Pure Physics Nov 2012_TYS ANSWERS

Nov 2012_Suggested ANSWERS_Aug 2023 Page 1 of 6 – Nov 2012 Catholic High School | O-Level Physics NOT IN SYLLABUS: 5059 Nov 2012 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 A 11 A 21 A 31 A 2 A 12 D 22 D 32 D 3 A 13 D 23 B 33 B 4 B 14 B 24 C 34 A 5 C 15 B 25 B 35 C 6 B 16 B 26 C 36 D 7 B 17 D 27 B 37 B 8 A 18 B 28 D 38 C 9 B 19 C 29 D 39 D 10 C 20 A 30 A 40 A *Q. 6: (B) The weight of the rocket must not be neglected. In F = ma, F is the resultant force. Thus F is not the thrust alone but F = thrust - weight. (D is incorrect.) *Q. 13: (D) There is kinetic energy at the top of the hill. (A is incorrect.) *Q. 22: (D) The image cannot be formed at the mirror surface unless the object is at the mirror surface. (B is incorrect.) *Q. 23: (B) There is partial reflection at the boundary. (A and D are incorrect.) *Q. 24: (C) The ray from O enters the lens normal to the surface PQ because O is the centre of the circle. Bending of the ray cannot occur at surface PQ, so D cannot be correct. (D is incorrect.) *Q. 34: (A) The flat part of the curve in B suggests that the resistance has become constant, which is inconsistent with as it is stated that ‘the resistance of the thermistor decreases as its temperature increases’ in the question. (B is incorrect.) *Q. 40: (A) The question asked for the current in the primary coil. (C is incorrect.) Paper 2 [80 marks] 1 a a = v - u t = 25−20 10−6 = 5 4 = 1.25 m s-2 (3 s.f.) 2 1 b The shaded area under the speed -time graph represents the distance travelled by the car, when it had a constant deceleration between 10 – 16 s. 1 1 ci The average speed of the car is the total distance travelled divided by the total time taken. [Note: Not ‘total distance per unit time.’!] 1 * cii The speed of the car between 6 – 16 s is always higher than 20 m s -1. 1

Nov 2012_Suggested ANSWERS_Aug 2023 Page 2 of 6 – Nov 2012 Between 0 – 20 s, the speed of the car is 20 m s-1 for half the journey. [Or: The time between 6 – 16 s is half the time taken during the journey. However the distance travelled is more than half.] [Or (longer method): Average speed between 6 – 16 s = Total distance Total time = 1 2(4)(20+25)+ 1 2(6)(25+20) 10 = 225 10 = 22.5 = 23 m s-1 (2 s.f.) Average speed between 0 – 20 s = Total distance Total time = 225+(10)(20) 20 = 425 20 = 21.25 = 21 m s-1 (2 s.f.) < 23 m s-1] [Note: Need to show comparison between 6 – 16 s and 0 – 20 s.] 2 a Chemical potential energy of swimmer to kinetic energy of swimmer. 1 b K.E. = ½mv2 = ½(60)(0.80)2 = 19.2 J (3 s.f.) 1 1 * ci The swimmer does work on the water by exerting a force backwards on the water, pushing the water backwards. [Note: Need to demonstrate ‘work done’, by showing the idea of ‘a force applied’ and ‘a distance moved’ → work done by swimmer or on the water.] 1 cii The work done by the swimmer on the water eventually