Cambridge A Level Physics 9702 — 2018 May/June Paper 2 · Variant 1

1 (a) State what is meant by a scalar quantity and by a vector quantity. scalar: ........................................................................................................................................ ................................................................................................................................................... vector: ........................................................................................................................................ ................................................................................................................................................... [2] (b) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar. quantity vector or scalar power temperature momentum Fig. 1.1 [2] (c) An aircraft is travelling in wind. Fig. 1.2 shows the velocities for the aircraft in still air and for the wind. west 65° aircraft velocity in still air 95 m s–1 wind velocity 28 m s–1 Fig. 1.2 The velocity of the aircraft in still air is 95 m s–1 to the west. The velocity of the wind is 28 m s–1 from 65° south of east. (i) On Fig. 1.2, draw an arrow, labelled R, in the direction of the resultant velocity of the aircraft. [1] (ii) Determine the magnitude of the resultant velocity of the aircraft. magnitude of velocity = ................................................. m s–1 [2] [Total: 7]

Mark scheme: 1(a) a scalar has magnitude (only) B1 a vector has magnitude and direction B1 1(b) power: scalar temperature: scalar momentum: vector (two correct 1 mark, all three correct 2 marks) B2 1(c)(i) arrow labelled R in a direction from 5° to 20° north of west B1 1(c)(ii) v2 = 282 + 952 – (2 × 28 × 95 × cos 115°) or v2 = [(95 + 28 cos 65°)2 + (28 sin 65°)2] C1 v = 110 ms–1 (109.8 ms–1) A1 or (scale diagram method) triangle of velocities drawn (C1) v = 110 m s–1 (allow 108–112 m s–1) (A1)

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

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

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

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

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4 (a) For a progressive wave, state what is meant by (i) the period, ........................................................................................................................................... .......................................................................................................................................[1] (ii) the wavelength. ........................................................................................................................................... .......................................................................................................................................[1] (b) Fig. 4.1 shows the variation with time t of the displacement x of two progressive waves P and Q passing the same point. 4.0 3.0 x / mm wave P 2.0 1.0 0 0 0.20 0.40 0.60 0.80 t / s –1.0 wave Q –2.0 –3.0 –4.0 Fig. 4.1 The speed of the waves is 20 cm s–1. (i) Calculate the wavelength of the waves. wavelength = .................................................... cm [2] (ii) Determine the phase difference between the two waves. phase difference = ....................................................... ° [1] (iii) Calculate the ratio intensity of wave Q . intensity of wave P ratio = .......................................................... [2] (iv) The two waves superpose as they pass the same point. Use Fig. 4.1 to determine the resultant displacement at time t = 0.45 s. displacement = ................................................... mm [1] [Total: 8]

Mark scheme: 4(a)(i) time for one oscillation/one vibration/one cycle or time between adjacent wavefronts/points in phase or shortest time between two wavefronts/points in phase B1 4(a)(ii) distance moved by wavefront/energy during one cycle/oscillation/period (of source) or minimum distance between two wavefronts or distance between two adjacent wavefronts or minimum distance between two points having the same displacement and moving in the same direction B1 4(b)(i) v = λ / T or v = fλ and f = 1 / T C1 λ = 20 × 0.60 = 12 cm A1 4(b)(ii) phase difference = 360° × (0.20 / 0.60) or 360° × (0.40 / 0.60) = 120° or 240° A1 4(b)(iii) I ∝ A2 C1 IQ / IP = AQ 2 / AP 2 = 2.02 / 3.02 = 0.44 A1 4(b)(iv) displacement = 1.00 – 3.00 = –2.00 mm A1

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5 (a) When monochromatic light is incident normally on a diffraction grating, the emergent light waves have been diffracted and are coherent. Explain what is meant by (i) diffracted waves, ........................................................................................................................................... .......................................................................................................................................[1] (ii) coherent waves. ........................................................................................................................................... .......................................................................................................................................[1] (b) Light consisting of only two wavelengths λ1 and λ2 is incident normally on a diffraction grating. The third order diffraction maximum of the light of wavelength λ1 and the fourth order θ to the direction of diffraction maximum of the light of wavelength λ2 are at the same angle the incident light. (i) Show that the ratio λ2 is 0.75. λ1 Explain your working. [2] (ii) The difference between the two wavelengths is 170 nm. Determine wavelength λ1. λ1 = .................................................... nm [1] [Total: 5]

Mark scheme: 5(a)(i) waves spread at (each) slit/gap B1 5(a)(ii) constant phase difference (between (each of) the waves) B1 5(b)(i) nλ = d sin θ B1 d sin θ is the same and 3λ1 = 4λ2 so λ2 / λ1 = 0.75 A1 5(b)(ii) λ2 / λ1 = 0.75 and λ1 – λ2 = 170 λ1 = 680 nm A1

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6 (a) Define the volt. ...............................................................................................................................................[1] (b) A battery of electromotive force (e.m.f.) 4.5 V and negligible internal resistance is connected to two filament lamps P and Q and a resistor R, as shown in Fig. 6.1. 4.5 V R P Q Fig. 6.1 The current in lamp P is 0.15 A. The I–V characteristics of the filament lamps are shown in Fig. 6.2. 0.20 P I / A 0.15 Q 0.10 0.05 0 0 1.0 2.0 3.0 4.0 V / V Fig. 6.2 (i) Use Fig. 6.2 to determine the current in the battery. Explain your working. current = ....................................................... A [2] (ii) Calculate the resistance of resistor R. resistance = ...................................................... Ω [2] (iii) The filament wires of the two lamps are made from material with the same resistivity at their operating temperature in the circuit. The diameter of the wire of lamp P is twice the diameter of the wire of lamp Q. Determine the ratio length of filament wire of lamp P length of filament wire of lamp Q. ratio = .......................................................... [3] (iv) The filament wire of lamp Q breaks and stops conducting. State and explain, qualitatively, the effect on the resistance of lamp P. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] [Total: 10]

Mark scheme: 6(a) joule / coulomb B1 6(b)(i) lamps have same p.d./lamps have p.d. of 2.7 V B1 current = 0.15 + 0.090 = 0.24 A A1 6(b)(ii) R = (4.5 – 2.7) / 0.24 or RP = 18 (Ω) and RQ = 30 (Ω) I / RT = 1 / 18 + 1 / 30 and so RT = 11.25 4.5 = 0.24 × (R + 11.25) C1 R = 7.5 Ω A1 Question Answer Marks 6(b)(iii) R = ρl / A C1 RP / RQ = [(2.7 / 0.15) / (2.7 / 0.09)] (= 0.60) C1 ratio = 0.60 × 22 = 2.4 A1 6(b)(iv) less p.d. across resistor/greater p.d. across P B1 greater current through P and so resistance (of P) increases B1

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7 A β– particle from a radioactive source is travelling in a vacuum with kinetic energy 460 eV. The particle enters a uniform electric field at a right-angle and follows the path shown in Fig. 7.1. path of β– particle β– particle kinetic energy 460 eV uniform electric field in the plane of the paper Fig. 7.1 (a) The direction of the electric field is in the plane of the paper. On Fig. 7.1, draw an arrow to show the direction of the electric field. [1] (b) Calculate the speed of the β– particle before it enters the electric field. speed = ................................................. m s–1 [3] (c) Other β– particles from the same radioactive source travel outside the electric field along the same incident path as that shown in Fig. 7.1. State and briefly explain whether those β– particles will all follow the same path inside the electric field. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 6]

Mark scheme: 7(a) arrow pointing vertically down the page B1 7(b) E = ½mv2 C1 E = 460 × 1.60 × 10–19 (= 7.36 × 10–17 (J)) C1 v = [(2 × 460 × 1.60 × 10–19) / (9.11 × 10–31)]½ = 1.3 × 107 m s–1 A1 7(c) β– particles have range of/different/various speeds/velocities/momenta/energies M1 so they follow different paths A1

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