Cambridge A Level Physics 9702 — 2025 May/June Paper 2 · Variant 2

1 (a) Table 1.1 lists some physical quantities. Identify with ticks (✓) which quantities are vectors and which are scalars. Table 1.1 quantity scalar vector acceleration displacement gravitational potential energy speed temperature [2] (b) A constant resultant force F acts on a car of mass m. The car moves from rest with constant acceleration a along horizontal ground. When the car has displacement s, the speed of the car is v. (i) Using the concept of work done on the car, show that the kinetic energy EK of the car is given by the equation 1 EK = mv2. 2 [3] (ii) The mass of the car is 920 kg. At time t = 0, the car is at rest. At time t = 5.8 s, its velocity is 17 m s–1. Calculate the kinetic energy of the car at time t = 5.8 s. kinetic energy = ....................................................... J [1] (iii) Between time t = 0 and time t = 5.8 s, the work done against resistive forces is 4.7 × 104 J. Determine the average output power of the car during this time. power = ..................................................... W [3] (iv) At time t = 5.8 s, the speed of the car becomes constant. State and explain whether the output power of the car is greater than, less than or the same as the output power just before t = 5.8 s. ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 10]

Mark scheme: Question Answer Marks 1(a) acceleration and displacement identified as vectors (and no others) B1 speed, temperature and gravitational potential energy identified as scalars (and no others) B1 1(b)(i) W = Fs or W = mas B1 s = v2 / 2a or a = v2 / 2s or as = v2 / 2 B1 W = ma(v2 / 2a) or W = m(v2 / 2s)s or W = m(v2 / 2) B1 and (so EK )= ½mv2 OR (B1) W = Fs or W = mas F = mv / t and s = ½vt (B1) W = mv / t  ½vt and (so EK )= ½mv 2 (B1) OR (B1) W = Fs or W = mas a = v / t and s = ½vt (B1) W = m(v / t)(½vt) and (so EK )= ½mv 2 (B1) OR (B1) W = Fs or W = mas a = v / t and s = ½at2 (B1) W = m(v / t)(½  (v / t)  t2) and (so EK )= ½mv 2 (B1) 1(b)(ii) kinetic energy = ½ mv2 A1 = ½  920  172 = 1.3  105 J 1(b)(iii) P = W / t C1 = (4.7  104 + 1.3  105) / 5.8 C1 = 3.1  104 W A1 1(b)(iv) (at/after t = 5.8 s) B1 the kinetic energy (of the car) does not change / work is done only against resistive forces / no work is done to accelerate (the car) so (power output is) less

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2 (a) Define the moment of a force about a point. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A tree of mass 270 kg grows out of sloping ground and is supported by a post, as shown in Fig. 2.1. centre of gravity of tree P F, 1800 N support post θ ground Q 1.2 m 1.6 m Fig. 2.1 (not to scale) The ground applies a total force R on the tree at point Q. The centre of gravity of the tree is a horizontal distance of 1.2 m from Q. The post applies a force F of 1800 N perpendicular to the line PQ. The line of action of F passes through point P at an angle θ to the vertical. P is a horizontal distance of 1.6 m from Q. The tree is in equilibrium and all forces act on the tree in the same plane. (i) By taking moments about point Q, show that θ is 25°. [3] (ii) On Fig. 2.2, draw a labelled scale vector triangle to represent the forces acting on the tree. The weight of the tree has been drawn to scale. weight Fig. 2.2 [2] (iii) The tree exerts a pressure of 150 kPa on the top of the post. Determine the surface area of the tree in contact with the post. area = .................................................... m2 [2] [Total: 8]

Mark scheme: 2(a) force  perpendicular distance (of line of action of force to / from the point) B1 2(b)(i) (moment due to weight =) 1.2  270  9.81 B1 (moment due to post =) 1800  (1.6 / cos) B1 1.2  270  9.81 = 1800  (1.6 / cos) so = 25(°) or B1 1.2  270  9.81 – 1800  (1.6 / cos) = 0 so = 25(°) 2(b)(ii) A closed tip-to-tail vector triangle M1 vector labelled F at an angle of 25°  3° anticlockwise from vertical and A1 vector labelled R at an angle of 37°  3° clockwise from vertical 2(b)(iii) A = F / p C1 = 1800 / (150  103) = 0.012 m2 A1

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3 Two progressive water waves X and Y travel along a straight line from point A to point B. The variation of displacement of the waves with distance from A at an instant in time is shown in Fig. 3.1. 20 displacement / cm 10 wave X 0 0 0.2 0.4 0.6 0.8 1.0 distance from A / m –10 wave Y –20 Fig. 3.1 (a) State the amplitude of wave X. amplitude = ................................................... cm [1] (b) Both waves have frequency 16 Hz. (i) Determine the speed of wave X. speed = ................................................ m s–1 [2] (ii) State and explain whether X and Y are coherent. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (c) Wave X and wave Y superpose to form a resultant wave. On Fig. 3.2, sketch the variation of displacement of the resultant wave with distance from A at the instant of time shown in Fig. 3.1. 20 displacement / cm 10 0 0 0.2 0.4 0.6 0.8 1.0 distance from A / m –10 –20 Fig. 3.2 [2] (d) The intensity of wave X is IX. The intensity of wave Y is IY. IX Use Fig. 3.1 to determine the ratio . IY ratio = ......................................................... [2] [Total: 8]

Mark scheme: 3(a) 10.0 cm A1 3(b)(i) v = f C1 = 16  0.40 = 6.4 m s–1 A1 3(b)(ii) (X and Y have a) constant phase difference (of 180°) so (they are) coherent B1 3(c) A single wave of amplitude 10.0 cm B1 A single negative sine wave of wavelength 0.40 m B1 3(d) I  A2 C1 I X = 102 / 202 IY ratio = 0.25 A1

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4 A small ball is dropped from rest from height h1 above the ground and falls vertically downwards. The ball collides with the ground and bounces back vertically upwards, reaching a maximum height h2. Fig. 4.1 shows the ball just before and just after hitting the ground. ball, mass 0.25 kg speed 3.6 m s–1 ground speed 5.2 m s–1 before hitting ground after hitting ground Fig. 4.1 The ball has mass 0.25 kg and is in contact with the ground for a time of 0.18 s. Just before the ball hits the ground, it has speed 5.2 m s–1. Just after it leaves the ground, it has speed 3.6 m s–1. Air resistance acting on the ball is negligible. (a) State and explain whether the collision is elastic or inelastic. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [1] (b) (i) Calculate the change in momentum of the ball during the collision with the ground. change in momentum = ............................................ kg m s–1 [2] (ii) Determine the average force on the ball during the collision with the ground. force = ...................................................... N [2] h2 (c) Calculate the ratio . h1 ratio = ......................................................... [3] [Total: 8]

Mark scheme: 4(a) The (total) kinetic energy changes / decreases so (the collision is) inelastic B1 OR (B1) (relative) speed of approach not equal to / greater than (relative) speed of separation so (collision is) inelastic 4(b)(i) p = mv or 0.25  3.6 or 0.25  5.2 C1 p = 0.25  (3.6 + 5.2) = 2.2 kg m s–1 A1 4(b)(ii) F = p / ()t C1 = 2.2 / 0.18 = 12 N A1 OR (C1) F = ma and a = (v–u) / t = mv / t = 0.25  (3.6 + 5.2) / 0.18 = 12 N (A1) 4(c) ½ mv2 = mg()h C1 ½  0.25  5.22 = 0.25  g  h1 h1 = 5.22 / 2g h1 = 1.38 ½  0.25  3.62 = 0.25  g  h2 C1 h2 = 3.62/2g h2 = 0.66 h2 / h1 = 0.66 / 1.38 ratio = 0.48 A1

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5 (a) Define the Young modulus. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A wire of unstretched length 0.81 m is made of a metal with Young modulus 95 GPa. The wire obeys Hooke’s law and has a constant cross-sectional area. Fig. 5.1 shows the force–extension graph for the wire. 500 400 force / N 300 200 100 0 0 1 2 3 4 5 extension / 10–3 m Fig. 5.1 (i) Determine the cross-sectional area of the wire. area = .................................................... m2 [3] (ii) The extension of the wire is initially 2.0 × 10–3 m. Determine the work done to increase the extension of the wire to 3.0 × 10–3 m. work done = ....................................................... J [3] [Total: 7]

Mark scheme: 5(a) the ratio of stress to strain B1 5(b)(i) A = FL / Ex C1 A = e.g. (500  0.81) / (95  109  4.0  10–3) C1 A = 1.1  10–6 m2 A1 OR (C1) A = kL / E or A = gradient  L / E k = e.g. 500 / 4.0  10–3 k = 1.25  105 A = 1.25  105  0.81 / 95  109 (C1) A = 1.1  10–6 m2 (A1) 5(b)(ii) E = ½ kx2 or E = ½ Fx or E = area (under graph) C1 ()E = ½  1.25  105  ((3.0  10–3)2 – (2.0  10–3)2) C1 or ()E =(½  375  3.0  10–3) – (½  250  2.0  10–3) or ()E = ½  (375 + 250)  1.0  10–3 work done = 0.31 J A1

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6 (a) Define electric potential difference across a component. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A circuit contains four resistors and a battery of electromotive force (e.m.f.) 8.0 V with negligible internal resistance. When the variable resistor has resistance R, the currents in the circuit are 0.030 A, I1 and I2, as shown in Fig. 6.1. 8.0 V 0.030 A I2 210 Ω R I1 430 Ω 240 Ω Fig. 6.1 (i) Determine the charge passing through the battery in a time of 4.0 minutes. charge = ...................................................... C [2] (ii) Calculate I1. I1 = ...................................................... A [2] (iii) Calculate I2. I2 = ...................................................... A [1] (iv) Determine R. R = ...................................................... Ω [2] (c) The variable resistor in (b) is fitted with a scale so that its resistance can be accurately determined. The resistor of resistance 240 Ω is now replaced by a new resistor X of unknown resistance. A galvanometer is connected as shown in Fig. 6.2. 8.0 V 210 Ω 430 Ω X Fig. 6.2 With reference to ratios of resistances, explain how this circuit can be used to determine the resistance of X. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 10]

Mark scheme: 6(a) energy transferred (to the component) per (unit) charge B1 6(b)(i) Q = It C1 = 0.030  4.0  60 = 7.2 C A1 6(b)(ii) I = V / R C1 I1 = 8.0 / (430+240) = 0.012 A A1 6(b)(iii) I2 = 0.030 – I1 A1 = 0.030 – 0.012 = 0.018 A 6(b)(iv) R = V / I2 C1 = (8.0 – (0.018  210)) / 0.018 = 230  A1 OR (C1) resistance of top branch = 8.0 / 0.018 R = 8.0 / 0.018 – 210 = 230  (A1) OR (C1) total circuit resistance = 8.0 / 0.030 = 267 1 / 267 = 1 / (210 + R) + 1 / (430 + 240) 1 / 267 – 1 / 670 = 1 / (210 +R) 210 + R = 443 R = 230  (A1) 6(c) (When) the galvanometer reads 0 (A) M1 The ratio of the resistances in the top branch will equal the ratio of the resistances in the bottom branch (so the resistance A1 of X can be determined) OR (A1) The ratio of the left pair of resistances will equal the ratio of the right pair of resistances

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7 (a) State what is meant by a fundamental particle. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A nucleus X has 14 nucleons and p protons. The ratio of charge to mass for nucleus X is 4.1 × 107 C kg–1. (i) Determine p. p = ......................................................... [3] (ii) Nucleus X undergoes β– decay to form nucleus Z. Complete the equation representing this decay. 14 ...... ...... ...... ......X ...... Z + ...... ..... + ...... ..... [3] (c) A sample of a radioactive substance emits particles that are positively charged and have a continuous range of kinetic energies. State and explain whether the nuclei in the sample are undergoing α-decay, β+ decay or β– decay. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 9]

Mark scheme: 7(a) (a particle that) cannot be divided/subdivided (into smaller particles) B1 7(b)(i) (p  e) / (14u) = 4.1  107 C1 p = (4.1  107  14  1.66  10–27) / (1.60  10–19) C1 p = 6 (answer should be an integer) A1 7(b)(ii) 14 6 X → 147 Z B1 0 ( − ) 0 ( − ) B1 − 1e or −1 00v ( e ) B1 7(c) (the nuclei are undergoing) + decay B1 A correct explanation in terms of charge and a correct explanation in terms of energy B1 Explanations in terms of charge: • (particles / decay) positively charged so cannot be – • (particles / decay) positively charged so could be / is + • – (particles / decay) are negatively charged • + (particles / decay) are positively charged Explanations in terms of energy: • range of energies so not  (particles / decay) • range of energies so is  (particles / decay) •  (particles / decay) have a range of energies •  (particles / decay) have discrete energies / not range of energies

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