Cambridge A Level Physics 9702 — 2014 Oct/Nov Paper 2 · Variant 3

1 (a) The kilogram, metre and second are SI base units. State two other base units. 1. ............................................................................................................................................... 2. ............................................................................................................................................... [2] (b) Determine the SI base units of (i) stress, SI base units ...........................................................[2] (ii) the Young modulus. SI base units ...........................................................[1]

Mark scheme: 1 (a) ampere B1 kelvin B1 [2] (allow mole and candela) (b) (i) stress: N m–2 C1 kg m s–2 / m2 = kg m–1 s–2 A1 [2] (ii) Young modulus = stress / strain and strain has no units hence units: kg m–1 s–2 B1 [1]

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2 A microphone detects a musical note of frequency f. The microphone is connected to a cathode- ray oscilloscope (c.r.o.). The signal from the microphone is observed on the c.r.o. as illustrated in Fig. 2.1. 1.0 cm 1.0 cm Fig. 2.1 The time-base setting of the c.r.o. is 0.50 ms cm–1. The Y-plate setting is 2.5 mV cm–1. (a) Use Fig. 2.1 to determine (i) the amplitude of the signal, amplitude = ................................................... mV [2] (ii) the frequency f, f = .................................................... Hz [3] (iii) the actual uncertainty in f caused by reading the scale on the c.r.o. actual uncertainty = .................................................... Hz [2] (b) State f with its actual uncertainty. f = ................................ ± ................................ Hz [1]

Mark scheme: 2 (a) (i) amplitude scale reading 2.2 (cm) C1 amplitude = 2.2 × 2.5 = 5.5 mV A1 [2] (ii) time period scale reading = 3.8 (cm) C1 time period = 3.8 × 0.5 × 10–3 = 0.0019 (s) C1 frequency f = 1 / 0.0019 = 530 (526) Hz A1 [3] (iii) uncertainty in reading = ± 0.2 in 3.8 (cm) or 5.3% or 0.2 in 7.6 (cm) or 2.6% [allow other variations of the distance on the x-axis] M1 actual uncertainty = 5.3% of 526 = 27.7 or 28 Hz or 2.6% of 526 = 13 or 14 A1 [2] (b) frequency = 530 ± 30 Hz or 530 ± 10 Hz A1 [1]

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3 (a) Force is a vector quantity. State three other vector quantities. 1. ............................................................................................................................................... 2. ............................................................................................................................................... 3. ............................................................................................................................................... [2] (b) Three coplanar forces X, Y and Z act on an object, as shown in Fig. 3.1. Y object θ X Z Fig. 3.1 The force Z is vertical and X is horizontal. The force Y is at an angle θ to the horizontal. The force Z is kept constant at 70 N. In an experiment, the magnitude of force X is varied. The magnitude and direction of force Y are adjusted so that the object remains in equilibrium. Fig. 3.2 shows the variation of the magnitude of force Y with the magnitude of force X. 130 Y / N 110 90 70 50 0 20 40 60 80 100 120 X / N Fig. 3.2 (i) Use Fig. 3.2 to estimate the magnitude of Y for X = 0. Y = ...................................................... N [1] (ii) State and explain the value of θ for X = 0. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (iii) The magnitude of X is increased to 160 N. Use resolution of forces to calculate the value of 1. angle θ, θ = ........................................................ ° [2] 2. the magnitude of force Y. Y = ...................................................... N [2] (c) The angle θ decreases as X increases. Explain why the object cannot be in equilibrium for θ = 0. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[1]

Mark scheme: 3 (a) displacement / velocity / acceleration / momentum / etc. three correct (none wrong) 2, two correct (none or one wrong) 1 A2 [2] (b) (i) Y = 70 N [allow 71 N as +½ small square on graph] A1 [1] (ii) θ = 90° M1 (for equilibrium) the direction of Y must be opposite to Z or using Y sin θ = Z, hence sin θ = 70 / 70 = 1, θ = 90° A1 [2] (iii) 1. Y cos θ = 160 and Y sin θ = 70 C1 tan θ = 70 / 160 hence θ = 23.6° (24°) A1 [2] 2. Y = 160 / cos 23.6° or 70 / sin 23.6° C1 = 174.6 or 175 or 170 N A1 [2] or: 1602 + 702 = Y2 (C1) Y = 174.6 or 175 or 170 N (A1) (c) (equilibrium not possible as) there is no vertical component from Y to balance Z B1 [1]

More questions on Momentum and Newton’s laws of motion

4 (a) State the principle of conservation of momentum. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A ball X and a ball Y are travelling along the same straight line in the same direction, as shown in Fig. 4.1. X Y 400 g 0.65 m s–1 600 g 0.45 m s–1 Fig. 4.1 Ball X has mass 400 g and horizontal velocity 0.65 m s–1. Ball Y has mass 600 g and horizontal velocity 0.45 m s–1. Ball X catches up and collides with ball Y. After the collision, X has horizontal velocity 0.41 m s–1 and Y has horizontal velocity v, as shown in Fig. 4.2. X Y 400 g 0.41 m s–1 600 g v Fig. 4.2 Calculate (i) the total initial momentum of the two balls, momentum = .................................................... N s [3] (ii) the velocity v, v = ................................................ m s–1 [2] (iii) the total initial kinetic energy of the two balls. kinetic energy = ....................................................... J [3] (c) Explain how you would check whether the collision is elastic. ................................................................................................................................................... ...............................................................................................................................................[1] (d) Use Newton’s third law to explain why, during the collision, the change in momentum of X is equal and opposite to the change in momentum of Y. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2]

Mark scheme: 4 (a) for a system (of interacting bodies) the total momentum remains constant M1 provided there is no resultant force acting (on the system) A1 [2] (b) (i) total momentum = m1v1 + m2v2 C1 = 0.4 × 0.65 + 0.6 × 0.45 C1 = 0.26 + 0.27 = 0.53 N s A1 [3] (ii) 0.53 = 0.4 × 0.41 + 0.6 × v C1 v = 0.366 / 0.6 = 0.61 m s–1 A1 [2] (iii) KE = ½ mv2 C1 total initial KE = ½ × 0.4 × (0.65)2 + ½ × 0.6 × (0.45)2 C1 = 0.0845 + 0.06075 = 0.15 (0.145) J A1 [3] (c) check relative speed of approach equals relative speed of separation or: total final kinetic energy equals the total initial kinetic energy B1 [1] (d) the forces on the two bodies (or on X and Y) are equal and opposite B1 time same for both forces and force is change in momentum / time B1 [2]

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5 Distinguish between evaporation and boiling. evaporation: ...................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... boiling: .............................................................................................................................................. .......................................................................................................................................................... .......................................................................................................................................................... [4]

Mark scheme: 5 evaporation: molecules escape from the surface B1 at all temperatures B1 boiling: takes place throughout / in the liquid B1 at the boiling point / at specific temperatures B1 [4]

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6 (a) A wire has length 100 cm and diameter 0.38 mm. The metal of the wire has resistivity 4.5 × 10–7 Ω m. Show that the resistance of the wire is 4.0 Ω. [3] (b) The ends B and D of the wire in (a) are connected to a cell X, as shown in Fig. 6.1. 2.0 V cell X 1.0 Ω l B C D 1.5 V metal wire 0.50 Ω cell Y Fig. 6.1 The cell X has electromotive force (e.m.f.) 2.0 V and internal resistance 1.0 Ω. A cell Y of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to the wire at points B and C, as shown in Fig. 6.1. The point C is distance l from point B. The current in cell Y is zero. Calculate (i) the current in cell X, current = ...................................................... A [2] (ii) the potential difference (p.d.) across the wire BD, p.d. = ...................................................... V [1] (iii) the distance l. l = .................................................... cm [2] (c) The connection at C is moved so that l is increased. Explain why the e.m.f. of cell Y is less than its terminal p.d. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2]

Mark scheme: 6 (a) R = ρl / A C1 A = [π × (0.38 × 10–3)2] / 4 (= 0.113 × 10–6 m2) C1 R = (4.5 × 10–7 × 1.00) / ( [π × (0.38 × 10–3)2] / 4 ) = 4.0 (3.97) Ω M1 [3] (b) (i) І = V / R C1 = 2.0 / 5.0 = 0.4(0) A A1 [2] (ii) p.d. across BD = 4 × 0.4 = 1.6 V A1 [1] (iii) p.d. across BC (l) = 1.5 (V) C1 BC (l) = (1.5 / 1.6) × 100 = 94 (93.75) cm A1 [2] (c) p.d. across wire not balancing e.m.f. of cell OR cell Y has current B1 energy lost or lost volts due to internal resistance B1 [2]

More questions on Kirchhoff’s laws

7 (a) (i) Explain what is meant by a progressive transverse wave. progressive: ....................................................................................................................... ........................................................................................................................................... transverse: ......................................................................................................................... ........................................................................................................................................... [2] (ii) Define frequency. ........................................................................................................................................... .......................................................................................................................................[1] (b) The variation with distance x of displacement y for a transverse wave is shown in Fig. 7.1. 2.0 R y / cm 1.0 Q S 0 0 0.4 0.8 1.2 1.6 2.0 x / cm –1.0 P T –2.0 Fig. 7.1 On Fig. 7.1, five points are labelled. Use Fig. 7.1 to state any two points having a phase difference of (i) zero, .......................................................................................................................................[1] (ii) 270°. .......................................................................................................................................[1] (c) The frequency of the wave in (b) is 15 Hz. Calculate the speed of the wave in (b). speed = ................................................ m s–1 [3] (d) Two waves of the same frequency have amplitudes 1.4 cm and 2.1 cm. Calculate the ratio intensity of wave of amplitude 1.4 cm . intensity of wave of amplitude 2.1 cm ratio = .......................................................... [2]

Mark scheme: 7 (a) (i) progressive: energy is moved / transferred / propagated from one place to another (without the bulk movement of the medium) B1 transverse: (particles) oscillate / vibrate at right angles to the direction of travel of the energy / wavefront B1 [2] (ii) number of oscillations per unit time / number of wavefronts passing a point per unit time B1 [1] (b) (i) P and T B1 [1] (ii) P and S or Q and T B1 [1] (c) λ = 1.2 × 10–2 (m) C1 v = fλ = 15 × 1.2 × 10–2 C1 = 0.18 m s–1 A1 [3] (d) ratio = (1.4)2 / (2.1)2 C1 = 0.44 A1 [2]

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