = 4.0 s. change in momentum = … N s [2] (iii) Determine the weight W of the rock on this planet. W = … N [2] (c) In practice, the planet in (b) does have an atmosphere that causes a viscous force to act on the moving rock. State and explain the variation, if any, in the resultant force acting on the rock as it moves vertically upwards. … … … … [2] [Total: 10]

10 marks

Mark scheme: 1(a) change in displacement / time (taken) B1 1(b)(i) (displacement =) area under graph C1 (at t = 4.0 s) v = (–) 2.4 C1 height = ½  2.5  4.0 – ½  1.5  2.4 = 3.2 m A1 1(b)(ii) change in momentum = 7.5 (–4.0 – 2.4) C1 = (–) 48 N s A1 1(b)(iii) W = ∆p / (∆)t or ∆mv / (∆)t C1 = 48 / 4.0 = 12 N A1 or W = ma or mg or m(v – u) / t (C1) = 7.5  1.6 or 7.5  (4 + 2.4) / 4.0 = 12 N (A1) 1(c) speed/velocity decreases so viscous force decreases B1 viscous force decreases (and weight constant) so resultant force decreases B1

This question in 9702/21 May/June 2022

2 A sphere is attached by a metal wire to the horizontal surface at the bottom of a river, as shown in Fig. 2.1. sphere direction of flow of water water wire horizontal 68° surface Fig. 2.1 (not to scale) The sphere is fully submerged and in equilibrium, with the wire at an angle of 68° to the horizontal surface. The weight of the sphere is 32 N. The upthrust acting on the sphere is 280 N. The density of the water is 1.0 × 103 kg m–3. Assume that the force on the sphere due to the water flow is in a horizontal direction. (a) By considering the components of force in the vertical direction, determine the tension in the wire. tension = … N [2] (b) For the sphere, calculate: (i) the volume volume = … m3 [1] (ii) the density. density = … kg m–3 [2] (c) The centre of the sphere is initially at a height of 6.2 m above the horizontal surface. The speed of the water then increases, causing the sphere to move to a different position. This movement of the sphere causes its gravitational potential energy to decrease by 77 J. Calculate the final height of the centre of the sphere above the horizontal surface. height = … m [3] (d) The extension of the wire increases when the sphere changes position as described in (c). The wire obeys Hooke’s law. (i) State a symbol equation that gives the relationship between the tension T in the wire and its extension x. Identify any other symbol that you use. … … [1] (ii) Before the sphere changed position, the initial elastic potential energy of the wire was 0.65 J. The change in position of the sphere causes the extension of the wire to double. Calculate the final elastic potential energy of the wire after the sphere has changed position. final elastic potential energy = … J [2] [Total: 11]

11 marks

Mark scheme: 2(a) T sin 68° + 32 = 280 C1 T = 270 N A1 2(b)(i) F = gV V = 280 / (1.0  103  9.81) = 0.029 m3 A1 2(b)(ii)  = (32 / 9.81) / 0.029 C1 = 110 kg m–3 A1 Question Answer Marks 2(c) (∆)E = mg(∆)h or (∆)E = W(∆)h C1 (∆)h = (–) 77 / 32 C1 (∆)h = (–) 2.4 final height = 6.2 – 2.4 = 3.8 m A1 2(d)(i) T = kx where k is a constant or T = (EA / L)x where A is (cross-sectional) area, E is Young modulus, L is (original/unstretched) length B1 2(d)(ii)  2 1 2 E kx or 1 2 E Fx and F = kx C1 E = 0.65  22 = 2.6 J A1 or 1 2 E Fx    1 0.65 270 2 x and so x = 4.8  10–3 m k = F / x = 270 / 4.8  10–3 = 5.6  104 (C1) Question Answer Marks 2(d)(ii) final      3 1 540 9.6 10 2 E or E =          3 1 270 2 4.8 10 2 2 or E =        2 4 3 1 5.6 10 9.6 10 2 = 2.6 J (A1)

This question in 9702/22 May/June 2022

3 A man standing on a wall throws a small ball vertically upwards with a velocity of 5.6 m s–1. The ball leaves his hand when it is at a height of 3.1 m above the ground, as shown in Fig. 3.1. ball velocity 5.6 m s–1 man 3.1 m wall ground Fig. 3.1 (not to scale) Assume that air resistance is negligible. (a) Show that the ball reaches a maximum height above the ground of 4.7 m. [2] (b) The man does not catch the ball as it falls. Calculate the time taken for the ball to fall from its maximum height to the ground. time taken = … s [2] (c) The ball leaves the man’s hand at time t = 0 and hits the ground at time t = T. On Fig. 3.2, sketch a graph to show the variation of the velocity v of the ball with time t from t = 0 to t = T. Numerical values of v and t are not required. Assume that v is positive in the upward direction. v 0 0 T t Fig. 3.2 [3] (d) State what is represented by the gradient of the graph in (c). … [1] (e) The man now throws a second ball with the same velocity and from the same height as the first ball. The mass of the second ball is greater than that of the first ball. Assume that air resistance is still negligible. For the first and second balls, compare: (i) the magnitudes of their accelerations … [1] (ii) the speeds with which they hit the ground. … [1] [Total: 10]

10 marks

Mark scheme: 3(a) v2 = u2 + 2as s = 5.62 / (2  9.81) (max height =) 3.1 + 5.62 / (2  9.81) = 4.7 (m) A1 3(b)   2 1 2 s ut at    2 1 4.7 9.81 2 t C1 t = 0.98 s A1 3(c) line drawn from a non-zero speed at t = 0 to a greater speed at t = T B1 a single sloping straight line drawn from t = 0 to t = T B1 line starts with a positive non-zero value of v and ends with a negative non-zero value of v B1 3(d) acceleration (of the ball) B1 3(e)(i) (magnitudes of accelerations are) equal / same B1 3(e)(ii) (speeds are) equal / same B1

This question in 9702/22 May/June 2022

4 (a) State the principle of conservation of momentum. … … … [2] (b) Two balls, X and Y, move along a horizontal frictionless surface, as shown from above in Fig. 4.1. 3.0 kg X 4.0 m s–1 θ 3.7 m s–1 A B A X Y B θ 4.8 m s–1 2.5 kg Y before collision after collision Fig. 4.1 (not to scale) Fig. 4.2 (not to scale) Ball X has a mass of 3.0 kg and a velocity of 4.0 m s–1 in a direction at angle θ to a line AB. Ball Y has a mass of 2.5 kg and a velocity of 4.8 m s–1 in a direction at angle θ to the line AB. The balls collide and stick together. After colliding, the balls have a velocity of 3.7 m s–1 along the line AB on the horizontal surface, as shown in Fig. 4.2. (i) By considering the components of the momenta along the line AB, calculate θ. θ = … ° [3] (ii) By calculation of kinetic energies, state and explain whether the collision of the balls is inelastic or perfectly elastic. … … [2] [Total: 7]

7 marks

Mark scheme: 4(a) sum/total momentum before = sum/total momentum after or sum/total momentum (of a system of objects) is constant M1 if no (resultant) external force/for a closed system A1 4(b)(i) (3.0  4.0  cos) or (2.5  4.8  cos) or (5.5  3.7) C1 (3.0  4.0  cos) + (2.5  4.8  cos) = (5.5  3.7) C1  = 32° A1 4(b)(ii) (initial EK = 1 2  3.0  4.02 + 1 2  2.5  4.82 =) 53 (J) or (final EK = 1 2  5.5  3.72 =) 38 (J) C1 values of initial EK and final EK both correct and inelastic stated A1

This question in 9702/22 May/June 2022

3 (a) Define velocity. … … [1] (b) A constant driving force of 2400 N acts on a car of mass 1200 kg. The car accelerates from rest in a straight line along a horizontal road. Assume that the resistive forces acting on the car are negligible. (i) Calculate the acceleration of the car. acceleration = … m s–2 [1] (ii) On Fig. 3.1, sketch a graph showing the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line A. 50 40 v / m s–1 30 20 10 0 0 4 8 12 16 20 t / s Fig. 3.1 [2] (c) In reality, a resistive force due to air resistance acts on the car in (b). This resistive force increases with speed until it becomes equal in magnitude to the driving force at time t = 12 s. (i) On Fig. 3.1, sketch a second line to show the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line B. [3] (ii) At time t = 20 s, the driving force is increased to 3000 N and remains constant at this value. Describe how the velocity of the car changes due to this increase in the driving force. … … … … [2] [Total: 9]

9 marks

Mark scheme: 3(a) (velocity =) change in displacement / time B1 3(b)(i) a = 2400 / 1200 = 2.0 m s−2 A1 3(b)(ii) straight line from the origin with positive gradient (labelled A) M1 ending at (20, 40) A1 3(c)(i) line starting at origin (with the same gradient as A) and beneath A at all points B1 gradient decreasing to zero B1 straight horizontal line from t = 12 s and ending at t = 20 s (and labelled B) B1 3(c)(ii) the velocity/speed will increase B1 to a new terminal/constant/maximum velocity/speed B1 or the car has an acceleration (B1) to a new (higher) terminal/constant/maximum velocity/speed (B1)

This question in 9702/23 May/June 2022

2 A spherical balloon is filled with a fixed mass of gas. A small block is connected by a string to the balloon, as shown in Fig. 2.1. balloon string block ground Fig. 2.1 (not to scale) The block is held on the ground by an external force so that the string is vertical. The density of the air surrounding the balloon is 1.2 kg m–3. The upthrust acting on the balloon is 0.071 N. The upthrust acting on the string and block is negligible. (a) By using Archimedes’ principle, calculate the radius r of the balloon. r = … m [2] (b) The total weight of the balloon, string and block is 0.053 N. The external force holding the block on the ground is removed so that the released block is lifted vertically upwards by the balloon. Calculate the acceleration of the block immediately after it is released. acceleration = … m s–2 [3] (c) The balloon continues to lift the block. The string breaks as the block is moving vertically upwards with a speed of 1.4 m s–1. After the string breaks, the detached block briefly continues moving upwards before falling vertically downwards to the ground. The block hits the ground with a speed of 3.6 m s–1. Assume that the air resistance on the block is negligible. (i) By considering the motion of the block after the string breaks, calculate the height of the block above the ground when the string breaks. height = … m [2] (ii) The string breaks at time t = 0 and the block hits the ground at time t = T. On Fig. 2.2, sketch a graph to show the variation of the velocity v of the block with time t from t = 0 to t = T. Numerical values of t are not required. Assume that v is positive in the upward direction. 4 3 v / m s–1 2 1 0 0 T t – 1 – 2 – 3 – 4 Fig. 2.2 [2] [Total: 9]

9 marks

Mark scheme: 2(a) F = gV C1 0.071 = 1.2  9.81  V (V = 6.03  10–3 m3) A1 4/3    r 3 = 6.03  10–3 r = 0.11 m 2(b) m = 0.053 / 9.81 C1 ( = 5.4  10–3 kg) F = 0.071 – 0.053 C1 ( = 0.018 N) a = (0.071 – 0.053) / (0.053 / 9.81) A1 = 3.3 m s–2 2(c)(i) v 2 = u 2 + 2as C1 3.62 = (–1.4)2 + 2  9.81  s or 3.62 = 1.42 + 2  9.81  s s = 0.56 m A1 2(c)(ii) single straight line from any positive non-zero value of v at t = 0 to any negative non-zero value of v at t = T B1 line starting at (0, 1.4) and ending at (T, –3.6) B1

This question in 9702/22 Oct/Nov 2022

4 A horizontal spring is fixed at one end. A block is pushed against the other end of the spring so that the spring is compressed, as shown in Fig. 4.1. compressed spring block frictionless surface Fig. 4.1 The block is released and accelerates along a horizontal frictionless surface as the spring returns to its original length. The block leaves the end of the spring with a speed of 2.3 m s–1, as shown in Fig. 4.2. speed 2.3 m s–1 spring block leaving the spring Fig. 4.2 The block has a mass of 250 g and the spring has a spring constant of 420 N m–1. Assume that the spring always obeys Hooke’s law and that all the elastic potential energy of the spring is transferred to the kinetic energy of the block. (a) Calculate the kinetic energy of the block as it leaves the spring. kinetic energy = … J [2] (b) Calculate the compression of the spring immediately before the block is released. compression = … m [2] (c) After leaving the spring, the block moves along the surface until it hits a barrier at a speed of 2.3 m s–1. The block then rebounds at a speed of 1.5 m s–1 and moves back along its original path. The block is in contact with the barrier for a time of 0.086 s. Calculate: (i) the change in momentum of the block during the collision change in momentum = … N s [2] (ii) the average resultant force exerted on the block during the collision. average resultant force = … N [1] (d) The maximum compression x of the spring is now varied in order to vary the kinetic energy EK of the block as it leaves the spring. Assume that all the elastic potential energy in the spring is always transferred to the kinetic energy of the block. On Fig. 4.3, sketch a graph to show the variation with x of EK. EK 00 x Fig. 4.3 [1] [Total: 8]

8 marks

Mark scheme: 4(a) E = ½mv2 C1 = ½  0.25  2.32 A1 = 0.66 J 4(b) E = ½kx2 C1 or E = ½Fx and F = kx x = [(2  0.66) / 420]0.5 A1 = 0.056 m or E = ½kx2 (C1) ½mv2 = ½kx2 (A1) x = (0.25  2.32 / 420)0.5 = 0.056 m 4(c)(i) (p =) 0.25  2.3 or 0.25  1.5 C1 change in momentum = 0.25 (2.3 + 1.5) A1 = 0.95 N s 4(c)(ii) resultant force = 0.95 / 0.086 or 0.25  (2.3 + 1.5) / 0.086 A1 = 11 N 4(d) curved line from the origin with an increasing gradient B1

This question in 9702/22 Oct/Nov 2022

2 The engine of a toy rocket pushes gases vertically downwards and this results in the rocket accelerating vertically upwards from the ground. The rocket starts to move from rest at time t = 0. The variation with time t of the vertical velocity v of the rocket for the first 0.30 s of the flight is shown in Fig. 2.1. 20 v / m s–1 15 10 5 0 0 0.05 0.10 0.15 0.20 0.25 0.30 t / s Fig. 2.1 As the rocket moves, the thrust force T provided by the rocket engine is 16 N. Assume that the mass of the rocket is constant for this part of its flight. Assume that air resistance is negligible. (a) For this part of the rocket’s flight: (i) show that the acceleration of the rocket is 55 m s–2 [1] (ii) state an expression for the resultant force F experienced by the rocket in terms of the thrust force T and the weight W of the rocket [1] (iii) calculate the mass of the rocket. mass = … kg [2] (b) At time t = 0.30 s, a small piece of metal separates from the rocket. Calculate: (i) the height of the rocket above the ground at t = 0.30 s height = … m [2] (ii) the speed at which the piece of metal strikes the ground. speed = … m s–1 [3] [Total: 9]

9 marks

Mark scheme: 2(a)(i) (e.g. a =) [16.5 – 0] / [0.30 – 0] = 55 (m s–2) A1 2(a)(ii) (F =) T – W A1 2(a)(iii) ma = T – mg C1 m = 16 / (55 + 9.81) m = 0.25 kg A1 2(b)(i) s = ut + ½at 2 or v2 = u2 + 2as or s = vt –½at 2 or s = ½(u + v)t or s = area under graph C1 s = ½  55  0.302 or s = 16.52 / (2  55) or s = 16.5  0.30 – ½  55  0.302 or s = ½  16.5  0.30 s = 2.5 m A1 2(b)(ii) u = 16.5 (m s–1) C1 v2 = u2 + 2as C1 v2 = 16.52 + 2  9.81  2.5 v = 18 m s–1 A1

This question in 9702/23 Oct/Nov 2022

2 A motor uses a wire to raise a block, as illustrated in Fig. 2.1. motor Z wire Y block, weight 1.4 × 104 N X Fig. 2.1 (not to scale) The base of the block takes a time of 0.49 s to move vertically upwards from level X to level Y at a constant speed of 0.64 m s–1. During this time the wire has a strain of 0.0012. The wire is made of metal of Young modulus 2.2 × 1011 Pa and has a uniform cross-section. The block has a weight of 1.4 × 104 N. Assume that the weight of the wire is negligible. (a) Calculate: (i) the cross-sectional area A of the wire A = … m2 [2] (ii) the increase in the gravitational potential energy of the block for the movement of its base from X to Y. increase in gravitational potential energy = … J [3] (b) The motor has an efficiency of 56%. Calculate the input power to the motor as the base of the block moves from X to Y. input power = … W [3] (c) The base of the block now has a uniform deceleration of magnitude 1.3 m s–2 from level Y until the base of the block stops at level Z. Calculate the tension T in the wire as the base of the block moves from Y to Z. T = … N [3] (d) The base of the block is at levels X, Y and Z at times tX, tY and tZ respectively. On Fig. 2.2, sketch a graph to show the variation with time t of the distance d of the base of the block from level X. Numerical values of d and t are not required. d 0 tX tY tZ t Fig. 2.2 [2] [Total: 13]

13 marks

Mark scheme: 2(a)(i) E =  /  or E = F / A C1 A = 1.4  104 / (2.2  1011  0.0012) A1 = 5.3  10–5 m2 2(a)(ii) (∆)h = 0.64  0.49 (= 0.3136) C1 (∆)E = mg(∆)h or W(∆)h C1 = 1.4  104  0.64  0.49 A1 = 4.4  103 J 2(b) P = Fv or W / t C1 = (1.4  104  0.64) / 0.56 or (4.4  103 / 0.49) / 0.56 C1 = 1.6  104 W A1 2(c) m = 1.4  104 / 9.81 C1 ( = 1427 kg) (resultant) F = (1.4  104 / 9.81)  1.3 C1 ( = 1855 N) T = 1.4  104 – 1855 or (1.4104 / 9.81)  (9.81 – 1.3) A1 = 1.2  104 N 2(d) upward sloping straight line from (tX, 0) to tY B1 from tY to tZ: an upward sloping curve with decreasing magnitude of gradient (that is horizontal at tZ) B1

This question in 9702/22 Feb/March 2023

4 Two blocks slide directly towards each other along a frictionless horizontal surface, as shown in Fig. 4.1. The blocks collide and then move as shown in Fig. 4.2. 0.37 kg m s–1 0.65 kg m s–1 0.13 kg m s–1 X Y X Y BEFORE COLLISION AFTER COLLISION Fig. 4.1 Fig. 4.2 Block X initially moves to the right with a momentum of 0.37 kg m s–1. Block Y initially moves to the left with a momentum of 0.65 kg m s–1. After the blocks collide, block X moves to the left back along its original path with a momentum of 0.13 kg m s–1. Block Y also moves to the left after the collision. (a) Block X has an initial kinetic energy of 0.30 J. Calculate the mass of block X. mass = … kg [3] (b) Determine the magnitude of the momentum of block Y after the collision. momentum = … kg m s–1 [1] (c) Block X exerts an average force of 7.7 N on block Y during the collision. Calculate the time that the blocks are in contact with each other. time = … s [2] [Total: 6]

6 marks

Mark scheme: 4(a) E = ½mv2 C1 p = mv C1 m = 0.372 / (2  0.30) or 0.37 / 1.6 or (0.30  2) / 1.62 A1 = 0.23 kg 4(b) 0.37 – 0.65 = –0.13 – p A1 p = 0.15 kg m s–1 4(c) 7.7 = (0.13 + 0.37) / (∆)t C1 or 7.7 = (0.65 – 0.15) / (∆)t time = 0.065 s A1

This question in 9702/22 Feb/March 2023

3 A block is pulled in a straight line along a rough horizontal surface by a varying force X, as shown in Fig. 3.1. momentum p block force X rough horizontal surface Fig. 3.1 Air resistance is negligible. Assume that the frictional force exerted on the block by the surface is constant and has magnitude 2.0 N. The variation with time t of the momentum p of the block is shown in Fig. 3.2. 7 6 p / kg m s–1 5 4 3 2 1 0 0 1 2 3 4 5 6 t / s Fig. 3.2 (a) State Newton’s second law of motion. … … [1] (b) Use Fig. 3.2 to determine, for the block at time t = 2.0 s, the magnitude of: (i) the resultant force on the block resultant force = … N [1] (ii) the force X. X = … N [1] (c) On Fig. 3.3, sketch a graph to show the variation of force X with time t from t = 0 to t = 6.0 s. 4.0 3.5 3.0 X / N 2.5 2.0 1.5 1.0 0.5 0 0 1 2 3 4 5 6 t / s Fig. 3.3 [3] [Total: 6]

6 marks

Mark scheme: 3(a) (resultant) force (on an object) is proportional to / equal to the rate of change of momentum B1 3(b)(i) resultant force = e.g. 6.0 / 4.0 = 1.5 N A1 3(b)(ii) force X = 1.5 + 2.0 = 3.5 N A1 3(c) from t = 0 to t = 4.0 s: horizontal line at any non-zero value of X B1 from t = 0 to t = 4.0 s: horizontal line at X = 3.5 N B1 from t = 4.0 s to t = 6.0 s: horizontal line at X = 2.0 N B1

This question in 9702/21 May/June 2023

3 A block is pulled by a force X in a straight line along a rough horizontal surface, as shown in Fig. 3.1. velocity total resistive force 0.80 N X horizontal surface Fig. 3.1 Assume that the total resistive force opposing the motion of the block is 0.80 N at all speeds of the block. The variation with time t of the magnitude of the force X is shown in Fig. 3.2. 2.0 X / N 1.5 1.0 0.5 0 0 1 2 3 4 5 6 t / s Fig. 3.2 (a) (i) Define force. … … [1] (ii) Determine the change in momentum of the block from time t = 0 to time t = 3.0 s. change in momentum = … kg m s–1 [2] (b) (i) Describe and explain the motion of the block between time t = 3.0 s and time t = 6.0 s. … … … … [2] (ii) Force X produces a total power of 2.0 W when moving the block between time t = 3.0 s and time t = 6.0 s. Calculate the distance moved by the block during this time interval. distance = … m [3] (c) The block is at rest at time t = 0. On Fig. 3.3, sketch a graph to show the variation of the momentum of the block with time t from t = 0 to t = 6.0 s. Numerical values of momentum are not required. momentum 0 0 1 2 3 4 5 6 t / s Fig. 3.3 [2] [Total: 10]

10 marks

Mark scheme: 3(a)(i) rate of change of momentum B1 3(a)(ii) change in momentum = (1.4 – 0.80)  3.0 C1 = 1.8 kg m s–1 A1 3(b)(i) resultant force (on block) is zero B1 (so) velocity is constant B1 3(b)(ii) P = Fv or P = Fs / t C1 v = 2.0 / 0.80 (= 2.5 m s–1) C1 distance = 2.5  3.0 = 7.5 m A1 or P = W / t or P = Fs / t (C1) W = 2.0  3.0 (= 6.0 J) (C1) distance = 6.0 / 0.80 = 7.5 m (A1) 3(c) 0 to 3.0 s: upward sloping straight line from the origin. B1 3.0 to 6.0 s: horizontal line at non-zero value of momentum with no ‘step change’ in momentum at 3.0 s B1

This question in 9702/22 May/June 2023

3 A trolley A moves along a horizontal surface at a constant velocity towards another trolley B which is moving at a lower constant speed in the same direction. Fig. 3.1 shows the trolleys at time t = 0. A B horizontal surface Fig. 3.1 Table 3.1 shows data for the trolleys. Table 3.1 trolley mass / kg initial speed / m s–1 A 0.25 0.48 B 0.75 0.12 The two trolleys collide elastically and then separate. Resistive forces are negligible. Fig. 3.2 shows the variation with time t of the velocity v for trolley B. 0.5 v / m s–1 0.4 0.3 B 0.2 0.1 0 / s 0 0.1 0.2 0.3 0.4 0.5t –0.1 –0.2 –0.3 –0.4 –0.5 Fig. 3.2 (a) State what is represented by the area under a velocity–time graph. … [1] (b) Use Table 3.1 and Fig. 3.2 to determine: (i) the acceleration of trolley B during the collision acceleration of B = … m s–2 [2] (ii) the magnitude and direction of the final velocity of trolley A. magnitude = … m s–1 direction … [3] (c) On Fig. 3.2, sketch the variation of the velocity of trolley A with time t from t = 0 to t = 0.50 s. [3] [Total: 9]

9 marks

Mark scheme: 3(a) displacement A1 3(b)(i) a = gradient or a = v / ()t or a = (v – u) / t C1 e.g. a = (0.30 – 0.12) / (0.35 – 0.15) A1 a = 0.90 m s–2 3(b)(ii) (0.25  0.48) + (0.75  0.12) = (0.25 v) + (0.75  0.30) C1 or (0.48 – 0.12) = (0.30 – v) or (½  0.25  0.482) + (½  0.75  0.122) = (½  0.25 × v 2) + (½ × 0.75  0.302) v = (–)0.060 m s–1 A1 direction: to the left / from the right / opposite to (its) initial velocity / opposite to (initial / final) velocity of B B1 3(c) sketch: horizontal line from (0, 0.48) to (0.15, 0.48) B1 horizontal line from (0.35, –0.06) to (0.5, –0.06) B1 straight line between (0.15, 0.48) and (0.35, –0.06) B1

This question in 9702/21 Oct/Nov 2023

3 (a) State the conditions for a system to be in equilibrium. … … … [2] (b) Fig. 3.1 shows an airship in flight. The airship is propelled by identical fans that can be angled to control the motion of the airship. airship fans Fig. 3.1 The upthrust on the airship is 93 000 N. The density of the surrounding air is 1.2 kg m–3. (i) Calculate the volume of air displaced by the airship. volume = … m3 [1] (ii) When fully loaded, the weight of the airship is greater than the upthrust. To maintain horizontal flight, the fans provide a total vertical force of 3.0 × 103 N upwards on the airship. Calculate the mass of the airship. mass = … kg [2] (c) At a certain time, the airship in (b) is stationary. The thrust force exerted by a fan on the airship is 2800 N. To produce this force, a mass of 64 kg of air is propelled through the blades of the fan in a time of 0.50 s. Assume that this air is initially stationary at the entrance to the fan. Calculate: (i) the change in momentum ∆p of the air propelled through the fan blades in this time ∆p = … kg m s–1 [2] (ii) the speed of the air as it leaves the fan speed = … m s–1 [2] (iii) the total kinetic energy of this air due to its movement through the fan. kinetic energy = … J [2] [Total: 11]

11 marks

Mark scheme: 3(a) resultant force (in any direction) is zero B1 resultant moment / torque (about any point) is zero B1 3(b)(i) F = Vg A1 V = 93 000 / (1.2  9.81) = 7900 m3 3(b)(ii) weight = 93 000 + 3(.0)  103 C1 m = (93 000 + 3.0  103) / 9.81 A1 = 9800 kg 3(c)(i) (p) = F()t or F = ∆p / ()t or F = rate of change of momentum or F = m (v – u) / t C1 ∆p = 2800  0.50 A1 = 1400 kg m s–1 3(c)(ii) (p =) mv C1 v = 1400 / 64 A1 = 22 m s–1 3(c)(iii) EK = ½mv2 C1 = ½  64  222 A1 = 15 000 J or EK = p2 / 2m (C1) = 14002 / (2  64) (A1) = 15 000 J

This question in 9702/23 Oct/Nov 2023

3 (a) State the principle of conservation of momentum. … … … [2] (b) An object of mass 2m is travelling at a speed of 5.0 m s–1 in a straight line. It collides with an object of mass 3m which is initially stationary, as shown in Fig. 3.1. 5.0 m s–1 object, mass 2m object, mass 3m Fig. 3.1 After the collision, the object of mass 2m moves with velocity v at an angle of 30° to its original direction of motion. The object of mass 3m moves with velocity w also at an angle of 30°, as shown in Fig. 3.2. object, mass 2m v original path 30° 30° w object, mass 3m Fig. 3.2 By considering the conservation of momentum in two dimensions, calculate the magnitudes of v and w. v = … m s–1 w = … m s–1 [4] (c) An object of mass 4.2 kg is travelling in a straight line at a speed of 6.0 m s–1. The object is brought to rest in a distance of 0.050 m by a constant force. Calculate the magnitude of this force. force = … N [3] [Total: 9]

9 marks

Mark scheme: 3(a) sum / total momentum before (a collision) = sum / total momentum after (a collision) or sum / total momentum (of a system) is constant M1 if no (resultant) external force (acts) / for an isolated system A1 3(b) along direction of motion: 10m = 2mv cos 30° + 3mw cos 30° C1 perpendicular to direction of motion: 2mv cos 60° = 3mw cos 60° (v = 3w / 2) C1 v = 2.9 m s–1 A1 w = 1.9 m s–1 A1 3(c) EK = ½  m  v2 ( = ½  4.2  6.02) ( = 76 J) C1 force = work done / distance C1 force = 76 / 0.050 = 1500 N A1 Question Answer Marks 3(c) or a = (–)u2 / 2s = (–)6.02 / (2  0.050) ( = (–)360 m s–2) (C1) F = ma (C1) F = 4.2  360 = 1500 N (A1) or a = (–)u2 / 2s = (–)6.02 / (2  0.050) ( = (–)360 m s–2) (C1) t = –u / a = –6.0 / –360 (= 0.017 s) F = p / t (C1) F = (0 – 4.2  6) / 0.017 = 1500 N (A1)

This question in 9702/21 May/June 2024

2 A skydiver jumps from an aircraft at time t = 0 and falls vertically downwards. The variation with t of her velocity v is shown in Fig. 2.1. 45 40 v / m s–1 35 30 25 20 15 10 5 0 0 5 10 15 20 25 30 t / s Fig. 2.1 (a) (i) Using Fig. 2.1, state the terminal velocity of the skydiver. terminal velocity = … m s–1 [1] (ii) By drawing a suitable line on Fig. 2.1, determine the acceleration of the skydiver at time t = 9.0 s. acceleration = … m s–2 [2] (b) The mass of the skydiver and her equipment is 68 kg. The upthrust on the skydiver is negligible. After reaching terminal velocity, the skydiver opens her parachute at time t1. A total drag force of 1800 N acts on the skydiver. Determine the magnitude and direction of the acceleration of the skydiver at time t1. acceleration = … m s–2 direction = … [3] (c) The parachute is fully open at time t2. At a later time t3 the skydiver reaches a constant velocity of 5.7 m s–1. (i) Describe and explain the variation with time of the magnitude of her acceleration between time t2 and time t3. … … … … … [2] (ii) Calculate the change in momentum of the skydiver between time t1 and time t3. change in momentum = … N s [2] [Total: 10]

10 marks

Mark scheme: 2(a)(i) 39 m s–1 A1 2(a)(ii) tangent line to curve drawn on Fig. 2.1 C1 a = gradient of tangent line = v / t e.g. = (44 – 26) / (18 – 0) 0.9 ⩽ a ⩽ 1.1 m s–2 A1 2(b) ()F = 68  9.81 – 1800 ( = –1133 N) C1 a = ()F / m = – 1133 / 68 = (–)17 m s–2 A1 upwards B1 2c(i) drag force decreases (as speed decreases) B1 (as speed decreases) resultant force decreases so (magnitude of) acceleration decreases (to zero) B1 2(c)(ii) ()p = mv or ()p = m(v – u) C1 = 68  (5.7 – 39) = (–)2300 N s A1

This question in 9702/22 May/June 2024

2 (a) Define linear momentum. … … [1] (b) A car of mass 1800 kg is moving in a straight line. Fig. 2.1 shows the variation with time t of the momentum p of the car. 8 p / 104 N s 6 4 2 0 0 2 4 6 8 10 12 t / s Fig. 2.1 (i) Calculate the maximum speed reached by the car. maximum speed = … m s–1 [1] (ii) Calculate the maximum kinetic energy of the car. maximum kinetic energy = … J [2] (iii) Show that the acceleration of the car at time t = 4.0 s is 5.0 m s–2. [2] (iv) Determine the distance travelled by the car between times t = 0 and t = 12.0 s. distance = … m [2] (c) On Fig. 2.2, sketch the variation with time t of the acceleration a of the car in (b) from t = 0 to t = 12.0 s. 10 a / m s−2 5 0 0 2 4 6 8 10 12 t / s −5 −10 Fig. 2.2 [3] [Total: 11]

11 marks

Mark scheme: 2(a) product of mass and velocity B1 2(b)(i) maximum speed = 7.2  104 / 1800 A1 = 40 m s–1 2(b)(ii) kinetic energy = ½ mv2 C1 = ½  1800  402 A1 = 1.4  106 J 2(b)(iii) a = gradient of line / mass or a = (v – u) / t or a = v / t C1 a = e.g. (3.6  104) / (4.0  1800) = 5.0 m s–2 A1 or a = e.g. 20 (– 0) / 4 = 5.0 m s–2 or F = e.g. 3.6  104 / 4.0 = 9.0  103 and a = 9.0  103 / 1800 = 5.0 m s–2 2(b)(iv) distance = ½  40  (8 + 4) C1 or distance = ½ × 72 000  (8 + 4) / 1800 or distance = [402 / (2  5)] + [402 / (2  10)] or distance = ½  5  82 + (40  4 – ½  10  42) distance = 240 m A1 2(c) stepped shape, showing one constant value up to 8.0 s then stepping to a different constant value from 8.0 s with no time B1 delay horizontal straight line from (0, 5.0) to (8.0, 5.0) B1 horizontal straight line from (8.0, –10.0) to (12.0, –10.0) B1

This question in 9702/21 Oct/Nov 2024

2 (a) Define momentum. … … [1] (b) A child stands on a scooter on horizontal ground. The combined mass of the child and the scooter is 16 kg. The child starts from rest and pushes once on the ground with her foot which causes her to accelerate. The push lasts for a time of 1.1 s. The speed of the child and the scooter after the push is 0.60 m s–1. Determine the average resultant force acting horizontally on the child and the scooter during the push. average force = … N [2] (c) Later, the child in (b) travels down a slope at a constant angle to the horizontal, as shown in Fig. 2.1. A B x Fig. 2.1 (not to scale) At point A her speed is 0.60 m s–1. She has a constant acceleration of 0.85 m s–2 parallel to the slope. After a time of 3.7 s, she reaches point B. Calculate the distance x travelled by the child along the slope from A to B. x = … m [2] (d) At point B, the child in (c) applies the brake with a constant force to maintain a constant velocity. Point C is 18 m from point B, as shown in Fig. 2.2. 18 m A B C Fig. 2.2 (not to scale) The work done by the braking force between B and C is 250 J. (i) Determine the magnitude of the braking force. force = … N [2] (ii) On Fig. 2.3, sketch the variation of the kinetic energy of the child and scooter with distance travelled from point A to point C. Numerical values for kinetic energy are not required. kinetic energy 0 0 x x + 18 m distance from A Fig. 2.3 [3] [Total: 10]

10 marks

Mark scheme: 2(a) product of mass and velocity B1 2(b) F = mv / t C1 = 16  0.60 / 1.1 = 8.7 N A1 2(c) x = ut + ½at2 C1 x = 0.60  3.7 + ½  0.85  3.72 or v = 0.60 + 0.85  3.7 (= 3.75 m s–1) (C1) x = 3.75  3.7 – 0.5  0.85  3.72 or x = ½  (0.60 + 3.75)  3.7 or x = (3.752 – 0.602) / (2  0.85) x = 8.0 m A1 2(d)(i) F = W / s C1 = 250 / 18 A1 = 14 N 2(d)(ii) any line starting at distance = 0 and a positive non-zero value of kinetic energy B1 a straight line from distance = 0 to distance = x with positive gradient B1 a straight horizontal line at a non-zero value of kinetic energy starting at distance = x and ending at distance = x + 18 m that B1 is continuous with the previous line

This question in 9702/22 Oct/Nov 2024

2 (a) State the principle of conservation of momentum. … … … [2] (b) A ball X has mass 240 g and moves in a straight line on a horizontal frictionless surface with an initial speed of 16 m s–1. The ball collides with a stationary ball Y that has mass 480 g. After the collision, ball X is stationary, as shown in Fig. 2.1. ball X, ball Y, ball X, ball Y, mass 240 g mass 480 g mass 240 g mass 480 g 16 m s–1 v surface surface BEFORE AFTER Fig. 2.1 (i) Show that the speed v of ball Y after the collision is 8.0 m s–1. [1] (ii) Calculate the change in the total kinetic energy ∆EK of the balls due to the collision. ∆EK = … J [3] (c) The collision in (b) lasts for a time of 2.0 ms. Assume that the contact force between the balls is constant during this time. (i) Determine the magnitude and direction of the force exerted on ball X by ball Y during the collision. magnitude = … N direction … [3] (ii) Compare the magnitude and direction of the force exerted on ball Y by ball X during the collision with the answers in (c)(i). No further calculations are required. … … … [2] [Total: 11]

11 marks

Mark scheme: 2(a) sum / total momentum (of a system of bodies) is constant M1 or sum / total momentum before = sum / total momentum after for an isolated system / no (resultant) external force A1 2(b)(i) 240  16 = 480v and so v = 8.0 m s–1 A1 or (initial momentum =) 240  16 (= 3840 g m s–1) and v = 3840 / 480 = 8.0 m s–1 2(b)(ii) (EK =) ½ mv2 C1 EK = ½ [(0.24  162) – (0.48  8.02)] C1 = 15 J A1 2(c)(i) F = (0.24  16) / (2.0  10–3) or F = (0.48  8) / (2.0  10–3) C1 = 1900 N A1 direction: to the left B1 2(c)(ii) equal (magnitude) B1 opposite (direction) B1

This question in 9702/23 Oct/Nov 2024