EJC 2019 J1H2 Promo P1 Solutions

©EJC 2021 9749/J1H2Promo/2019 EUNOIA JUNIOR COLLEGE JC1 Promotional Examination 2019 General Certificate of Education Advanced Level Higher 2 PHYSICS MARK SCHEME 9749 Sep/Oct 2019 Paper 1 Multiple Choice Q Key Q Key Q Key Q Key Q Key Q Key 1 B 6 C 11 C 16 B 21 C 26 B 2 C 7 B 12 B 17 D 22 D 27 A 3 D 8 C 13 C 18 B 23 A 28 B 4 A 9 D 14 C 19 D 24 C 29 A 5 D 10 B 15 D 20 A 25 A 30 C 1 radius of Earth ~ 4x of Moon so select item ~ 3 cm radius: tennis ball 2 [unorthodox question, can afford to ignore] the uncertainty of 1% applies on the extreme values obtained       min min max max max min A 1 3 04 0 03 A100 A 1 3 3 06 3 04 0 03 3 01 3 08 0 03 08 0 03 A100 A 1 0 052 3 06 0 05 3 11 A .. .. . .. . . . . . . .                   I I I I I I I I I 3 using 22 2uav s                 22 2 2 2 2 22 2 1 2 2 2 4 4 2 16 32 15 32 9 81 3 06 32 9 81 3 06 15 8 0 m s A A A A AB A A A B A A B A A A v u g y u g y v g y g y v v g y v . . ..v v v y .              4 acceleration from X so eliminate C and D (constant initial speed) constant speed thereafter so eliminate B 5 constant horizontal speed so eliminate AB for horizontal displacement with time loss in KE is gain in GPE so negative gradient of straight line for KE vs s y expected even at highest point there is KE associated with horizontal velocity so eliminate Graph 2

2 ©EJC 2021 9749/J1H2Promo/2019 6 Consider 2 kg & 3 kg as combined mass   2 3 2 10 N F ma    OR Consider total mass 2 3 4 2 18 9 N F ma FFa F      Consider 2 kg mass:   btwn 2 and 3 9 N2 4 ma F T    Consider 3 kg mass:   btwn 3 and 4 btwn 2 and 3 btwn 3 and 4 32 41 N0 aTT T m    7 Free-body diagram: vertical equilibrium sinT mg  horizontal acceleration cos tan 2 63 4 Ta . m g a         angle wrt vertical is 26.6 8 momentum so think vectorially     final initial sin 30 sin 30 p p m p p v p p       9 A: climbs at constant speed, nett force is zero, can be in equilibrium B: constant altitude so vertical equilibrium, if horizontal speed is constant then glider can be in equilibrium C: if ship is constant velocity, nett force is zero, can be in equilibrium D: changing direction is changing acceleration implies net force, not in equilibrium 10 Recall “origin of upthrust” 11 think of both translational & rotational equilibrium translational eqm: 16 9 kN25W    By POM, pivot about front wheels, sum of  moments = sum of  moments,     0 54 9 1 5 9 2 m1 5 5 Wx . . .x    CG is 0.54 after front wheels so 1.14 from front. 12 EPE = KE left + w.d. against friction (no change in GPE) 21KE 2 kx fd  T W = mg θ 30 30 pfinal