Cambridge A Level Physics 9702 — 2007 May/June Paper 2 · Variant 1

1 The uncalibrated scale and the pointer of a meter are shown in Fig. 1.1. uncalibrated scale Fig. 1.1 The pointer is shown in the zero position. The meter is to be used to indicate the volume of fuel in the tank of a car. A known volume V of fuel is poured into the tank and the deflection θ of the pointer is noted. Fig. 1.2 shows the variation with θ of V. 80 V /103 cm3 60 40 20 0 0 20 40 60 80 100 θ/° Fig. 1.2 Examiner’s Use (a) On Fig. 1.1, (i) calibrate the scale at 20 × 103 cm3 intervals, [2] (ii) mark a possible position for a volume of 1.0 × 105 cm3. [1] (b) Suggest one advantage of this scale, as compared with a uniform scale, for measuring fuel volumes in the tank of the car. .......................................................................................................................................... ......................................................................................................................................[1]

Mark scheme: 1 (a) (i) all positions (accept 20, 40, 60, 80) marked to within ±5° B2 positions are 40°, 70°, 90° and 102° (-1 for each error or omission) (ii) allow 107° → 113° B1 [3] (b) e.g. more sensitive at low volumes B1 [1] (do not allow reference to ‘accuracy’)

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2 (a) Define electric field strength. .......................................................................................................................................... ......................................................................................................................................[1] (b) Two flat parallel metal plates, each of length 12.0 cm, are separated by a distance of 1.5 cm, as shown in Fig. 2.1. +210 V electron 1.5 cm speed 5.0 x 107 m s–1 12.0 cm Fig. 2.1 The space between the plates is a vacuum. The potential difference between the plates is 210 V. The electric field may be assumed to be uniform in the region between the plates and zero outside this region. Calculate the magnitude of the electric field strength between the plates. field strength = ........................................N C–1 [1] Examiner’s Use (c) An electron initially travels parallel to the plates along a line mid-way between the plates, as shown in Fig. 2.1. The speed of the electron is 5.0 × 107 m s–1. For the electron between the plates, (i) determine the magnitude and direction of its acceleration, acceleration = .............................................. m s–2 direction ...................................................[4] (ii) calculate the time for the electron to travel a horizontal distance equal to the length of the plates. time = ............................................... s [1] (d) Use your answers in (c) to determine whether the electron will hit one of the plates or emerge from between the plates. [3]

Mark scheme: 2 (a) force per unit positive charge (on a small test charge) B1 [1] (b) field strength = (210/{1.5 × 10-2} =) 1.4 ×104 N C-1 A1 [1] (c) (i) acceleration = Eq / m C1 = (1.4 × 104 × 1.6 × 10-19) / (9.1 × 10-31) C1 = 2.5 × 1015 m s-2 (2.46 × 1015) A1 towards positive plate / upwards (and normal to plate) B1 [4] (ii) time = 2.4 × 10-9 s A1 [1] (d) either vertical displacement after acceleration for 2.4 × 10-9 s = ½ × 2.46 × 1015 × (2.4 × 10-9)2 C1 = 7.1 × 10-3 m A1 (0.71 cm < 0.75 cm and) so will pass between plates A1 [3] i.e. valid conclusion based on a numerical value or 0.75 × 10-2 = ½ × 2.46 × 1015 × t2 (C1) t is time to travel ‘half-way across’ plates = 2.47 × 10-9 s (A1) (2.4 ns < 2.47 ns) so will pass between plates (A1) i.e. valid conclusion based on a numerical value

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3 (a) Define density. ......................................................................................................................................[1] (b) Liquid of density ρ fills a container to a depth h, as illustrated in Fig. 3.1. h area A Fig. 3.1 The container has vertical sides and a base of area A. (i) State, in terms of A, h and ρ, the mass of liquid in the container. ..............................................................................................................................[1] (ii) Hence derive an expression for the pressure p exerted by the liquid on the base of the container. Explain your working. [2] Examiner’s Use (c) The density of liquid water is 1.0 g cm–3. The density of water vapour at atmospheric 1 pressure is approximately g cm–3. 1600 Determine the ratio volume of water vapour (i) , volume of equal mass of liquid water ratio = ..................................................[1] mean separation of molecules in water vapour (ii) . mean separation of molecules in liquid water ratio = ...................................................[2] (d) State the evidence for (i) the molecules in solids and liquids having approximately the same separation, .................................................................................................................................. ..............................................................................................................................[1] (ii) strong rigid forces between molecules in solids. strong: ...................................................................................................................... rigid: .....................................................................................................................[2]

Mark scheme: 3 (a) mass / volume (ratio idea essential) B1 [1] (b) (i) mass = Ahρ B1 [1] (ii) pressure = force/area B1 weight (of liquid)/force (on base) = Ahρg B1 pressure = hρg A0 [2] (c) (i) ratio = 1600 or 1600:1 A1 [1] (ii) ratio = 3√1600 C1 = 11.7 (allow 12) A1 [2] GCE A/AS LEVEL – May/June 2007 9702 2 (d) (i) density of solids and liquids are (about) equal B1 [1] (ii) strong forces: fixed volume B1 rigid forces: retains shape / does not flow / little deformation B1 [2] (allow 1 mark for fixed volume, fixed shape)

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4 (a) A stone of mass 56 g is thrown horizontally from the top of a cliff with a speed of 18 m s–1, as illustrated in Fig. 4.1. 18 m s–1 16 m sea level Fig. 4.1 The initial height of the stone above the level of the sea is 16 m. Air resistance may be neglected. (i) Calculate the change in gravitational potential energy of the stone as a result of falling through 16 m. change = ............................................... J [2] (ii) Calculate the total kinetic energy of the stone as it reaches the sea. kinetic energy = .............................................. J [3] Examiner’s Use (b) Use your answer in (a)(ii) to show that the speed of the stone as it hits the water is approximately 25 m s–1. [1] (c) State the horizontal velocity of the stone as it hits the water. horizontal velocity = .........................................m s–1 [1] (d) (i) On the grid of Fig. 4.2, draw a vector diagram to represent the horizontal velocity and the resultant velocity of the stone as it hits the water. [1] Fig. 4.2 (ii) Use your vector diagram to determine the angle with the horizontal at which the stone hits the water. angle = .............................................. ° [2]

Mark scheme: 4 (a) (i) (change in) potential energy = mgh C1 = 0.056 × 9.8 × 16 = 8.78 J (allow 8.8) A1 [2] (ii) (initial) kinetic energy = ½mv2 C1 = ½ × 0.056 × 182 = 9.07 J (allow 9.1) C1 total kinetic energy = 8.78 + 9.07 = 17.9 J A1 [3] (b) kinetic energy = ½mv2 17.9 = ½ × 0.056 × v2 and v = 25(.3) m s-1 B1 [1] (c) horizontal velocity = 18 m s-1 B1 [1] (d) (i) correct shape of diagram (two sides of right-angled triangle with correct orientation) B1 (ii) angle = 41° → 48° (allow trig. solution based on diagram) A2 [3] (for angle 38°→ 41° or 48°→ 51°, allow 1 mark)

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5 Light reflected from the surface of smooth water may be described as a polarised transverse wave. (a) By reference to the direction of propagation of energy, explain what is meant by (i) a transverse wave, .................................................................................................................................. ..............................................................................................................................[1] (ii) polarisation. .................................................................................................................................. ..............................................................................................................................[1] (b) A glass tube, closed at one end, has fine dust sprinkled along its length. A sound source is placed near the open end of the tube, as shown in Fig. 5.1. dust heap tube sound source 39.0 cm Fig. 5.1 The frequency of the sound emitted by the source is varied and, at one frequency, the dust forms small heaps in the tube. (i) Explain, by reference to the properties of stationary waves, why the heaps of dust are formed. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3] Examiner’s Use (ii) One frequency at which heaps are formed is 2.14 kHz. The distance between six heaps, as shown in Fig. 5.1, is 39.0 cm. Calculate the speed of sound in the tube. speed = .........................................m s–1 [3] (c) The wave in the tube is a stationary wave. Explain, by reference to the formation of a stationary wave, what is meant by the speed calculated in (b)(ii). .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3]

Mark scheme: 5 (a) (i) vibrations (in plane) normal to direction of energy propagation B1 [1] (ii) vibrations in one direction (normal to direction of propagation) B1 [1] (b) (i) at (displacement) antinodes / where there are no heaps, wave has maximum amplitude (of vibration) B1 at (displacement) nodes/where there are heaps, amplitude of vibration is zero/minimum B1 dust is pushed to / settles at (displacement) nodes B1 [3] (ii) 2.5λ = 39 cm C1 v = fλ C1 v = 2.14 × 103 × 15.6 × 10-2 = 334 m s-1 (allow 330, not 340) A1 [3] (c) Stationary wave formed by interference / superposition / overlap of B1 either wave travelling down tube and its reflection or two waves of same (type and) frequency travelling in opposite directions B1 speed is the speed of the incident / reflected waves B1 [3] GCE A/AS LEVEL – May/June 2007 9702 2

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Use 6 A car battery has an internal resistance of 0.060 Ω. It is re-charged using a battery charger having an e.m.f. of 14 V and an internal resistance of 0.10 Ω, as shown in Fig. 6.1. car 0.10 Ω 0.060 Ω battery battery + charger 14 V – E Fig. 6.1 (a) At the beginning of the re-charging process, the current in the circuit is 42 A and the e.m.f. of the battery is E (measured in volts). (i) For the circuit of Fig. 6.1, state 1. the magnitude of the total resistance, resistance = ............................................. Ω 2. the total e.m.f. in the circuit. Give your answer in terms of E. e.m.f. = .............................................. V [2] (ii) Use your answers to (i) and data from the question to determine the e.m.f. of the car battery at the beginning of the re-charging process. e.m.f. = ...............................................V [2] Examiner’s Use (b) For the majority of the charging time of the car battery, the e.m.f. of the car battery is 12 V and the charging current is 12.5 A. The battery is charged at this current for 4.0 hours. Calculate, for this charging time, (i) the charge that passes through the battery, charge = .............................................. C [2] (ii) the energy supplied from the battery charger, energy = ............................................... J [2] (iii) the total energy dissipated in the internal resistance of the battery charger and the car battery. energy = ............................................... J [2] (c) Use your answers in (b) to calculate the percentage efficiency of transfer of energy from the battery charger to stored energy in the car battery. efficiency = ..............................................% [2]

Mark scheme: 6 (a) (i) 1 total resistance = 0.16 Ω A1 2 e.m.f. = either (14 – E) or (E – 14) A1 [2] (ii) either 14 – E = 42 × 0.16 or (E – 14) = -42 × 0.16 C1 E = 7.3 V A1 [2] (b) (i) charge = It C1 = 12.5 × 4 × 60 × 60 = 1.8 × 105 C A1 [2] (ii) either energy = EQ or energy = Eit C1 either energy = 14 × 1.8 × 105 or energy = 14 × 12.5 × 4 × 3600 = 2.52 × 106 J A1 [2] (iii) energy = I2Rt or Vit and V = IR C1 = 12.52 × 0.16 × 4 × 3600 = 3.6 × 105 J A1 [2] (c) efficiency = (2.52 × 106 – 3.6 × 105)/(2.52 × 106) C1 = 86% A1 [2]

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7 The radioactive decay of a strontium (Sr) nucleus is represented in Fig. 7.1. 92 nucleon 91 number Sr 90 89 88 36 37 38 39 40 proton number Fig. 7.1 (a) State whether Fig. 7.1 represents α-decay, β-decay or γ-decay. ......................................................................................................................................[1] (b) One type of radioactive decay cannot be represented on Fig. 7.1. Identify this decay and explain why it cannot be represented. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

Mark scheme: 7 (a) β(-decay) B1 [1] (b) γ(-decay) B1 either any two of Z, N and A do not change or it is loss of energy only or it is an electromagnetic wave B1 [2] Allow ‘α(-decay) as change of 4 in the nucleon number cannot be shown on the diagram’ (B2) Do not give credit for a ‘bald’ α(-decay)

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