Cambridge A Level Physics 9702 — 2012 Oct/Nov Paper 2 · Variant 3
9702/23/O/N/12 · 6 questions · 60 marks · ≈68 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper12 pages












Mark scheme4 pages
Answers below. Sit the paper first if you are practising.




Questions as text
Q1 · The spacing between two atoms in a crystal is 3.8 × 10–10 m
1 (a) The spacing between two atoms in a crystal is 3.8 × 10–10 m. State this distance in pm. spacing = .......................................... pm [1] (b) Calculate the time of one day in Ms. time = .......................................... Ms [1] (c) The distance from the Earth to the Sun is 0.15 Tm. Calculate the time in minutes for light to travel from the Sun to the Earth. time = ......................................... min [2] (d) Underline all the vector quantities in the list below. distance energy momentum weight work [1] (e) The velocity vector diagram for an aircraft heading due north is shown to scale in For Fig. 1.1. There is a wind blowing from the north-west. Examiner’s Use wind 45° aircraft Fig. 1.1 The speed of the wind is 36 m s–1 and the speed of the aircraft is 250 m s–1. (i) Draw an arrow on Fig. 1.1 to show the direction of the resultant velocity of the aircraft. [1] (ii) Determine the magnitude of the resultant velocity of the aircraft. resultant velocity = ...................................... m s–1 [2]
Mark scheme: 1 (a) spacing = 380 or 3.8 × 102 pm B1 [1] (b) time = 24 × 3600 time = 0.086 (0.0864) Ms B1 [1] 1.5 × 10 11 (c) time = distance / speed = C1 3 × 10 8 = 500 (s) = 8.3 min A1 [2] (d) momentum and weight B1 [1] (e) (i) arrow to the right of plane direction (about 4° to 24°) B1 [1] (ii) scale diagram drawn or use of cosine formula v2 = 2502 + 362 – 2 × 250 × 36 × cos 45° or resolving v = [(36 cos 45°)2 + (250 – 36 sin 45°)2]1/2 C1 resultant velocity = 226 (220 – 240 for scale diagram) m s–1 allow one mark for values 210 to 219 or 241 to 250 m s–1 or use of formula (v2 = 51068) v = 230 (226) m s–1 A1 [2]
Q2 · Two planks of wood AB and BC are inclined at an angle of 15° to the horizontal
2 Two planks of wood AB and BC are inclined at an angle of 15° to the horizontal. The two For wooden planks are joined at point B, as shown in Fig. 2.1. Examiner’s Use M C A 0.26 m 0.26 m 15° B 15° Fig. 2.1 A small block of metal M is released from rest at point A. It slides down the slope to B and up the opposite side to C. Points A and C are 0.26 m above B. Assume frictional forces are negligible. (a) (i) Describe and explain the acceleration of M as it travels from A to B and from B to C. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3] (ii) Calculate the time taken for M to travel from A to B. time = ............................................. s [3] (iii) Calculate the speed of M at B. speed = ...................................... m s–1 [2] (b) The plank BC is adjusted so that the angle it makes with the horizontal is 30°. M is released from rest at point A and slides down the slope to B. It then slides a distance along the plank from B towards C. Use the law of conservation of energy to calculate this distance. Explain your working. distance = ............................................ m [2]
Mark scheme: 2 (a) (i) accelerations (A to B and B to C) are same magnitude B1 accelerations (A to B and B to C) are opposite directions or both accelerations are toward B B1 (A to B and B to C) the component of the weight down the slope provides the acceleration B1 [3] (ii) acceleration = g sin15 ° C1 s = 0 + ½ at2 s = 0.26 / sin 15 ° = 1.0 C1 2 1 . 0 × 2 t = t = 0.89 s A1 [3] 9 . 8 × sin15 ° (iii) v = 0 + g sin15t or v2 = 0 + 2g sin15 × 1.0 C1 v = 2.26 m s–1 A1 [2] (using loss of GPE = gain KE can score full marks) (b) loss of GPE at A = gain in GPE at C or loss of KE at B = gain in GPE at C B1 h1 = h2 = 0.26 m or ½ mv2 = mgh h2 = 0.5 × (2.26)2 / 9.81 = 0.26 m x = 0.26 / sin 30° = 0.52 m A1 [2]
Question 3
3 (a) Define power. For Examiner’s .......................................................................................................................................... Use .................................................................................................................................... [1] (b) A cyclist travels along a horizontal road. The variation with time t of speed v is shown in Fig. 3.1. 12.0 10.0 8.0 v / m s–1 6.0 4.0 2.0 0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 t / s Fig. 3.1 The cyclist maintains a constant power and after some time reaches a constant speed of 12 m s–1. (i) Describe and explain the motion of the cyclist. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3] (ii) When the cyclist is moving at a constant speed of 12 m s–1 the resistive force is For 48 N. Show that the power of the cyclist is about 600 W. Explain your working. Examiner’s Use [2] (iii) Use Fig. 3.1 to show that the acceleration of the cyclist when his speed is 8.0 m s–1 is about 0.5 m s–2. [2] (iv) The total mass of the cyclist and bicycle is 80 kg. Calculate the resistive force R acting on the cyclist when his speed is 8.0 m s–1. Use the value for the acceleration given in (iii). R = ............................................ N [3] (v) Use the information given in (ii) and your answer to (iv) to show that, in this situation, the resistive force R is proportional to the speed v of the cyclist. [1]
Mark scheme: 3 (a) power is the rate of doing work or power = work done / time (taken) or power = energy transferred / time (taken) B1 [1] (b) (i) as the speed increases drag / air resistance increases B1 resultant force reduces hence acceleration is less B1 constant speed when resultant force is zero B1 [3] (allow one mark for speed increases and acceleration decreases) GCE AS/A LEVEL – October/November 2012 9702 23 (ii) force from cyclist = drag force / resistive force B1 P = 12 × 48 M1 P = 576 W A0 [2] (iii) tangent drawn at speed = 8.0 m s–1 M1 gradient values that show acceleration between 0.44 to 0.48 m s–2 A1 [2] (iv) F – R = ma C1 600 / 8 – R = 80 × 0.5 [using P = 576] 576 / 8 – R = 80 × 0.5 C1 R = 75 – 40 = 35 N R = 72 – 40 = 32 N A1 [3] (v) at 12 m s–1 drag is 48 N, at 8 m s–1 drag is 35 or 32 N R / v calculated as 4 and 4 or 4.4 and consistent response for whether R is proportional to v or not B1 [1]
Q4 · A circuit used to measure the power transfer from a battery is shown in Fig
4 A circuit used to measure the power transfer from a battery is shown in Fig. 4.1. The power is For transferred to a variable resistor of resistance R. Examiner’s Use E r A I R V Fig. 4.1 The battery has an electromotive force (e.m.f.) E and an internal resistance r. There is a potential difference (p.d.) V across R. The current in the circuit is І. (a) By reference to the circuit shown in Fig. 4.1, distinguish between the definitions of e.m.f. and p.d. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [3] (b) Using Kirchhoff’s second law, determine an expression for the current І in the circuit. [1] (c) The variation with current І of the p.d. V across R is shown in Fig. 4.2. For Examiner’s 6.0 Use 4.0 V / V 2.0 0 0 1.0 2.0 3.0 4.0 I /A Fig. 4.2 Use Fig. 4.2 to determine (i) the e.m.f. E, E = ............................................ V [1] (ii) the internal resistance r. r = ............................................ Ω [2] (d) (i) Using data from Fig. 4.2, calculate the power transferred to R for a current of 1.6 A. power = ........................................... W [2] (ii) Use your answers from (c)(i) and (d)(i) to calculate the efficiency of the battery for a current of 1.6 A. efficiency = ........................................... % [2]
Mark scheme: 4 (a) e.m.f. = chemical energy to electrical energy M1 p.d. = electrical energy to thermal energy M1 idea of per unit charge A1 [3] (b) E = I (R +r) or I = E / (R +r) (any subject) B1 [1] (c) (i) E = 5.8 V B1 [1] (ii) evidence of gradient calculation or calculation with values from graph e.g. 5.8 = 4 + 1.0 × r C1 r = 1.8 Ω A1 [2] (d) (i) P = VI C1 P = 2.9 × 1.6 = 4.6 (4.64) W A1 [2] (ii) power from battery = 1.6 × 5.8 = 9.28 or efficiency = VI / EI C1 efficiency = (4.64 / 9.28) × 100 = 50 % or (2.9 / 5.8) × 100 = 50% A1 [2]
Q5 · State one property of electromagnetic waves that is not common to other transverse For…
5 (a) State one property of electromagnetic waves that is not common to other transverse For waves. Examiner’s Use .................................................................................................................................... [1] (b) The seven regions of the electromagnetic spectrum are represented by blocks labelled A to G in Fig. 5.1. visible region A B C D E F G wavelength decreasing Fig. 5.1 A typical wavelength for the visible region D is 500 nm. (i) Name the principal radiations and give a typical wavelength for each of the regions B, E and F. B: name: ............................................ wavelength: ............................................. m E: name: ............................................ wavelength: ............................................. m F: name: ............................................ wavelength: ............................................. m [3] (ii) Calculate the frequency corresponding to a wavelength of 500 nm. frequency = .......................................... Hz [2] (c) All the waves in the spectrum shown in Fig. 5.1 can be polarised. Explain the meaning of the term polarised. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2]
Mark scheme: 5 (a) travel through a vacuum / free space B1 [1] (b) (i) B : name: microwaves wavelength: 10– 4 to 10–1 m B1 C : name: ultra-violet / UV wavelength: 10–7 to 10–9 m B1 F : name: X –rays wavelength: 10–9 to 10–12 m B1 [3] 3 × 10 8 (ii) f = C1 500 × 10 − 9 f = 6(.0) × 1014 Hz A1 [2] GCE AS/A LEVEL – October/November 2012 9702 23 (c) vibrations are in one direction M1 perpendicular to direction of propagation / energy transfer or good sketch showing this A1 [2]
Q6 · Β-radiation is emitted during the spontaneous radioactive decay of an unstable nucleus
6 (a) β-radiation is emitted during the spontaneous radioactive decay of an unstable nucleus. For Examiner’s (i) State the nature of a β-particle. Use ............................................................................................................................ [1] (ii) State two properties of β-radiation. 1. ............................................................................................................................... 2. ............................................................................................................................... [2] (iii) Explain the meaning of spontaneous radioactive decay. .................................................................................................................................. ............................................................................................................................ [1] (b) The following equation represents the decay of a nucleus of hydrogen-3 by the emission of a β-particle. Complete the equation. ...... ...... 31H He + β [2] ...... ...... (c) The β-particle is emitted with an energy of 5.7 × 103 eV. Calculate the speed of the β-particle. speed = ...................................... m s–1 [3] (d) A different isotope of hydrogen is hydrogen-2 (deuterium). Describe the similarities and differences between the atoms of hydrogen-2 and hydrogen-3. .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2]
Mark scheme: 6 (a) (i) electron B1 [1] (ii) any two: can be deflected by electric and magnetic fields or negatively charged / absorbed by few (1 – 4) mm of aluminum / 0.5 to 2 m or metres for range in air / speed up to 0.99c / range of speeds / energies B2 [2] (iii) decay occurs and cannot be affected by external / environmental factors or two stated factors such as chemical / pressure / temperature / humidity B1 [1] (b) 3 and 0 for superscript numbers B1 2 and –1 for subscript numbers B1 [2] (c) energy = 5.7 × 103 × 1.6 × 10–19 (= 9.12 × 10–16 J) C1 2 × 9.12 × 10 −16 v2 = C1 9.11 × 10 − 31 v = 4.5 × 107 m s–1 A1 [3] (d) both have 1 proton and 1 electron B1 1 neutron in hydrogen-2 and 2 neutrons in hydrogen-3 B1 [2] (special case: for one mark ‘same number of protons / atomic number different number of neutrons’)
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Cambridge’s own grade thresholds for 2012 Oct/Nov, Paper 2 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.