Equations of motion — Paper 2 · A Level Physics 9702

3 A stone on a string is made to travel along a horizontal circular path, as shown in Fig. 3.1. For Examiner’s Use path of stone stone Fig. 3.1 The stone has a constant speed. (a) Define acceleration. … … [1] (b) Use your definition to explain whether the stone is accelerating. … … … [2] (c) The stone has a weight of 5.0 N. When the string makes an angle of 35° to the vertical, For the tension in the string is 6.1 N, as illustrated in Fig. 3.2. Examiner’s Use 6.1 N 35° 5.0 N Fig. 3.2 Determine the resultant force acting on the stone in the position shown. magnitude of force = ……………..……………………. N direction of force….………………..………………….. [4]

7 marks

Mark scheme: 3 (a) change in velocity/time (taken) B1 [1] (b) velocity is a vector/velocity has magnitude & direction B1 direction changing so must be accelerating B1 [2] (c) either 6.1 × cos35 = 4.99 N or scale shown B1 so no resultant vertical force triangle of correct shape B1 6.1 sin35 = 3.5 N resultant = 3.5 ± 0.2 N B1 horizontally horizontal ± 3° B1 [4] allow answer based on centripetal force: resultant is centripetal force (which is horizontal) (B1) resultant is horizontal component of tension (B1) 6.1 sin35 = 3.5 N (B1) horizontally (B1)

This question in 9702/21 Oct/Nov 2005

4 A trolley of mass 930 g is held on a horizontal surface by means of two springs, as shown in For Fig. 4.1. Examiner’s Use trolley spring Fig. 4.1 The variation with time t of the speed v of the trolley for the first 0.60 s of its motion is shown in Fig. 4.2. 8.0 v / cm s – 1 6.0 4.0 2.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s Fig. 4.2 (a) Use Fig. 4.2 to determine (i) the initial acceleration of the trolley, acceleration = ………………………. m s–2 [2] (ii) the distance moved during the first 0.60 s of its motion. For Examiner’s Use distance = ………….………………… m [3] (b) (i) Use your answer to (a)(i) to determine the resultant force acting on the trolley at time t = 0. force = …………………………….. N [2] (ii) Describe qualitatively the variation with time of the resultant force acting on the trolley during the first 0.60 s of its motion. … … … … [3]

10 marks

Mark scheme: 4 (a) (i) use of tangent at time t = 0 B1 acceleration = 42 ± 4 cm s-2 A1 [2] (ii) use of area of loop B1 distance = 0.031 ± 0.001 m B2 [3] allow 1 mark if 0.031 ± 0.002 m) (b) (i) F = ma C1 = 0.93 × 0.42 {allow e.c.f. from (a)(i)} = 0.39 N A1 [2] (ii) force reduces to zero in first 0.3 s B1 then increases again in next 0.3 s M1 in the opposite direction A1 [3] GCE A/AS LEVEL – November 2005 9702 2

This question in 9702/21 Oct/Nov 2005

8 (a) Explain the concept of work. For Examiner’s … Use … … [2] (b) A table tennis ball falls vertically through air. Fig. 8.1 shows the variation of the kinetic energy EK of the ball with distance h fallen. The ball reaches the ground after falling through a distance h0. energy E K 0 0 h h 0 Fig. 8.1 (i) Describe the motion of the ball. … … … … … [3] (ii) On Fig. 8.1, draw a line to show the variation with h of the gravitational potential energy EP of the ball. At h = h0, the potential energy is zero. [3]

8 marks

Mark scheme: 8 (a) product of force and distance M1 moved in the direction of the force A1 [2] (b) (i) falls from rest B1 decreasing acceleration B1 reaches a constant speed B1 [3] (ii) straight line with negative gradient B1 y-axis intercept above maximum EK B1 reasonable gradient (same magnitude as that for EK initially) B1 [3]

This question in 9702/21 Oct/Nov 2005

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 car is travelling along a straight road at speed v. A hazard suddenly appears in front of For the car. In the time interval between the hazard appearing and the brakes on the car coming Examiner’s into operation, the car moves forward a distance of 29.3 m. With the brakes applied, the front Use wheels of the car leave skid marks on the road that are 12.8 m long, as illustrated in Fig. 2.1. position of car skid mark when hazard appears 29.3 m 12.8 m Fig. 2.1 It is estimated that, during the skid, the magnitude of the deceleration of the car is 0.85 g, where g is the acceleration of free fall. (a) Determine (i) the speed v of the car before the brakes are applied, v = …………………… m s–1 [2] (ii) the time interval between the hazard appearing and the brakes being applied. time = ……………..…………. s [2] (b) The legal speed limit on the road is 60 km per hour. For Use both of your answers in (a) to comment on the standard of the driving of the car. Examiner’s Use … … … … [3]

7 marks

Mark scheme: 2 (a) (i) v2 = 2as v2 = 2 × 0.85 × 9.8 × 12.8 C1 v = 14.6 m s-1 A1 [2] (ii) time = 29.3 / 14.6 C1 = 2.0 s A1 [2] (any acceleration scores 0 marks; allow 1 s.f.) (b) either 60 km h–1 = 16.7 m s–1 or 14.6 m s–1 = 53 km h–1 or 22.1 m s–1 = 79.6 km h–1 M1 so driving within speed limit A1 but reaction time is too long / too slow B1 [3]

This question in 9702/21 Oct/Nov 2008

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 small ball is thrown horizontally with a speed of 4.0 m s–1. It falls through a vertical height of For 1.96 m before bouncing off a horizontal plate, as illustrated in Fig. 3.1. Examiner’s Use 4.0 m s–1 1.96 m 0.98 m plate Fig. 3.1 Air resistance is negligible. (a) For the ball, as it hits the horizontal plate, (i) state the magnitude of the horizontal component of its velocity, horizontal velocity = … m s–1 [1] (ii) show that the vertical component of the velocity is 6.2 m s–1. [1] (b) The components of the velocity in (a) are both vectors. For Examiner’s Complete Fig. 3.2 to draw a vector diagram, to scale, to determine the velocity of the Use ball as it hits the horizontal plate. Fig. 3.2 velocity = … m s–1] at … ° to the vertical [3] (c) After bouncing on the plate, the ball rises to a vertical height of 0.98 m. (i) Calculate the vertical component of the velocity of the ball as it leaves the plate. vertical velocity = … m s–1 [2] (ii) The ball of mass 34 g is in contact with the plate for a time of 0.12 s. For Examiner’s Use your answer in (c)(i) and the data in (a)(ii) to calculate, for the ball as it bounces Use on the plate, 1. the change in momentum, change = … kg m s–1 [3] 2. the magnitude of the average force exerted by the plate on the ball due to this momentum change. force = … N [2]

12 marks

Mark scheme: 3 (a) (i) speed = 4.0 m s-1 …(allow 1 s.f.) … A1 [1] (ii) v2 = 2gh = 2 × 9.8 × 1.96 … M1 v = 6.2 m s-1 … A0 [1] (use of g = 10 m s-2 loses the mark) (b) correct basic shape with correct directions for vectors … M1 speed = (7.4 ± 0.2) m s-1 … A1 at (33 ± 2)° to the vertical … A1 [3] (for credit to be awarded, speed and angle must be correct on the diagram – not calculated) GCE A/AS LEVEL – October/November 2009 9702 22 (c) (i) either v2 = 2 × 9.8 × 0.98 or v = 6.2 / √2 … C1 speed = 4.4 m s-1 … A1 [2] (allow calculation of t = 0.447 s, then v = 4.4 m s-1) (ii) 1 momentum = mv … C1 change in momentum = 0.034 (6.2 + 4.4) … C1 = 0.36 kg m s-1 … A1 [3] (use of 0.034 (6.2 - 4.4) loses last two marks) 2 force = ∆p / ∆t …….(however expressed) … C1 0.36 = 0.12 = 3.0 N ……(allow 1 s.f.) … A1 [2] [Total: 12]

This question in 9702/22 Oct/Nov 2009

2 (a) Complete Fig. 2.1 to show whether each of the quantities listed is a vector or a scalar. For Examiner’s Use vector / scalar distance moved … speed … acceleration … Fig. 2.1 [3] (b) A ball falls vertically in air from rest. The variation with time t of the distance d moved by the ball is shown in Fig. 2.2. 5 4 d /m 3 2 1 0 0 0.2 0.4 0.6 0.8 1.0 1.2 t /s Fig. 2.2 (i) By reference to Fig. 2.2, explain how it can be deduced that For Examiner’s 1. the ball is initially at rest, Use … … … [2] 2. air resistance is not negligible. … … [1] (ii) Use Fig. 2.2 to determine the speed of the ball at a time of 0.40 s after it has been released. speed = … m s–1 [2] (iii) On Fig. 2.2, sketch a graph to show the variation with time t of the distance d moved by the ball for negligible air resistance. You are not expected to carry out any further calculations. [3]

11 marks

Mark scheme: 2 (a) scalar …………………………………………………………..………………………… B1 scalar …………………………………………………………..………………………… B1 vector …………………………………………………………..………………………… B1 [3] (b) (i) 1 gradient (of graph) is the speed/velocity (can be scored here or in 2)………. B1 initial gradient is zero …………………………………………………………… B1 [2] 2 gradient (of line/graph) becomes constant ……….……..…………………… B1 [1] (ii) speed = (2.8 ± 0.1) m s–1 ……… ………………………………………………… A2 [2] (if answer > ±0.1 but ≤ ±0.2, then award 1 mark) (iii) curved line never below given line and starts from zero …..………………….. B1 continuous curve with increasing gradient …………………..…………………. B1 line never vertical or straight ………………………………..……………………. B1 [3]

This question in 9702/21 May/June 2010

2 (a) The distance s moved by an object in time t may be given by the expression For Examiner’s s = 12at 2 Use where a is the acceleration of the object. State two conditions for this expression to apply to the motion of the object. 1. … … 2. … … [2] (b) A student takes a photograph of a steel ball of radius 5.0 cm as it falls from rest. The image of the ball is blurred, as illustrated in Fig. 2.1. The image is blurred because the ball is moving while the photograph is being taken. initial position 80 of ball in photograph cm final position 90 of ball in photograph cm 100 cm Fig. 2.1 The scale shows the distance fallen from rest by the ball. At time t = 0, the top of the ball is level with the zero mark on the scale. Air resistance is negligible. Calculate, to an appropriate number of significant figures, For Examiner’s (i) the time the ball falls before the photograph is taken, Use time = … s [3] (ii) the time interval during which the photograph is taken. time interval = … s [3] (c) The student in (b) takes a second photograph starting at the same position on the scale. The ball has the same radius but is less dense, so that air resistance is not negligible. State and explain the changes that will occur in the photograph. … … … … [2]

10 marks

Mark scheme: 2 (a) e.g. initial speed is zero constant acceleration straight line motion (any two, one mark each) ……………………………………………………………….B2 [2] (b) (i) s = ½a t 2 0.79 = ½ × 9.8 × t 2 ………………………………………………………….. C1 t = 0.40 s allow 1 SF or greater ……………………………………………. A1 2 or 3 SF answer ……………………………………………………….. A1 [3] (ii) distance travelled by end of time interval = 90 cm ………………………. C1 0.90 = ½ × 9.8 × t 2 t = 0.43 s allow 2 SF or greater ……………………………………………. C1 time interval = 0.03 s ………………………………………………………... A1 [3] (c) (air resistance) means ball’s speed/acceleration is less ……………………… M1 length of image is shorter ………………………………………………………… A1 [2]

This question in 9702/22 May/June 2010

3 A cyclist is moving up a slope that has a constant gradient. The cyclist takes 8.0 s to climb For the slope. Examiner’s The variation with time t of the speed v of the cyclist is shown in Fig. 3.1. Use 8 v / m s–1 6 4 2 0 0 2 4 6 8 t / s Fig. 3.1 (a) Use Fig. 3.1 to determine the total distance moved up the slope. distance = … m [3] (b) The bicycle and cyclist have a combined mass of 92 kg. For The vertical height through which the cyclist moves is 1.3 m. Examiner’s Use (i) For the movement of the bicycle and cyclist between t = 0 and t = 8.0 s, 1. use Fig. 3.1 to calculate the change in kinetic energy, change = … J [2] 2. calculate the change in gravitational potential energy. change = … J [2] (ii) The cyclist pedals continuously so that the useful power delivered to the bicycle is 75 W. Calculate the useful work done by the cyclist climbing up the slope. work done = … J [2] (c) Some energy is used in overcoming frictional forces. For Examiner’s (i) Use your answers in (b) to show that the total energy converted in overcoming Use frictional forces is approximately 670 J. [1] (ii) Determine the average magnitude of the frictional forces. average force = … N [1] (d) Suggest why the magnitude of the total resistive force would not be constant. … … … [2]

13 marks

Mark scheme: 3 (a) evidence of use of area below the line B1 distance = 39 m (allow ±0.5 m) A2 [3] (if > ±0.5 m but ≤ 1.0 m, then allow 1 mark) (b) (i) 1 EK = ½mv 2 C1 ∆ EK = ½ × 92 × (62 – 32) = 1240 J A1 [2] 2 EP = mgh C1 ∆ EP = 92 × 9.8 × 1.3 = 1170 J A1 [2] (ii) E = Pt C1 E = 75 × 8 = 600 J A1 [2] (c) (i) energy = (1240 + 600) – 1170 M1 = 670 J A0 [1] (ii) force = 670/39 = 17 N A1 [1] (d) frictional forces include air resistance B1 air resistance decreases with decrease of speed B1 [2] GCE AS/A LEVEL – May/June 2010 9702 23

This question in 9702/23 May/June 2010

2 A ball is thrown horizontally from the top of a building, as shown in Fig. 2.1. For Examiner’s Use 8.2 m s–1 60° P x Fig. 2.1 The ball is thrown with a horizontal speed of 8.2 m s–1. The side of the building is vertical. At point P on the path of the ball, the ball is distance x from the building and is moving at an angle of 60° to the horizontal. Air resistance is negligible. (a) For the ball at point P, (i) show that the vertical component of its velocity is 14.2 m s–1, [2] (ii) determine the vertical distance through which the ball has fallen, distance = … m [2] (iii) determine the horizontal distance x. For Examiner’s Use x = … m [2] (b) The path of the ball in (a), with an initial horizontal speed of 8.2 m s–1, is shown again in Fig. 2.2. 8.2 m s–1 Fig. 2.2 On Fig. 2.2, sketch the new path of the ball for the ball having an initial horizontal speed (i) greater than 8.2 m s–1 and with negligible air resistance (label this path G), [2] (ii) equal to 8.2 m s–1 but with air resistance (label this path A). [2]

10 marks

Mark scheme: 2 (a) (i) horizontal speed constant at 8.2 m s–1 C1 vertical component of speed = 8.2 tan 60° M1 = 14.2 m s–1 A0 [2] (ii) 14.22 = 2 × 9.8 × h (using g = 10 then –1) C1 vertical distance = 10.3 m A1 [2] (iii) time of descent = 14.2 / 9.8 = 1.45 s C1 x = 1.45 × 8.2 = 11.9 m A1 [2] (b) (i) smooth path curved and above given path M1 hits ground at more acute angle A1 [2] (ii) smooth path curved and below given path M1 hits ground at steeper angle A1 [2]

This question in 9702/21 Oct/Nov 2010

2 A ball is thrown from a point P, which is at ground level, as illustrated in Fig. 2.1. For Examiner’s Use wall path of ball h P 36° Fig. 2.1 The initial velocity of the ball is 12.4 m s–1 at an angle of 36° to the horizontal. The ball just passes over a wall of height h. The ball reaches the wall 0.17 s after it has been thrown. (a) Assuming air resistance to be negligible, calculate (i) the horizontal distance of point P from the wall, distance = … m [2] (ii) the height h of the wall. For Examiner’s Use h = … m [3] (b) A second ball is thrown from point P with the same velocity as the ball in (a). For this ball, air resistance is not negligible. This ball hits the wall and rebounds. On Fig. 2.1, sketch the path of this ball between point P and the point where it first hits the ground. [2]

7 marks

Mark scheme: 2 (a) (i) VH = 12.4 cos 36° (= 10.0 m s–1) C1 distance = 10.0 × 0.17 = 1.7 m A1 [2] (ii) VV = 12.4 sin 36° (= 7.29 m s–1) C1 h = 7.29 × 0.17 – ½ × 9.81 × 0.172 C1 = 1.1 m A1 [3] (b) smooth curve with ball hitting wall below original B1 smooth curve showing rebound to ground with correct reflection at wall B1 [2]

This question in 9702/22 Oct/Nov 2010

4 A student takes measurements to determine a value for the acceleration of free fall. Some of For the apparatus used is illustrated in Fig. 4.1. Examiner’s Use electromagnet iron ball d bench Fig. 4.1 The student measures the vertical distance d between the base of the electromagnet and the bench. The time t for an iron ball to fall from the electromagnet to the bench is also measured. Corresponding values of t2 and d are shown in Fig. 4.2. 60 50 d /cm 40 30 20 10 0 0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 t 2 / s2 Fig. 4.2 (a) On Fig. 4.2, draw the line of best fit for the points. [1] For Examiner’s (b) State and explain why there is a non-zero intercept on the graph of Fig. 4.2. Use … … … [2] (c) Determine the student’s value for (i) the diameter of the ball, diameter = … cm [1] (ii) the acceleration of free fall. acceleration = … m s–2 [3]

7 marks

Mark scheme: 4 (a) acceptable straight line drawn (touching every point) B1 [1] (b) the distance fallen is not d C1 d is the distance fallen plus the diameter of the ball A1 [2] (‘d is not measured to the bottom of the ball’ scores 2/2) (c) (i) diameter: allow 1.5 ± 0.5 cm (accept one SF) A1 [1] no ecf from (a) (ii) gradient = 4.76, ± 0.1 with evidence that origin has not been used C1 gradient = g / 2 C1 g = 9.5 m s–2 A1 [3] GCE A LEVEL – October/November 2010 9702 23

This question in 9702/23 Oct/Nov 2010

1 (a) Distinguish between scalar quantities and vector quantities. … … … [2] (b) In the following list, underline all the scalar quantities. acceleration force kinetic energy mass power weight [1] (c) A stone is thrown with a horizontal velocity of 20 m s–1 from the top of a cliff 15 m high. The path of the stone is shown in Fig. 1.1. 20 m s–1 cliff 15 m ground Fig. 1.1 Air resistance is negligible. For this stone, (i) calculate the time to fall 15 m, time = … s [2] (ii) calculate the magnitude of the resultant velocity after falling 15 m, resultant velocity = … m s–1 [3] (iii) describe the difference between the displacement of the stone and the distance For that it travels. Examiner’s Use … … … [2]

10 marks

Mark scheme: 1 (a) scalar has only magnitude B1 vector has magnitude and direction B1 [2] (b) kinetic energy, mass, power all three underlined B1 [1] (c) (i) s = ut + ½ at2 15 = 0.5 × 9.81 × t2 C1 T = 1.7 s A1 [2] if g = 10 is used then –1 but only once on paper (ii) vertical component vv: vv2 = u2 + 2as = 0 + 2 × 9.81 × 15 or vv = u + at = 9.81 × 1.7(5) vv =17.16 C1 resultant velocity: v2 = (17.16)2 + (20)2 C1 v = 26 m s–1 A1 [3] If u = 20 is used instead of u = 0 then 0/3 Allow the solution using: initial (potential energy + kinetic energy) = final kinetic energy (iii) distance is the actual path travelled B1 displacement is the straight line distance between start and finish points (in that direction) / minimum distance B1 [2]

This question in 9702/22 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

3 A ball is thrown against a vertical wall. The path of the ball is shown in Fig. 3.1. For Examiner’s Use P 15.0 m s–1 wall 60.0° S F 6.15 m 9.95 m Fig. 3.1 (not to scale) The ball is thrown from S with an initial velocity of 15.0 m s–1 at 60.0° to the horizontal. Assume that air resistance is negligible. (a) For the ball at S, calculate (i) its horizontal component of velocity, horizontal component of velocity = … m s–1 [1] (ii) its vertical component of velocity. vertical component of velocity = … m s–1 [1] (b) The horizontal distance from S to the wall is 9.95 m. The ball hits the wall at P with a velocity that is at right angles to the wall. The ball rebounds to a point F that is 6.15 m from the wall. Using your answers in (a), (i) calculate the vertical height gained by the ball when it travels from S to P, height = … m [1] (ii) show that the time taken for the ball to travel from S to P is 1.33 s, For Examiner’s Use [1] (iii) show that the velocity of the ball immediately after rebounding from the wall is about 4.6 m s–1. [1] (c) The mass of the ball is 60 × 10–3 kg. (i) Calculate the change in momentum of the ball as it rebounds from the wall. change in momentum = … N s [2] (ii) State and explain whether the collision is elastic or inelastic. … … … [1]

8 marks

Mark scheme: 3 (a) (i) horizontal velocity = 15 cos 60° = 7.5 m s–1 A1 [1] (ii) vertical velocity = 15 sin 60° = 13 m s–1 A1 [1] (b) (i) v2 = u2 + 2as s = (13)2 / (2 × 9.81) = 8.6(1) m A1 [1] using g = 10 then max. 1 (ii) t = 13 / 9.81 = 1.326 s or t = 9.95 / 7.5 = 1.327 s A1 [1] (iii) velocity = 6.15 / 1.33 M1 = 4.6 m s–1 A0 [1] (c) (i) change in momentum = 60 × 10–3 [–4.6 – 7.5] C1 = (–)0.73 N s A1 [2] (ii) final velocity / kinetic energy is less after the collision or relative speed of separation < relative speed of approach M1 hence inelastic A0 [1] GCE AS/A LEVEL – October/November 2011 9702 21

This question in 9702/21 Oct/Nov 2011

1 The variation with time t of the displacement s for a car is shown in Fig. 1.1. 600 500 s / m 400 300 200 100 0 0 20 40 60 80 100 t / s Fig. 1.1 (a) Determine the magnitude of the average velocity between the times 5.0 s and 35.0 s. average velocity = … m s–1 [2] (b) On Fig. 1.2, sketch the variation with time t of the velocity v for the car. v / m s–1 0 0 20 40 60 80 100 t / s Fig. 1.2 [4]

6 marks

Mark scheme: 1 (a) average velocity = 540 / 30 C1 = 18 m s–1 A1 [2] (b) velocity zero at time t = 0 B1 positive value and horizontal line for time t = 5 s to 35 s B1 line / curve through v = 0 at t = 45 s to negative velocity B1 negative horizontal line from 53 s with magnitude less than positive value and horizontal line to time = 100 s B1 [4]

This question in 9702/22 Oct/Nov 2011

2 The variation with time t of velocity v of a car is shown in Fig. 2.1. For Examiner’s Use stage 1 stage 2 20.0 15.0 v / m s–1 10.0 5.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 t / s Fig. 2.1 At time t = 0, the driver sees an obstacle in the road. A short time later, the driver applies the brakes. The car travels in two stages, as shown in Fig. 2.1. (a) Use Fig. 2.1 to describe the velocity of the car in 1. stage 1, … … [1] 2. stage 2. … … [1] (b) (i) Calculate the distance travelled by the car from t = 0 to t = 3.5 s. total distance = … m [2] (ii) The car has a total mass of 1250 kg. Determine the total resistive force acting on For the car in stage 2. Examiner’s Use force = … N [3] (c) For safety reasons drivers are asked to travel at lower speeds. For each stage, describe and explain the effect on the distance travelled for the same car and driver travelling at half the initial speed shown in Fig. 2.1. (i) stage 1: … … … [1] (ii) stage 2: … … … … [2]

10 marks

Mark scheme: 2 (a) 1. constant velocity / speed B1 [1] 2. either constant / uniform decrease (in velocity/speed) or constant rate of decrease (in velocity/speed) B1 [1] (b) (i) distance is area under graph for both stages C1 stage 1: distance (18 × 0.65) = 11.7 (m) stage 2: distance = (9 × [3.5 – 0.65]) = 25.7 (m) total distance = 37.(4) m A1 [2] (–1 for misreading graph) {for stage 2, allow calculation of acceleration (6.32 m s–2) and then s = (18 × 2.85) + ½ × 6.32 (2.85)2 = 25.7 m} (ii) either F = ma or EK = ½mv2 C1 a = (18 – 0)/(3.5 – 0.65) EK = ½ × 1250 × (18)2 C1 F = 1250 × 6.3 =7900 N or F = ½ × 1250 × (18)2 / 25.7 = 7900 N A1 [3] or initial momentum = 1250 × 18 (C1) F = change in momentum / time taken (C1) F = (1250 × 18) / 2.85 = 7900 (A1) (c) (i) stage 1: either half / less distance as speed is half / less or half distance as the time is the same or sensible discussion of reaction time B1 [1] (ii) stage 2: either same acceleration and s = v2 / 2a or v2 is ¼ B1 ¼ of the distance B1 [2] GCE AS/A LEVEL – October/November 2011 9702 23

This question in 9702/23 Oct/Nov 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

2 (a) A ball is thrown vertically down towards the ground and rebounds as illustrated in For Fig. 2.1. Examiner’s Use ball passing point A A 8.4 m s–1 ball at maximum height after rebound 5.0 m B h Fig. 2.1 As the ball passes A, it has a speed of 8.4 m s–1. The height of A is 5.0 m above the ground. The ball hits the ground and rebounds to B. Assume that air resistance is negligible. (i) Calculate the speed of the ball as it hits the ground. speed = … m s–1 [2] (ii) Show that the time taken for the ball to reach the ground is 0.47 s. [1] (b) The ball rebounds vertically with a speed of 4.2 m s–1 as it leaves the ground. The time For the ball is in contact with the ground is 20 ms. The ball rebounds to a maximum height h. Examiner’s Use The ball passes A at time t = 0. On Fig. 2.2, plot a graph to show the variation with time t of the velocity v of the ball. Continue the graph until the ball has rebounded from the ground and reaches B. v / m s–1 0 0 t / s Fig. 2.2 [3] (c) The ball has a mass of 0.050 kg. It moves from A and reaches B after rebounding. (i) For this motion, calculate the change in 1. kinetic energy, change in kinetic energy = … J [2] 2. gravitational potential energy. change in potential energy = … J [3] (ii) State and explain the total change in energy of the ball for this motion. For Examiner’s … Use … … … [2]

13 marks

Mark scheme: 2 (a) (i) v2 = u2 + 2as = (8.4)2 + 2 × 9.81 × 5 C1 = 12.99 m s–1 (allow 13 to 2 s.f. but not 12.9) A1 [2] (ii) t = (v – u) / a or s = ut + ½at2 = (12.99 – 8.4) / 9.81 or 5 = 8.4t + ½ × 9.81t2 M1 t = 0.468 s A0 [1] (b) reasonable shape M1 suitable scale A1 correctly plotted 1st and last points at (0,8.4) and (0.88 – 0.96,0) with non-vertical line at 0.47 s A1 [3] (c) (i) 1. kinetic energy at end is zero so ∆KE = ½ mv2 or ∆KE = ½ mu2 – ½ mv2 C1 = ½ × 0.05 × (8.4)2 = (–) 1.8 J A1 [2] 2. final maximum height = (4.2)2 / (2 × 9.8) = (0.9 (m)) change in PE = mgh2 – mgh1 C1 = 0.05 × 9.8 × (0.9 – 5) C1 = (–) 2.0 J A1 [3] (ii) change is – 3.8 (J) B1 energy lost to ground (on impact) / energy of deformation of the ball / thermal energy in ball B1 [2]

This question in 9702/22 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 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) 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

2 (a) A student walks from A to B along the path shown in Fig. 2.1. For Examiner’s Use A B Fig. 2.1 The student takes time t to walk from A to B. (i) State the quantity, apart from t, that must be measured in order to determine the average value of 1. speed, … … [1] 2. velocity. … … [1] (ii) Define acceleration. … [1] (b) A girl falls vertically onto a trampoline, as shown in Fig. 2.2. For Examiner’s Use springy material Fig. 2.2 The trampoline consists of a central section supported by springy material. At time t = 0 the girl starts to fall. The girl hits the trampoline and rebounds vertically. The variation with time t of velocity v of the girl is illustrated in Fig. 2.3. 10.0 8.0 6.0 v / m s–1 4.0 2.0 0 0 0.5 1.0 1.5 2.0 t / s – 2.0 – 4.0 rebound – 6.0 time – 8.0 Fig. 2.3 For the motion of the girl, calculate (i) the distance fallen between time t = 0 and when she hits the trampoline, distance = … m [2] (ii) the average acceleration during the rebound. For Examiner’s Use acceleration = … m s–2 [2] (c) (i) Use Fig. 2.3 to compare, without calculation, the accelerations of the girl before and after the rebound. Explain your answer. … … … [2] (ii) Use Fig. 2.3 to compare, without calculation, the potential energy of the girl at t = 0 and t = 1.85 s. Explain your answer. … … … [2]

11 marks

Mark scheme: 2 (a) (i) 1. distance of path / along line AB B1 [1] 2. shortest distance between AB / distance in straight line between AB or displacement from A to B B1 [1] (ii) acceleration = rate of change of velocity A1 [1] (b) (i) distance = area under line or (v/2)t or s = (8.8)2 / (2 × 9.81) C1 = 8.8 / 2 × 0.90 = 3.96 m or s = 3.95 m = 4(.0) m A1 [2] (ii) acceleration = (– 4.4 – 8.8) / 0.50 C1 = (–) 26(.4) m s–2 A1 [2] (c) (i) the accelerations are constant as straight lines B1 the accelerations are the same as same gradient or no air resistance as acceleration is constant or change of speed in opposite directions (one speeds up one slows down) B1 [2] (ii) area under the lines represents height or KE at trampoline equals PE at maximum height B1 second area is smaller / velocity after rebound smaller hence KE less B1 hence less height means loss in potential energy A0 [2]

This question in 9702/23 May/June 2013

1 (a) (i) Define velocity. … … [1] (ii) Distinguish between speed and velocity. … … [2] (b) A car of mass 1500 kg moves along a straight, horizontal road. The variation with time t of the velocity v for the car is shown in Fig. 1.1. 40 30 v / m s–1 20 10 0 0 1.0 2.0 3.0 4.0 5.0 6.0 t / s Fig. 1.1 The brakes of the car are applied from t = 1.0 s to t = 3.5 s. For the time when the brakes are applied, (i) calculate the distance moved by the car, distance = … m [3] (ii) calculate the magnitude of the resultant force on the car. resultant force = … N [3] (c) The direction of motion of the car in (b) at time t = 2.0 s is shown in Fig. 1.2. direction of motion Fig. 1.2 On Fig. 1.2, show with arrows the directions of the acceleration (label this arrow A) and the resultant force (label this arrow F). [1]

10 marks

Mark scheme: 1 (a) (i) either rate of change of displacement or (change in) displacement / time (taken) B1 [1] (ii) speed has magnitude only B1 velocity has magnitude and direction B1 [2] (u + v ) (b) (i) idea of area under graph / use of s = × t C1 2 (18 + 32) s = × 2.5 C1 2 = 62.5 m A1 [3] (ii) a = (18 – 32) / 2.5 (= –5.6) C1 F = ma C1 F = 1500 × (–) 5.6 = (–) 8400 N A1 [3] (c) arrow labelled A and arrow labelled F both to the left B1 [1]

This question in 9702/21 May/June 2014

2 (a) Explain what is meant by a scalar quantity and by a vector quantity. scalar: … … vector: … … [2] (b) A ball leaves point P at the top of a cliff with a horizontal velocity of 15 m s–1, as shown in Fig. 2.1. ball P 15 m s–1 path of ball 25 m cliff Q ground Fig. 2.1 The height of the cliff is 25 m. The ball hits the ground at point Q. Air resistance is negligible. (i) Calculate the vertical velocity of the ball just before it makes impact with the ground at Q. vertical velocity = … m s–1 [2] (ii) Show that the time taken for the ball to fall to the ground is 2.3 s. [1] (iii) Calculate the magnitude of the displacement of the ball at point Q from point P. displacement = … m [4] (iv) Explain why the distance travelled by the ball is different from the magnitude of the displacement of the ball. … … … [2]

11 marks

Mark scheme: 2 (a) scalar has magnitude only B1 vector has magnitude and direction B1 [2] 1 (b) (i) v2 = 0 + 2 × 9.81 × 25 (or using m v2 = mgh) C1 2 v = 22(.1) m s–1 A1 [2] 1 (ii) 22.1 = 0 + 9.81 × t (or 25 = × 9.81 × t 2) M1 2 t (=22.1 / 9.81) = 2.26 s or t [=(5.097)1/2] = 2.26 s A0 [1] (iii) horizontal distance = 15 × t = 15 × 2.257 = 33.86 (allow 15 × 2.3 = 34.5) C1 (displacement)2 = (horizontal distance)2 + (vertical distance)2 C1 = (25)2 + (33.86)2 C1 displacement = 42 (42.08) m (allow 43 (42.6) m, allow 2 or more s.f.) A1 [4] (iv) distance is the actual (curved) path followed by ball B1 displacement is the straight line / minimum distance P to Q B1 [2]

This question in 9702/23 May/June 2014

4 A trolley moves down a slope, as shown in Fig. 4.1. trolley v 25° horizontal Fig. 4.1 The slope makes an angle of 25° with the horizontal. A constant resistive force FR acts up the slope on the trolley. At time t = 0, the trolley has velocity v = 0.50 m s−1 down the slope. At time t = 4.0 s, v = 12 m s−1 down the slope. (a) (i) Show that the acceleration of the trolley down the slope is approximately 3 m s−2. [2] (ii) Calculate the distance x moved by the trolley down the slope from time t = 0 to t = 4.0 s. x = … m [2] (iii) On Fig. 4.2, sketch the variation with time t of distance x moved by the trolley. x 0 0 4.0 t / s Fig. 4.2 [2] (b) The mass of the trolley is 2.0 kg. (i) Show that the component of the weight of the trolley down the slope is 8.3 N. [1] (ii) Calculate the resistive force FR. FR = … N [2]

9 marks

Mark scheme: 4 (a) (i) acceleration = (v – u) / t or (12 – 0.5) / 4 C1 = (12 – 0.5) / 4 = 2.9 (2.875) (= approximately 3 m s–2) M1 [2] (ii) x = (u + v) t / 2 = [(12 + 0.5) × 4] / 2 C1 = 25 m A1 [2] (iii) line with increasing gradient M1 non-zero gradient at origin A1 [2] (b) (i) weight down slope = 2 × 9.81 × sin 25° = 8.29 / 8.3 M1 [1] (ii) (F = ma) 8.3 – FR = 2 × 2.9 C1 FR = 2.5 (2.3 if 3 used for a) N A1 [2] 2

This question in 9702/21 Oct/Nov 2014

2 A ball is thrown from A to B as shown in Fig. 2.1. V 60° A B Fig. 2.1 The ball is thrown with an initial velocity V at 60° to the horizontal. The variation with time t of the vertical component Vv of the velocity of the ball from t = 0 to t = 0.60 s is shown in Fig. 2.2. 6.0 Vv 4.0 2.0 velocity / m s–1 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 t / s –2.0 –4.0 –6.0 Fig. 2.2 Assume air resistance is negligible. (a) (i) Complete Fig. 2.2 for the time until the ball reaches B. [2] (ii) Calculate the maximum height reached by the ball. height = … m [2] (iii) Calculate the horizontal component Vh of the velocity of the ball at time t = 0. Vh = … m s−1 [2] (iv) On Fig. 2.2, sketch the variation with t of Vh. Label this sketch Vh. [1] (b) The ball has mass 0.65 kg. Calculate, for the ball, (i) the maximum kinetic energy, maximum kinetic energy = … J [3] (ii) the maximum potential energy above the ground. maximum potential energy = … J [2]

12 marks

Mark scheme: 2 (a) (i) straight line from t = 0.60 s to t = 1.2 s and |Vv| = 5.9 at t = 1.2 s M1 Vv = – 5.9 at t = 1.2 s i.e. line is for negative values of Vv A1 [2] (ii) s = 0 + ½ × 9.81 × (0.6)2 or area of graph = (5.9 × 0.6) / 2 C1 = 1.8 (1.77) m = 1.8 (1.77) m A1 [2] (iii) Vh = V cos 60° and Vv = V sin 60° or Vh = 5.9 / tan 60° or Vh = 5.9 tan 30° C1 Vh = 3.4 m s−1 A1 [2] (iv) horizontal line at 3.4 from t = 0 to t = 1.2 s [to half a small square] B1 [1] (b) (i) KE = ½ mv2 C1 = ½ × 0.65 × (6.81)2 [allow if valid method to find v] C1 = 15 (15.1) J A1 [3] (ii) PE = 0.65 × 9.81 × 1.77 C1 = 11(11.3) J A1 [2]

This question in 9702/22 Oct/Nov 2014

2 (a) Define speed and velocity and use these definitions to explain why one of these quantities is a scalar and the other is a vector. speed: … velocity: … … … [2] (b) A ball is released from rest and falls vertically. The ball hits the ground and rebounds vertically, as shown in Fig. 2.1. initial position ball rebound ground Fig. 2.1 The variation with time t of the velocity v of the ball is shown in Fig. 2.2. 12.0 10.0 8.0 v / m s–1 6.0 4.0 2.0 0 0 1.0 2.0 3.0 t / s – 2.0 – 4.0 – 6.0 – 8.0 – 10.0 Fig. 2.2 Air resistance is negligible. (i) Without calculation, use Fig. 2.2 to describe the variation with time t of the velocity of the ball from t = 0 to t = 2.1 s. … … … … … … [3] (ii) Calculate the acceleration of the ball after it rebounds from the ground. Show your working. acceleration = … m s–2 [3] (iii) Calculate, for the ball, from t = 0 to t = 2.1 s, 1. the distance moved, distance = … m [3] 2. the displacement from the initial position. displacement = … m [2] (iv) On Fig. 2.3, sketch the variation with t of the speed of the ball. 12.0 10.0 8.0 speed / m s–1 6.0 4.0 2.0 0 0 1.0 2.0 3.0 t / s – 2.0 – 4.0 – 6.0 – 8.0 – 10.0 Fig. 2.3 [2]

15 marks

Mark scheme: 2 (a) speed = distance / time and velocity = displacement / time B1 speed is a scalar as distance has no direction and velocity is a vector as displacement has direction B1 [2] (b) (i) constant acceleration or linear/uniform increase in velocity until 1.1 s B1 rebounds or bounces or changes direction B1 decelerates to zero velocity at the same acceleration as initial value B1 [3] (ii) a = (v – u) / t or use of gradient implied C1 = (8.8 + 8.8) / 1.8 or appropriate values from line or = (8.6 + 8.6) / 1.8 B1 = 9.8 (9.78) m s–2 or = 9.6 m s–2 A1 [3] (iii) 1. distance = first area above graph + second area below graph C1 = (1.1 × 10.8) / 2 + (0.9 × 8.8) / 2 (= 5.94 + 3.96) C1 = 9.9 m A1 [3] 2. displacement = first area above graph – second area below graph C1 = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2 = 2.0 (1.98) m A1 [2] (iv) correct shape with straight lines and all lines above the time axis or all below M1 correct times for zero speeds (0.0, 1.15 s, 2.1 s) and peak speeds (10.8 m s–1 at 1.1 s and 8.8 m s–1 at 1.2 s and 3.0 s) A1 [2]

This question in 9702/21 May/June 2015

2 A stone is thrown vertically upwards. The variation with time t of the displacement s of the stone is shown in Fig. 2.1. s 0 0 1.0 2.0 3.0 t / s Fig. 2.1 (a) Use Fig. 2.1 to describe, without calculation, the speed of the stone from t = 0 to t = 3.0 s. … … … [2] (b) Assume air resistance is negligible and therefore the stone has constant acceleration. Calculate, for the stone, (i) the speed at 3.0 s, speed = … m s−1 [3] (ii) the distance travelled from t = 0 to t = 3.0 s, distance = … m [3] (iii) the displacement from t = 0 to t = 3.0 s. displacement = … m direction … [2] (c) On Fig. 2.2, draw the variation with time t of the velocity v of the stone from t = 0 to t = 3.0 s. v / m s–1 0 0 1.0 2.0 3.0 t / s Fig. 2.2 [3]

13 marks

Mark scheme: 2 (a) speed decreases/stone decelerates to rest/zero at 1.25 s B1 speed then increases/stone accelerates (in opposite direction) B1 [2] (b) (i) v = u + at (or s = ut + ½at2 and v2 = u2 + 2as) C1 = 0 + (3.00 – 1.25) × 9.81 C1 = 17.2 (17.17) m s–1 A1 [3] (ii) s = ut + ½at2 s = ½ × 9.81 × (1.25)2 [= 7.66] C1 s = ½ × 9.81 × (1.75)2 [= 15.02] C1 (distance = 7.66 + 15.02) [v = u + at = 0 + 9.81 × (2.50 – 1.25) = 12.26 m s–1] or s = ½ × 9.81 × (1.25)2 [= 7.66] (C1) s = 12.26 × 0.50 + ½ × 9.81 × (3.00 – 2.50)2 [= 7.36] (C1) (distance = 2 × 7.66 + 7.36) Example alternative method: s = (v2 – u2) / 2a = (12.262 – 0) / 2 × 9.81 [= 7.66] (C1) s = (v2 – u2) / 2a = (17.172 – 12.262) / 2 × 9.81 [= 7.36] (C1) (distance = 2 × 7.66 + 7.36) 22.7 (22.69 or 23) m A1 [3] (iii) (s = 15.02 – 7.66 =) 7.4 (7.36) m (ignore sign in answer) A1 down A1 [2] (c) straight line from positive value of v to t axis M1 same straight line crosses t axis at t = 1.25 s A1 same straight line continues with same gradient to t = 3.0 s A1 [3]

This question in 9702/22 May/June 2015

1 (a) The distance between the Sun and the Earth is 1.5 × 1011 m. State this distance in Gm. distance = … Gm [1] (b) The distance from the centre of the Earth to a satellite above the equator is 42.3 Mm. The radius of the Earth is 6380 km. A microwave signal is sent from a point on the Earth directly below the satellite. Calculate the time taken for the microwave signal to travel to the satellite and back. time = … s [2] (c) The speed v of a sound wave through a gas of density ρ and pressure P is given by CP v = ρ where C is a constant. Show that C has no unit. [3] (d) Underline all the scalar quantities in the list below. acceleration energy momentum power weight [1] (e) A boat travels across a river in which the water is moving at a speed of 1.8 m s–1. The velocity vectors for the boat and the river water are shown to scale in Fig. 1.1. water velocity 1.8 m s–1 river boat velocity 3.0 m s–1 60° river bank Fig. 1.1 (shown to scale) In still water the speed of the boat is 3.0 m s–1. The boat is directed at an angle of 60° to the river bank. (i) On Fig. 1.1, draw a vector triangle or a scale diagram to show the resultant velocity of the boat. [2] (ii) Determine the magnitude of the resultant velocity of the boat. resultant velocity = … m s–1 [2]

11 marks

Mark scheme: 1 (a) 150 or 1.5 × 102 Gm A1 [1] (b) distance = 2 × (42.3 – 6.38) × 106 (= 7.184 × 107 m) C1 (time =) 7.184 × 107 / (3.0 × 108) = 0.24 (0.239) s A1 [2] (c) units of pressure P: kg m s–2 / m2 = kg m–1 s–2 M1 units of density ρ: kg m–3 and speed v: m s–1 M1 simplification for units of C: C = v2 ρ / P units: (m2 s–2 kg m–3) / kg m–1 s–2 and cancelling to give no units for C A1 [3] (d) energy and power (both underlined and no others) A1 [1] (e) (i) vector triangle of correct orientation M1 three arrows for the velocities in the correct directions A1 [2] (ii) length measured from scale diagram 5.2 ± 0.2 cm or components of boat speed determined parallel and perpendicular to river flow C1 velocity = 2.6 m s–1 (allow ± 0.1 m s–1) A1 [2]

This question in 9702/23 May/June 2015

2 The variation with time t of the velocity v of a ball is shown in Fig. 2.1. 5 v / m s–1 0 0 2 4 6 8 10 12 14 16 t / s ï Fig. 2.1 The ball moves in a straight line from a point P at t = 0. The mass of the ball is 400 g. (a) Use Fig. 2.1 to describe, without calculation, the velocity of the ball from t = 0 to t = 16 s. … … … … … [2] (b) Use Fig. 2.1 to calculate, for the ball, (i) the displacement from P at t = 10 s, displacement = … m [2] (ii) the acceleration at t = 10 s, acceleration = … m s–2 [2] (iii) the maximum kinetic energy. kinetic energy = … J [2] (c) Use your answers in (b)(i) and (b)(ii) to determine the time from t = 0 for the ball to return to P. time = … s [2]

10 marks

Mark scheme: 2 (a) constant rate of increase in velocity/acceleration from t = 0 to t = 8 s B1 constant deceleration from t = 8 s to t = 16 s or constant rate of increase in velocity in the opposite direction from t = 10 s to t = 16 s B1 [2] (b) (i) area under lines to 10 s C1 (displacement =) (5.0 × 8.0) / 2 + (5.0 × 2.0) / 2 = 25 m or ½ (10.0 × 5.0) = 25 m A1 [2] (ii) a = (v – u) / t or gradient of line C1 = (–15.0 –5.0) / 8.0 = (–) 2.5 m s–2 A1 [2] (iii) KE = ½ m v2 C1 = 0.5 × 0.4 × (15.0)2 = 45 J A1 [2] (c) (distance =) 25 (m) (= ut + ½ at 2) = 0 + ½ × 2.5 × t 2 C1 (t = 4.5 (4.47) s therefore) time to return = 14.5 s A1 [2]

This question in 9702/23 May/June 2015

3 (a) An object is moved from point P to point R either by a direct path or by the path P to Q to R, as shown in Fig. 3.1. R vertical object horizontal P Q Fig. 3.1 P and Q are on the same horizontal level. R is vertically above Q. Explain whether the work done moving the object against the gravitational field is the same or different along paths PR and PQR. … … … … [2] (b) A ball is thrown with an initial velocity V at an angle θ to the horizontal, as shown in Fig. 3.2. path of ball V e horizontal Fig. 3.2 (not to scale) The variation with time t of the height h of the ball is shown in Fig. 3.3. 12.0 10.0 h / m 8.0 6.0 4.0 2.0 0 0 1.00 2.00 3.00 t / s Fig. 3.3 Air resistance is negligible. (i) Use the time to reach maximum height to determine the vertical component Vv of the velocity of the ball for time t = 0. Vv = … m s–1 [2] (ii) The horizontal displacement of the ball at t = 3.00 s is 25.5 m. On Fig. 3.4, draw the variation with t of the horizontal displacement x of the ball. 30 x / m 20 10 0 0 1.00 2.00 3.00 t / s Fig. 3.4 [1] (iii) For the ball at maximum height, calculate the ratio potential energy of the ball . kinetic energy of the ball ratio = … [3] (iv) In practice, air resistance is not negligible. State and explain the effect of air resistance on the time taken for the ball to reach maximum height. … … … … [2]

10 marks

Mark scheme: 3 (a) work done is force × distance moved in direction of force or no work done along PQ as no displacement/distance moved in direction of force B1 work done is same in vertical direction as same distance moved in direction of force B1 [2] (b) (i) at maximum height t = 1.5 (s) or s = ½(u + v) t, s = 11 m and t = 1.5 s C1 Vv = 0 + 9.81 × 1.5 Vv = (11 × 2) / 1.5 = 15 (14.7) m s–1 A1 [2] (ii) straight line from (0,0) to (3.00, 25.5) B1 [1] (iii) at maximum height Vh = 25.5 / 3 (= 8.5 m s–1) B1 ratio = mgh / ½ mv2 C1 = (2 × 9.81 × 11.0) / (8.5)2 = 3.0 (2.99) A1 [3] (iv) deceleration is greater/resultant force (weight and friction force) is greater M1 time is less A1 [2]

This question in 9702/21 Oct/Nov 2015

2 Fig. 2.1 shows an object M on a slope. 3.6 m s–1 Q M P R Fig. 2.1 M moves up the slope, comes to rest at point Q and then moves back down the slope to point R. M has a constant acceleration of 3.0 m s–2 down the slope at all times. At time t = 0, M is at point P and has a velocity of 3.6 m s–1 up the slope. The total distance from P to Q and then to R is 6.0 m. (a) Calculate, for the motion of M from P to Q, (i) the time taken, time = … s [2] (ii) the distance travelled. distance = … m [1] (b) Show that the speed of M at R is 4.8 m s–1. [2] (c) On Fig. 2.2, draw the variation with time t of the velocity v of M for the motion P to Q to R. 6.0 4.0 v / m s–1 2.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 t / s –2.0 –4.0 –6.0 Fig. 2.2 [3] (d) The mass of M is 450 g. Calculate the difference in the kinetic energy of M at P and at R. difference in kinetic energy = … J [2]

10 marks

Mark scheme: 2 (a) (i) v = u + at C1 0 = 3.6 – 3.0t t (= 3.6 / 3.0) = 1.2 s A1 [2] (ii) (distance to rest from P = (3.6 × 1.2) / 2 =) 2.2 (2.16) m A1 [1] or [0 – (3.6)2] / [2 × (–3.0)] = 2.2 (2.16) m or 3.6 × 1.2 – ½ × 3.0 × (1.2)2 = 2.2 (2.16) m or 0 + ½ × 3.0 × (1.2)2 = 2.2 (2.16) m (b) distance = 6.0 – 2.16 (= 3.84) C1 v2 = u2 + 2as = 2 × 3.0 × 3.84 (= 23.04) M1 or x + 2 × 2.16 = 6.0 gives x = 1.68 (m) (C1) v2 = 3.62 + 2 × 1.68 × 3.0 (= 23.04) (M1) or correct method with intermediate time calculated (t = 1.6 s from Q to R) v = 4.8 m s–1 A0 [2] (c) straight line from v = 3.6 m s–1 to v = 0 at t = 1.2 s B1 straight line continues with the same gradient as v changes sign B1 straight line from v = 0 intercept to v = –4.8 m s–1 B1 [3] (d) difference in KE = ½m(v2 – u2) = 0.5 × 0.45 (4.82 – 3.62) [= 5.184 – 2.916] C1 = 2.3 (2.27) J A1 [2]

This question in 9702/22 Oct/Nov 2015

3 A steel ball falls from a platform on a tower to the ground below, as shown in Fig. 3.1. ball platform path of tower 192 m ball ground Fig. 3.1 The ball falls from rest through a vertical distance of 192 m. The mass of the ball is 270 g. (a) Assume air resistance is negligible. (i) Calculate 1. the time taken for the ball to fall to the ground, time taken = … s [2] 2. the maximum kinetic energy of the ball. maximum kinetic energy = … J [2] (ii) State and explain the variation of the velocity of the ball with time as the ball falls to the ground. … … [1] (iii) Show that the velocity of the ball on reaching the ground is approximately 60 m s–1. [1] (b) In practice, air resistance is not negligible. The variation of the air resistance R with the velocity v of the ball is shown in Fig. 3.2. 4.0 3.0 R / N 2.0 1.0 0 0 20 40 60 80 100 v / m s–1 Fig. 3.2 (i) Use Fig. 3.2 to state and explain qualitatively the variation of the acceleration of the ball with the distance fallen by the ball. … … … … [3] (ii) The speed of the ball reaches 40 m s–1. Calculate its acceleration at this speed. acceleration = … m s–2 [2] (iii) Use information from (a)(iii) and Fig. 3.2 to state and explain whether the ball reaches terminal velocity. … … … [2]

13 marks

Mark scheme: 3 (a) (i) 1. s = ut + ½ at2 192 = ½ × 9.81 × t2 C1 t = 6.3 (6.26) s A1 [2] 2. max Ek (= mgh) = 0.27 × 9.81 × 192 C1 or calculation of v (= 61.4) and use of EK (= ½ mv2) = ½ × 0.27 × (61.4)2 (C1) max Ek = 510 (509) J A1 [2] (ii) velocity is proportional to time or velocity increases at a constant rate as acceleration is constant or resultant force is constant B1 [1] (iii) use of v = at or v2 = 2as or E = ½ mv2 to give v = 61(.4) m s–1 B1 [1] (b) (i) R increases with velocity B1 resultant force is mg – R or resultant force decreases B1 acceleration decreases B1 [3] (ii) at v = 40 m s–1, R = 0.6 (N) C1 0.27 × 9.8 – 0.6 = 0.27 × a a = 7.6 (7.58) m s–2 A1 [2] (iii) R = weight for terminal velocity B1 either weight requires velocity to be about 80 m s–1 or at 60 m s–1, R is less than weight so does not reach terminal velocity B1 [2]

This question in 9702/23 Oct/Nov 2015

2 (a) Define acceleration. … … [1] (b) A ball is kicked from horizontal ground towards the top of a vertical wall, as shown in Fig. 2.1. path of ball v wall ball 28° horizontal ground 24 m Fig. 2.1 (not to scale) The horizontal distance between the initial position of the ball and the base of the wall is 24 m. The ball is kicked with an initial velocity v at an angle of 28° to the horizontal. The ball hits the top of the wall after a time of 1.5 s. Air resistance may be assumed to be negligible. (i) Calculate the initial horizontal component vX of the velocity of the ball. vX = … m s−1 [1] (ii) Show that the initial vertical component vY of the velocity of the ball is 8.5 m s−1. [2] (iii) Calculate the time taken for the ball to reach its maximum height above the ground. time = … s [2] (iv) The ball is kicked at time t = 0. On Fig. 2.2, sketch the variation with time t of the vertical component vY of the velocity of the ball until it hits the wall. It may be assumed that velocity is positive when in the upwards direction. 10.0 vY / m s–1 5.0 0 00 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 t / s –5.0 –10.0 Fig. 2.2 [2] (c) (i) Use the information in (b) to determine the maximum height of the ball above the ground. maximum height = … m [2] (ii) The maximum gravitational potential energy of the ball above the ground is 22 J. Calculate the mass of the ball. mass = … kg [2] (d) A ball of greater mass is kicked with the same velocity as the ball in (b). State and explain the effect, if any, of the increased mass on the maximum height reached by the ball. Air resistance is still assumed to be negligible. … … [1] [Total: 13]

13 marks

Mark scheme: 2 (a) change in velocity / time (taken) or rate of change of velocity B1 (b) (i) vX = (24 / 1.5) = 16 (m s–1) A1 (ii) tan 28° = vY / vX or vX = v cos 28° and vY = v sin 28° C1 vY = 16 tan 28° or vY = 16 × (sin 28° / cos 28°) so vY = 8.5 (m s–1) A1 (iii) v = u + at C1 t = (0 – 8.5) / (–9.81) = 0.87 (s) A1 (iv) straight line from positive vY at t = 0 to negative vY at t =1.5 s M1 line starts at (0, 8.5) and crosses t-axis at (0.87, 0) and does not go beyond t = 1.5 s. A1 (c) (i) (v 2 = u 2 + 2as) 0 = 8.52 + 2(–9.81)s or (s = ut + ½at 2) s = 8.5×0.87 + ½ × (–9.81) × 0.872 or (s = vt – ½at 2) s = 0 – ½×(–9.81)×0.872 or (s = ½(u + v)t or area under graph) s = 0.5 × 8.5 × 0.87 C1 s = 3.7 (m) A1 (ii) ∆EP = mg∆h (allow E = mgh) C1 m = 22 / (9.81 × 3.7) = 0.61 (kg) A1 (d) acceleration (of freefall) is unchanged / not dependent on mass, and so no effect (on maximum height) or explanation in terms of energy: (initial) KE ∝ mass, (∆)KE = (∆)PE, (max) PE ∝ mass, and so no effect (on maximum height) B1

This question in 9702/22 Feb/March 2016

2 A ball is thrown from a point P with an initial velocity u of 12 m s–1 at 50° to the horizontal, as illustrated in Fig. 2.1. path of ball Q X =12 m s–1 50° P horizontal Fig. 2.1 The ball reaches maximum height at Q. Air resistance is negligible. (a) Calculate (i) the horizontal component of u, horizontal component = … m s–1 [1] (ii) the vertical component of u. vertical component = … m s–1 [1] (b) Show that the maximum height reached by the ball is 4.3 m. [2] (c) Determine the magnitude of the displacement PQ. displacement = … m [4] [Total: 8]

8 marks

Mark scheme: 2 (a) (i) horizontal component (= 12 cos 50°) = 7.7 m s–1 A1 [1] (ii) vertical component (= 12 sin 50° or 7.7 tan 50°) = 9.2 m s–1 A1 [1] (b) v2 = u2 + 2as and v = 0 or mgh = ½mv2 or s = v2 sin2θ / 2g C1 9.22 = 2 × 9.81 × h hence h = 4.3 (4.31) m A1 [2] alternative methods using time to maximum height of 0.94 s: s = ut + ½at2 and t = 0.94 (s) (C1) s = 9.2 × 0.94 – ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = vt – ½at2 and t = 0.94 (s) (C1) s = ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = ½(u + v)t and t = 0.94 (s) (C1) s = ½ × 9.2 × 0.94 hence s = 4.3 m (A1) (c) t (= 9.2 / 9.81) = 0.94 (0.938) s C1 horizontal distance = 0.938 × 7.7 (= 7.23 m) C1 displacement = [4.32 + 7.232]1/2 C1 = 8.4 m A1 [4]

This question in 9702/21 May/June 2016

1 (a) Define acceleration. … … [1] (b) A man travels on a toboggan down a slope covered with snow from point A to point B and then to point C. The path is illustrated in Fig. 1.1. man toboggan, at rest A 40° horizontal B 20° horizontal C Fig. 1.1 (not to scale) The slope AB makes an angle of 40° with the horizontal and the slope BC makes an angle of 20° with the horizontal. Friction is not negligible. The man and toboggan have a combined mass of 95 kg. The man starts from rest at A and has constant acceleration between A and B. The man takes 19 s to reach B. His speed is 36 m s–1 at B. (i) Calculate the acceleration from A to B. acceleration = … m s–2 [2] (ii) Show that the distance moved from A to B is 340 m. [1] (iii) For the man and toboggan moving from A to B, calculate 1. the change in kinetic energy, change in kinetic energy = … J [2] 2. the change in potential energy. change in potential energy = … J [2] (iv) Use your answers in (iii) to determine the average frictional force that acts on the toboggan between A and B. frictional force = … N [2] (v) A parachute opens on the toboggan as it passes point B. There is a constant deceleration of 3.0 m s–2 from B to C. Calculate the frictional force that produces this deceleration between B and C. frictional force = … N [2] [Total: 12]

12 marks

Mark scheme: 1 (a) acceleration = change in velocity / time (taken) or rate of change of velocity B1 [1] (b) (i) v = 0 + at or v = at C1 (a = 36 / 19 =) 1.9 (1.8947) m s–2 A1 [2] (ii) s = ½(u + v)t or s = v2 / 2a or s = ½at2 = ½ × 36 × 19 = 362 / (2 × 1.89) = ½ × 1.89 × 192 = 340 m (342 m / 343 m / 341 m) M1 [1] (iii) 1. (∆KE =) ½ × 95 × (36)2 C1 = 62 000 (61 560) J A1 [2] 2. (∆PE =) 95 × 9.81 × 340 sin 40° or 95 × 9.81 × 218.5 C1 = 200 000 J A1 [2] (iv) work done (by frictional force) = ∆PE – ∆KE or work done = 200 000 – 62 000 (values from 1b(iii) 1. and 2.) C1 (frictional force = 138 000 / 340 =) 410 (406) N [420 N if full figures used] A1 [2] (v) –ma = mg sin 20° – f or ma = –mg sin 20° + f C1 –95 × 3.0 = 95 × 3.36 – f f = 600 (604) N A1 [2]

This question in 9702/22 May/June 2016

3 (a) Explain what is meant by gravitational potential energy and by kinetic energy. gravitational potential energy: … … kinetic energy: … … [2] (b) A motion sensor is used to measure the velocity of a ball falling vertically towards the ground, as illustrated in Fig. 3.1. motion sensor v A B ground Fig. 3.1 The ball passes through points A and B as it falls. The ball has a mass of 1.5 kg. The variation with time t of the velocity v of the ball as it falls from A to B is shown in Fig. 3.2. 8.0 7.0 6.0 v / m s–1 5.0 4.0 3.0 0.40 0.60 0.80 t / s ball at position A ball at position B Fig. 3.2 Use Fig. 3.2 to calculate, for the ball falling from A to B, (i) the displacement, displacement = … m [3] (ii) the acceleration, acceleration = … m s–2 [2] (iii) the change in kinetic energy. change in kinetic energy = … J [3] (c) Show that the work done by the gravitational field on the ball in (b) as it moves from A to B is equal to the change in kinetic energy. [2] [Total: 12]

12 marks

Mark scheme: 3 (a) (gravitational potential energy is) the energy/ability to do work of a mass that it has or is stored due to its position/height in a gravitational field B1 kinetic energy is energy/ability to do work a object/body/mass has due to its speed/velocity/motion/movement B1 [2] (b) (i) s = [(u + v) t] / 2 or acceleration = 9.8/9.75 (using gradient) C1 = [(7.8 + 3.9) × 0.4] / 2 or s = 3.9 × 0.4 + 21 × 9.75 × (0.4)2 C1 s = 2.3(4) m A1 [3] (ii) a = (v – u) / t or gradient of line C1 = (7.8 – 3.9) / 0.4 = 9.8 (9.75) m s–2 (allow ± 21 small square in readings) A1 [2] (iii) KE = 21 mv2 C1 change in kinetic energy = 21 mv2 – 21 mu2 = 21 × 1.5 × (7.82 – 3.92) C1 = 34 (34.22) J A1 [3] (c) work done = force × distance (moved) or Fd or Fx or mgh or mgd or mgx M1 = 1.5 × 9.8 × 2.3 = 34 (33.8) J (equals the change in KE) A1 [2]

This question in 9702/23 May/June 2016

2 (a) Define electric field strength. … … [1] (b) A potential difference of 2.5 kV is applied across a pair of horizontal metal plates in a vacuum, as shown in Fig. 2.1. metal plate \ 2.0 cm electron B + 2.5 kV velocity A – 2.0 cm 3.7 × 107 m s–1 metal 5.9 cm plate Fig. 2.1 (not to scale) Each plate has a length of 5.9 cm. The separation of the plates is 4.0 cm. The arrangement produces a uniform electric field between the plates. Assume the field does not extend beyond the edges of the plates. An electron enters the field at point A with horizontal velocity 3.7 × 107 m s–1 along a line mid-way between the plates. The electron leaves the field at point B. (i) Calculate the time taken for the electron to move from A to B. time taken = … s [1] (ii) Calculate the magnitude of the electric field strength. field strength = … N C–1 [2] (iii) Show that the acceleration of the electron in the field is 1.1 × 1016 m s–2. [2] (iv) Use the acceleration given in (iii) and your answer in (i) to determine the vertical distance y between point B and the upper plate. y = … cm [3] (v) Explain why the calculation in (iv) does not need to include the gravitational effects on the electron. … … [1] (vi) The electron enters the field at time t = 0. On Fig. 2.2, sketch graphs to show the variation with time t of 1. the horizontal component vX of the velocity of the electron, 2. the vertical component v of the velocity of the electron. Y Numerical values are not required. vX vY 0 0 0 0 t t Fig. 2.2 [2] [Total: 12]

12 marks

Mark scheme: 2 (a) force per unit positive charge B1 [1] (b) (i) time = 5.9 × 10–2 / 3.7 × 107 = 1.6 × 10–9 s (1.59 × 10–9 s) A1 [1] (ii) E = V / d C1 = 2500 / 4.0 × 10–2 = 6.3 × 104 N C–1 (6.25 × 104 or 62500 N C–1) A1 [2] (iii) a = Eq / m or F = ma and F = Eq C1 = (6.3 × 104 × 1.60 × 10–19) / 9.11 × 10–31 = 1.1 × 1016 m s–2 A1 [2] (iv) s = ut + ½at 2 = ½ × 1.1 × 1016 × (1.6 × 10–9)2 C1 = 1.4 × 10–2 (m) C1 distance from plate = 2.0 – 1.4 = 0.6 cm (allow 1 or more s.f.) A1 [3] (v) electric force ≫ gravitational force (on electron)/weight or acceleration due to electric field ≫ acceleration due to gravitational field B1 [1] (vi) vX–t graph: horizontal line at a non-zero value of vX B1 vY–t graph: straight line through the origin with positive gradient B1 [2]

This question in 9702/21 Oct/Nov 2016

2 (a) Define electric field strength. … … [1] (b) A potential difference of 2.5 kV is applied across a pair of horizontal metal plates in a vacuum, as shown in Fig. 2.1. metal plate y 2.0 cm electron B + 2.5 kV velocity A – 2.0 cm 3.7 × 107 m s–1 metal 5.9 cm plate Fig. 2.1 (not to scale) Each plate has a length of 5.9 cm. The separation of the plates is 4.0 cm. The arrangement produces a uniform electric field between the plates. Assume the field does not extend beyond the edges of the plates. An electron enters the field at point A with horizontal velocity 3.7 × 107 m s–1 along a line mid-way between the plates. The electron leaves the field at point B. (i) Calculate the time taken for the electron to move from A to B. time taken = … s [1] (ii) Calculate the magnitude of the electric field strength. field strength = … N C–1 [2] (iii) Show that the acceleration of the electron in the field is 1.1 × 1016 m s–2. [2] (iv) Use the acceleration given in (iii) and your answer in (i) to determine the vertical distance y between point B and the upper plate. y = … cm [3] (v) Explain why the calculation in (iv) does not need to include the gravitational effects on the electron. … … [1] (vi) The electron enters the field at time t = 0. On Fig. 2.2, sketch graphs to show the variation with time t of 1. the horizontal component vX of the velocity of the electron, 2. the vertical component v of the velocity of the electron. Y Numerical values are not required. vX vY 0 0 0 0 t t Fig. 2.2 [2] [Total: 12]

12 marks

Mark scheme: 2 (a) force per unit positive charge B1 [1] (b) (i) time = 5.9 × 10–2 / 3.7 × 107 = 1.6 × 10–9 s (1.59 × 10–9 s) A1 [1] (ii) E = V / d C1 = 2500 / 4.0 × 10–2 = 6.3 × 104 N C–1 (6.25 × 104 or 62500 N C–1) A1 [2] (iii) a = Eq / m or F = ma and F = Eq C1 = (6.3 × 104 × 1.60 × 10–19) / 9.11 × 10–31 = 1.1 × 1016 m s–2 A1 [2] (iv) s = ut + ½at 2 = ½ × 1.1 × 1016 × (1.6 × 10–9)2 C1 = 1.4 × 10–2 (m) C1 distance from plate = 2.0 – 1.4 = 0.6 cm (allow 1 or more s.f.) A1 [3] (v) electric force ≫ gravitational force (on electron)/weight or acceleration due to electric field ≫ acceleration due to gravitational field B1 [1] (vi) vX–t graph: horizontal line at a non-zero value of vX B1 vY–t graph: straight line through the origin with positive gradient B1 [2]

This question in 9702/23 Oct/Nov 2016

5 An electron is travelling in a straight line through a vacuum with a constant speed of 1.5 × 107 m s–1. The electron enters a uniform electric field at point A, as shown in Fig. 5.1. uniform electric field 2.0 cm electron speed A B 1.5 × 107 m s–1 Fig. 5.1 The electron continues to move in the same direction until it is brought to rest by the electric field at point B. Distance AB is 2.0 cm. (a) State the direction of the electric field. … [1] (b) Calculate the magnitude of the deceleration of the electron in the field. deceleration = … m s–2 [2] (c) Calculate the electric field strength. electric field strength = … V m–1 [3] (d) The electron is at point A at time t = 0. On Fig. 5.2, sketch the variation with time t of the velocity v of the electron until it reaches point B. Numerical values of v and t do not need to be shown. v 0 0 t Fig. 5.2 [1] [Total: 7]

7 marks

Mark scheme: 5(a) to the right / from the left / from A to B / in the same direction as electron velocity B1 5(b) v 2 = u 2 + 2as a = (1.5 × 107)2 / (2 × 2.0 × 10–2) Other alternative calculations for the C1 mark: e.g. a = 1.5×107 / 2.67×10–9 e.g. a = [(1.5×107 × 2.67×10–9) – 2.0×10–2] × [2 / (2.67×10–9)2] e.g. a = (2.0×10–2 × 2) / (2.67×10–9)2 C1 = 5.6 × 1015 m s–2 A1 5(c) E = F / Q C1 = (9.1 × 10–31 × 5.6 × 1015) / 1.6 × 10–19 C1 = 3.2 × 104 V m–1 A1 5(d) straight line with negative gradient starting at an intercept on the v-axis and ending at an intercept on the t-axis. B1

This question in 9702/22 Feb/March 2017

2 (a) State the two conditions for a system to be in equilibrium. 1. … … 2. … … [2] (b) A paraglider P of mass 95 kg is pulled by a wire attached to a boat, as shown in Fig. 2.1. parachute paraglider P wire boat horizontal 25° water Fig. 2.1 The wire makes an angle of 25° with the horizontal water surface. P moves in a straight line parallel to the surface of the water. The variation with time t of the velocity v of P is shown in Fig. 2.2. 10.0 8.0 v / m s–1 6.0 4.0 2.0 0 0 2.0 4.0 6.0 8.0 t / s Fig. 2.2 (i) Show that the acceleration of P is 1.4 m s–2 at time t = 5.0 s. [2] (ii) Calculate the total distance moved by P from time t = 0 to t = 7.0 s. distance = … m [2] (iii) Calculate the change in kinetic energy of P from time t = 0 to t = 7.0 s. change in kinetic energy = … J [2] (iv) The tension in the wire at time t = 5.0 s is 280 N. Calculate, for the horizontal motion, 1. the vertical lift force F supporting P, F = … N [3] 2. the force R due to air resistance acting on P in the horizontal direction. R = … N [3] [Total: 14]

14 marks

Mark scheme: 2(a) resultant force (in any direction) is zero B1 resultant torque/moment (about any point) is zero B1 2(b)(i) a = (v − u) / t or gradient or ∆v / (∆)t C1 e.g. a = (8.8 − 4.6) / (7.0 – 4.0) = 1.4 m s–2 A1 2(b)(ii) s = 4.6 × 4 + [(8.8 + 4.6) / 2] × 3 C1 = 18.4 + 20.1 = 39 (38.5) m A1 2(b)(iii) ∆E = ½ × 95 [(8.8)2 − (4.6)2] C1 = 3678 – 1005 = 2700 (2673) J A1 2(b)(iv)1. weight = 95 × 9.81 (= 932 N) C1 vertical tension force = 280 sin 25° or 280 cos 65° (=118.3 N) C1 F = 932 + 118 = 1100 (1050) N A1 2(b)(iv)2. horizontal tension force = 280 cos 25° or 280 sin 65° (= 253.8 N) C1 resultant force = 95 × 1.4 (= 133 N) C1 133 = 253.8 – R R = 120 (120.8) N A1

This question in 9702/21 May/June 2017

2 (a) Define velocity. … … [1] (b) A ball of mass 0.45 kg leaves the edge of a table with a horizontal velocity v, as shown in Fig. 2.1. ball v path of ball 1.25 m table 1.50 m floor horizontal Fig. 2.1 The height of the table is 1.25 m. The ball travels a distance of 1.50 m horizontally before hitting the floor. Air resistance is negligible. Calculate, for the ball, (i) the horizontal velocity v as it leaves the table, v = … m s–1 [3] (ii) the velocity just as it hits the floor, magnitude of velocity = … m s–1 angle to the horizontal = … ° [4] (iii) the kinetic energy just as it hits the floor, kinetic energy = … J [2] (iv) the loss in gravitational potential energy as it falls from the table to the floor. loss in potential energy = … J [2] (c) Explain why the kinetic energy of the ball in (b)(iii) does not equal the loss of gravitational potential energy in (b)(iv). … … [1] [Total: 13]

13 marks

Mark scheme: 2(a) rate of change of displacement or change in displacement/time taken B1 2(b)(i) s = ut + ½at2 C1 t = [(2 × 1.25) / 9.81]1/2 (= 0.5048 s) C1 or v2 = u2 + 2as vvert = (2 × 9.81 × 1.25)1/2 (= 4.95) (C1) t = [2s / (u + v)] = 2 × 1.25 / 4.95 (= 0.5048 s) (C1) v = d / t = 1.5 / 0.50(48) = 3.0 (2.97) m s–1 A1 2(b)(ii) vertical velocity = at = 9.81 × 0.5048 (= 4.95) [using t = 0.50 gives 4.9] C1 velocity = [(vh)2 + (vv)2]1/2 C1 = [(2.97)2 + (4.95)2]1/2 = 5.8 (5.79) [using t = 0.50 leads to 5.7] A1 direction (= tan–1 4.95/2.97) = 59° A1 2(b)(iii) kinetic energy = ½mv2 C1 = ½ × 0.45 × (5.8)2 = 7.6 (7.57) J [using t = 0.50 leads to 7.3 J] A1 Question Answer Marks 2(b)(iv) potential energy = mgh C1 = (0.45 × 9.81 × 1.25) = 5.5 (5.52) J A1 2(c) there is KE of the ball at the start/leaving table or the ball has an initial/constant horizontal velocity or the ball has velocity at start/leaving table B1

This question in 9702/22 May/June 2017

5 (a) Define electric field strength. … … [1] (b) Two parallel metal plates in a vacuum are separated by 0.045 m. A potential difference V is applied between the plates, as shown in Fig. 5.1. metal plate – 0.045 m V + metal plate proton Fig. 5.1 A proton is initially at rest on the surface of the positive plate. The proton in the uniform electric field takes a time of 1.5 × 10–7 s to reach the negative plate. (i) Show that the acceleration of the proton is 4.0 × 1012 m s–2. [2] (ii) Calculate the electric force on the proton. force = … N [1] (iii) Use your answer in (ii) to determine 1. the electric field strength, field strength = … N C–1 [2] 2. the potential difference V between the plates. V = … V [2] (c) An α particle is now accelerated between the two metal plates in (b) by the electric field. Calculate the ratio acceleration of α particle . acceleration of proton ratio = … [2] [Total: 10]

10 marks

Mark scheme: 5(a) force per unit positive charge B1 5(b)(i) s = ½at 2 C1 a = (2 × 0.045) / (1.5 × 10–7)2 = 4(.0) × 1012 m s–2 A1 5(b)(ii) F = 1.67 × 10–27 × 4.0 × 1012 = 6.7 (6.68) × 10–15N A1 5(b)(iii) 1. E = F / Q C1 = 6.68 × 10–15 / 1.6 × 10–19 = 4.2 (4.18) × 104 N C–1 A1 2. E = V / d C1 V = 4.18 × 104 × 0.045 = 1.9 × 103 V A1 Question Answer Marks 5(c) a = Eq / m or F = ma and F = Eq C1 ratio = − − − − × × × × × × × × 19 27 19 27 (2 1.6 10 ) (1.67 10 ) (1.6 10 ) (4 1.66 10 ) or × × 2 1 1 4 = 0.50 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

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

1 (a) Define (i) displacement, … … [1] (ii) acceleration. … … [1] (b) A remote-controlled toy car moves up a ramp and travels across a gap to land on another ramp, as illustrated in Fig. 1.1. path of car 5.5 m s–1 car ramp P ramp Q d ground θ Fig. 1.1 The car leaves ramp P with a velocity of 5.5 m s–1 at an angle θ to the horizontal. The horizontal component of the car’s velocity as it leaves the ramp is 4.6 m s–1. The car lands at the top of ramp Q. The tops of both ramps are at the same height and are distance d apart. Air resistance is negligible. (i) Show that the car leaves ramp P with a vertical component of velocity of 3.0 m s–1. [1] (ii) Determine the time taken for the car to travel between the ramps. time taken = … s [2] (iii) Calculate the horizontal distance d between the tops of the ramps. d = … m [1] (iv) Calculate the ratio kinetic energy of the car at its maximum height . kinetic energy of the car as it leaves ramp P ratio = … [3] (c) Ramp Q is removed. The car again leaves ramp P as in (b) and now lands directly on the ground. The car leaves ramp P at time t = 0 and lands on the ground at time t = T. On Fig. 1.2, sketch the variation with time t of the vertical component vy of the car’s velocity from t = 0 to t = T. Numerical values of vy and t are not required. vy 0 0 TT t t Fig. 1.2 [2] [Total: 11]

11 marks

Mark scheme: 1(a)(i) distance in a specified direction (from a point) B1 1(a)(ii) change in velocity / time (taken) B1 1(b)(i) vertical component of velocity = (5.52 – 4.62)1/2 = 3.0 (m s–1) or 5.5 cos θ = 4.6 (so θ = 33.2°) and 5.5 sin 33.2° = 3.0 (m s–1) A1 1(b)(ii) s = ut + ½at 2 0 = (3.0 × t) – (½ × 9.81 × t 2) or v = u + at –3.0 = 3.0 – 9.81t C1 t = 0.61 s A1 1(b)(iii) d = 4.6 × 0.61 = 2.8 m A1 1(b)(iv) E = ½mv2 C1 ratio = (½ × m × 4.62) / (½ × m × 5.52) or ratio = (½ × m × 5.52 – m × 9.81 × 0.459) / (½ × m × 5.52) C1 ratio = 0.70 A1 1(c) straight line from positive value of vy at t = 0 to negative value of vy M1 straight line ends at t = T and final magnitude of vy greater than initial magnitude of vy A1

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

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

2 (a) A resultant force F moves an object of mass m through distance s in a straight line. The force gives the object an acceleration a so that its speed changes from initial speed u to final speed v. (i) State an expression for: 1. the work W done by the force, in terms of a, m and s W = … [1] 2. the distance s, in terms of a, u and v. s = … [1] (ii) Use your answers in (i) to show that the kinetic energy of the object is given by 1 kinetic energy = × mass × (speed)2. 2 Explain your working. [2] (b) A ball of mass 0.040 kg is projected into the air from horizontal ground, as illustrated in Fig. 2.1. Y path of ball h ball, mass 0.040 kg X ground Fig. 2.1 The ball is launched from a point X with a kinetic energy of 4.5 J. At point Y, the ball has a speed of 9.5 m s−1. Air resistance is negligible. (i) For the movement of the ball from X to Y, draw a solid line on Fig. 2.1 to show: 1. the distance moved (label this line D) 2. the displacement (label this line S). [2] (ii) By consideration of energy transfer, determine the height h of point Y above the ground. h = … m [3] (iii) On Fig. 2.2, sketch the variation of the kinetic energy of the ball with its vertical height above the ground for the movement of the ball from X to Y. Numerical values are not required. kinetic energy 0 0 h height Fig. 2.2 [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) 1. W = mas B1 2. s = (v 2 – u 2) / 2a B1 2(a)(ii) W/work equals energy transferred/gain or change in kinetic energy B1 W (= mas) = ma(v 2 – u 2) / 2a leading to W = m(v 2 – u 2) / 2 (so KE = ½mv 2) B1 2(b)(i) 1. solid curved line drawn from X to Y along path of ball and labelled D B1 2. solid straight line drawn from X to Y and labelled S B1 2(b)(ii) (∆)E = mg(∆)h C1 4.5 = (0.040 × 9.81 × h) + (½ × 0.040 × 9.52) C1 h = 6.9 m A1 2(b)(iii) line with a negative gradient starting from a non-zero value of kinetic energy when the vertical height is zero M1 straight line ends at a non-zero value of kinetic energy when the vertical height is h A1

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

2 A small charged glass bead of weight 5.4 × 10–5 N is initially at rest at point A in a vacuum. The bead then falls through a uniform horizontal electric field as it moves in a straight line to point B, as illustrated in Fig. 2.1. vertical glass bead weight 5.4 × 10–5 N A horizontal charge –3.7 × 10–9 C uniform horizontal path of the electric field, × 104 V m–1 falling bead field strength 1.3 B side view Fig. 2.1 (not to scale) The electric field strength is 1.3 × 104 V m–1. The charge on the bead is –3.7 × 10–9 C. (a) Describe how two metal plates could be used to produce the electric field. Numerical values are not required. … … … [2] (b) Determine the magnitude of the electric force acting on the bead. electric force = … N [2] (c) Use your answer in (b) and the weight of the bead to show that the resultant force acting on it is 7.2 × 10–5 N. [1] (d) Explain why the resultant force on the bead of 7.2 × 10–5 N is constant as the bead moves along path AB. … … … … [2] (e) (i) Calculate the magnitude of the acceleration of the bead along the path AB. acceleration = … m s–2 [2] (ii) The path AB has length 0.58 m. Use your answer in (i) to determine the speed of the bead at point B. speed = … m s–1 [2] [Total: 11]

11 marks

Mark scheme: 2(a) the (two) plates are vertical (and separated) B1 left plate positively charged and right plate negatively charged/earthed or right plate negatively charged and left plate positively charged/earthed B1 2(b) F = Eq C1 = 1.3 × 104 × 3.7 × 10–9 = 4.8 × 10–5 N A1 2(c) F2 = (4.8 × 10–5)2 + (5.4 × 10–5)2 so F = 7.2 × 10–5 N or F = [(4.8 × 10–5)2 + (5.4 × 10–5)2]0.5 so F = 7.2 × 10–5 N A1 2(d) electric force is constant (because field strength/E is constant) B1 weight is constant (and so resultant force constant) B1 2(e)(i) m = 5.4 × 10–5 / 9.81 (= 5.5 × 10–6) C1 a = 7.2 × 10–5 / (5.5 × 10–6) =13 m s–2 A1 2(e)(ii) v2 = u2 + 2as v2 = 2 × 13 × 0.58 C1 v = 3.9 m s–1 A1

This question in 9702/21 Oct/Nov 2019

2 (a) Define acceleration. … [1] (b) A steel ball of diameter 0.080 m is released from rest and falls vertically in air, as illustrated in Fig. 2.1. position of ball steel ball of when released diameter 0.080 m 0.280 m horizontal beam of light of position P negligible width of ball Fig. 2.1 (not to scale) A horizontal beam of light of negligible width is a vertical distance of 0.280 m below the bottom of the ball when it is released. The ball falls through and breaks the beam of light. (i) Explain why the force due to air resistance acting on the ball may be neglected when calculating the time taken for the ball to reach the beam of light. … … [1] (ii) Calculate the time taken for the ball to fall from rest to position P where the bottom of the ball touches the beam of light. time taken = … s [2] (iii) Determine the time interval during which the beam of light is broken by the ball. time interval = … s [2] (c) A different ball is released from the same position as the steel ball in (b). This ball has the same diameter but a much lower density. For this ball, the force due to air resistance cannot be neglected as the ball falls. State and explain the change, if any, to the time interval during which the beam of light is broken by the ball. … … … … [2] [Total: 8]

8 marks

Mark scheme: 2(a) change in velocity / time (taken) A1 2(b)(i) weight ≫ (force due to) air resistance or (force due to) air resistance is negligible compared to weight B1 2(b)(ii) s = ut + ½at 2 0.280 = ½ × 9.81 × t 2 C1 t = 0.24 s A1 Question Answer Marks 2(b)(iii) total distance fallen = 0.280 + 0.080 = 0.360 0.360 = ½ × 9.81 × t 2 t = 0.27 s C1 time taken = 0.27 – 0.24 = 0.03 s A1 or v = 9.81 × 0.239 or (2 × 9.81 × 0.280)0.5 or (2 × 0.280) / 0.239 v = 2.34 (m s–1) (C1) 0.080 = 2.34t + ½ × 9.81 × t 2 solving quadratic equation gives t = 0.03 s allow any correct method using equations of uniform accelerated motion (A1) 2(c) (average) resultant force/acceleration/speed/velocity (of low-density ball) is less B1 (so) time interval is longer B1

This question in 9702/22 Oct/Nov 2019

2 A dolphin is swimming under water at a constant speed of 4.50 m s–1. (a) The dolphin emits a sound as it swims directly towards a stationary submerged diver. The frequency of the sound heard by the diver is 9560 Hz. The speed of sound in the water is 1510 m s–1. Determine the frequency, to three significant figures, of the sound emitted by the dolphin. frequency = … Hz [2] (b) The dolphin strikes the bottom of a floating ball so that the ball rises vertically upwards from the surface of the water, as illustrated in Fig. 2.1. path of ball height of ball above ball surface surface of water speed 5.6 m s–1 Fig. 2.1 The ball leaves the water surface with speed 5.6 m s–1. Assume that air resistance is negligible. (i) Calculate the maximum height reached by the ball above the surface of the water. height = … m [2] (ii) The ball leaves the water at time t = 0 and reaches its maximum height 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 time t Fig. 2.2 [1] (iii) The mass of the ball is 0.45 kg. Use your answer in (b)(i) to calculate the change in gravitational potential energy of the ball as it rises from the surface of the water to its maximum height. change in gravitational potential energy = … J [2] (iv) State and explain the variation in the magnitude of the acceleration of the ball as it falls back towards the surface of the water if air resistance is not negligible. … … … … … [2] [Total: 9]

9 marks

Mark scheme: 2(a) f0 = fS v / (v – vS) 9560 = f × 1510 / (1510 – 4.50) C1 f = 9530 Hz A1 2(b)(i) v 2 = u 2 + 2as height = 5.62 / (2 × 9.81) C1 = 1.6 m A1 2(b)(ii) downward sloping straight line starting from a point on the speed axis and ending at point (T, 0) B1 Question Answer Marks 2(b)(iii) (Δ)E = mg(Δ)h = 0.45 × 9.81 × 1.6 C1 = 7.1 J A1 2(b)(iv) air resistance increases (and weight constant) B1 (resultant force decreases so) acceleration decreases B1

This question in 9702/22 Feb/March 2020

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

2 (a) Fig. 2.1 shows the velocity–time graph for an object moving in a straight line. v velocity u 0 0 t time Fig. 2.1 (i) Determine an expression, in terms of u, v and t, for the area under the graph. area = … [1] (ii) State the name of the quantity represented by the area under the graph. … [1] (b) A ball is kicked with a velocity of 15 m s–1 at an angle of 60° to horizontal ground. The ball then strikes a vertical wall at the instant when the path of the ball becomes horizontal, as shown in Fig. 2.2. path of ball vertical velocity wall 15 m s–1 ball 60° horizontal ground Fig. 2.2 (not to scale) Assume that air resistance is negligible. (i) By considering the vertical motion of the ball, calculate the time it takes to reach the wall. time = … s [3] (ii) Explain why the horizontal component of the velocity of the ball remains constant as it moves to the wall. … … [1] (iii) Show that the ball strikes the wall with a horizontal velocity of 7.5 m s–1. [1] (c) The mass of the ball in (b) is 0.40 kg. It is in contact with the wall for a time of 0.12 s and rebounds horizontally with a speed of 4.3 m s–1. (i) Use the information from (b)(iii) to calculate the change in momentum of the ball due to the collision. change in momentum = … kg m s–1 [2] (ii) Calculate the magnitude of the average force exerted on the ball by the wall. average force = … N [1] [Total: 10]

10 marks

Mark scheme: 2(a)(i) area = ut + ½(v – u)t or area = vt – ½(v – u)t or area = ½(u + v)t 2(a)(ii) displacement A1 2(b)(i) u = 15 sin 60° (= 13 m s–1) C1 t = 15 sin 60° / 9.81 C1 = 1.3 s A1 2(b)(ii) the force in the horizontal direction is zero B1 2(b)(iii) (velocity =) 15 cos 60° = 7.5 (m s–1) or (velocity =) 15 sin 30° = 7.5 (m s–1) A1 2(c)(i) p = mv or 0.40 × 7.5 or 0.40 × 4.3 C1 Δp = 0.40 (7.5 + 4.3) = 4.7 kg m s–1 A1 2(c)(ii) force = 4.7 / 0.12 or 0.40 × [(7.5 + 4.3) / 0.12] = 39 N A1

This question in 9702/22 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

1 (a) Complete Table 1.1 by putting a tick (3) in the appropriate column to indicate whether the listed quantities are scalars or vectors. Table 1.1 quantity scalar vector acceleration density temperature momentum [2] (b) A toy train moves along a straight section of track. Fig. 1.1 shows the variation with time t of the distance d moved by the train. 0.6 0.5 d / m 0.4 0.3 0.2 0.1 0 0 1 2 3 t / s Fig. 1.1 (i) Describe qualitatively the motion of the train between time t = 0 and time t = 1.0 s. … … [1] (ii) Determine the speed of the train at time t = 2.0 s. speed = … m s−1 [2] (c) The straight section of track in (b) is part of the loop of track shown in Fig. 1.2. track Fig. 1.2 The train completes exactly one lap of the loop. State and explain the average velocity of the train over the one complete lap. … … … [1] [Total: 6]

6 marks

Mark scheme: 1(a) density and temperature indicated as scalars B1 acceleration and momentum indicated as vectors B1 1(b)(i) decelerates or speed/velocity decreases B1 1(b)(ii) speed = (Δ)d / (Δ)t or gradient C1 = e.g. (0.56 – 0.20) / 1.5 = 0.24 m s–1 A1 1(c) displacement is zero (so) average velocity is zero B1

This question in 9702/22 Oct/Nov 2020

3 A ball is fired horizontally with a speed of 41.0 m s–1 from a stationary cannon at the top of a hill. The ball lands on horizontal ground that is a vertical distance of 57 m below the cannon, as shown in Fig. 3.1. ball, initial speed cannon 41.0 m s–1 path of ball 57 m horizontal ground Fig. 3.1 (not to scale) Assume air resistance is negligible. (a) Show that the time taken for the ball to reach the ground, after being fired, is 3.4 s. [2] (b) Calculate the horizontal distance of the ball from the cannon at the point where the ball lands on the ground. horizontal distance = … m [1] (c) Determine the magnitude of the displacement of the ball from the cannon at the point where the ball lands on the ground. displacement = … m [2] (d) The ball leaves the cannon at time t = 0. On Fig. 3.2, sketch a graph to show the variation of the magnitude v of the vertical component of the velocity of the ball with time t from t = 0 to t = 3.4 s. Numerical values are not required. v 0 0 3.4 t / s Fig. 3.2 [1] (e) The cannon recoils horizontally with a speed of 0.340 m s–1 when it fires the ball. The total mass of the ball and the cannon is 1480 kg. Assume that no external horizontal forces act on the ball-cannon system. Determine, to three significant figures, the mass of the ball. mass = … kg [2] (f) The cannon now fires a ball of smaller mass. Assume that air resistance is still negligible. State and explain the change, if any, to the graph in Fig. 3.2 due to the decreased mass of the ball. … … … [2] [Total: 10]

10 marks

Mark scheme: 3(a) s = ½at 2 C1 57 = ½ × 9.81 × t 2 and t = 3.4 (s) A1 3(b) horizontal distance = 41 × 3.4 = 140 m A1 3(c) (displacement)2 = 572 + 1402 C1 displacement = ( 572 + 1402)0.5 = 150 m A1 3(d) straight line from the origin with positive gradient B1 3(e) (1480 – m) × 0.340 = m × 41.0 C1 m = 12.2 kg A1 or mc 0.34 = mb 41 and mc + mb = 1480 (C1) mc = (41 / 0.34)mb (41 / 0.34)mb + mb = 1480 mb = 12.2 kg (A1) 3(f) acceleration (of free fall) is unchanged/is not dependent on mass M1 (so) no change (to the graph) A1

This question in 9702/23 Oct/Nov 2020

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

3 (a) Define power. … … [1] (b) A car of mass 1700 kg moves in a straight line along a slope that is at an angle θ to the horizontal, as shown in Fig. 3.1. B 25 m car, slope A θ mass 1700 kg horizontal Fig. 3.1 (not to scale) The car moves at constant velocity for a distance of 25 m from point A to point B. Air resistance and friction provide a total resistive force of 440 N that opposes the motion of the car. For the movement of the car from A to B: (i) state the change in the kinetic energy change in kinetic energy = … J [1] (ii) calculate the work done against the total resistive force. work done = … J [1] (c) The movement of the car in (b) from A to B causes its gravitational potential energy to increase by 4.8 × 104 J. Calculate: (i) the increase in vertical height h of the car for its movement from A to B h = … m [2] (ii) angle θ. θ = … ° [1] (d) The engine of the car in (b) produces an output power of 1.7 × 104 W to move the car along the slope. Calculate the time taken for the car to move from A to B. time = … s [2] [Total: 8]

8 marks

Mark scheme: 3(a) work (done) / time (taken) B1 3(b)(i) zero / 0 J A1 3(b)(ii) work done = 440 × 25 = 1.1 × 104 J A1 3(c)(i) (Δ)E(P) = mg(Δ)h C1 h = 4.8 × 104 / (1700 × 9.81) = 2.9 m A1 3(c)(ii) θ = sin–1 (2.9 / 25) = 6.7° A1 3(d) work done = 4.8 × 104 + 1.1 × 104 (= 5.9 × 104 J) C1 time = 5.9 × 104 / 1.7 × 104 = 3.5 s A1

This question in 9702/22 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

2 Water leaves the end of a hose pipe at point P with a horizontal velocity of 6.6 m s–1, as shown in Fig. 2.1. hose pipe P 6.6 m s–1 path of water h Q ground 3.5 m Fig. 2.1 (not to scale) Point P is at height h above the ground. The water hits the ground at point Q. The horizontal distance from P to Q is 3.5 m. Air resistance is negligible. Assume that the water between P and Q consists of non-interacting droplets of water and that the only force acting on each droplet is its weight. (a) Explain, briefly, why the horizontal component of the velocity of a droplet of water remains constant as it moves from P to Q. … … [1] (b) Show that the time taken for a droplet of water to move from P to Q is 0.53 s. [1] (c) Calculate height h. h = … m [2] (d) For the movement of a droplet of water from P to Q, state and explain whether the displacement of the droplet is less than, more than or the same as the distance along its path. … … … [1] (e) Calculate the magnitude of the displacement of a droplet of water that moves from P to Q. displacement = … m [2] [Total: 7]

7 marks

Mark scheme: 2(a) force (on droplet of water) in horizontal direction is zero. B1 2(b) (time taken =) 3.5 / 6.6 = 0.53 (s) A1 2(c) s = ut + ½at 2 s = ½ × 9.81 × 0.532 C1 h = 1.4 m A1 Question Answer Marks 2(d) displacement is straight-line distance (from P to Q) so less (than distance along path) or displacement is the shortest distance (from P to Q). B1 2(e) (displacement)2 = 3.52 + 1.42 C1 displacement = 3.8 m 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

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

2 An archer releases an arrow towards a target at a velocity of 65.0 m s–1 at an angle of 4.30° above the horizontal, as shown in Fig. 2.1. arrow, speed 65.0 m s–1 4.30° centre of target target archer 1.66 m 70.0 m ground Fig. 2.1 (not to scale) When released, the tip of the arrow is a horizontal distance of 70.0 m from the target and 1.66 m above the horizontal ground. The arrow hits the centre of the target. Assume that air resistance is negligible and that all the mass of the arrow is at its tip. (a) Show that the time taken for the arrow to reach the target is 1.08 s. [2] (b) Calculate the height of the centre of the target above the ground. height above ground = … m [3] (c) By considering energy changes, state and explain how the final kinetic energy of the arrow as it hits the target compares with its initial kinetic energy immediately after release. A numerical calculation is not required. … … … … [2] [Total: 7]

7 marks

Mark scheme: 2(a) (time =) displacement / velocity C1 (time =) 70.0 / 65.0 cos 4.30° = 1.08 (s) A1 2(b) s = ut + ½at2 = (65  sin 4.30°  1.08) – (0.5  9.81  1.082) C1 s = −0.46 (m) C1 height above ground = 1.66 − 0.46 = 1.2 m A1 2(c) GPE has decreased M1 (total energy conserved so) KE has increased A1

This question in 9702/23 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 steel ball is projected horizontally from the top of a table, as shown in Fig. 2.1. ball table 4.9 m s–1 path of ball edge of table ground 180 cm Fig. 2.1 (not to scale) The ball is projected horizontally at a speed of 4.9 m s–1. The ball lands on the ground a horizontal distance of 180 cm from the edge of the table. Assume that air resistance is negligible. (a) (i) Calculate the time taken for the ball to reach the ground. time = … s [1] (ii) Calculate the vertical component of the velocity of the ball as it hits the ground. velocity = … m s–1 [2] (iii) Determine the magnitude and the angle to the horizontal of the velocity of the ball as it hits the ground. magnitude of velocity = … m s–1 angle to the horizontal = … ° [3] (b) The ball is projected by means of a compressed spring which is attached to a fixed block as shown in Fig. 2.2. ball x0 frictionless fixed track block spring Fig. 2.2 The ball is placed on a frictionless track in front of the spring. The ball is then pulled back so that the spring has compression x0. When the spring is released, the ball is projected horizontally as shown in Fig. 2.3. ball spring Fig. 2.3 The variation with compression x of the applied force F for the spring is shown in Fig. 2.4. 8 F / N 6 4 2 0 0 2 4 6 8 10 x / cm Fig. 2.4 The ball is a uniform sphere of steel of diameter 0.016 m and mass 0.017 kg. (i) Calculate the density of the steel. density = … kg m–3 [3] (ii) All of the elastic potential energy in the spring is converted into kinetic energy of the ball. The speed of the ball as it leaves the spring is 4.9 m s–1. Show that the maximum elastic potential energy of the spring is 0.20 J. [2] (iii) Use Fig. 2.4 to determine the spring constant k of the spring. k = … N m–1 [2] (iv) Use your answer in (b)(iii) and the value of energy given in (b)(ii) to determine the compression x0 of the spring. x0 = … m [2] (c) The steel ball is replaced by a polystyrene ball of the same diameter but of much lower mass. The spring is given compression x0 and is then released. Air resistance on this ball is not negligible after it leaves the spring. Explain: (i) why this ball leaves the spring with a greater speed than that of the steel ball … … … [1] (ii) why this ball takes a longer time to reach the ground than the steel ball. … … … [1] [Total: 17]

17 marks

Mark scheme: 2(a)(i) t = 1.8 / 4.9 A1 = 0.37 s 2(a)(ii) v = u + at C1 = 9.81  0.37 = 3.6 m s–1 A1 2(a)(iii) v 2 = 3.62 + 4.92 C1 v = 6.1 m s–1 A1 = tan–1 (3.6 / 4.9) A1 = 36° 2(b)(i) = m / V C1 4 C1 V = r3 3 4 A1 = 0.017 / [   (0.016 / 2)3 ] 3 = 7900 kg m–3 2(b)(ii) (E =) ½mv2 C1 (E =) ½  0.017  4.92 = 0.20 (J) A1 2(b)(iii) k = F / x or k = gradient C1 e.g. k = 6.4 / 10  10–2 A1 = 64 N m–1 (allow 63–65 N m–1) 2(b)(iv) E = ½kx2 C1 or E = ½Fx and F = kx x0 = [(2  0.20) / 64]0.5 A1 = 0.079 m or 0.080 m 2(c)(i) same elastic potential energy / same (initial) kinetic energy and (polystyrene ball has) smaller mass (so greater speed) B1 or same (average) force and (polystyrene ball has) smaller mass, (so greater average acceleration so greater speed) 2(c)(ii) (for the polystyrene ball there is) B1 less (average vertical) acceleration / smaller (average vertical component of) resultant force (so takes longer time to reach ground)

This question in 9702/21 Oct/Nov 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

1 (a) Underline all the SI base units in the following list. ampere coulomb current kelvin newton [1] (b) A toy car moves in a horizontal straight line. The displacement s of the car is given by the equation v 2 s = 2a where a is the acceleration of the car and v is its final velocity. State two conditions that apply to the motion of the car in order for the above equation to be valid. 1 … 2 … [2] (c) An experiment is performed to determine the acceleration of the car in (b). The following measurements are obtained: s = 3.89 m ± 0.5% v = 2.75 m s–1 ± 0.8%. (i) Calculate the acceleration a of the car. a = … m s–2 [1] (ii) Determine the percentage uncertainty, to two significant figures, in a. percentage uncertainty = … % [2] (iii) Use your answers in (c)(i) and (c)(ii) to determine the absolute uncertainty in the calculated value of a. absolute uncertainty = … m s–2 [1] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) only ampere and kelvin underlined B1 1(b) initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1 1(c)(i) a = 2.752 / (2  3.89) A1 = 0.97 m s–2 1(c)(ii) percentage uncertainty = (2  0.8) + 0.5 C1 = 2.1% A1 1(c)(iii) absolute uncertainty = (2.1 / 100)  0.97 A1 = 0.02 m s–2

This question in 9702/22 Feb/March 2023

1 A well has a depth of 36 m from ground level to the surface of the water in the well, as shown in Fig. 1.1. ground 36 m well surface of water Fig. 1.1 (not to scale) A student wishes to find the depth of the well. The student plans to drop a stone down the well and record the time taken from releasing the stone to hearing the splash made by the stone as it enters the water. (a) Assume that air resistance is negligible and that the stone is released from rest. Calculate the time taken for the stone to fall from ground level to the surface of the water. time = … s [2] (b) The time recorded by the student using a stop-watch is not equal to the time in (a). Suggest three possible reasons, other than the effect of air resistance, for this difference. 1 … … 2 … … 3 … … [3] (c) The student repeats the experiment three times and uses the results to calculate the depth of the well. The values are shown in Table 1.1. Table 1.1 1st experiment 2nd experiment 3rd experiment depth / m 54.4 53.9 54.1 The true depth of the well is 36.0 m. Explain why these results may be described as precise but not accurate. … … … … [2] [Total: 7]

7 marks

Mark scheme: 1(a) t = √(2s / g) = √[(2  36) / 9.81] C1 = 2.7 s A1 1(b)  reaction time between hearing the splash and stopping the stop-watch  the sound (of the splash) takes time to reach the student or the stone hits the water at a different time to the sound being heard or the sound (of the splash) has to travel to the student  the student might not let go of the stone from ground level  the student might not let go of the stone and start the stop-watch at the same time  stop-watch may not be properly calibrated / has a zero error  (local value of) g is not (exactly) 9.81 (m s2)  stone given initial velocity / initial velocity not zero  stone does not fall (exactly) vertically / in a straight line Any three points, 1 mark each B3 1(c) precise: results are close together / have little scatter B1 not accurate: the values are not close to / 50% different / (very) different from the true value B1

This question in 9702/23 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

1 (a) In the following list, underline all quantities that are SI base quantities. charge electric current force time [1] (b) Under certain conditions, the distance s moved in a straight line by an object in time t is given by s = 12at 2 where a is the acceleration of the object. State two conditions under which the above expression applies to the motion of the object. 1 … 2 … [2] (c) The variation with time t of the velocity v of a car that is moving in a straight line is shown in Fig. 1.1. 25 20 15 v / m s–1 10 5 0 0 2 4 6 8 10 12 14 16 t / s –5 –10 –15 Fig. 1.1 (i) Compare, qualitatively, the acceleration of the car at time t = 8.0 s and at time t = 14.0 s in terms of: ● magnitude … … ● direction. … … [2] (ii) Determine the magnitude of the acceleration of the car at time t = 4.0 s. acceleration = … m s–2 [2] (iii) The car is at point X at time t = 0. Determine the magnitude of the displacement of the car from X at time t = 12.0 s. displacement = … m [2] [Total: 9]

9 marks

Mark scheme: Question Answer Marks 1(a) only electric current and time underlined B1 1(b) initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1 1(c)(i) • magnitude of acceleration at t = 8.0 s is less than that at t = 14.0 s B1 • direction of acceleration at t = 8.0 s is opposite to that at t = 14.0 s B1 1(c)(ii) a = gradient or a = (v – u) / t or a = ∆v / (∆)t C1 a = e.g. (20 + 10) / 12 or (0 + 10) / 4 or (20 – 0) / (12 – 4) A1 a = 2.5 m s–2 1(c)(iii) displacement = [½  (12 – 4)  20] – [½  4  10] C1 or displacement = (–10  12) + (½  2.5  122) or displacement = (20  12) – (½  2.5  122) or displacement = ½  (20 – 10)  12 displacement = 60 m A1

This question in 9702/22 Oct/Nov 2023

2 A high‑altitude balloon is stationary in still air. A solid sphere is suspended from the balloon by a string, as shown in Fig. 2.1. balloon string sphere Fig. 2.1 (not to scale) The volume of the balloon is 7.5 m3. The total weight of the balloon, string and sphere is 65 N. The upthrust acting on the string and sphere is negligible. (a) Calculate the density of the air surrounding the balloon. density = … kg m–3 [2] (b) The string breaks, releasing the sphere. (i) State the magnitude of the acceleration of the sphere immediately after the string breaks. acceleration = … m s–2 [1] (ii) State and explain the variation, if any, in the magnitude of the acceleration of the sphere when it is moving downwards before it reaches terminal (constant) velocity. … … … … … … … [3] (c) The sphere has a mass of 4.0 kg. Calculate the total resistive force acting on the sphere at the instant when its acceleration is 1.9 m s–2. resistive force = … N [2] [Total: 8]

8 marks

Mark scheme: 2(a) 65 = gV C1 or 65 = mg and m = V  = 65 / (9.81  7.5) A1 = 0.88 kg m–3 2(b)(i) acceleration = 9.8 m s–2 A1 2(b)(ii) air resistance (acting on sphere) increases B1 resultant force (on sphere) decreases B1 (magnitude of) acceleration decreases B1 2(c) F = ma C1 = 4.0  1.9 ( = 7.6 N) resistive force = (4.0  9.81) – (4.0  1.9) A1 = 32 N

This question in 9702/22 Oct/Nov 2023

2 A ball on horizontal ground is kicked towards a vertical wall. Fig. 2.1 shows the path of the ball. path of ball horizontal h u ground wall 38° 9.0 m ball Fig. 2.1 (not to scale) The ball has an initial velocity u at an angle of 38° to the ground. The ball travels a horizontal distance of 9.0 m before striking the wall at a height h above the ground. The horizontal component uH of the initial velocity of the ball is 9.5 m s–1. Air resistance is negligible. (a) (i) Show that the time t for the ball to reach the wall is 0.95 s. [1] (ii) Calculate the vertical component uV of the initial velocity of the ball. uV = … m s–1 [2] (iii) Determine h. h = … m [2] (b) The speed of the ball just after striking the wall is less than its speed just before striking the wall. State what this indicates about the nature of the collision of the ball with the wall. … … [1] [Total: 6]

6 marks

Mark scheme: 2(a)(i) (time / t =) 9(.0) / 9.5 = 0.95 (s) A1 2(a)(ii) (uV) = 9.5 tan 38° or 9.5 / tan 52° C1 or 9.5 = u cos 38° and uV = u sin 38° or 9.5 = u cos 38° and uV = (u2 – 9.52)½ uV = 7.4 m s–1 A1 2(a)(iii) s = ut + ½at 2 C1 (h =) 7.4  0.95 – ½  9.81  0.952 h = 2.6 m A1 2(b) (collision is) inelastic B1

This question in 9702/23 Oct/Nov 2023

2 (a) Define acceleration. … … [1] (b) An Olympic diver stands on a platform above a pool of water, as shown in Fig. 2.1. 5.9 m s–1 diver 60° horizontal platform 9.0 m surface of water 1.2 m Fig. 2.1 (not to scale) When the diver is on the platform his centre of gravity is a vertical height of 9.0 m above the surface of the water. The diver jumps from the platform with a velocity of 5.9 m s–1 at an angle of 60° to the horizontal. Air resistance is negligible. When the diver hits the surface of the water, his centre of gravity is a vertical height of 1.2 m above the surface of the water. Calculate the speed of the diver at the instant he hits the surface of the water. speed = … m s–1 [3] (c) The diver in (b) enters the water and decelerates. (i) Describe and explain the variation of the viscous drag force acting on the diver in the water as he moves downwards. … … … … [2] (ii) The diver has a volume of 7.5 × 10–2 m3 . The density of the water is 1.0 × 103 kg m–3 . Show that the upthrust acting on the diver when he is entirely underwater is 740 N. [1] (iii) At a particular instant when the diver is entirely underwater his horizontal velocity is zero. The viscous drag force acting on him at this instant is 950 N vertically upwards. The diver has mass 78 kg. Determine the magnitude and direction of the acceleration of the diver. acceleration = … m s–2 direction … [4] [Total: 11]

11 marks

Mark scheme: 2(a) rate of change of velocity B1 2(b) ½ m()v2= mg()h C1 v2 = 5.92 + 2  9.81  7.8 C1 v2 = 188 v = 14 m s–1 A1 or by resolving components (C1) Vertically: v2 = u2 + 2as v2 = (5.9sin60)2 +2  –9.81  (1.2–9.0) vv = 13.4 horizontally: (C1) vh = 5.9cos60 vh = 2.95 resultant velocity = √(13.42 + 2.952) (A1) = 14 m s–1 2(c)(i) (As the diver moves down their) speed decreases B1 (So) viscous force / drag (force) decreases B1 2(c)(ii) (F =) gV A1 = 1000  9.81  7.5  10–2 = 740 (N)

This question in 9702/22 Feb/March 2024

2 (a) Define velocity. … … [1] (b) A student throws a ball over a vertical wall of height h, as shown in Fig. 2.1. path of ball wall 22 m s–1 ball 40° horizontal h ground 1.2 m 36 m Fig. 2.1 (not to scale) The ball leaves the hand of the student at a height of 1.2 m above the horizontal ground. The ball has an initial velocity of 22 m s–1 at an angle of 40° to the horizontal. The wall is a horizontal distance of 36 m from where the student releases the ball. Air resistance is negligible. (i) Determine the time taken for the ball to reach the wall. time taken = … s [2] (ii) Calculate the vertical component u of the initial velocity of the ball. u = … m s–1 [1] (iii) The ball just goes over the wall. Calculate the height h of the wall. h = … m [3] [Total: 7]

7 marks

Mark scheme: 2(a) change in displacement / time (taken) B1 2(b)(i) horizontal velocity = 22  cos 40° C1 time taken = 36 / (22  cos 40°) = 2.1 s A1 2(b)(ii) u = 22  sin 40° = 14 m s−1 A1 2(b)(iii) s = ut + ½ at2 = (14  2.1) + (½  −9.81  2.12) C1 = 7.8 (m) C1 (therefore) height of wall = 7.8 + 1.2 = 9.0 m A1 Question Answer Marks 2(b)(iii) or other methods possible e.g. time to maximum height = (0 – 14) / –9.81 (= 1.43 s) time from maximum height to wall = 2.1 – 1.43 (= 0.67 s) maximum height above release = (14  1.43) + (½  −9.81  1.432) = 9.99 (m) (C1) height from maximum to wall = 0.5  9.81  0.672 (= 2.2 m) height above release = 9.99 – 2.2 = 7.8 (m) (C1) height of wall = 1.2 + 7.8 = 9.0 m (A1)

This question in 9702/21 May/June 2024

2 (a) Define displacement from a point. … … [1] (b) An object is projected horizontally at a speed of 6.0 m s–1 from a slope, as shown in Fig. 2.1. 6.0 m s–1 object slope θ horizontal Fig. 2.1 (not to scale) The slope is at an angle θ to the horizontal. Air resistance is negligible. The object lands on the slope a time of 0.71 s later and stops without rolling or bouncing. (i) Determine the horizontal distance travelled by the object. distance = … m [1] (ii) Determine the vertical distance travelled by the object. distance = … m [2] (iii) Use your answers in (b)(i) and (b)(ii) to calculate θ. θ = … ° [2] (iv) Determine the magnitude of the displacement of the object from its original position. displacement = … m [2] (v) By considering energy, calculate the speed of the object just before it lands. speed = … m s–1 [3] [Total: 11]

11 marks

Mark scheme: 2(a) distance (from the point) in a straight line in a given direction B1 2(b)(i) distance = speed  time = 6.0  0.71 = 4.3 m A1 2(b)(ii) s = ut + ½ at2 = ½  9.81  0.712 C1 = 2.5 m A1 2(b)(iii) tan  = 2.5 / 4.3 or hypotenuse = √(4.32 + 2.52) ( = 4.97 m) cos  = 4.3 / 4.97 or sin  = 2.5 / 4.97 C1  = 30° A1 Question Answer Marks 2(b)(iv) displacement = √(4.32 + 2.52) C1 = 4.9 m or 5.0 m A1 or displacement = 2.5 / sin 30° or displacement = 4.3 / cos 30° (C1) = 5.0 m (A1) 2(b)(v) KE = ½mv2 or GPE = mgh C1 initial KE + loss in GPE = final KE (½  m  6.02) + (m  9.81  2.5) = (½  m  v2) C1 v = 9.2 m s–1 A1

This question in 9702/23 May/June 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

1 (a) Define acceleration. … … [1] (b) A small aircraft is flying horizontally at a speed of 42 m s–1 at a height of 63 m above horizontal ground, as shown in Fig. 1.1. speed 42 m s–1 63 m ground Fig. 1.1 The aircraft drops a small parcel. The parcel is released from the aircraft at the instant shown in Fig. 1.1. Air resistance is negligible. (i) On Fig. 1.1, draw a line to show the path of the parcel as it falls from the aircraft to the ground. [1] (ii) Calculate the time taken from the instant of release to the instant the parcel reaches the ground. time = … s [2] (iii) Calculate the vertical component of the velocity of the parcel immediately before it reaches the ground. vertical component of velocity = … m s–1 [1] (iv) Determine the speed at which the parcel reaches the ground. speed = … m s–1 [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) rate of change of velocity B1 1(b)(i) curved path from aircraft to ground, starting horizontal at aircraft and then with increasing negative gradient as it moves B1 towards the ground 1(b)(ii) s = ut + ½at2 C1 63 = ½  9.81  t2 time = 3.6 s A1 1(b)(iii) v2 = 2  9.81  63 A1 or v = 0 + (9.81  3.6) or 63 = (v  3.6) – (½  9.81  3.62) or 63 = ½  (0 + v)  3.6 v = 35 m s–1 1(b)(iv) speed2 = 352 + 422 C1 speed = 55 m s–1 A1

This question in 9702/23 Oct/Nov 2024

3 (a) A truck R of mass 9400 kg moves with constant acceleration in a straight line down a slope, as illustrated in Fig. 3.1. R A 180 m B Fig. 3.1 At point A the speed of the truck is 13 m s–1 and at point B the speed of the truck is 22 m s–1. A and B are a distance of 180 m apart. (i) Calculate the acceleration of the truck between A and B. acceleration = … m s–2 [2] (ii) Determine the gain in kinetic energy of the truck between A and B. gain in kinetic energy = … J [3] (b) A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes. Fig. 3.2 shows the variation with time of the momentum of the truck. 25 20 momentum / 104 kg m s–1 15 10 5 0 0 5 10 15 20 25 time / s Fig. 3.2 (i) Define force. … … [1] (ii) Show that the average resultant force F acting on truck R between time t = 0 and t = 15 s is –1.2 × 104 N. [1] (iii) An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force F in (b)(ii). State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R. … … … … … … … [3] [Total: 10]

10 marks

Mark scheme: 3(a)(i) a = (v2 – u2) / 2s C1 = (222 – 132) / (2  180) = 0.88 m s–2 A1 OR (C1) [t = (180  2) / (22 + 13) = 10.3] 180 = 13  10.3 + ½ a  10.32 or 180 = 22  10.3 – ½ a  10.3 2 or 22 = 13 + a  10.3 a = 0.88 m s–2 (A1) 3(a)(ii) ()E = ½m()v2 C1 gain in KE = ½m(v2 – u2) C1 = ½  9400  (222 – 132) = 1.5  106 J A1 OR (C1) W = Fs = ma  d gain in KE = 9400  0.88  180 (C1) = 1.5  106 J (A1) 3(b)(i) rate of change of momentum B1 3(b)(ii) (Force =) (2.5  104 – 21  104) / 15 = – 1.2  104 (N) A1 3(b)(iii) The change of momentum (for S and R to come to rest) is the same B1 (Average) force on R (to come to rest) is B1 (–21  104 / 23 =) 0.91  104 N or (Average) force on R (to come to rest) is less than the force on S / less than F (S will come to rest in) less time (than R). B1 OR (B1) The change of momentum (for S and R to come to rest) is the same Time for S to come to rest is (–21  104 / 1.2  104 =) 17.5 s (and time for R to come to rest is 23 s) (B1) (S comes to rest in) less time (than R). (B1)

This question in 9702/22 Feb/March 2025

1 (a) Define velocity. … … [1] (b) In an experiment, two objects A and B are released from the side of a building, as shown in Fig. 1.1. building A 10.0 m 3.0 m s–1 B h ground Fig. 1.1 (not to scale) Object A is released from rest at a height of 10.0 m above horizontal ground. Object B is released with an initial upward velocity of 3.0 m s−1 at a height h above the ground. Both objects take the same time to reach the ground and they do not collide with each other. Air resistance is negligible. Calculate h. h = … m [3] (c) In a second experiment, object B is released from the same height as in (b) but with a speed of 6.0 m s−1 at an angle of 60° to the vertical, as shown in Fig. 1.2. 60° building 6.0 m s–1 B Fig. 1.2 (i) State and explain whether the time taken for object B to reach the ground is less than, the same as, or greater than the time taken in the first experiment. … … … [2] (ii) By considering energy, state and explain whether the speed at which object B reaches the ground is less than, the same as, or greater than in the first experiment. … … … [2] [Total: 8]

8 marks

Mark scheme: Question Answer Marks 1(a) Rate of change of displacement B1 1(b) For object A: C1 1 2 s = ut + at 2 2  10 so t = 9.81 =1.4 Then for object B: C1 1 2 s = ut + at 2 h = −(3 1.4) + (0.5  9.81  1.42) = 5.7 m A1 OR (C1) h = −(3 1.4) + 10 = 5.7 m (A1) 1(c)(i) time taken (to reach the ground is) same B1 The initial vertical (component of the) velocity is the same (as in part (1b)) B1 1(c)(ii) The (total) initial energy is greater (than in part (1b)) B1 change in gravitational potential energy is same, so speed is greater B1

This question in 9702/23 May/June 2025

1 (a) Define acceleration. … … [1] (b) A rocket is launched vertically from the surface of the Earth. Fig. 1.1 shows the variation of the velocity of the rocket with time for the first 20 s after its launch. 400 velocity / m s–1 200 0 0 5 10 15 20 time / s Fig. 1.1 (i) Determine the acceleration of the rocket. acceleration = … m s–2 [1] (ii) Show that the height of the rocket above the surface of the Earth at a time of 20 s after launch is 3.2 km. [2] (c) The mass of the rocket in (b) is 2.9 × 106 kg. Assume that this mass remains constant. For this rocket, from launch to its height at a time of 20 s after launch: (i) calculate the gain in gravitational potential energy ΔEP ΔEP = … J [2] (ii) calculate the gain in kinetic energy ΔEK ΔEK = … J [2] (iii) determine the average power output of the rocket engines. Assume that resistive forces are negligible. power = … W [2] [Total: 10]

10 marks

Mark scheme: Question Answer Marks 1(a) rate of change of velocity B1 1(b)(i) acceleration = 320 / 20 A1 = 16 m s–2 1(b)(ii) 1 C1 s = ((u) + v)t or distance = area under graph 2 1 A1 (height) =  20  320 = 3200 m or 3.2 km 2 1(c)(i) (EP) = mgh C1 = 2.9  106  9.81  3200 A1 = 9.1  1010 J 1(c)(ii) 1 C1 (EK) = m(v2) 2 1 A1 =  2.9  106  3202 2 = 1.5  1011 J 1(c)(iii) (power =) work done / time C1 power = ((1.5 + 0.91)  1011) / 20 A1 = 1.2  1010 W

This question in 9702/23 Oct/Nov 2025