Pure Physics Nov 2015 TYS ANSWERS

Nov 2015_Suggested ANSWERS_as at July 2023 Page 1 of 7 – Nov 2015 Catholic High School | O-Level Physics NOT IN SYLLABUS: 5059 Nov 2015 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 C 11 D 21 C 31 C 2 D 12 B 22 A 32 B 3 B 13 A 23 C 33 C 4 D 14 B 24 C 34 D 5 B 15 A 25 A 35 B 6 C 16 B 26 B 36 D 7 D 17 A 27 B 37 C 8 C 18 A 28 A 38 D 9 C 19 B 29 A 39 A 10 D 20 A 30 A 40 D *Q. 1: C Most appeared to be aware of the order of magnitude of the diameter of an atom. Only a significantly smaller proportion were aware of the order of magnitude of the diameter of the Earth. (Both A and B are incorrect.) *Q. 5: B Those who chose D were clearly aware of the effect of air resistance on the velocity of the falling box, but did not convert the increasing magnitude of the velocity to a displacement-time graph of increasing positive gradient. (D is incorrect.) *Q. 7: D In Newton’s Second Law of Motion equation FR = ma, FR is resultant force. The resultant force acting on the car is FR = driving force - frictional force, i.e. FR = driving force - 1200 ma = driving force - 1200 (800)(2.5) = driving force - 1200 Hence, driving force = 2000 + 1200 = 3200 N (A is incorrect.) *Q. 26: B For a converging lens, the rays will bend away from its original direction, towards the principal axis. Also, after passing through the converging, the rays do not necessarily converge to meet (think of magnifying glass). (C is incorrect.) *Q. 34: D When the amount of light increases, the resistance across the LDR decreases. Hence the p.d. across the LDR decreases and the reading on the voltmeter across the resistor increases. (C is incorrect.)

Nov 2015_Suggested ANSWERS_as at July 2023 Page 2 of 7 – Nov 2015 Paper 2 [80 marks] 1 a 1. Using a stopwatch, record t1 and t2: timing for first and second set of 20 complete movements. 2. Calculate tav = 𝑡1+ 𝑡2 2 : average time for 20 complete movements. 3. Calculate t = 𝑡𝑎𝑣 20 : average time for 1 complete movement. 1 1 1 bi The distance moved by the athlete along the track. 1 bii 1. Work done by athlete = (force)(distance along track) (i.e. W = Fd) 2. Power of the athlete = work done by athlete average time for 1 complete movement (i.e. P = 𝑊 𝑡 ) 1 1 2 a When an object is in equilibrium, the sum of clockwise moments about the pivot (/ any point) is equal to sum of anti-clockwise moments about the same pivot (/ same point). 1 1 bi With a 30 N force applied vertically (upwards) at A, the perpendicular distance between the line of action of the force and the pivot is the largest. 1 bii M = Fd = (30)(20) = 600 Ncm or 6.0 Nm ) 1 1 biii Taking moments about the pivot, Clockwise moment by brake cable = Anticlockwise moment by hand-grip 600 = F(1.2) F = M ÷ d = 600 ÷ 1.2 = 500 N (2 or 3 s.f.) 1 3 ai1 p = 𝐹 𝐴 Force that oil exerts on piston P = pA = (3.0 × 105)(5.0 × 10-5) = 15.0 N (3 s.f.) 1 1 ai2 Force that atmosphere exerts on piston P = pA = (1.0 × 105)(5.0 × 10-5) = 5.00 N (3 s.f.) 1 ai3 By equilibrium of forces, F + force by atmosphere = force by oil F + 5.0 = 15 F = 15 - 5.0 = 10.0 N (3 s.f.) 1 aii Since the pressure transmitted throughout the oil is the same, as piston Q has a larger cross-sectional area (than piston P), a larger force will be exerted on the metal plate at R by piston Q. 1 bi The air molecules are in constant, random motion. In Fig. 3.2b, the number of air molecules per unit volume has increased. The air molecules hence collide more frequently with the surface of the air bubble, hence exerting a larger force per unit area. 1 1 bii Air is compressible, so any pressure exerted on the hydraulic press would be used to compress the bubbles of air first. 1 force by atmosphere force by oil

Nov 2015_Suggested ANSWERS_as at July 2023 Page 3 of 7 – Nov 2015 4 a Gravitational force acting per unit mass. 1 b Increase in weight = Wfinal - Winitial = m(gfinal – ginitial) = (80)(9.8 – 9.7) = 8.00 N (3 s.f.) 1 c Initial G.P.E. at height of 39 km (at start of free-fall) = mgh = (80)(9.7)(39 000) = 30 264 000 J Final G.P.E. at height of 3.0 km (at end of free-fall) = mgh = (80)(9.8)(3000) = 2 352 000 J Loss in GPE during free-fall = 30 264 000 - 2 352 000 = 27 912 000 = 2.79 × 107 J (3 s.f.) 1 1 d Gravitational potential energy (G.P.E.) of diver at the start of jump is converted to kinetic energy (K.E.) of diver at the end of jump. By Conservation of Energy, Loss of G.P.E. = Gain in K.E. Since mgh = ½mv2  v = √2𝑔ℎ, speed of the diver does not depend on mass. 1 1 5 a Insert the bulb of the thermometer completely into a beaker of pure, melting ice. The lower fixed point is marked on the thermometer when the level of the mercury is steady. [NOTE: the lower fixed point is 0 °C, not -10 °C.] 1 b physical property that changes substance or object involved electromotive force, e.m.f. thermocouple electrical resistance metal pressure gas 1 any one set c Heat loss by water = Heat gain by thermometer (mc)water = (Co)thermometer (m)(4.2)(85 - 80) = (2.5)(80 - 20) m = 150 ÷ 21 = 7.14 = 7.14 g (3 s.f.) 2 1

Nov 2015_Suggested ANSWERS_as at July 2023 Page 4 of 7 – Nov 2015 6 ai 1 aii Total internal reflection occurs as: 1. the angle of incidence of at P is greater than the critical angle of glass, and 2. light is travelling from optically denser medium (glass) towards an optically less dense medium (air). 1 bi n = sin 𝑖 sin 𝑟 (from air to glass) 1.5 = sin 45° sin 𝑟 sin r = sin 45° 1.5 r = sin-1 ( sin 45° 1.5 ) = 28.1 (3 s.f.) 1 1 2 biii If using mirrors, due to light being refracted (besides reflection), there is a loss of intensity of light. OR: If using mirrors, due to the weak reflected ray, there is another weak reflected image (besides the refracted image). As result of 2 images being seen, the image is not as clear. [NOTE: If using prisms, all the light is total internally reflected and there is no refraction. There is hence only 1 image.] 1 7 a As the fuel passes through the hose, it becomes charged by friction as it loses electrons to the hose. 1

Nov 2015_Suggested ANSWERS_as at July 2023 Page 5 of 7 – Nov 2015 b As the fuel becomes more and more charged, a spark can occur to ignite the fuel. 1 c Connecting the wire allows electrons to flow from the ground to the aircraft to neutralise any excess positive charges on the aircraft. 1 d Lightning / Damage to electronic equipment (such as hard drives). 1 8 ai V = IR I = 𝑉 𝑅 = 2.9 3.2 = 0.906 A (3 s.f.) 2 aii R = 𝑉 𝐼 = 12 0.906 = 13.2  (3 s.f.) 1 bi 1 𝑅 = 1 𝑅1 + 1 𝑅2 R of XB and lamp = (𝑅1)(𝑅2) 𝑅1+ 𝑅2 = (3.2)(3.2) 3.2 + 3.2 = 1.6  Resistance of AX = 10 - 3.2 = 6.8  Total resistance = 1.6 + 6.8 = 8.4  1 1 bii at A: 12.0 V at B: 0.0 V 1 biii The circuit in Fig. 8.2 allows the lamp to be dimmed completely (when the slider is at B and the p.d. across the lamp is 0 V). In Fig. 8.1, when the resistance of R1 is maximum, there is still a p.d. of 2.9 V across the lamp, so it cannot be dimmed completely. [NOTE: both lamps can be at the same maximum brightness.] 1 9 ai1 The p.d. across X is proportional to the current in X. 1 ai2 The p.d. across X increases at an increasing rate when the current is increased. 1 aii The longer the length of the wire, the smaller the current that causes it to melt. 1 aiii The two wires have different lengths, and hence different resistances (R =  𝐿 𝐴). Hence using V = IR, with the same current but different resistances, the p.d. across the two wires is different. 1 1 bi With a strong wind, the temperature of the wires is reduced, resulting in a lower resistance and hence a lowe