Cambridge A Level Physics 9702 — 2012 Oct/Nov Paper 2 · Variant 1

1 (a) (i) Define acceleration. .................................................................................................................................. ............................................................................................................................. [1] (ii) State Newton’s first law of motion. .................................................................................................................................. ............................................................................................................................. [1] (b) The variation with time t of vertical speed v of a parachutist falling from an aircraft is shown in Fig. 1.1. 60 B C 50 v / m s–1 40 30 20 D 10 E A 0 0 10 20 30 t / s Fig. 1.1 (i) Calculate the distance travelled by the parachutist in the first 3.0 s of the motion. For Examiner’s Use distance = ............................................ m [2] (ii) Explain the variation of the resultant force acting on the parachutist from t = 0 (point A) to t = 15 s (point C). .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [3] (iii) Describe the changes to the frictional force on the parachutist 1. at t = 15 s (point C), .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [1] 2. between t = 15 s (point C) and t = 22 s (point E). .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [1] (iv) The mass of the parachutist is 95 kg. For Examiner’s Calculate, for the parachutist between t = 15 s (point C) and t = 17 s (point D), Use 1. the average acceleration, acceleration = ....................................... m s–2 [2] 2. the average frictional force. frictional force = ............................................. N [3] Please turn over for Question 2.

Mark scheme: 1 (a) (i) acceleration = change in velocity / time (taken) or acceleration = rate of change of velocity B1 [1] (ii) a body continues at constant velocity unless acted on by a resultant force B1 [1] (b) (i) distance is represented by the area under graph C1 distance = ½ × 29.5 × 3 = 44.3 m (accept 43.5 m for 29 to 45 m for 30) A1 [2] (ii) resultant force = weight – frictional force B1 frictional force increases with speed B1 at start frictional force = 0 / at end weight = frictional force B1 [3] (iii) 1. frictional force increases B1 [1] 2. frictional force (constant) and then decreases B1 [1] (iv) 1. acceleration = (v2 – v1) / t = (20 – 50) / (17 – 15) C1 = (–) 15 m s–2 A1 [2] 2. W – F = ma C1 W = 95 × 9.81 (= 932) C1 F = (95 × 15) + 932 = 2400 (2360) (2357) N A1 [3]

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2 (a) Define electrical resistance. For Examiner’s .......................................................................................................................................... Use ..................................................................................................................................... [1] (b) A circuit is set up to measure the resistance R of a metal wire. The potential difference (p.d.) V across the wire and the current І in the wire are to be measured. (i) Draw a circuit diagram of the apparatus that could be used to make these measurements. [3] (ii) Readings for p.d. V and the corresponding current І are obtained. These are shown in Fig. 2.1. 0.30 0.25 0.20 I / A 0.15 0.10 0.05 0 0 1.0 2.0 3.0 4.0 5.0 V / V Fig. 2.1 Explain how Fig. 2.1 indicates that the readings are subject to For Examiner’s 1. a systematic uncertainty, Use .................................................................................................................................. ............................................................................................................................. [1] 2. random uncertainties. .................................................................................................................................. ............................................................................................................................. [1] (iii) Use data from Fig. 2.1 to determine R. Explain your working. R = ............................................. Ω [3] (c) In another experiment, a value of R is determined from the following data: Current І = 0.64 ± 0.01 A and p.d. V = 6.8 ± 0.1 V. Calculate the value of R, together with its uncertainty. Give your answer to an appropriate number of significant figures. R = ..................... ± .................... Ω [3]

Mark scheme: 2 (a) resistance = potential difference / current B1 [1] (b) (i) metal wire in series with power supply and ammeter B1 voltmeter in parallel with metal wire B1 rheostat in series with power supply or potential divider arrangement or variable power supply B1 [3] (ii) 1. intercept on graph B1 [1] 2. scatter of readings about the best fit line B1 [1] (iii) correction for zero error explained B1 use of V and corrected І values from graph C1 resistance = V / І = 22.(2) Ω [e.g. 4.0 / 0.18] A1 [3] (c) R = 6.8 / 0.64 = 10.625 C1 %R = %V + %І = (0.1 / 6.8) × 100 + (0.01 / 0.64) × 100 C1 = 1.47% + 1.56% ∆R = 0.0303 × 10.625 = 0.32 Ω R = 10.6 ± 0.3 Ω A1 [3] GCE AS/A LEVEL – October/November 2012 9702 21

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3 (a) Define pressure. For Examiner’s .......................................................................................................................................... Use ..................................................................................................................................... [1] (b) Explain, in terms of the air molecules, why the pressure at the top of a mountain is less than at sea level. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3] (c) Fig. 3.1 shows a liquid in a cylindrical container. container liquid 0.250 m Fig. 3.1 The cross-sectional area of the container is 0.450 m2. The height of the column of liquid is 0.250 m and the density of the liquid is 13 600 kg m–3. (i) Calculate the weight of the column of liquid. weight = ............................................ N [3] (ii) Calculate the pressure on the base of the container caused by the weight of the For liquid. Examiner’s Use pressure = ........................................... Pa [1] (iii) Explain why the pressure exerted on the base of the container is different from the value calculated in (ii). .................................................................................................................................. ............................................................................................................................. [1]

Mark scheme: 3 (a) pressure = force / area B1 [1] (b) molecules collide with object / surface and rebound B1 molecules have change in momentum hence force acts B1 fewer molecules per unit volume on top of mountain / temperature is less hence lower speed of molecules B1 hence less pressure A0 [3] (c) (i) ρ = m / V C1 W = Vρg = 0.25 × 0.45 × 9.81 × 13600 C1 = 15000 (15009) N A1 [3] (ii) p = W / A (or using p = ρgh) = 15009 / 0.45 = 3.3 × 104 Pa A1 [1] (iii) pressure will be greater due to the air pressure (acting on the surface of the liquid) B1 [1]

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4 (a) Describe the diffraction of monochromatic light as it passes through a diffraction grating. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ..................................................................................................................................... [2] (b) White light is incident on a diffraction grating, as shown in Fig. 4.1. spectrum (first order) white light white (zero order) diffraction spectrum (first order) grating screen Fig. 4.1 (not to scale) The diffraction pattern formed on the screen has white light, called zero order, and coloured spectra in other orders. (i) Describe how the principle of superposition is used to explain 1. white light at the zero order, .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2] 2. the difference in position of red and blue light in the first-order spectrum. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2] (ii) Light of wavelength 625 nm produces a second-order maximum at an angle of 61.0° For to the incident direction. Examiner’s Determine the number of lines per metre of the diffraction grating. Use number of lines = ......................................... m–1 [2] (iii) Calculate the wavelength of another part of the visible spectrum that gives a maximum for a different order at the same angle as in (ii). wavelength = ……………………..…….. nm [2]

Mark scheme: 4 (a) waves pass through the elements / gaps / slits in the grating M1 spread into geometric shadow A1 [2] (b) (i) 1. displacements add to give resultant displacement B1 each wavelength travels the same path difference or are in phase B1 hence produce a maximum A0 [2] 2. to obtain a maximum the path difference must be λ or phase difference 360° / 2π rad B1 λ of red and blue are different B1 hence maxima at different angles / positions A0 [2] (ii) nλ = d sin θ C1 N = sin 61° / (2 × 625 × 10–9) = 7.0 × 105 A1 [2] (iii) nλ = 2 × 625 is a constant (1250) C1 n = 1 → λ = 1250 outside visible n = 3 → λ = 417 in visible n = 4 → λ = 312.5 outside visible λ = 420 nm A1 [2]

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5 (a) Explain what is meant by plastic deformation. For Examiner’s .......................................................................................................................................... Use ..................................................................................................................................... [1] (b) A copper wire of uniform cross-sectional area 1.54 × 10–6 m2 and length 1.75 m has a breaking stress of 2.20 × 108 Pa. The Young modulus of copper is 1.20 × 1011 Pa. (i) Calculate the breaking force of the wire. breaking force = ............................................. N [2] (ii) A stress of 9.0 × 107 Pa is applied to the wire. Calculate the extension. extension = ............................................ m [2] (c) Explain why it is not appropriate to use the Young modulus to determine the extension when the breaking force is applied. .......................................................................................................................................... ..................................................................................................................................... [1]

Mark scheme: 5 (a) when the load is removed then the wire / body object does not return to its original shape / length B1 [1] (b) (i) stress = force / area C1 F = 220 × 106 × 1.54 × 10–6 = 340 (338.8) N A1 [2] (ii) E = (F × l) / (A × e) C1 e = (90 × 106) × 1.75 / (1.2 × 1011) = 1.31 × 10–3 m A1 [2] (c) the stress is no longer proportional to the extension B1 [1] GCE AS/A LEVEL – October/November 2012 9702 21

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6 (a) Describe the structure of an atom of the nuclide 23592U. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) The deflection of α-particles by a thin metal foil is investigated with the arrangement shown in Fig. 6.1. All the apparatus is enclosed in a vacuum. vacuum detector of _-particles D W _ source path of deflected X _-particles Y Fig. 6.1 The detector of α-particles, D, is moved around the path labelled WXY. (i) Explain why the apparatus is enclosed in a vacuum. .................................................................................................................................. ............................................................................................................................. [1] (ii) State and explain the readings detected by D when it is moved along WXY. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [3] Question 6 continues on page 16. (c) A beam of α-particles produces a current of 1.5 pA. Calculate the number of α-particles For per second passing a point in the beam. Examiner’s Use number = ........................................... s–1 [3]

Mark scheme: 6 (a) 92 protons in the nucleus and 92 electrons around nucleus B1 143 neutrons (in the nucleus) B1 [2] (b) (i) α-particle travels short distance in air B1 [1] (ii) very small proportion in backwards direction / large angles B1 majority pass through with no /small deflections B1 either most of mass is in very small volume (nucleus) and is charged or most of atom is empty space B1 [3] (c) I = Q / t C1 n / t = (1.5 × 10–12) /( 2 × 1.6 × 10–19) C1 n / t = 4.7 × 106 s–1 A1 [3]

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