Cambridge A Level Physics 9702 — 2013 Oct/Nov Paper 2 · Variant 2

1 (a) State two SI base units other than the kilogram, metre and second. Use 1. ...................................................................................................................................... 2. ...................................................................................................................................... [2] (b) A metal wire has original length l0. It is then suspended and hangs vertically as shown in Fig. 1.1. wire Fig. 1.1 The weight of the wire causes it to stretch. The elastic potential energy stored in the wire is E. (i) Show that the SI base units of E are kg m2 s–2. [2] (ii) The elastic potential energy E is given by For Examiner’s E = Cρ2g 2Al03 Use where ρ is the density of the metal, g is the acceleration of free fall, A is the cross-sectional area of the wire and C is a constant. Determine the SI base units of C. SI base units of C ..................................................[3]

Mark scheme: 1 (a) kelvin / K B1 ampere / amp / A B1 [2] [allow mole / mol and candela / Cd] (b) (i) energy OR work = force × distance [allow any energy expression] C1 units: kg m s–2 × m OR kg (m s–1)2 for ½ mv2 or mc2 M1 (ignore any numerical factor) units: = kg m2 s–2 A0 [2] (ii) units: ρ: kg m–3 g: m s–2 A: m2 l0: m C1 C: kg m2 s–2 / kg2 m–6 m2 s–4 m2 m3 [any subject] C1 = kg–1 m s2 (allow m s2 / kg) A1 [3]

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2 A source of radio waves sends a pulse towards a reflector. The pulse returns from the For reflector and is detected at the same point as the source. The emitted and reflected pulses Examiner’s are recorded on a cathode-ray oscilloscope (c.r.o.) as shown in Fig. 2.1. Use 1 cm 1 cm Fig. 2.1 The time-base setting is 0.20 μs cm–1. (a) Using Fig. 2.1, determine the distance between the source and the reflector. distance = ............................................. m [4] (b) Determine the time-base setting required to produce the same separation of pulses on the c.r.o. when sound waves are used instead of radio waves. The speed of sound is 300 m s–1. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3]

Mark scheme: 2 (a) d = v × t C1 t = 0.2 × 4 (allow t = 0.2 × 2) C1 d = 3 × 108 × 0.8 × 10–6 OR 3 × 108 × 0.4 × 10–6 C1 d = 240 m hence distance from source to reflector = 120 m A1 [4] (b) speed of sound 300 cf speed of light 3 × 108 OR time = 240 / 300 (= 0.8) OR time = 120 / 300 (= 0.4) C1 sound slower by factor of 106 OR time for one division 0.8 / 4 OR time for one division 0.4 / 2 C1 time base setting 0.2 s cm–1 [unit required] A1 [3]

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3 (a) State what is meant by work done. For Examiner’s .......................................................................................................................................... Use ......................................................................................................................................[1] (b) A trolley of mass 400 g is moving at a constant velocity of 2.5 m s–1 to the right as shown in Fig. 3.1. 2.5 m s–1 trolley 400 g Fig. 3.1 Show that the kinetic energy of the trolley is 1.3 J. [2] (c) The trolley in (b) moves to point P as shown in Fig. 3.2. 2.5 m s–1 trolley F 400 g P Q x Fig. 3.2 At point P the speed of the trolley is 2.5 m s–1. A variable force F acts to the left on the trolley as it moves between points P and Q. The variation of F with displacement x from P is shown in Fig. 3.3. 20 F / N 10 0 0 x P Q Fig. 3.3 The trolley comes to rest at point Q. For Examiner’s (i) Calculate the distance PQ. Use distance PQ = ............................................. m [3] (ii) On Fig. 3.4, sketch the variation with x of velocity v for the trolley moving between P and Q. 2.5 v / m s–1 0 P Q x Fig. 3.4 [2]

Mark scheme: 3 (a) (work =) force × distance moved / displacement in the direction of the force OR when a force moves in the direction of the force work is done B1 [1] (b) kinetic energy = ½ mv2 C1 kinetic energy = ½ 0.4 (2.5)2 = 1.25 / 1.3 J A1 [2] (c) (i) area under graph is work done / work done = ½ Fx C1 1.25 = (14 x) / 2 C1 x = 0.18 (0.179) m [allow x = 0.19 m using kinetic energy = 1.3 J] A1 [3] (ii) smooth curve from v = 2.5 at x = 0 to v = 0 at Q M1 curve with increasing gradient A1 [2] GCE AS/A LEVEL – October/November 2013 9702 22

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4 (a) Define the torque of a couple. For Examiner’s .......................................................................................................................................... Use ......................................................................................................................................[2] (b) A wheel is supported by a pin P at its centre of gravity, as shown in Fig. 4.1. 25 cm 35 N P 35 N Fig. 4.1 The plane of the wheel is vertical. The wheel has radius 25 cm. Two parallel forces each of 35 N act on the edge of the wheel in the vertical directions shown in Fig. 4.1. Friction between the pin and the wheel is negligible. (i) List two other forces that act on the wheel. State the direction of these forces and where they act. 1. ............................................................................................................................... 2. ............................................................................................................................... [2] (ii) Calculate the torque of the couple acting on the wheel. torque = .......................................... N m [2] (iii) The resultant force on the wheel is zero. Explain, by reference to the four forces acting on the wheel, how it is possible that the resultant force is zero. .................................................................................................................................. ..............................................................................................................................[1] (iv) State and explain whether the wheel is in equilibrium. ..............................................................................................................................[1]

Mark scheme: 4 (a) torque of a couple = one of the forces / a force × distance M1 multiplied by the perpendicular distance between the forces A1 [2] (b) (i) weight at P (vertically) down B1 normal reaction OR contact force at (point of contact with the pin) P (vertically) up B1 [2] (ii) torque = 35 × 0.25 (or 25) × 2 C1 torque = 18 (17.5) N m A1 [2] (iii) the two 35 N forces are equal and opposite and the weight and the upward / contact / reaction force are equal and opposite B1 [1] (iv) not in equilibrium as the (resultant) torque is not zero B1 [1]

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5 A long rope is held under tension between two points A and B. Point A is made to vibrate For vertically and a wave is sent down the rope towards B as shown in Fig. 5.1. Examiner’s Use direction of travel of wave B A Fig. 5.1 (not to scale) The time for one oscillation of point A on the rope is 0.20 s. The point A moves a distance of 80 mm during one oscillation. The wave on the rope has a wavelength of 1.5 m. (a) (i) Explain the term displacement for the wave on the rope. .................................................................................................................................. ..............................................................................................................................[1] (ii) Calculate, for the wave on the rope, 1. the amplitude, amplitude = .......................................... mm [1] 2. the speed. speed = ........................................ m s–1 [3] (b) On Fig. 5.1, draw the wave pattern on the rope at a time 0.050 s later than that shown. [2] (c) State and explain whether the waves on the rope are (i) progressive or stationary, .................................................................................................................................. ..............................................................................................................................[1] (ii) longitudinal or transverse. .................................................................................................................................. ..............................................................................................................................[1]

Mark scheme: 5 (a) (i) displacement is the distance the rope / particles are (above or below) from the equilibrium / mean / rest / undisturbed position (not ‘distance moved’) B1 [1] (ii) 1. amplitude (= 80 / 4) = 20 mm B1 [1] 2. v = fλ or v = λ / T C1 f = 1 / T = 1 / 0.2 (5 Hz) C1 v = 5 × 1.5 = 7.5 m s–1 A1 [3] (b) point A of rope shown at equilibrium position B1 same wavelength, shape, peaks / wave moved ¼λ to right B1 [2] (c) (i) progressive as energy OR peaks OR troughs is/are transferred/moved /propagated (by the waves) B1 [1] (ii) transverse as particles/rope movement is perpendicular to direction of travel /propagation of the energy/wave velocity B1 [1]

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6 (a) Define potential difference (p.d.). For Examiner’s ......................................................................................................................................[1] Use (b) A power supply of e.m.f. 240 V and zero internal resistance is connected to a heater as shown in Fig. 6.1. 240 V Fig. 6.1 The wires used to connect the heater to the power supply each have length 75 m. The wires have a cross-sectional area 2.5 mm2 and resistivity 18 nΩ m. The heater has a constant resistance of 38 Ω. (i) Show that the resistance of each wire is 0.54 Ω. [3] (ii) Calculate the current in the wires. current = .............................................. A [3] (iii) Calculate the power loss in the wires. power = ............................................. W [3] (c) The wires to the heater are replaced by wires of the same length and material but For having a cross-sectional area of 0.50 mm2. Without further calculation, state and explain Examiner’s the effect on the power loss in the wires. Use .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

Mark scheme: 6 (a) p.d. = work (done) / charge OR energy transferred from (electrical to other forms) / (unit) charge B1 [1] (b) (i) R = ρl / A C1 ρ = 18 × 10–9 C1 R = (18 × 10–9 × 75) / 2.5 × 10–6 = 0.54 Ω A1 [3] (ii) V = IR C1 R = 38 + (2 × 0.54) C1 I = 240 / 39.08 = 6.1 (6.14) A A1 [3] GCE AS/A LEVEL – October/November 2013 9702 22 (iii) P = I 2R or P = VI and V = IR or P = V2/R and V = IR C1 P = (6.14)2 × 2 × 0.54 C1 P = 41 (40.7) W A1 [3] (c) area of wire is less (1/5) hence resistance greater (×5) M1 OR R is ∝ 1/A therefore R is greater p.d. across wires greater so power loss in cables increases A1 [2]

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7 (a) An electric field is set up between two parallel metal plates in a vacuum. The deflection For of α-particles as they pass between the plates is shown in Fig. 7.1. Examiner’s Use metal plate path of _-particles electric field metal plate Fig. 7.1 The electric field strength between the plates is reduced. The α-particles are replaced by β-particles. The deflection of β-particles is shown in Fig. 7.2. metal plate path of `-particles electric field metal plate Fig. 7.2 (i) State one similarity of the electric fields shown in Fig. 7.1 and Fig. 7.2. .................................................................................................................................. ..............................................................................................................................[1] (ii) The electric field strength in Fig. 7.2 is less than that in Fig. 7.1. State two methods of reducing this electric field strength. 1. ............................................................................................................................... 2. ............................................................................................................................... [2] (iii) By reference to the properties of α-particles and β-particles, suggest three reasons For for the differences in the deflections shown in Fig. 7.1 and Fig. 7.2. Examiner’s Use 1. ............................................................................................................................... .................................................................................................................................. 2. ............................................................................................................................... .................................................................................................................................. 3. ............................................................................................................................... .................................................................................................................................. [3] (b) A source of α-particles is uranium-238. The nuclear reaction for the emission of α-particles is represented by 23892U WXQ + YZ α. State the values of W ............... X ............... Y ............... Z ............... [2] (c) A source of β-particles is phosphorus-32. The nuclear reaction for the emission of β-particles is represented by 3215P ABR + CDβ. State the values of A ............... B ............... C ............... D ............... [1]

Mark scheme: 7 (a) (i) the direction of the fields is the same OR fields are uniform OR constant electric field strength OR E = V / d with symbols explained B1 [1] (ii) reduce p.d. across plates B1 increase separation of plates B1 [2] (iii) α opposite charge to β (as deflection in opposite direction) B1 β has a range of velocities OR energies (as different deflections) and α all have same velocity OR energy (as constant deflection) B1 α are more massive (as deflection is less for greater field strength) B1 [3] (b) W = 234 and X = 90 B1 Y = 4 and Z = 2 B1 [2] (c) A = 32 and B = 16 and C = 0 and D = –1 B1 [1]

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