Cambridge A Level Physics 9702 — 2015 May/June Paper 2 · Variant 2
9702/22/M/J/15 · 7 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 paper16 pages
Mark scheme5 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · Use the definition of work done to show that the SI base units of energy are kg m2 s−2
1 (a) Use the definition of work done to show that the SI base units of energy are kg m2 s−2. [2] (b) Define potential difference. ................................................................................................................................................... .............................................................................................................................................. [1] (c) Determine the SI base units of resistance. Show your working. units .......................................................... [3]
Mark scheme: 1 (a) (work =) force × distance or force × displacement or (W =) F × d M1 units of work: kg m s–2 × m = kg m2 s–2 A1 [2] work (done) or energy (transform ed) (from electrical to other forms) (b) (p.d. = ) B1 [1] charge (c) R = V / I B1 units of V: kg m2 s–2 / A s and units of I: A C1 or R = P / I2 [or P = VI and V = IR] (B1) units of P: kg m2 s–3 and units of I: A (C1) or R = V2 / P (B1) units of V: kg m2 s–2 / A s and units of P: kg m2 s–3 (C1) units of R: (kg m2 s–2 / A2 s =) kg m2 s–3 A–2 A1 [3]
Q2 · A stone is thrown vertically upwards
2 A stone is thrown vertically upwards. The variation with time t of the displacement s of the stone is shown in Fig. 2.1. s 0 0 1.0 2.0 3.0 t / s Fig. 2.1 (a) Use Fig. 2.1 to describe, without calculation, the speed of the stone from t = 0 to t = 3.0 s. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Assume air resistance is negligible and therefore the stone has constant acceleration. Calculate, for the stone, (i) the speed at 3.0 s, speed = ............................................... m s−1 [3] (ii) the distance travelled from t = 0 to t = 3.0 s, distance = ..................................................... m [3] (iii) the displacement from t = 0 to t = 3.0 s. displacement = ........................................................... m direction ............................................................... [2] (c) On Fig. 2.2, draw the variation with time t of the velocity v of the stone from t = 0 to t = 3.0 s. v / m s–1 0 0 1.0 2.0 3.0 t / s Fig. 2.2 [3]
Mark scheme: 2 (a) speed decreases/stone decelerates to rest/zero at 1.25 s B1 speed then increases/stone accelerates (in opposite direction) B1 [2] (b) (i) v = u + at (or s = ut + ½at2 and v2 = u2 + 2as) C1 = 0 + (3.00 – 1.25) × 9.81 C1 = 17.2 (17.17) m s–1 A1 [3] (ii) s = ut + ½at2 s = ½ × 9.81 × (1.25)2 [= 7.66] C1 s = ½ × 9.81 × (1.75)2 [= 15.02] C1 (distance = 7.66 + 15.02) [v = u + at = 0 + 9.81 × (2.50 – 1.25) = 12.26 m s–1] or s = ½ × 9.81 × (1.25)2 [= 7.66] (C1) s = 12.26 × 0.50 + ½ × 9.81 × (3.00 – 2.50)2 [= 7.36] (C1) (distance = 2 × 7.66 + 7.36) Example alternative method: s = (v2 – u2) / 2a = (12.262 – 0) / 2 × 9.81 [= 7.66] (C1) s = (v2 – u2) / 2a = (17.172 – 12.262) / 2 × 9.81 [= 7.36] (C1) (distance = 2 × 7.66 + 7.36) 22.7 (22.69 or 23) m A1 [3] (iii) (s = 15.02 – 7.66 =) 7.4 (7.36) m (ignore sign in answer) A1 down A1 [2] (c) straight line from positive value of v to t axis M1 same straight line crosses t axis at t = 1.25 s A1 same straight line continues with same gradient to t = 3.0 s A1 [3]
Q3 · A rod PQ is attached at P to a vertical wall, as shown in Fig
3 A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1. R wire wall F 0.64 m 0.96 m 30° Q P rod W Fig. 3.1 The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P. The rod is kept horizontal and in equilibrium by a wire attached to Q and to the wall at R. The wire provides a force F on the rod of 44 N at 30° to the horizontal. (a) Determine (i) the vertical component of F, vertical component = ...................................................... N [1] (ii) the horizontal component of F. horizontal component = ...................................................... N [1] (b) By taking moments about P, determine the weight W of the rod. W = ...................................................... N [2] (c) Explain why the wall must exert a force on the rod at P. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [1] (d) On Fig. 3.1, draw an arrow to represent the force acting on the rod at P. Label your arrow with the letter S. [1]
Mark scheme: 3 (a) (i) (vertical component = 44 sin 30° =) 22 N A1 [1] (ii) (horizontal component = 44 cos 30° =) 38(.1) N A1 [1] (b) W × 0.64 = 22 × 1.60 C1 (W =) 55 N A1 [2] (c) F has a horizontal component (not balanced by W) or F has 38 N acting horizontally or 38 N acts on wall or vertical component of F does not balance W or F and W do not make a closed triangle of forces B1 [1] (d) line from P in direction towards point on wire vertically above W and direction up B1 [1]
Q4 · A gas molecule has a mass of 6.64 × 10−27 kg and a speed of 1250 m s−1
4 (a) A gas molecule has a mass of 6.64 × 10−27 kg and a speed of 1250 m s−1. The molecule collides normally with a flat surface and rebounds with the same speed, as shown in Fig. 4.1. flat surface flat surface molecule molecule before collision after collision Fig. 4.1 Calculate the change in momentum of the molecule. change in momentum = ................................................... N s [2] (b) (i) Use the kinetic model to explain the pressure exerted by gases. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) Explain the effect of an increase in density, at constant temperature, on the pressure of a gas. ........................................................................................................................................... ...................................................................................................................................... [1]
Mark scheme: 4 (a) (p =) mv C1 ∆p (= – 6.64 × 10–27 × 1250 – 6.64 × 10–27 × 1250) = 1.66 × 10–23 N s A1 [2] (b) (i) molecule collides with wall/container and there is a change in momentum B1 change in momentum / time is force or ∆p = Ft B1 many/all/sum of molecular collisions over surface/area of container produces pressure B1 [3] (ii) more collisions per unit time so greater pressure B1 [1]
Question 5
5 (a) On Fig. 5.1, sketch the temperature characteristic of a thermistor. resistance 0 0 100 temperature / °C Fig. 5.1 [2] (b) A potential divider circuit is shown in Fig. 5.2. X Y 12 V Z Fig. 5.2 The battery of electromotive force (e.m.f.) 12 V and negligible internal resistance is connected in series with resistors X and Y and thermistor Z. The resistance of Y is 15 kΩ and the resistance of Z at a particular temperature is 3.0 kΩ. The potential difference (p.d.) across Y is 8.0 V. (i) Explain why the power transformed in the battery equals the total power transformed in X, Y and Z. ...................................................................................................................................... [1] (ii) Calculate the current in the circuit. current = ...................................................... A [2] (iii) Calculate the resistance of X. resistance = ...................................................... Ω [3] (iv) The temperature of Z is increased. State and explain the effect on the potential difference across Z. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]
Mark scheme: 5 (a) curved line showing decreasing gradient with temperature rise M1 smooth line not touching temperature axis, not horizontal or vertical anywhere A1 [2] (b) (i) (no energy lost in battery because) no/negligible internal resistance B1 [1] (ii) I = V / R = 8 / 15 × 103 or 1.6 / 3.0 × 103 or 2.4 / 4.5 × 103 or 12 / 22.5 × 103 C1 = 0.53 × 10–3 A A1 [2] (iii) p.d. across X = 12 – 8.0 – 3.0 × 103 × 0.53 × 10–3 (= 2.4 V) C1 RX = 2.4 / (0.53 × 10–3) C1 or Rtot = 12 / 0.53 × 10–3 (= 22.5 × 103 Ω) (C1) RX = (22.5 – 15.0 – 3.0) × 103 (C1) 4.5(2) × 103 Ω A1 [3] (iv) resistance decreases hence current (in circuit) is greater M1 p.d. across X and Y is greater hence p.d across Z decreases A1 or explanation in terms of potential divider: RZ decreases so RZ / (RX + RY + RZ) is less (M1) therefore p.d. across Z decreases (A1) [2]
Q6 · State two differences between progressive waves and stationary waves
6 (a) State two differences between progressive waves and stationary waves. 1. .............................................................................................................................................. ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... [2] (b) A source S of microwaves is placed in front of a metal reflector R, as shown in Fig. 6.1. metal reflector R microwave detector D microwave source S meter Fig. 6.1 A microwave detector D is placed between R and S. Describe (i) how stationary waves are formed between R and S, ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) how D is used to show that stationary waves are formed between R and S, ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) how the wavelength of the microwaves may be determined using the apparatus in Fig. 6.1. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (c) The wavelength of the microwaves in (b) is 2.8 cm. Calculate the frequency, in GHz, of the microwaves. frequency = ................................................. GHz [3] Please turn over for Question 7.
Mark scheme: 6 (a) progressive waves transfer/propagate energy and stationary waves do not B1 amplitude constant for progressive wave and varies (from max/antinode to min/zero/node) for stationary wave B1 adjacent particles in phase for stationary wave and out of phase for progressive wave (B1) [2] (b) (i) wave / microwave from source/S reflects at reflector/R B1 reflected and (further) incident waves overlap/meet/superpose B1 waves have same frequency/wavelength/period and speed (so stationary waves formed) B1 [3] (ii) detector/D is moved between reflector/R and source/S (or v.v.) B1 maximum, minimum/zero, (maximum… etc.) observed on meter/deflections/readings/measurements/recordings B1 [2] (iii) determine/measure the distance between adjacent minima/nodes or maxima/antinodes or across specific number of nodes/antinodes B1 wavelength is twice distance between adjacent nodes/minima or maxima/ antinodes (or other correct method of calculation of wavelength from measurement) B1 [2] (c) v = fλ C1 f = 3.0 × 108 / (2.8 × 10–2) [= 1.07 × 1010 Hz] C1 11 (10.7) GHz A1 [3]
Q7 · A uranium-235 nucleus absorbs a neutron and then splits into two nuclei
7 A uranium-235 nucleus absorbs a neutron and then splits into two nuclei. A possible nuclear reaction is given by 23592U + abn 9337Rb + dXc + 2abn + energy. (a) State the constituent particles of the uranium-235 nucleus. .............................................................................................................................................. [1] (b) Complete Fig. 7.1 for this reaction. value a b c d [3] Fig. 7.1 (c) Suggest a possible form of energy released in this reaction. .............................................................................................................................................. [1] (d) Explain, using the law of mass-energy conservation, how energy is released in this reaction. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]
Mark scheme: 7 (a) 92 protons and 143 neutrons B1 [1] (b) value a 1 b 0 (a and b both required) B1 c 141 B1 d 55 B1 [3] (c) kinetic energy (of products) or gamma/γ (radiation or photon) B1 [1] (d) (total) mass on left-hand side/reactants is greater than (total) mass on right-hand side/products M1 difference in mass is (converted to) energy A1 [2]
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