Cambridge A Level Physics 9702 — 2005 Oct/Nov Paper 4 · Variant 1

1 The Earth may be considered to be a sphere of radius 6.4 ×106 m with its mass of 6.0 ×1024 kg concentrated at its centre. A satellite of mass 650 kg is to be launched from the Equator and put into geostationary orbit. (a) Show that the radius of the geostationary orbit is 4.2 ×107m. [3] (b) Determine the increase in gravitational potential energy of the satellite during its launch from the Earth’s surface to the geostationary orbit. energy = ………………………………... J [4] (c) Suggest one advantage of launching satellites from the Equator in the direction of rotation of the Earth. .......................................................................................................................................... ......................................................................................................................................[1]

Mark scheme: 1 (a) GM / R2 = Rω2 …………….…………………...…..………………….. C1 ω = 2π / (24 × 3600) ………………………………..……..…………… C1 6.67 × 10–11 × 6.0 × 1024 = R3 × ω2

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2 The air in a car tyre has a constant volume of 3.1 ×10–2m3. The pressure of this air is 2.9 ×105Pa at a temperature of 17 °C. The air may be considered to be an ideal gas. (a) State what is meant by an ideal gas. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) Calculate the amount of air, in mol, in the tyre. amount = ……………………………. mol [2] (c) The pressure in the tyre is to be increased using a pump. On each stroke of the pump, 0.012 mol of air is forced into the tyre. Calculate the number of strokes of the pump required to increase the pressure to 3.4 ×105Pa at a temperature of 27 °C. number = ……………………………. [3]

Mark scheme: 2 (a) obeys the law pV = constant × T ………………………..……………….. M1 at all values of p, V and T ………………………………………………. A1 [2] (b) n = (2.9 × 105 × 3.1 × 10–2) / (8.31 × 290) …..………………..………... C1 = 3.73 mol ………………………………………………………………. A1 [2] 3 . 4 290 (c) at new pressure, nn = 3.73 × × 2 . 9 300 = 4.23 mol ….………………………………………. C1 change = 0.50 mol ……………………………………………………….… C1 number of strokes = 0.50 / 0.012 = 42 (must round up for mark) ……. A1 [3]

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3 (a) State the first law of thermodynamics in terms of the increase in internal energy ∆U, the heating q of the system and the work w done on the system. .......................................................................................................................................... ..................................................................................................................................... [1] (b) The volume occupied by 1.00 mol of liquid water at 100 °C is 1.87 ×10–5m3. When the water is vaporised at an atmospheric pressure of 1.03 ×105 Pa, the water vapour has a volume of 2.96 ×10–2m3. The latent heat required to vaporise 1.00 mol of water at 100 °C and 1.03 ×105 Pa is 4.05 ×104 J. Determine, for this change of state, (i) the work w done on the system, w = ……………………………. J [2] (ii) the heating q of the system, q = ……………………………. J [1] (iii) the increase in internal energy ∆U of the system. ∆U = ……………………………. J [1] Use (c) Using your answer to (b)(iii), estimate the binding energy per molecule in liquid water. energy = ………………………………. J [2]

Mark scheme: 3 (a) correct statement, words or symbols …..…………………………...….. B1 [1] (b) (i) w = p∆V ………………………………………………………………….. C1 = 1.03 × 105 × (2.96 × 10–2 – 1.87 × 10–5) = (–) 3050 J …………………..……………….…………..…………… A1 [2] (ii) q = 4.05 × 104 J …………………………………………………………. B1 [1] (iii) ∆U = 4.05 × 104 – 3050 = 37500 J …no e.c.f. from (a)………………… A1 [1] penalise 2 sig.fig. once only (c) number of molecules = NA ………………………………………………. C1 energy = 37500 / (6.02 × 1023) = 6.2 × 10–20 J (accept 1 sig.fig.) …………………………..…. A1 [2]

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4 The centre of the cone of a loudspeaker is oscillating with simple harmonic motion of frequency 1400 Hz and amplitude 0.080 mm. (a) Calculate, to two significant figures, (i) the angular frequency ω of the oscillations, ω = ………………………………. rad s–1 [2] (ii) the maximum acceleration, in m s–2, of the centre of the cone. acceleration = ……………………………….. m s–2 [2] (b) On the axes of Fig. 4.1, sketch a graph to show the variation with displacement x of the acceleration a of the centre of the cone. a 0 0 x [2] Fig. 4.1 Use (c) (i) State the value of the displacement x at which the speed of the centre of the cone is a maximum. x = ……………………………… mm [1] (ii) Calculate, in m s–1, this maximum speed. speed = ……………………………. m s–1 [2]

Mark scheme: 4 (a) (i) ω = 2πf ………………………………………………………....………….. C1 = 2π × 1400 = 8800 rad s–1 ………………………………………………………….. A1 [2] (ii) a0 = (–)ω2x0 ……………………………..………………………………… C1 = (8800)2 × 0.080 × 10–3 = 6200 m s–2 …………………….……………………………………. A1 [2] (b) straight line through origin with negative gradient …….…………….... M1 end points of line correctly labelled …………………………………….. A1 [2] (c) (i) zero displacement ………………………………………………………… B1 [1] (ii) v = ωx0 ……………………………………………………………………. C1 = 8800 × 0.080 × 10–3 = 0.70 m s–1 ……………………………………………………………. A1 [2] A LEVEL – NOVEMBER 2005 9702 4 2

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5 (a) An electron is accelerated from rest in a vacuum through a potential difference of 1.2 ×104V. Show that the final speed of the electron is 6.5 ×107m s–1. [2] (b) The accelerated electron now enters a region of uniform magnetic field acting into the plane of the paper, as illustrated in Fig. 5.1. magnetic field into plane of paper + + + path of + + + electron + + + Fig. 5.1 (i) Describe the path of the electron as it passes through, and beyond, the region of the magnetic field. You may draw on Fig. 5.1 if you wish. path within field: ........................................................................................................ ................................................................................................................................... path beyond field: .................................................................................................... .............................................................................................................................. [3] Use (ii) State and explain the effect on the magnitude of the deflection of the electron in the magnetic field if, separately, 1. the potential difference accelerating the electron is reduced, ........................................................................................................................... ........................................................................................................................... ...................................................................................................................... [2] 2. the magnetic field strength is increased. ........................................................................................................................... ........................................................................................................................... ...................................................................................................................... [2]

Mark scheme: 5 (a) ½mv2 = qV ……(or some verbal explanation) …..……………..…… B1 ½ × 9.11 × 10-31 × v2 = 1.6 × 10-19 × 1.2 × 104 ………………………… B1 v = 6.49 × 107 m s–1 ………….………………………………………… A0 [2] (b) (i) within field: circular arc ………………..………………..…………….. B1 in ‘downward’ direction ……………..………………….. B1 beyond field: straight, with no ‘kink’ on leaving field ………………… B1 [3] (ii) 1. v is smaller …………………………………………………………………. M1 deflection is larger ………………………………………………………… A1 [2] 2. (magnetic) force is larger ………………………………………………… M1 deflection is larger ……………………………………………………….. A1 [2]

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6 (a) Define magnetic flux density. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3] (b) A flat coil consists of N turns of wire and has area A. The coil is placed so that its plane is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig. 6.1. flat coil area A θ magnetic field flux density B θ Fig. 6.1 Using the symbols A, B, N and θ and making reference to the magnetic flux in the coil, derive an expression for the magnetic flux linkage through the coil. [2] Use (c) (i) State Faraday’s law of electromagnetic induction. ................................................................................................................................... ................................................................................................................................... .............................................................................................................................. [2] (ii) The magnetic flux density B in the coil is now made to vary with time t as shown in Fig. 6.2. B 0 0 t T 2T 3T Fig. 6.2 E 0 0 t T 2T 3T Fig. 6.3 On Fig. 6.3, sketch the variation with time t of the e.m.f. E induced in the coil. [3]

Mark scheme: 6 (a) (numerically equal to) force per unit length …………………….…….… M1 on straight conductor carrying unit current ……………………………. A1 normal to the field ………………………………………………………… A1 [3] (b) flux through coil = BA sinθ ……………………………………………….. B1 flux linkage = BAN sinθ ………………..………………………………… B1 [2] (c) (i) (induced) e.m.f. proportional to ………………………………..…..…..… M1 rate of change of flux (linkage) …………………….……………………. A1 [2] (ii) graph: two square sections in correct positions, zero elsewhere ….. B1 pulses in opposite directions …………………………………… B1 amplitude of second about twice amplitude of first ………….. B1 [3]

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