Cambridge A Level Physics 9702 — 2005 Oct/Nov Paper 2 · Variant 1

1 (a) (i) Define pressure. Use ................................................................................................................................... .............................................................................................................................. [1] (ii) State the units of pressure in base units. .............................................................................................................................. [1] (b) The pressure p at a depth h in an incompressible fluid of density ρ is given by p = ρgh, where g is the acceleration of free fall. Use base units to check the homogeneity of this equation. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3]

Mark scheme: 1 (a) (i) force per unit area (ratio idea essential) B1 (ii) kg m-1 s-2 B1 [2] (b) ρ has base unit kg m-3 B1 g has base unit m s-2 B1 hρg has base unit m × kg m-3 × m s-2 M1 same as pressure QED A0 [3]

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2 (a) Explain what is meant by the centre of gravity of a body. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ..................................................................................................................................... [2] (b) An irregularly-shaped piece of cardboard is hung freely from one point near its edge, as shown in Fig. 2.1. pivot cardboard Fig. 2.1 Explain why the cardboard will come to rest with its centre of gravity vertically below the pivot. You may draw on Fig. 2.1 if you wish. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2]

Mark scheme: 2 (a) point where whole weight of body (allow mass) M1 may be considered to act (do not allow ‘acts’) A1 [2] (b) when CG below pivot, weight acts through the pivot B1 (so) weight has no turning effect about pivot B1 [2]

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3 A stone on a string is made to travel along a horizontal circular path, as shown in Fig. 3.1. For Examiner’s Use path of stone stone Fig. 3.1 The stone has a constant speed. (a) Define acceleration. .......................................................................................................................................... ..................................................................................................................................... [1] (b) Use your definition to explain whether the stone is accelerating. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (c) The stone has a weight of 5.0 N. When the string makes an angle of 35° to the vertical, For the tension in the string is 6.1 N, as illustrated in Fig. 3.2. Examiner’s Use 6.1 N 35° 5.0 N Fig. 3.2 Determine the resultant force acting on the stone in the position shown. magnitude of force = ……………..……………………. N direction of force….………………..………………….. [4]

Mark scheme: 3 (a) change in velocity/time (taken) B1 [1] (b) velocity is a vector/velocity has magnitude & direction B1 direction changing so must be accelerating B1 [2] (c) either 6.1 × cos35 = 4.99 N or scale shown B1 so no resultant vertical force triangle of correct shape B1 6.1 sin35 = 3.5 N resultant = 3.5 ± 0.2 N B1 horizontally horizontal ± 3° B1 [4] allow answer based on centripetal force: resultant is centripetal force (which is horizontal) (B1) resultant is horizontal component of tension (B1) 6.1 sin35 = 3.5 N (B1) horizontally (B1)

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4 A trolley of mass 930 g is held on a horizontal surface by means of two springs, as shown in For Fig. 4.1. Examiner’s Use trolley spring Fig. 4.1 The variation with time t of the speed v of the trolley for the first 0.60 s of its motion is shown in Fig. 4.2. 8.0 v / cm s – 1 6.0 4.0 2.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s Fig. 4.2 (a) Use Fig. 4.2 to determine (i) the initial acceleration of the trolley, acceleration = ………………………. m s–2 [2] (ii) the distance moved during the first 0.60 s of its motion. For Examiner’s Use distance = ………….………………… m [3] (b) (i) Use your answer to (a)(i) to determine the resultant force acting on the trolley at time t = 0. force = …………………………….. N [2] (ii) Describe qualitatively the variation with time of the resultant force acting on the trolley during the first 0.60 s of its motion. ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... .............................................................................................................................. [3]

Mark scheme: 4 (a) (i) use of tangent at time t = 0 B1 acceleration = 42 ± 4 cm s-2 A1 [2] (ii) use of area of loop B1 distance = 0.031 ± 0.001 m B2 [3] allow 1 mark if 0.031 ± 0.002 m) (b) (i) F = ma C1 = 0.93 × 0.42 {allow e.c.f. from (a)(i)} = 0.39 N A1 [2] (ii) force reduces to zero in first 0.3 s B1 then increases again in next 0.3 s M1 in the opposite direction A1 [3] GCE A/AS LEVEL – November 2005 9702 2

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5 Fig. 5.1 shows the variation with time t of the displacements xA and xB at a point P of two For sound waves A and B. Examiner’s Use wave A 3 xA / 10 – 4 cm 2 1 0 0 1 2 3 4 5 6 t / ms –1 –2 –3 wave B 2 xB / 10 – 4 cm 1 0 0 1 2 3 4 5 6 t / ms –1 –2 Fig. 5.1 (a) By reference to Fig. 5.1, state one similarity and one difference between these two waves. similarity: .......................................................................................................................... difference: ................................................................................................................... [2] (b) State, with a reason, whether the two waves are coherent. .......................................................................................................................................... ..................................................................................................................................... [1] (c) The intensity of wave A alone at point P is I. For Examiner’s (i) Show that the intensity of wave B alone at point P is 4 I. Use 9 [2] (ii) Calculate the resultant intensity, in terms of I, of the two waves at point P. resultant intensity = ……………………………… I [2] (d) Determine the resultant displacement for the two waves at point P (i) at time t = 3.0 ms, resultant displacement = ……………………………… cm [1] (ii) at time t = 4.0 ms. resultant displacement = ……………………………… cm [2]

Mark scheme: 5 (a) similarity: e.g. same wavelength/frequency/period, constant phase difference B1 difference: e.g. different amplitude/phase B1 [2] (do not allow a reference to phase for both similarity and difference) (b) constant phase difference so coherent B1 [1] (c) (i) intensity ∝ amplitude2 C1 I ∝ 32 and IB ∝ 22 leading to M1 4 IB = I A0 [2] 9 (ii) resultant amplitude = 1.0 × 10-4 cm C1 1 resultant intensity = I A1 [2] 9 (d) (i) displacement = 0 B1 [1] (ii) xA = -2.6 × 10-4 cm and xB = +1.7 × 10-4 cm C1 allow ± 0.5 × 10-4 cm) resultant displacement = (-) 0.9 × 10-4 cm A1 [2]

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6 Two horizontal metal plates X and Y are at a distance 0.75 cm apart. A positively charged For particle of mass 9.6 ×10–15kg is situated in a vacuum between the plates, as illustrated in Examiner’s Fig. 6.1. Use plate X + 0.75 cm plate Y Fig. 6.1 The potential difference between the plates is adjusted until the particle remains stationary. (a) State, with a reason, which plate, X or Y, is positively charged. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) The potential difference required for the particle to be stationary between the plates is found to be 630 V. Calculate (i) the electric field strength between the plates, field strength = …………………………….. N C–1 [2] (ii) the charge on the particle. For Examiner’s Use charge = …………………………….. C [3]

Mark scheme: 6 (a) force must be upwards (on positive charge) M1 so plate Y is positive A1 [2] (b) (i) E = V / d C1 = 630/(0.75 × 10-2) = 8.4 × 104 N C-1 A1 [2] (ii) qE = mg C1 q = (9.6 × 10-15 × 9.8) / (8.4 × 104) C1 = 1.12 × 10-18 C A1 [3]

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7 A battery of e.m.f. 4.50 V and negligible internal resistance is connected in series with a fixed For resistor of resistance 1200 Ωand a thermistor, as shown in Fig. 7.1. Examiner’s Use C 1200 Ω 4.50 V B A Fig. 7.1 (a) At room temperature, the thermistor has a resistance of 1800 Ω. Deduce that the potential difference across the thermistor (across AB) is 2.70 V. [2] (b) A uniform resistance wire PQ of length 1.00 m is now connected in parallel with the resistor and the thermistor, as shown in Fig. 7.2. C P 1200 Ω 4.50 V B V 1.00 m M A Q Fig. 7.2 A sensitive voltmeter is connected between point B and a moveable contact M on the For wire. Examiner’s Use (i) Explain why, for constant current in the wire, the potential difference between any two points on the wire is proportional to the distance between the points. ................................................................................................................................... ................................................................................................................................... ...............................................................................................................................[2] (ii) The contact M is moved along PQ until the voltmeter shows zero reading. 1. State the potential difference between the contact at M and the point Q. potential difference = …………………………. V [1] 2. Calculate the length of wire between M and Q. length = ………………………….. cm [2] (iii) The thermistor is warmed slightly. State and explain the effect on the length of wire between M and Q for the voltmeter to remain at zero deflection. ................................................................................................................................... ................................................................................................................................... ...............................................................................................................................[2]

Mark scheme: 7 (a) either V = E R1 / (R1 + R2) or I = E / (R1 + R2) C1 1800 1800 = × 4.50 V = × 4.50 M1 3000 3000 = 2.70 V = 2.70 V A0 [2] (b) (i) for a wire, V = I x (ρL/A) M1 I, ρ and A are constant A1 so V ∝ L A0 [2] GCE A/AS LEVEL – November 2005 9702 2 (ii) 1 2.70 V A1 [1] 2 L = 2 . 70 C1 100 4 . 50 L = 60.0 cm A1 [2] (iii) thermistor resistance decreases as temperature rises M1 so QM is shorter A1 [2]

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8 (a) Explain the concept of work. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ..................................................................................................................................... [2] (b) A table tennis ball falls vertically through air. Fig. 8.1 shows the variation of the kinetic energy EK of the ball with distance h fallen. The ball reaches the ground after falling through a distance h0. energy E K 0 0 h h 0 Fig. 8.1 (i) Describe the motion of the ball. ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... .............................................................................................................................. [3] (ii) On Fig. 8.1, draw a line to show the variation with h of the gravitational potential energy EP of the ball. At h = h0, the potential energy is zero. [3]

Mark scheme: 8 (a) product of force and distance M1 moved in the direction of the force A1 [2] (b) (i) falls from rest B1 decreasing acceleration B1 reaches a constant speed B1 [3] (ii) straight line with negative gradient B1 y-axis intercept above maximum EK B1 reasonable gradient (same magnitude as that for EK initially) B1 [3]

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