Cambridge A Level Physics 9702 — 2009 May/June Paper 2 · Variant 1

1 (a) State the most appropriate instrument, or instruments, for the measurement of the following. (i) the diameter of a wire of diameter about 1 mm ............................................................................................................................ [1] (ii) the resistance of a filament lamp ............................................................................................................................ [1] (iii) the peak value of an alternating voltage ............................................................................................................................ [1] (b) The mass of a cube of aluminium is found to be 580 g with an uncertainty in the measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm. Calculate the density of aluminium with its uncertainty. Express your answer to an appropriate number of significant figures. density = ................ ± ................ g cm–3 [5]

Mark scheme: 1 (a) (i) micrometer (screw gauge) / travelling microscope ...................... B1 [1] (ii) either ohm-meter or voltmeter and ammeter or multimeter/avo on ohm setting ................................................ B1 [1] (iii) either (calibrated) c.r.o. or a.c. voltmeter and × √2 ...................... B1 [1] (b) density = mass / volume .................................................................... C1 = 580 / 63 = 2.685 g cm-3 …(allow 2.68, 2.69, 2.7) ............. A1 % uncertainty in mass = (10 / 580) × 100 = 1.7% .............................. C1 % uncertainty in volume = 3 × (0.1 / 6) × 100 = 5.0% ........................ C1 uncertainty in density = 0.18 g cm-3 density = 2.7 ± 0.2 g cm-3 ................................................................... A1 [5] (answer 2.69 ± 0.09 g cm-3 scores 4 marks)

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2 A ball B of mass 1.2 kg travelling at constant velocity collides head-on with a stationary ball S For of mass 3.6 kg, as shown in Fig. 2.1. Examiner’s Use v ball B ball S mass 1.2 kg mass 3.6 kg Fig. 2.1 Frictional forces are negligible. The variation with time t of the velocity v of ball B before, during and after colliding with ball S is shown in Fig. 2.2. +4 +3 v / m s–1 +2 +1 0 0 0.1 0.2 0.3 0.4 0.5 t / s –1 –2 Fig. 2.2 (a) State the significance of positive and negative values for v in Fig. 2.2. .......................................................................................................................................... .................................................................................................................................... [1] (b) Use Fig. 2.2 to determine, for ball B during the collision with ball S, For Examiner’s (i) the change in momentum of ball B, Use change in momentum = .......................................... N s [3] (ii) the magnitude of the force acting on ball B. force = ............................................. N [3] (c) Calculate the speed of ball S after the collision. speed = ....................................... m s–1 [2] (d) Using your answer in (c) and information from Fig. 2.2, deduce quantitatively whether For the collision is elastic or inelastic. Examiner’s Use .......................................................................................................................................... .................................................................................................................................... [2]

Mark scheme: 2 (a) ball moving in opposite direction (after collision) ................................ B1 [1] (b) (i) change in momentum = 1.2 (4.0 + 0.8) ....................................... C2 (correct values, 1 mark; correct sign {values added}, 1 mark ) = 5.76 N s …(allow 5.8) ......................... A1 [3] (ii) force = ∆p / ∆t or m∆v / ∆t ................................................... C1 = 5.76 / 0.08 or 1.2 × 4.8 / 0.08 ........................................ C1 = 72 N ............................................................................... A1 [3] (c) 5.76 = 3.6 × V ..................................................................................... C1 V = 1.6 m s-1 ....................................................................................... A1 [2] (d) either speed of approach = 4.0 m s-1 and speed of separation = 2.4 m s-1 .............................................. M1 not equal and so inelastic ....................................................... A1 or kinetic energy before = 9.6 J and kinetic energy after collision = 4.99 J ...................................... M1 kinetic energy after is less / not conserved so inelastic .......... A1 [2]

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3 (a) Define the torque of a couple. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... .................................................................................................................................... [2] (b) A torque wrench is a type of spanner for tightening a nut and bolt to a particular torque, as illustrated in Fig. 3.1. force F nut torque scale C 45 cm Fig. 3.1 The wrench is put on the nut and a force is applied to the handle. A scale indicates the torque applied. The wheel nuts on a particular car must be tightened to a torque of 130 N m. This is achieved by applying a force F to the wrench at a distance of 45 cm from its centre of rotation C. This force F may be applied at any angle to the axis of the handle, as shown in Fig. 3.1. For the minimum value of F to achieve this torque, (i) state the magnitude of the angle that should be used, = .............................................. ° [1] (ii) calculate the magnitude of F. F = ............................................. N [2]

Mark scheme: 3 (a) product of (magnitude of one) force and distance between forces ..... M1 reference to either perpendicular distance between forces or line of action of forces and perpendicular distance .... A1 [2] (b) (i) 90° ............................................................................................... B1 [1] (ii) 130 = F × 0.45 (allow e.c.f. for angle in (i)) ................................ C1 F = 290 N ..................................................................................... A1 [2] (allow 1 mark only if angle stated in (i) is not used in (ii)) GCE A/AS LEVEL – May/June 2009 9702 21

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4 A spring having spring constant k hangs vertically from a fixed point. A load of weight L, when For hung from the spring, causes an extension e. The elastic limit of the spring is not exceeded. Examiner’s Use (a) State (i) what is meant by an elastic deformation, .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2] (ii) the relation between k, L and e. ............................................................................................................................ [1] Question 4 continues on page 10 (b) Some identical springs, each with spring constant k, are arranged as shown in Fig. 4.1. For Examiner’s Use arrangement total extension spring constant of arrangement L L L Fig. 4.1 The load on each of the arrangements is L. For each arrangement in Fig. 4.1, complete the table by determining (i) the total extension in terms of e, (ii) the spring constant in terms of k. [5]

Mark scheme: 4 (a) (i) change of shape / size / length / dimension ................................ C1 when (deforming) force is removed, returns to original shape / size A1 [2] (ii) L = ke ........................................................................................... B1 [1] (b) 2e ....................................................................................................... B1 ½k …(allow e.c.f. from extension) .................................................... B1 ½e and 2k ........................................................................................... B1 3 e …(allow e.c.f. from extension in part 2) ...................................... B1 2 2 k …(allow e.c.f. from extension) .................................................... B1 [5] 3

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5 Two sources S1 and S2 of sound are situated 80 cm apart in air, as shown in Fig. 5.1. For Examiner’s Use 100 cm S1 M 80 cm S2 Fig. 5.1 The frequency of vibration can be varied. The two sources always vibrate in phase but have different amplitudes of vibration. A microphone M is situated a distance 100 cm from S1 along a line that is normal to S1S2. As the frequency of S1 and S2 is gradually increased, the microphone M detects maxima and minima of intensity of sound. (a) State the two conditions that must be satisfied for the intensity of sound at M to be zero. 1. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... [2] (b) The speed of sound in air is 330 m s–1. The frequency of the sound from S1 and S2 is increased. Determine the number of minima that will be detected at M as the frequency is increased from 1.0 kHz to 4.0 kHz. number = ................................................ [4]

Mark scheme: 5 (a) either phase difference is π rad / 180° or path difference (between waves from S1 and S2) is ½λ / (n + ½)λ . B1 either same amplitude / intensity at M or ratio of amplitudes is 1.28 / ratio of intensities is 1.282 .................. B1 [2] (b) path difference between waves from S1 and S2 = 28 cm ................. B1 wavelength changes from 33 cm to 8.25 cm ...................................... B1 minimum when λ = (56 cm,) 18.7 cm, 11.2 cm, (8.0 cm) ................ B1 so two minima .................................................................................... B1 [4]

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6 Two vertical parallel metal plates are situated 2.50 cm apart in a vacuum. The potential For difference between the plates is 350 V, as shown in Fig. 6.1. Examiner’s Use 350 V electron + – 2.50 cm Fig. 6.1 An electron is initially at rest close to the negative plate and in the uniform electric field between the plates. (a) (i) Calculate the magnitude of the electric field between the plates. electric field strength = ....................................... N C–1 [2] (ii) Show that the force on the electron due to the electric field is 2.24 × 10–15 N. [2] (b) The electron accelerates horizontally across the space between the plates. Determine For Examiner’s (i) the horizontal acceleration of the electron, Use acceleration = ....................................... m s–2 [2] (ii) the time to travel the horizontal distance of 2.50 cm between the plates. time = ............................................. s [2] (c) Explain why gravitational effects on the electron need not be taken into consideration in your calculation in (b). .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2]

Mark scheme: 6 (a) (i) E = V / d ....................................................................................... C1 = 350 / (2.5 × 10-2) = 1.4 × 104 N C-1 .......................................................................... A1 [2] (ii) force = Eq .................................................................................... C1 = 1.4 × 104 × 1.6 × 10-19 ............................................................... M1 = 2.24 × 10-15 ............................................................................... A0 [2] (b) (i) F = ma ......................................................................................... C1 a = (2.24 × 10-15) / (9.1 × 10-31) = 2.46 × 1015 m s-2 …(allow 2.5 × 105) ...................................... A1 [2] (ii) s = ½at2 ....................................................................................... C1 2.5 × 10-2 = ½ × 2.46 × 1015 × t2 t = 4.5 × 10-9 s ............................................................................. A1 [2] (c) either gravitational force is normal to electric force or electric force horizontal, gravitational force vertical ................ B2 [2] special case: force/acceleration due to electric field >> force/acceleration due to gravitational field, allow 1 mark GCE A/AS LEVEL – May/June 2009 9702 21

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7 A network of resistors, each of resistance R, is shown in Fig. 7.1. For Examiner’s Use A Z R R R B Y R C X Fig. 7.1 (a) Calculate the total resistance, in terms of R, between points (i) A and C, resistance = ................................................ [1] (ii) B and X, resistance = ................................................ [1] (iii) A and Z. resistance = ................................................ [1] (b) Two cells of e.m.f. E1 and E2 and negligible internal resistance are connected into the For network in (a), as shown in Fig. 7.2. Examiner’s Use E1 A Z R R I1 I3 R B Y I2 R C X E2 Fig. 7.2 The currents in the network are as indicated in Fig. 7.2. Use Kirchhoff’s laws to state the relation (i) between currents I1, I2 and I3, ............................................................................................................................ [1] (ii) between E2, R, I2 and I3 in loop BCXYB, ............................................................................................................................ [1] (iii) between E1, E2, R, I1 and I2 in loop ABCXYZA. ............................................................................................................................ [1]

Mark scheme: 7 (a) (i) R .................................................................................................. B1 [1] (ii) 0.5R ............................................................................................. B1 [1] (iii) 2.5R …(allow e.c.f. from (ii)) ..................................................... B1 [1] (b) (i) I1 + I2 = I3 .................................................................................... B1 [1] (ii) E2 = I3R + I2R .............................................................................. B1 [1] (iii) E1 – E2 = 2I1R – I2R .................................................................... B1 [1]

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