2019 9749 H2 Physics MS EJC Suggested_ver Aug2022

©EJC 2021 9749/H2 PHYSICS 2019 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2019 Paper 1 Multiple Choice Question Key Question Key Question Key 1 D 6 B 11 D 2 D 7 A 12 A 3 D 8 D 13 C 4 C 9 D 14 C 5 A 10 B 15 B 16 B 21 B 26 C 17 B 22 B 27 B 18 C 23 C 28 C 19 D 24 D 29 B 20 C 25 B 30 A Notes: Q5, 13, 23, 24 and questions on parallel D.C. circuits were not well done. Question 5 As the total momentum of the system is zero before the explosion, the centre of gravity cannot move throughout. If it does, this would mean that an external resultant force must have been applied to the system. Question 8 Those who incorrectly selected option A thought that KE was dropping, and missed out on the fact that there was a force exerted to the body continually.

2 ©EJC 2021 9749/H2 PHYSICS 2019 Question 13 Those who chose option A (43 m s−1) did not convert the mass of the helium into kilograms. Since the r.m.s. speed of a typical air molecule is about 500 m s−1, a helium atom with the same KE but smaller mass should have a higher speed. Question 15 Those who chose option C did not consider the direction of heat transfer. Question 17 Option A - the times of minimum and maximum disturbance Option C – the times of maximum disturbance Question 18 Question 21 A charged particle in a uniform electric field undergoes constant acceleration, and the gradient of the v-t graph is constant. Question 23 Possible misconception – if two components are connected in parallel and one of them is not conducting, then the other will not conduct. Question 24 Those who chose C did not realise that when the value of the variable resistor is 0, the resistance of the left-hand side of the circuit is 0, not 1000 Ω.

©EJC 2021 9749/H2 PHYSICS 2019 Paper 2 Structured Questions Notes: Singaporean candidates have to be good at the maths and the presentation of maths. Weaker answers often did not make use of scientific terms appropriately. Weaker answers for definitions were often imprecise. Better answers tend to be short and direct. Qns Marks 1(a) resultant force (in any direction) is zero B1 resultant torque/moment (about any point) on the body is zero Notes: must refer to resultant/net/sum of… B1 1(b) 12 sum of forces along vertical: sin sin 10 7 020 00TT °+ °− = M1 12 sum of forces along horizon cos20 tal: cos10 0TT °− °= M1 1 2 solving simultaneousl 0 y N N 1400 13 0 T T = = OR 1 1 70 4 0 80 30 1 00 sin sin N T T = =°° 2 2 700 70 30 1300 sin sin N T T = =°° A1 A1 (M1) (A1) (M1) (A1) 1(c) let angle that T makes with pole be θ 16 tan 12 53 1o . . . θ θ = = B1 take moments about base of pole, sum of clockwise moments = sum anti-clockwise moments ( ) ( )( ) ( ) ( )1 2 sin53.1 1 2 0 60 150 cos 10o.T . .= + ° M1 T = 280 N A1