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

1 (a) Show that the SІ base units of power are kg m2 s–3. [3] Q (b) The rate of flow of thermal energy in a material is given by t Q CAT = t x where A is the cross-sectional area of the material, T is the temperature difference across the thickness of the material, x is the thickness of the material, C is a constant. Determine the SІ base units of C. base units .......................................................... [4]

Mark scheme: 1 (a) power = energy / time or work done / time B1 force: kg m s–2 (including from mg in mgh or Fv) 1 or kinetic energy ( mv2): kg (m s–1)2 B1 2 (distance: m and (time) –1: s–1) and hence power: kg m s–2 m s–1 = kg m2 s–3 B1 [3] (b) Q / t : kg m2 s–3 C1 A: m2 and x: m and T: K C1 correct substitution into C = (Qx) / tAT or equivalent, or with cancellation C1 units of C : kg m s–3 K–1 A1 [4]

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2 A coin is made in the shape of a thin cylinder, as shown in Fig. 2.1. diameter thickness Fig. 2.1 Fig. 2.2 shows the measurements made in order to determine the density ρ of the material used to make the coin. quantity measurement uncertainty mass 9.6 g ± 0.5 g thickness 2.00 mm ± 0.01 mm diameter 22.1 mm ± 0.1 mm Fig. 2.2 (a) Calculate the density ρ in kg m–3. ρ = ...............................................kg m–3 [3] (b) (i) Calculate the percentage uncertainty in ρ. percentage uncertainty = ......................................................... [3] (ii) State the value of ρ with its actual uncertainty. ρ = ........................................................ ± ........................................... kg m–3 [1]

Mark scheme: 2 (a) ρ = m / V C1 V = (π d 2 / 4) × t = 7.67 × 10–7 m3 ρ = (9.6 × 10–3) / [π(22.1/2 × 10–3)2 × 2.00 × 10–3] C1 ρ = 12513 kg m–3 (allow 2 or more s.f.) A1 [3] (b) (i) ∆ρ / ρ = ∆m / m + ∆t / t + 2∆d / d C1 = 5.21% + 0.50% + 0.905% [or correct fractional uncertainties] C1 = 6.6% (6.61%) A1 [3] (ii) ρ = 12 500 ± 800 kg m–3 A1 [1]

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3 (a) State Newton’s first law of motion. ................................................................................................................................................... .............................................................................................................................................. [1] (b) A box slides down a slope, as shown in Fig. 3.1. v box 20° horizontal Fig. 3.1 The angle of the slope to the horizontal is 20°. The box has a mass of 65 kg. The total resistive force R acting on the box is constant as it slides down the slope. (i) State the names and directions of the other two forces acting on the box. 1. ........................................................................................................................................ 2. ........................................................................................................................................ [2] (ii) The variation with time t of the velocity v of the box as it moves down the slope is shown in Fig. 3.2. 8.0 6.0 v / m s–1 4.0 2.0 0 0 1.0 2.0 t / s Fig. 3.2 1. Use data from Fig. 3.2 to show that the acceleration of the box is 2.6 m s–2. [2] 2. Calculate the resultant force on the box. resultant force = ...................................................... N [1] 3. Determine the resistive force R on the box. R = ...................................................... N [3]

Mark scheme: 3 (a) a body / mass / object continues (at rest or) at constant / uniform velocity unless acted on by a resultant force B1 [1] (b) (i) weight vertically down B1 normal / reaction / contact (force) perpendicular / normal to the slope B1 [2] (ii) 1. acceleration = gradient or (v – u) / t or ∆v / t C1 = (6.0 – 0.8) / (2.0 – 0.0) = 2.6 m s–2 M1 [2] 2. F = ma = 65 × 2.6 = 169 N (allow to 2 or 3 s.f.) A1 [1] 3. weight component seen: mg sinθ (218 N) C1 218 – R = 169 C1 R = 49 N (require 2 s.f.) A1 [3] GCE AS/A LEVEL – May/June 2014 9702 22

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4 (a) Explain what is meant by gravitational potential energy and kinetic energy. gravitational potential energy: .................................................................................... .............. .…................................................................................................................................ .............. kinetic energy: ........................................................................................................................... ................................................................................................................................................... [2] (b) A ball of mass 400 g is thrown with an initial velocity of 30.0 m s–1 at an angle of 45.0° to the horizontal, as shown in Fig. 4.1. path of ball 30.0 m s–1 H ball 45.0° Fig. 4.1 Air resistance is negligible. The ball reaches a maximum height H after a time of 2.16 s. (i) Calculate 1. the initial kinetic energy of the ball, kinetic energy = ............................................... J [3] 2. the maximum height H of the ball, H = ..............................................m [2] 3. the gravitational potential energy of the ball at height H. potential energy = ....................................................... J [2] (ii) 1. Determine the kinetic energy of the ball at its maximum height. kinetic energy = ....................................................... J [1] 2. Explain why the kinetic energy of the ball at maximum height is not zero. ...................................................................................................................................... ................................................................................................................................. [1]

Mark scheme: 4 (a) GPE: energy of a mass due to its position in a gravitational field B1 KE: energy (a mass has) due to its motion / speed / velocity B1 [2] 1 (b) (i) 1. KE = mv2 C1 2 1 = × 0.4 × (30)2 C1 2 = 180 J A1 [3] 1 2. s = 0 + × 9.81 × (2.16)2 or s = (30 sin 45°)2 / (2 × 9.81) C1 2 = 22.88 (22.9) m = 22.94 (22.9) m A1 [2] 3. GPE = mgh C1 = 0.4 × 9.81 × 22.88 = 89.8 (90) J A1 [2] (ii) 1. KE = initial KE – GPE = 180 – 90 = 90 J A1 [1] 2. (horizontal) velocity is not zero / (object) is still moving / answer explained in terms of conservation of energy B1 [1]

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5 (a) Define the Young modulus. ................................................................................................................................................... .............................................................................................................................................. [1] (b) Two wires P and Q of the same material and same original length l0 are fixed so that they hang vertically, as shown in Fig. 5.1. l 0 l 0 P Q F F Fig. 5.1 (not to scale) The diameter of P is d and the diameter of Q is 2d. The same force F is applied to the lower end of each wire. Show your working and determine the ratio stress in P (i) , stress in Q ratio = ......................................................... [2] strain in P (ii) . strain in Q ratio = ......................................................... [2]

Mark scheme: 5 (a) (Young modulus / E =) stress / strain B1 [1] (b) (i) stress = F / A or = F / (π d2/4) or = F / (π d2) M1 ratio = 4 (or 4:1) A1 [2] (ii) E is the same for both wires (as same material) [e.g. EP = EQ] M1 strain = stress / E ratio = 4 (or 4:1) [must be same as (i)] A1 [2]

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6 A battery is connected in series with resistors X and Y, as shown in Fig. 6.1. 24 V I X Y B A C 6.0 1 R Fig. 6.1 The resistance of X is constant. The resistance of Y is 6.0 Ω. The battery has electromotive force (e.m.f.) 24 V and zero internal resistance. A variable resistor of resistance R is connected in parallel with X. The current І from the battery is changed by varying R from 5.0 Ω to 20 Ω. The variation with R of І is shown in Fig. 6.2. 2.5 I / A 2.0 1.5 5 10 15 20 R / 1 Fig. 6.2 (a) Explain why the potential difference (p.d.) between points A and C is 24 V for all values of R. ................................................................................................................................................... .............................................................................................................................................. [1] (b) Use Fig. 6.2 to state and explain the variation of the p.d. across resistor Y as R is increased. Numerical values are not required. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (c) For R = 6.0 Ω, (i) show that the p.d. between points A and B is 9.6 V, [2] (ii) calculate the resistance of X, resistance = ...................................................... Ω [3] (iii) calculate the power provided by the battery. power = ..................................................... W [2] (d) State and explain qualitatively how the power provided by the battery changes as the resistance R is increased. ................................................................................................................................................... .............................................................................................................................................. [1]

Mark scheme: 6 (a) there are no lost volts / energy lost in the battery or there are no lost volts / energy lost in the internal resistance B1 [1] (b) the current / I decreases (as R increases) M1 p.d. decreases (as R increases) A1 or the parallel resistance (of X and R) increases M1 p.d. across parallel resistors increases, so p.d. (across Y) decreases A1 [2] GCE AS/A LEVEL – May/June 2014 9702 22 (c) (i) current = 2.4 (A) C1 p.d. across AB = 24 – 2.4 × 6 = 9.6 V M1 or total resistance = 10 Ω (= 24 V / 2.4 A) C1 (parallel resistance = 4 Ω), p.d. = 24 × (4 / 10) = 9.6 V M1 [2] (ii) R (AB) = 9.6 / 2.4 = 4.0 Ω C1 1 / 6 + 1 / X = 1 / 4 [must correctly substitute for R] C1 X = 12 Ω A1 or IR = 9.6 / 6.0 = 1.6 (A) (C1) IX = 2.4 – 1.6 = 0.8 (A) (C1) X (= 9.6 / 0.8) = 12 Ω (A1) [3] (iii) power = VI or EI or V 2 / R or E 2 / R or І 2R C1 = 24 × 2.4 or (24)2 / 10 or (2.4)2 × 10 = 57.6 W (allow 2 or more s.f.) A1 [2] (d) power decreases M0 e.m.f. constant or power = 24 × current, and current decreases or e.m.f. constant or power = 242 / resistance, and resistance increases A1 [1]

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7 A laser is placed in front of a double slit, as shown in Fig. 7.1. P double slit 12 mm laser Q bright fringes 2.8 m screen Fig. 7.1 (not to scale) The laser emits light of frequency 670 THz. Interference fringes are observed on the screen. (a) Explain how the interference fringes are formed. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) Show that the wavelength of the light is 450 nm. [2] (c) The separation of the maxima P and Q observed on the screen is 12 mm. The distance between the double slit and the screen is 2.8 m. Calculate the separation of the two slits. separation = ..................................................... m [3] (d) The laser is replaced by a laser emitting red light. State and explain the effect on the interference fringes seen on the screen. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]

Mark scheme: 7 (a) waves from the double slit are coherent / constant phase difference B1 waves (from each slit) overlap / superpose / meet (not interfere) B1 maximum / bright fringe where path difference is nλ or phase difference is n360U / 2πn rad 1 or minimum / dark fringe where path difference is (n + )λ 2 or phase difference is (2n + 1) 180U / (2n + 1)π rad B1 [3] (b) v = fλ C1 λ = (3 × 108) / 670 × 1012 = 448 (or 450) (nm) M1 [2] (c) w = 12 / 9 C1 a (= Dλ / w) = (2.8 × 450 × 10–9) / (12 / 9 × 10–3) [allow nm, mm] C1 = 9.5 × 10–4 m [9.4 × 10–4 m using λ = 448 nm] A1 [3] (d) (red light has) larger / higher / longer wavelength (must be comparison) M1 fringes further apart / larger separation A1 [2]

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