EJC 2017 J1H2 Promo P1 Solutions

©EJC 2021 9749/J1H2Promo/2017 EUNOIA JUNIOR COLLEGE JC1 Promotional Examination 2017 General Certificate of Education Advanced Level Higher 2 PHYSICS MARK SCHEME 9749 Sep/Oct 2017 Paper 1 Multiple Choice Q Key Q Key Q Key Q Key 1 A 6 D 11 B 16 C 2 A 7 B 12 B 17 A 3 C 8 A 13 C 18 C 4 D 9 D 14 A 19 C 5 void 10 C 15 B 20 D 1 Singapore population ~ 5 million Take average mass per pax ~ 60 kg    6 total 8 60 3 5 10 10 kg m   2 Set 1 and 2 has the same accuracy but Set 1 is less precise. Set 3 is precise but inaccurate. 3 Consider error propagation 05 2 2100 100 10 32 0 85 P m v t P m v t . .%           4 gradient of s-t graph must tend to zero at time t2 6 Consider horizontal force needed to move sand from no horizontal speed to constant speed.      final initial total 4 1 5 0 6 N 17 2 N63 mm v v vtt . F F                   7 constant speed so no net force             sin 1 38 sin 5 55 5 55 sin sin sin1 38 80 9 81 si at faster speed: 5 55 5 55 1138 207 N n5 r . mg f k . mg F k . f k . mg mg m . g. .               θ

2 ©EJC 2021 9749/J1H2Promo/2017 8 By PCE,             barrel man all barrel 2 man man barrel barrel man man barrel 1 2 loss in GPE gain in GPE gain in KE 1 2 9 80 120 5 94 2 120 80 m 1 s 98 m m m m m m g gh g h v v . h . mm                9 recall “origin of upthrust” 10 consider vertical equilibrium si si nnT HW TH TW    consider rotation equilibrium pivot at L/4 where cable attaches to door: sum of  moments = sum of  moments   si44 n so LLHW HW    11 consider half the extended cord: by conserving energy 1 2    2 25 20 1L.k       2 2 2222 0 5 m 80 0 5 0 2 10 0 56 .. x. mv x.     12 consider view from Earth’s north pole: (answer is not unique) the first possible answer is plane has flew one extra round to end up at same angular displacement         Earth Earth Earth plane plane plane plane 18 15 2 3 5 3 22 24 215 2 10 5 18 3 h t T . T T T. t . .t .                   θ L 0.5 m

3 ©EJC 2021 9749/J1H2Promo/2017 13 compare ratios:   2 1 120 4000 0 05 4000 3 16 3 00 ms a a v. . r r r.           14 the keyword is the “now” in “The spacecraft will now move” i.e. the behaviour of the spacecraft at the instant the engine is fired. Net force zero at that instant so by N1L the spacecraft will continue with its linear velocity tangential to the orbit. 15 gravitational force provides centripetal force EarthGM m 2r m 2v r   Earth new old old new new 1 6610 6890 6610 6890 1 7780 7620 m s vr vr v GMv r r        16 displacement is a sine graph so velocity is cosine graph and 21 2 mv resembles C 17 res