Cambridge A Level Physics 9702 — 2014 May/June Paper 2 · Variant 3

1 (a) Underline all the base quantities in the following list. ampere charge current mass second temperature weight [2] (b) The potential energy EP stored in a stretched wire is given by EP = ½Cσ2V where C is a constant, σ is the strain, V is the volume of the wire. Determine the SІ base units of C. base units ...........................................................[3]

Mark scheme: 1 (a) current, mass and temperature two correct 2/2, one omission or error 1/2 A2 [2] (b) σ : no units, V: m3 C1 EP: kg m2 s–2 C1 C: kg m2 s–2 × m–3 = kg m–1 s–2 A1 [3]

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2 (a) Explain what is meant by a scalar quantity and by a vector quantity. scalar: ....................................................................................................................................... ................................................................................................................................................... vector: ....................................................................................................................................... ................................................................................................................................................... [2] (b) A ball leaves point P at the top of a cliff with a horizontal velocity of 15 m s–1, as shown in Fig. 2.1. ball P 15 m s–1 path of ball 25 m cliff Q ground Fig. 2.1 The height of the cliff is 25 m. The ball hits the ground at point Q. Air resistance is negligible. (i) Calculate the vertical velocity of the ball just before it makes impact with the ground at Q. vertical velocity = ................................................. m s–1 [2] (ii) Show that the time taken for the ball to fall to the ground is 2.3 s. [1] (iii) Calculate the magnitude of the displacement of the ball at point Q from point P. displacement = ...................................................... m [4] (iv) Explain why the distance travelled by the ball is different from the magnitude of the displacement of the ball. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2]

Mark scheme: 2 (a) scalar has magnitude only B1 vector has magnitude and direction B1 [2] 1 (b) (i) v2 = 0 + 2 × 9.81 × 25 (or using m v2 = mgh) C1 2 v = 22(.1) m s–1 A1 [2] 1 (ii) 22.1 = 0 + 9.81 × t (or 25 = × 9.81 × t 2) M1 2 t (=22.1 / 9.81) = 2.26 s or t [=(5.097)1/2] = 2.26 s A0 [1] (iii) horizontal distance = 15 × t = 15 × 2.257 = 33.86 (allow 15 × 2.3 = 34.5) C1 (displacement)2 = (horizontal distance)2 + (vertical distance)2 C1 = (25)2 + (33.86)2 C1 displacement = 42 (42.08) m (allow 43 (42.6) m, allow 2 or more s.f.) A1 [4] (iv) distance is the actual (curved) path followed by ball B1 displacement is the straight line / minimum distance P to Q B1 [2]

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3 (a) Explain what is meant by work done. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A boy on a board B slides down a slope, as shown in Fig. 3.1. boy on board B 30° horizontal Fig. 3.1 The angle of the slope to the horizontal is 30°. The total resistive force F acting on B is constant. (i) State a word equation that links the work done by the force F on B to the changes in potential and kinetic energy. ........................................................................................................................................... .......................................................................................................................................[1] (ii) The boy on the board B moves with velocity v down the slope. The variation with time t of v is shown in Fig. 3.2. 8.0 6.0 v / m s–1 4.0 2.0 0 0 1.0 2.0 3.0 t / s Fig. 3.2 The total mass of B is 75 kg. For B, from t = 0 to t = 2.5 s, 1. show that the distance moved down the slope is 9.3 m, [2] 2. calculate the gain in kinetic energy, gain in kinetic energy = ....................................................... J [3] 3. calculate the loss in potential energy, loss in potential energy = ....................................................... J [3] 4. calculate the resistive force F. F = ...................................................... N [3]

Mark scheme: 3 (a) work done is the product of force and the distance moved in the direction of the force or product of force and displacement in the direction of the force B1 [1] GCE AS/A LEVEL – May/June 2014 9702 23 (b) (i) work done equals the decrease in GPE – gain in KE B1 [1] (ii) 1. distance = area under line C1 = (7.4 × 2.5) / 2 = 9.3 m (9.25 m) M1 [2] or acceleration from graph a = 7.4 / 2.5 (= 2.96) (C1) and equation of motion (7.4)2 = 2 × 2.96 × s gives s = 9.3 (9.25) m (A1) 1 2 2. kinetic energy = m v C1 2 1 = × 75 × (7.4)2 C1 2 = 2100 J A1 [3] 3. potential energy = mgh C1 h = 9.3 sin 30 ° C1 PE = 75 × 9.81 × 9.3 sin 30 ° = 3400 J A1 [3] 4. work done = energy loss C1 R = (3421 – 2054) / 9.3 C1 = 150 (147) N A1 [3]

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4 A spring hangs vertically from a point P, as shown in Fig. 4.1. P metre rule spring mass M reading x Fig. 4.1 A mass M is attached to the lower end of the spring. The reading x from the metre rule is taken, as shown in Fig. 4.1. Fig. 4.2 shows the relationship between x and M. 0.60 0.40 M / kg 0.20 0 20 22 24 26 28 30 32 x / cm Fig. 4.2 (a) Explain how the apparatus in Fig. 4.1 may be used to determine the load on the spring at the elastic limit. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) State and explain whether Fig. 4.2 suggests that the spring obeys Hooke’s law. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (c) Use Fig. 4.2 to determine the spring constant, in N m–1, of the spring. spring constant = ................................................ N m–1 [3]

Mark scheme: 4 (a) add small mass to cause extension then remove mass to see if spring returns to original length M1 repeat for larger masses and note maximum mass for which, when load is removed, the spring does return to original length A1 [2] (b) Hooke’s law requires force proportional to extension B1 graph shows a straight line, hence obeys Hooke’s law M1 [2] (c) k = force / extension C1 = (0.42 × 9.81) / [(30 – 21.2) × 10–2] C1 = 47 (46.8) N m–1 A1 [3]

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5 (a) Explain why the terminal potential difference (p.d.) of a cell with internal resistance may be less than the electromotive force (e.m.f.) of the cell. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A battery of e.m.f. 4.5 V and internal resistance r is connected in series with a resistor of resistance 6.0 Ω, as shown in Fig. 5.1. battery 4.5V r I 6.0 1 Fig. 5.1 The current I in the circuit is 0.65 A. Determine (i) the internal resistance r of the battery, r = ...................................................... Ω [2] (ii) the terminal p.d. of the battery, p.d. = ....................................................... V [2] (iii) the power dissipated in the resistor, power = ..................................................... W [2] (iv) the efficiency of the battery. efficiency = .......................................................... [2] (c) A second resistor of resistance 20 Ω is connected in parallel with the 6.0 Ω resistor in Fig. 5.1. Describe and explain qualitatively the change in the heating effect within the battery. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3]

Mark scheme: 5 (a) lost volts / energy used within the cell / internal resistance B1 when cell supplies a current B1 [2] GCE AS/A LEVEL – May/June 2014 9702 23 (b) (i) E = І(R + r) C1 4.5 = 0.65 (6.0 + r) r = 0.92 Ω A1 [2] (ii) І = 0.65 (A) and V = ІR C1 V = 0.65 × 6 = 3.9 V A1 [2] (iii) P = V 2 / R or P = І2R and P = ІV C1 = (3.9)2 / 6 = 2.5 W A1 [2] (iv) efficiency = power out / power in C1 = І 2R / І 2(R + r) = R / (R + r) = 6.0 / ( 6.0 + 0.92 ) = 0.87 A1 [2] (c) (circuit) resistance decreases B1 current increases M1 more heating effect A1 [3]

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6 A hollow tube is used to investigate stationary waves. The tube is closed at one end and open at the other end. A loudspeaker connected to a signal generator is placed near the open end of the tube, as shown in Fig. 6.1. L loudspeaker Q P signal generator hollow tube Fig. 6.1 The tube has length L. The frequency of the signal generator is adjusted so that the loudspeaker produces a progressive wave of frequency 440 Hz. A stationary wave is formed in the tube. A representation of this stationary wave is shown in Fig. 6.1. Two points P and Q on the stationary wave are labelled. (a) (i) Describe, in terms of energy transfer, the difference between a progressive wave and a stationary wave. ........................................................................................................................................... .......................................................................................................................................[1] (ii) Explain how the stationary wave is formed in the tube. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] (iii) State the direction of the oscillations of an air particle at point P. ........................................................................................................................................... .......................................................................................................................................[1] (b) On Fig. 6.1 label, with the letter N, the nodes of the stationary wave. [1] (c) State the phase difference between points P and Q on the stationary wave. phase difference = .......................................................... [1] (d) The speed of sound in the tube is 330 m s–1. Calculate (i) the wavelength of the sound wave, wavelength = ...................................................... m [2] (ii) the length L of the tube. length = ...................................................... m [2]

Mark scheme: 6 (a) (i) progressive wave transfers energy, stationary wave no transfer of energy / keeps energy within wave B1 [1] (ii) (progressive) wave / wave from loudspeaker reflects at end of tube B1 reflected wave overlaps (another) progressive wave B1 same frequency and speed hence stationary wave formed B1 [3] (iii) (side to side) along length of tube / along axis of tube B1 [1] (b) all three nodes clearly marked with N / clearly labelled at cross-over points B1 [1] (c) phase difference = 0 A1 [1] (d) (i) v = fλ C1 λ = 330 / 440 = 0.75 m A1 [2] (ii) L = 5/4 λ C1 = 5/4 × 0.75 = 0.94 m A1 [2]

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