Cambridge A Level Physics 9702 — 2008 May/June Paper 2 · Variant 1

1 Make reasonable estimates of the following quantities. (a) the frequency of an audible sound wave frequency = ........................................... Hz [1] (b) the wavelength, in nm, of ultraviolet radiation wavelength = ........................................... nm [1] (c) the mass of a plastic 30 cm ruler mass = .............................................. g [1] (d) the density of air at atmospheric pressure density = ..................................... kg m–3 [1]

Mark scheme: 1 (a) allow anything in range 20 Hz → 20 kHz B1 [1] (b) allow anything in range 10 nm → 400 nm B1 [1] (c) allow anything in range 10 g → 100 g B1 [1] (d) allow anything in range 0.1 kg m–3 → 10 kg m–3 B1 [1]

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2 A spring is placed on a flat surface and different weights are placed on it, as shown in For Fig. 2.1. Examiner’s Use weights spring Fig. 2.1 The variation with weight of the compression of the spring is shown in Fig. 2.2. 4 compression / cm 3 2 1 0 0 10 20 30 40 weight / N Fig. 2.2 The elastic limit of the spring has not been exceeded. (a) (i) Determine the spring constant k of the spring. k = ........................................... N m–1 [2] (ii) Deduce that the strain energy stored in the spring is 0.49 J for a compression of For 3.5 cm. Examiner’s Use [2] (b) Two trolleys, of masses 800 g and 2400 g, are free to move on a horizontal table. The spring in (a) is placed between the trolleys and the trolleys are tied together using thread so that the compression of the spring is 3.5 cm, as shown in Fig. 2.3. thread spring trolley trolley mass 800g mass 2400g Fig. 2.3 Initially, the trolleys are not moving. The thread is then cut and the trolleys move apart. (i) Deduce that the ratio speed of trolley of mass 800 g speed of trolley of mass 2400 g is equal to 3.0. [2] (ii) Use the answers in (a)(ii) and (b)(i) to calculate the speed of the trolley of mass For 800 g. Examiner’s Use speed = ........................................... m s–1 [3]

Mark scheme: 2 (a) (i) k is the reciprocal of the gradient of the graph C1 k = {32 / (4 × 10–2) = } 800 N m–1 A1 [2] (ii) either energy = average force × extension or ½kx2 or area under graph line C1 energy = ½ × 800 × (3.5 × 10–2)2 or ½ × 28 × 3.5 × 10–2 M1 energy = 0.49 J A0 [2] (b) (i) momentum before cutting thread = momentum after C1 0 = 2400 × V – 800 × v M1 v / V = 3.0 A0 [2] (ii) energy stored in spring = kinetic energy of trolleys C1 0.49 = ½ × 2.4 × ( 1 v)2 + ½ × 0.8 × v2 C1 3 v = 0.96 m s–1 A1 [3] (if only one trolley considered, or masses combined, allow max 1 mark) 2

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3 A shopping trolley and its contents have a total mass of 42 kg. The trolley is being pushed For along a horizontal surface at a speed of 1.2 m s–1. When the trolley is released, it travels a Examiner’s distance of 1.9 m before coming to rest. Use (a) Assuming that the total force opposing the motion of the trolley is constant, (i) calculate the deceleration of the trolley, deceleration = ........................................... m s–2 [2] (ii) show that the total force opposing the motion of the trolley is 16 N. [1] (b) Using the answer in (a)(ii), calculate the power required to overcome the total force opposing the motion of the trolley at a speed of 1.2 m s–1. power = ........................................... W [2] (c) The trolley now moves down a straight slope that is inclined at an angle of 2.8° to the For horizontal, as shown in Fig. 3.1. Examiner’s Use 2.8° Fig. 3.1 The constant force that opposes the motion of the trolley is 16 N. Calculate, for the trolley moving down the slope, (i) the component down the slope of the trolley’s weight, component of weight = ........................................... N [2] (ii) the time for the trolley to travel from rest a distance of 3.5 m along the length of the slope. time = ............................................ s [4] (d) Use your answer to (c)(ii) to explain why, for safety reasons, the slope is not made any steeper. .......................................................................................................................................... ......................................................................................................................................[1]

Mark scheme: 3 (a) (i) v2 = 2as 1.22 = 2 × a × 1.9 M1 a = 0.38 m s–2 A1 [2] (ii) F = ma = 42 × 0.38 M1 = 16 N A0 [1] (b) power = Fv C1 = 16 × 1.2 = 19 W A1 [2] (c) (i) component = 42 × 9.8 × sin2.8 C1 = 20.1 N A1 [2] (ii) accelerating force = 20.1 – 16 = 4.1 N C1 acceleration of trolley = 4.1 / 42 = 0.098 m s–2 C1 s = ½at2 3.5 = ½ × 0.098 × t2 C1 t = 8.5 s A1 [4] GCE A/AS LEVEL – May/June 2008 9702 02 (d) either allows plenty of time to stop runaway trolley or speed of trolley increases gradually or trolley will travel faster B1 [1] (answer must be unambiguous when read in conjunction with question)

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4 (a) (i) Define the terms For Examiner’s 1. tensile stress, Use .................................................................................................................................. ..............................................................................................................................[1] 2. tensile strain, .................................................................................................................................. ..............................................................................................................................[1] 3. the Young modulus. .................................................................................................................................. ..............................................................................................................................[1] (ii) Suggest why the Young modulus is not used to describe the deformation of a liquid or a gas. .................................................................................................................................. ..............................................................................................................................[1] (b) The change ∆V in the volume V of some water when the pressure on the water increases by ∆p is given by the expression ∆V ∆p = 2.2 × 109 , V where ∆p is measured in pascal. In many applications, water is assumed to be incompressible. By reference to the expression, justify this assumption. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (c) Normal atmospheric pressure is 1.01 × 105 Pa. For Examiner’s Divers in water of density 1.08 × 103 kg m–3 frequently use an approximation that every Use 10 m increase in depth of water is equivalent to one atmosphere increase in pressure. Determine the percentage error in this approximation. error = ........................................... % [3]

Mark scheme: 4 (a) (i) 1. stress = force / (cross-sectional) area B1 [1] 2. strain = extension / original length B1 [1] 3. Young modulus = stress / strain B1 [1] (ratios must be clear in each answer) (ii) either fluids cannot be deformed in one direction / cannot be stretched or fluids can only have volume change or no fixed shape B1 [1] (b) either unless ∆p is very large or 2.2 × 109 is a large number M1 ∆V is very small or ∆V/V is very small, (so ‘incompressible’) A1 [2] (c) ∆p = hρg 1.01 × 105 = h × 1.08 × 103 × 9.81 C1 h = 9.53 m C1 ∆h / h = 0.47 / 10 or 0.47 / 9.53 error = 4.7% or 4.9% or 5% A1 [3]

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5 (a) State what is meant by For Examiner’s (i) the frequency of a progressive wave, Use .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) the speed of a progressive wave. .................................................................................................................................. ..............................................................................................................................[1] (b) One end of a long string is attached to an oscillator. The string passes over a frictionless pulley and is kept taut by means of a weight, as shown in Fig. 5.1. string pulley oscillator weight Fig. 5.1 The frequency of oscillation is varied and, at one value of frequency, the wave formed on the string is as shown in Fig. 5.1. (i) Explain why the wave is said to be a stationary wave. .................................................................................................................................. ..............................................................................................................................[1] (ii) State what is meant by an antinode. .................................................................................................................................. ..............................................................................................................................[1] (iii) On Fig. 5.1, label the antinodes with the letter A. [1] (c) A weight of 4.00 N is hung from the string in (b) and the frequency of oscillation is For adjusted until a stationary wave is formed on the string. The separation of the antinodes Examiner’s on the string is 17.8 cm for a frequency of 125 Hz. Use The speed v of waves on a string is given by the expression T v = , m where T is the tension in the string and m is its mass per unit length. Determine the mass per unit length of the string. mass per unit length = ........................................... kg m–1 [5]

Mark scheme: 5 (a) (i) frequency: number of oscillations per unit time M1 of the source / of a point on the wave A1 [2] (ii) speed: speed at which energy is transferred / speed of wavefront B1 [1] (b) (i) does not transfer energy (along the wave) B1 [1] (ii) position (along wave) where amplitude of vibration is a maximum B1 [1] (iii) all three positions marked B1 [1] (c) wavelength = 2 × 17.8 = 35.6 cm C1 v = fλ C1 v = 125 × 0.356 = 44.5 m s–1 C1 44.52 = 4.00 / m C1 m = 2.0 × 10–3 kg m–1 A1 [5] GCE A/AS LEVEL – May/June 2008 9702 02 2

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6 An electric heater consists of three similar heating elements A, B and C, connected as shown For in Fig. 6.1. Examiner’s Use 240V S1 A B S2 C S3 Fig. 6.1 Each heating element is rated as 1.5 kW, 240 V and may be assumed to have constant resistance. The circuit is connected to a 240 V supply. (a) Calculate the resistance of one heating element. resistance = ……………….……….. Ω [2] (b) The switches S1, S2 and S3 may be either open or closed. For Examiner’s Complete Fig. 6.2 to show the total power dissipation of the heater for the switches in Use the positions indicated. S1 S2 S3 total power / kW open closed closed closed closed open closed closed closed closed open open closed open closed [5] Fig. 6.2

Mark scheme: 6 (a) either P = VI and V = IR or P = V2 / R C1 resistance = 38.4 Ω A1 [2] (b) zero B1 1.5 kW B1 3.0 kW B1 0.75 kW B1 2.25 kW B1 [5]

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7 Uranium-236 (23692U) and Uranium-237 (2392U)7 are both radioactive. For Uranium-236 is an α-emitter and Uranium-237 is a β-emitter. Examiner’s Use (a) Distinguish between an α-particle and a β-particle. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[4] (b) The grid of Fig. 7.1 shows some proton numbers Z on the x-axis and the number N of neutrons in the nucleus on the y-axis. 149 148 number of neutrons N 147 146 145 236U92U 144 143 142 141 140 88 89 90 91 92 93 94 95 96 97 proton number Z Fig. 7.1 The α-decay of Uranium-236 (23692U) is represented on the grid. This decay produces a For nucleus of thorium (Th). Examiner’s Use (i) Write down the nuclear equation for this α-decay. ..............................................................................................................................[2] (ii) On Fig. 7.1, mark the position for a nucleus of 1. Uranium-237 (mark this position with the letter U), 2. Neptunium, the nucleus produced by the β-decay of Uranium-237 (mark this position with the letters Np). [2]

Mark scheme: 7 (a) α-particle: either helium nucleus or contains 2 protons + 2 neutrons or 42 He B1 β-particle: either electron or −01 e B1 α speed < β speed (1) α discrete values of speed/energy, β continuous spectrum (1) either α ionising power >> β ionising power or α range << β range (1) α positive, β negative (only if first two B marks not scored) (1) α mass > β mass (only if first two B marks not scored) (1) (any two sensible pairs of statements relevant to differences, – do not allow statements relevant to only α or β, 1 each, max 2) B2 [4] (b) (i) 23692 U → 23290 Th M1 + 42 He A1 [2] (ii) 1. correct position for U at Z = 92, N = 145 B1 2. correct position for Np relative to U i.e. Z + 1 and N – 1 B1 [2]

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