Cambridge A Level Physics 9702 — 2018 Feb/March Paper 2 · Variant 2

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]

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

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2 (a) Explain what is meant by (i) work done, ........................................................................................................................................... .......................................................................................................................................[1] (ii) kinetic energy. ........................................................................................................................................... .......................................................................................................................................[1] (b) A leisure-park ride consists of a carriage that moves along a railed track. Part of the track lies in a vertical plane and follows an arc XY of a circle of radius 13 m, as shown in Fig. 2.1. 13 m Y 13 m carriage 22 m s–1 mass 580 kg track X Fig. 2.1 The mass of the carriage is 580 kg. At point X, the carriage has velocity 22 m s–1 in a horizontal direction. The velocity of the carriage then decreases to 12 m s–1 in a vertical direction at point Y. (i) For the carriage moving from X to Y 1. show that the decrease in kinetic energy is 9.9 × 104 J, [2] 2. calculate the gain in gravitational potential energy. gain in gravitational potential energy = ....................................................... J [2] (ii) Show that the length of the track from X to Y is 20 m. [1] (iii) Use your answers in (b)(i) and (b)(ii) to calculate the average resistive force acting on the carriage as it moves from X to Y. resistive force = ...................................................... N [2] (iv) Describe the change in the direction of the linear momentum of the carriage as it moves from X to Y. ........................................................................................................................................... .......................................................................................................................................[1] (v) Determine the magnitude of the change in linear momentum when the carriage moves from X to Y. change in momentum = .................................................... N s [3] [Total: 13]

Mark scheme: 2(a)(i) B1 2(a)(ii) energy (of a mass/body) due to motion / speed / velocity B1 2(b)(i) 1 E = ½mv 2 C1 (∆)E = ½ × 580 × (222 – 122) = 9.9 × 104 J A1 2 (∆)E = mg(∆)h ∆E = 580 × 9.81 × 13 C1 = 7.4 × 104 J A1 Question Answer Marks 2(b)(ii) length = (2π×13) / 4 or (π×26) / 4 or (π×13) / 2 = 20 m A1 2(b)(iii) work done against resistive force = 9.9 × 104 – 7.4 × 104 average resistive force = (9.9 × 104 – 7.4 × 104) / 20 C1 = 1300 N A1 2(b)(iv) from horizontal/right to vertical / up or 90° A1 2(b)(v) p = mv or (580 × 22) or (580 × 12) C1 ∆p = [ (580×12)2 + (580×22)2 ]0.5 C1 = 1.5 × 104 N s A1

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3 (a) For the deformation of a wire under tension, define (i) stress, ........................................................................................................................................... .......................................................................................................................................[1] (ii) strain. ........................................................................................................................................... .......................................................................................................................................[1] (b) A wire is fixed at one end so that it hangs vertically. The wire is given an extension x by suspending a load F from its free end. The variation of F with x is shown in Fig. 3.1. 8 F / N 7 6 5 4 3 2 1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 x / mm Fig. 3.1 The wire has cross-sectional area 9.4 × 10–8 m2 and original length 2.5 m. (i) Describe how measurements can be taken to determine accurately the cross-sectional area of the wire. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] (ii) Determine the Young modulus E of the material of the wire. E = .................................................... Pa [2] (iii) Use Fig. 3.1 to calculate the increase in the energy stored in the wire when the load is increased from 2.0 N to 4.0 N. increase in energy = ....................................................... J [2] (c) The wire in (b) is replaced by a new wire of the same material. The new wire has twice the length and twice the diameter of the old wire. The new wire also obeys Hooke’s law. On Fig. 3.1, sketch the variation with extension x of the load F for the new wire from x = 0 to x = 0.80 mm. [2] [Total: 11]

Mark scheme: 3(a)(i) force / (cross-sectional) area B1 3(a)(ii) extension / original length B1 3(b)(i) measure / determine / find diameter B1 using a micrometer / digital calipers B1 several measurements in different places / along the wire / around the circumference (and average them) B1 3(b)(ii) E = σ / ε or E = FL / Ax or E = gradient × (L / A) E = (4 × 2.5) / (0.8 × 10–3) × (9.4 × 10–8) C1 = 1.3 × 1011 Pa A1 Question Answer Marks 3(b)(iii) E = ½Fx or E = ½kx 2 or E = area under graph E = ½ × (2+4) × 0.4 × 10–3 or E = (½ × 4 × 0.8×10–3) – (½ × 2 × 0.4×10–3) or E = [ ½ × 5000 × (0.8×10–3)2 ] – [ ½ × 5000 × (0.4×10–3)2 ] C1 E = 1.2 × 10–3 J A1 3(c) straight line from the origin and above the original line M1 straight line passes through (0.80, 8.0) A1

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4 (a) State the conditions required for the formation of a stationary wave. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) The sound from a loudspeaker is detected by a microphone that is connected to a cathode-ray oscilloscope (c.r.o.). Fig. 4.1 shows the trace on the screen of the c.r.o. 1 cm 1 cm Fig. 4.1 In air, the sound wave has a speed of 330 m s–1 and a wavelength of 0.18 m. (i) Calculate the frequency of the sound wave. frequency = .................................................... Hz [2] (ii) Determine the time-base setting, in s cm–1, of the c.r.o. time-base setting = ............................................... s cm–1 [2] (iii) The intensity of the sound from the loudspeaker is now halved. The wavelength of the sound is unchanged. Assume that the amplitude of the trace is proportional to the amplitude of the sound wave. On Fig. 4.1, sketch the new trace shown on the screen of the c.r.o. [2] (c) The loudspeaker in (b) is held above a vertical tube of liquid, as shown in Fig. 4.2. loudspeaker liquid level A level A tube level B level B liquid tap Fig. 4.2 Fig. 4.3 A tap at the bottom of the tube is opened so that liquid drains out at a constant rate. The wavelength of the sound from the loudspeaker is 0.18 m. The sound that is heard first becomes much louder when the liquid surface reaches level A. The next time that the sound becomes much louder is when the liquid surface reaches level B, as shown in Fig. 4.3. (i) Calculate the vertical distance between level A and level B. distance = ...................................................... m [1] (ii) On Fig. 4.3, label with the letter N the positions of the nodes of the stationary wave that is formed in the air column when the liquid surface is at level B. [1] (iii) The mass of liquid leaving the tube per unit time is 6.7 g s–1. The tube has an internal cross-sectional area of 13 cm2. The density of the liquid is 0.79 g cm–3. Calculate the time taken for the liquid to move from level A to level B. time = ....................................................... s [2] [Total: 12]

Mark scheme: 4(a) (two) waves (travelling at same speed) in opposite directions overlap B1 (waves are same type and) have same frequency / wavelength B1 4(b)(i) v = fλ f = 330 / 0.18 C1 = 1800 Hz (1830 Hz) A1 4(b)(ii) T = 1 / 1800 (= 5.5 × 10–4) time-base setting = (1.5 × 5.5 ×10–4) / 8.0 or 1 / (1800 × 5.3) C1 = 1.0 × 10–4 s cm–1 A1 4(b)(iii) waveform drawn with same period as original waveform B1 waveform drawn with amplitude of 1.7 cm B1 4(c)(i) distance = λ / 2 = 0.18 / 2 = 0.090 m A1 Question Answer Marks 4(c)(ii) letter N shown at level B and at level A and not anywhere else. B1 4(c)(iii) m = ρAx = 0.79 × 13 × 9.0 (=92.4) or 790 × 13×10–4 × 0.090 (=0.0924) t = 92.4 / 6.7 or 0.0924 / 0.0067 C1 = 14 s A1

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5 (a) State Kirchhoff’s second law. ................................................................................................................................................... ...............................................................................................................................................[2] (b) Two batteries, each of electromotive force (e.m.f.) 6.0 V and negligible internal resistance, are connected in series with three resistors, as shown in Fig. 5.1. R 4.0 Ω X 6.0 V V 6.0 V Y 1.5 Ω I Fig. 5.1 Resistor X has resistance 4.0 Ω and resistor Y has resistance 1.5 Ω. (i) The resistance R of the variable resistor is changed until the voltmeter in the circuit reads zero. Calculate 1. the current I in the circuit, I = ....................................................... A [1] 2. the resistance R. R = ...................................................... Ω [2] (ii) Resistors X and Y are wires made from the same material. The diameter of the wire of X is twice the diameter of the wire of Y. Determine the ratio average drift speed of free electrons in X . average drift speed of free electrons in Y ratio = .......................................................... [2] (iii) The resistance R of the variable resistor is now increased. State and explain the effect of the increase in R on the power transformed by each of the batteries. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] [Total: 10]

Mark scheme: 5(a) sum of e.m.f.(s) = sum of p.d.(s) M1 around a loop / around a closed circuit A1 5(b)(i) 1 6.0 – 4.0I = 0 I = 1.5 A A1 2 6.0 + 6.0 = I (4.0 + R + 1.5) 12 = 1.5 (4.0 + R + 1.5) C1 R = 2.5 Ω A1 or 6.0 = I (R + 1.5) 6.0 = 1.5 (R + 1.5) (C1) R = 2.5 Ω (A1) or combines 6 = 4I and 6 = I(R + 1.5) to give 4 = R + 1.5 (C1) R = 2.5 Ω (A1) Question Answer Marks 5(b)(ii) I = Anvq ratio = 12 / 22 C1 = 0.25 A1 5(b)(iii) total (circuit) resistance increases B1 current / I decreases or P ∝ I or P ∝ 1 / (total resistance) M1 power (transformed) decreases A1

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6 A sample of a radioactive isotope emits a beam of β– radiation. (a) State the change, if any, to the number of neutrons in a nucleus of the sample that emits a β– particle. ...............................................................................................................................................[1] (b) The number of β– particles passing a fixed point in the beam in a time of 2.0 minutes is 9.8 × 1010. Calculate the current, in pA, produced by the beam of β– particles. current = ..................................................... pA [3] (c) Suggest why the β– particles are emitted with a range of kinetic energies. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 6]

Mark scheme: 6(a) –1 / decreases by 1 A1 6(b) I = Q / t or Ne / t C1 = (9.8×1010 × 1.6×10–19) / (2.0 × 60) = 1.3 × 10–10 (A) C1 = 130 pA A1 6(c) antineutrino(s) (emitted) / other particle(s) (emitted) C1 energy / momentum shared with antineutrino(s) A1

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