Cambridge A Level Physics 9702 — 2011 Oct/Nov Paper 2 · Variant 3

1 (a) Distinguish between scalars and vectors. .......................................................................................................................................... ......................................................................................................................................[1] (b) Underline all the vector quantities in the list below. acceleration kinetic energy momentum power weight [2] (c) A force of 7.5 N acts at 40° to the horizontal, as shown in Fig. 1.1. 7.5 N 40° horizontal Fig. 1.1 Calculate the component of the force that acts (i) horizontally, horizontal component = ............................................. N [1] (ii) vertically. vertical component = ............................................. N [1] (d) Two strings support a load of weight 7.5 N, as shown in Fig. 1.2. For Examiner’s Use T1 T2 50° 40° horizontal 7.5N Fig. 1.2 One string has a tension T1 and is at an angle 50° to the horizontal. The other string has a tension T2 and is at an angle 40° to the horizontal. The object is in equilibrium. Determine the values of T1 and T2 by using a vector triangle or by resolving forces. T1 = .................................................. N T2 = .................................................. N [4]

Mark scheme: 1 (a) scalar has magnitude/size, vector has magnitude/size and direction B1 [1] (b) acceleration, momentum, weight B2 [2] (–1 for each addition or omission but stop at zero) (c) (i) horizontally: 7.5 cos 40° / 7.5 sin 50° = 5.7(45) / 5.75 not 5.8 N A1 [1] (ii) vertically: 7.5 sin 40° / 7.5 cos 50° = 4.8(2) N A1 [1] (d) either correct shaped triangle M1 correct labelling of two forces, three arrows and two angles A1 or correct resolving: T2 cos 40° = T1 cos 50° (B1) T1 sin 50° + T2 sin 40° = 7.5 (B1) T1 = 5.7(45) (N) A1 T2 = 4.8 (N) A1 [4] (allow ± 0.2 N for scale diagram)

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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]

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

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3 (a) Define the terms For Examiner’s (i) power, Use ..............................................................................................................................[1] (ii) the Young modulus. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[1] (b) A crane is used to lift heavy objects, as shown in Fig. 3.1. motor cable object of mass 1800 kg ground Fig. 3.1 The motor in the crane lifts a total mass of 1800 kg from rest on the ground. The cable supporting the mass is made of steel of Young modulus 2.4 × 1011 Pa. The cross-sectional area of the cable is 1.3 × 10– 4 m2. As the mass leaves the ground, the strain in the cable is 0.0010. Assume the weight of the cable to be negligible. (i) 1. Use the Young Modulus of the steel to show that the tension in the cable is 3.1 × 104 N. [2] 2. Calculate the acceleration of the mass as it is lifted from the ground. acceleration = ....................................... m s–2 [3] (ii) The motor now lifts the mass through a height of 15 m at a constant speed. For Examiner’s Calculate Use 1. the tension in the lifting cable, tension = ............................................. N [1] 2. the gain in potential energy of the mass. gain in potential energy = ............................................. J [2] (iii) The motor of the crane is 30% efficient. Calculate the input power to the motor required to lift the mass at a constant speed of 0.55 m s–1. input power = ............................................ W [3]

Mark scheme: 3 (a) (i) power = work done per unit time / energy transferred per unit time / rate of work done B1 [1] (ii) Young modulus = stress / strain B1 [1] (b) (i) 1. E = T / (A × strain) (allow strain = ε) C1 T = E × A × strain = 2.4 × 1011 × 1.3 × 10–4 × 0.001 M1 = 3.12 × 104 N A0 [2] 2. T – W = ma C1 [3.12 × 104 – 1800 × 9.81] = 1800a C1 a = 7.52 m s–2 A1 [3] (ii) 1. T = 1800 × 9.81 = 1.8 × 104 N A1 [1] 2. potential energy gain = mgh C1 = 1800 × 9.81 × 15 = 2.7 × 105 J A1 [2] (iii) P = Fv C1 = 1800 × 9.81 × 0.55 C1 input power = 9712 × (100/30) = 32.4 × 103 W A1 [3]

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4 (a) Distinguish between potential difference (p.d.) and electromotive force (e.m.f.) in terms For of energy transformations. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) Two cells A and B are connected in series with a resistor R of resistance 5.5 Ω, as shown in Fig. 4.1. 4.4 V 2.3 Ω cell A R 5.5 Ω 2.1 V 1.8 Ω cell B Fig. 4.1 Cell A has e.m.f. 4.4 V and internal resistance 2.3 Ω. Cell B has e.m.f. 2.1 V and internal resistance 1.8 Ω. (i) State Kirchhoff’s second law. .................................................................................................................................. ..............................................................................................................................[1] (ii) Calculate the current in the circuit. current = ............................................. A [2] (iii) On Fig. 4.1, draw an arrow to show the direction of the current in the circuit. Label this arrow I. [1] (iv) Calculate For Examiner’s 1. the p.d. across resistor R, Use p.d. = ............................................. V [1] 2. the terminal p.d. across cell A, p.d. = ............................................. V [1] 3. the terminal p.d. across cell B. p.d. = ............................................. V [2]

Mark scheme: 4 (a) p.d. = energy transformed from electrical to other forms unit charge B1 e.m.f. = energy transformed from other forms to electrical unit charge B1 [2] (b) (i) sum of e.m.f.s (in a closed circuit) = sum of potential differences B1 [1] (ii) 4.4 – 2.1 = I × (1.8 + 5.5 + 2.3) M1 I = 0.24 A A1 [2] (iii) arrow (labelled) I shown anticlockwise A1 [1] (iv) 1. V = I × R = 0.24 × 5.5 = 1.3(2) V A1 [1] 2. VA = 4.4 – (I × 2.3) = 3.8(5) V A1 [1] 3. either VB = 2.1 + (I × 1.8) or VB = 3.8 – 1.3 C1 = 2.5(3) V A1 [2] GCE AS/A LEVEL – October/November 2011 9702 23

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5 (a) By reference to vibrations of the points on a wave and to its direction of energy transfer, For distinguish between transverse waves and longitudinal waves. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) Describe what is meant by a polarised wave. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (c) The variation with distance x of the displacement y of a transverse wave is shown in Fig. 5.1. 3.0 AA 2.0 y / cm 1.0 BB 0 0 0.2 0.4 0.6 0.8 1.0 1.2 x / m –1.0 –2.0 –3.0 Fig. 5.1 (i) Use Fig. 5.1 to determine 1. the amplitude of the wave, amplitude = .......................................... cm [1] 2. the phase difference between the points labelled A and B. phase difference = ..................................................[2] (ii) Determine the amplitude of a wave with twice the intensity of that shown in For Fig. 5.1. Examiner’s Use amplitude = .......................................... cm [1]

Mark scheme: 5 (a) transverse waves have vibrations that are perpendicular / normal to the direction of energy travel B1 longitudinal waves have vibrations that are parallel to the direction of energy travel B1 [2] (b) vibrations are in a single direction M1 either applies to transverse waves or normal to direction of wave energy travel or normal to direction of wave propagation A1 [2] (c) (i) 1. amplitude = 2.8 cm B1 [1] 2. phase difference = 135° or 0.75π rad or ¾π rad or 2.36 radians (three sf needed) numerical value M1 unit A1 [2] (ii) amplitude = 3.96 cm (4.0 cm) A1 [1]

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6 Two horizontal metal plates are separated by distance d in a vacuum. A potential difference V For is applied across the plates, as shown in Fig. 6.1. Examiner’s Use +V metal plate radioactive d source metal plate beam of α-particles 0V Fig. 6.1 A horizontal beam of α-particles from a radioactive source is made to pass between the plates. (a) State and explain the effect on the deflection of the α-particles for each of the following changes: (i) The magnitude of V is increased. .................................................................................................................................. ..............................................................................................................................[1] (ii) The separation d of the plates is decreased. .................................................................................................................................. ..............................................................................................................................[1] (b) The source of α-particles is replaced with a source of β-particles. For Compare, with a reason in each case, the effect of each of the following properties on Examiner’s the deflections of α- and β-particles in a uniform electric field: Use (i) charge .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) mass .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (iii) speed .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[1] (c) The electric field gives rise to an acceleration of the α-particles and the β-particles. Determine the ratio acceleration of the α-particles . acceleration of the β-particles ratio = ..................................................[3]

Mark scheme: 6 (a) (i) greater deflection M0 greater electric field / force on α-particle A1 [1] (ii) greater deflection M0 greater electric field / force on α-particle A1 [1] (b) (i) either deflections in opposite directions M1 because oppositely charged A1 or β less deflection (M1) β has smaller charge (A1) [2] (ii) α smaller deflection M1 because larger mass A1 [2] (iii) β less deflection because higher speed B1 [1] (c) either F = ma and F = Eq or a = Eq / m C1 ratio = either (2 × 1.6 × 10–19) × (9.11 × 10–31) (1.6 × 10–19) × 4 × (1.67 × 10–27) or [2e × 1 / 2000 u] / [e × 4u] C1 ratio = 1 /4000 or 2.5 × 10–4 or 2.7 × 10–4 A1 [3]

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