Cambridge A Level Physics 9702 — 2015 May/June Paper 2 · Variant 1

1 (a) Use the definition of power to show that the SI base units of power are kg m2 s–3. [2] (b) Use an expression for electrical power to determine the SI base units of potential difference. units ...........................................................[2]

Mark scheme: 1 (a) power = work / time or energy / time or (force × distance) / time B1 = kg m s–2 × m s–1 = kg m2 s–3 A1 [2] (b) power = VI [or V2 / R and V = IR or I 2R and V = IR] B1 (units of V:) kg m2 s–3 A–1 B1 [2]

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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]

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]

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

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

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4 A spring is kept horizontal by attaching it to points A and B, as shown in Fig. 4.1. slider spring cart, mass 1.7 kg v support A B Fig. 4.1 Point A is on a movable slider and point B is on a fixed support. A cart of mass 1.7 kg has horizontal velocity v towards the slider. The cart collides with the slider. The spring is compressed as the cart comes to rest. The variation of compression x of the spring with force F exerted on the spring is shown in Fig. 4.2. 4.5 3.5 F / N 2.5 1.5 0.5 1.0 1.5 2.0 x / cm Fig. 4.2 Fig. 4.2 shows the compression of the spring for F = 1.5 N to F = 4.5 N. The cart comes to rest when F is 4.5 N. (a) Use Fig. 4.2 to (i) show that the compression of the spring obeys Hooke’s law, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) determine the spring constant of the spring, spring constant = ................................................ N m–1 [2] (iii) determine the elastic potential energy EP stored in the spring due to the cart being brought to rest. EP = ....................................................... J [3] (b) Calculate the speed v of the cart as it makes contact with the slider. Assume that all the kinetic energy of the cart is converted to the elastic potential energy of the spring. speed = ................................................. m s–1 [2]

Mark scheme: 4 (a) (i) two sets of co-ordinates taken to determine a constant value (F / x) M1 F / x constant hence obeys Hooke’s law A1 [2] or gradient calculated and one point on line used (M1) to show no intercept hence obeys Hooke’s law (A1) (ii) gradient or one point on line used e.g. 4.5 / 1.8 × 10–2 C1 (k =) 250 N m–1 A1 [2] (iii) work done or EP = area under graph or ½Fx or ½kx2 C1 = 0.5 × 4.5 × 1.8 × 10–2 or 0.5 × 250 × (1.8 × 10–2)2 C1 = 0.041 (0.0405) J A1 [3] (b) KE = ½mv2 ½mv2 = 0.0405 or KE = 0.0405 (J) C1 (v = [2 × 0.0405 / 1.7]1/2 =) 0.22 (0.218) m s–1 A1 [2]

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5 The variation with potential difference (p.d.) V of current I for a semiconductor diode is shown in Fig. 5.1. 12.0 10.0 8.0 I / mA 6.0 4.0 2.0 0 – 0.5 0 0.5 1.0 V / V Fig. 5.1 (a) Use Fig. 5.1 to describe the variation of the resistance of the diode between V = −0.5 V and V = 0.8 V. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) On Fig. 5.2, sketch the variation with p.d. V of current I for a filament lamp. Numerical values are not required. I 0 0 V Fig. 5.2 [2] (c) Fig. 5.3 shows a power supply of electromotive force (e.m.f.) 12 V and internal resistance 0.50 Ω connected to a filament lamp and switch. 12 V 0.50 1 Fig. 5.3 The filament lamp has a power of 36 W when the p.d. across it is 12 V. (i) Calculate the resistance of the lamp when the p.d. across it is 12 V. resistance = ...................................................... Ω [1] (ii) The switch is closed and the current in the lamp is 2.8 A. Calculate the resistance of the lamp. resistance = ...................................................... Ω [3] (d) Explain how the two values of resistance calculated in (c) provide evidence for the shape of the sketch you have drawn in (b). ................................................................................................................................................... ...............................................................................................................................................[1]

Mark scheme: 5 (a) very high/infinite resistance for negative voltages up to about 0.4 V B1 resistance decreases from 0.4 V B1 [2] (b) initial straight line from (0,0) into curve with decreasing gradient but not to horizontal M1 repeated in negative quadrant A1 [2] (c) (i) R = 122 / 36 = 4.0 Ω A1 or I = P / V = 36 / 12 = 3.0 A and R = 12 / 3.0 = 4.0 Ω (A1) [1] (ii) lost volts = 0.5 × 2.8 = 1.4 (V) or E = 12 = 2.8 × (R + r) C1 R = V / I = (12 – 1.4) / 2.8 or (R + r) = 4.29 Ω C1 = 3.8 (3.79) Ω or R = 3.8 Ω A1 [3] (d) resistance of the lamp increases with increase of V or I B1 [1]

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6 (a) State what is meant by diffraction and by interference. diffraction: ................................................................................................................................. ................................................................................................................................................... interference: .............................................................................................................................. ................................................................................................................................................... [3] (b) Light from a source S1 is incident on a diffraction grating, as illustrated in Fig. 6.1. diffraction grating light S1 zero order Fig. 6.1 (not to scale) The light has a single frequency of 7.06 × 1014 Hz. The diffraction grating has 650 lines per millimetre. Calculate the number of orders of diffracted light produced by the grating. Do not include the zero order. Show your working. number = .......................................................... [3] (c) A second source S2 is used in place of S1. The light from S2 has a single frequency lower than that of the light from S1. State and explain whether more orders are seen with the light from S2. ................................................................................................................................................... ...............................................................................................................................................[1]

Mark scheme: 6 (a) diffraction is the spreading of a wave as it passes through a slit or past an edge B1 when two (or more) waves superpose/meet/overlap M1 resultant displacement is the sum of the displacement of each wave A1 [3] (b) nλ = d sin θ and v = fλ C1 max order number for θ = 90° hence n (= f / vN) = 7.06 × 1014 / (3 × 108 × 650 × 103) M1 n = 3.6 hence number of orders = 3 A1 [3] (c) greater wavelength so fewer orders seen A1 [1]

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7 (a) Explain what is meant by an electric field. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A uniform electric field is produced between two vertical metal plates AB and CD, as shown in Fig. 7.1. A C _-particle 16 mm B D 450 V + – Fig. 7.1 The potential difference between the plates is 450 V and the separation of the plates is 16 mm. An α-particle is accelerated from plate AB to plate CD. (i) On Fig. 7.1, draw lines to represent the electric field between the plates. [2] (ii) Calculate the electric field strength between the plates. electric field strength = ................................................ V m–1 [2] (iii) Calculate the work done by the electric field on the α-particle as it moves from AB to CD. work done = ....................................................... J [3] Question 7 continues on page 16. (iv) A β-particle moves from AB to CD. Calculate the ratio work done by the electric field on the α-particle work done by the electric field on the β-particle. Show your working. ratio = .......................................................... [1]

Mark scheme: 7 (a) a region/space/area where a (stationary) charge experiences an (electric) force B1 [1] (b) (i) at least four parallel equally spaced straight lines perpendicular to plates B1 consistent direction of an arrow on line(s) from left to right B1 [2] (ii) electric field strength E = V / d C1 E = (450 / 16 × 10–3) = 28 × 103 (28 125) V m–1 A1 [2] (iii) W = Eqd or Vq C1 q = 3.2 × 10–19 (C) C1 W = 28 125 × 3.2 × 10–19 × 16 × 10–3 or 450 × 3.2 × 10–19 = 1.4(4) × 10–16 J A1 [3] 450 × 3.2 × 10 −19 (iv) ratio = −19 (evidence of working required) 450 × − 1.6 × 10 = (–) 2 A1 [1]

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