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 pinitial pfinal − pinitial p

3 ©EJC 2021 9749/J1H2Promo/2019 13 max speed  no net force engine resistance engine resistance 2 engine engine 22 engine 3 3 3 10 0 110 21 Ff Ff kv P F vk vv P v         boat provides constant power from engine 3 engine,new 2 new engine, new new 33 2 3 10 1 110 15 110 110 15 21 46 0 10 15 N60 P Fv PF v FkF v              14 surfaces in contact have same linear speed because no slipping, eliminate B and D Rr vr r R       15 vertical circular motion that has PCE. At top, weight and tension provides centripetal force: WT   2 0 top 2 top 11 2 1 22 Wr m m r v gr v m   Conserve energy:   22 bottom top 11 2 1 22 5 2 5 2 mv mgh mv mg m r r r glv mg g     16 gravitational force provides centripetal force Eum Jupiter Eu2 2 Eu GM mr r m      2 2 3 Jupiter 3532 2 11 27 2 2 6 7 10 102 85 60 6 67 10 1 90 10 kg r T r TG . . . M             17 at mid point, net force is zero. 18 gravitational force provides centripetal force 2 2 2 1 2 1 2 EE E GM m GM m r GM mmv m r v r   2 mr  2 32 E r GM r   need to compare energy total 2 KE GPE 1 2 1 2 1KE or GP 1 2 E2 E EE E E GM m r GM m GM mv m rr GM m r         A: true. Sat_A is nearer so has less GPE so less total energy if the Sats have same mass. For Sat A to have same total energy, Sat_A is more massive. C: true. Sat_B has larger r so smaller ω D: E_total = − KE so same KE

4 ©EJC 2021 9749/J1H2Promo/2019 19 Consider change in GPE     final final 5 4 prove D false5 EE E GM GM RR G m mM R U m m              20 bouncing ball has constant acceleration of g = 9.81 m s -2 except during contact with ground, does not obey 2ax  21 Using 22 0 xvx    22 0 22 0 2 0 2 1 20 30 solve simultaneous 1 1 25 9 mm s 30 20 1300 1300 25 x v x x .              22 By PCE, max PE = max KE           22 max 0 max anytime 2 2 2 2 2 00 22 2 2 2 2 2 1 2 11 22 1 2 1141 22 4 2 4 1 4 KE PE KE KE 2 215 24 69 1 s mx m x m x x mx . m m mT . . . T T.                     23 reduce 30% means 70% transmitted 2 0 2 0 basic 07 cos cos 33 .         II I I 24 8000 5000vf   2 2 2 25 8 5 2/ s             25 Huygen’s principle (see Annex - single slit) 26 Grating X produces the horizontal dots   9 2 2 5 1 opposite hypotenuse 1.6 3 450 10 3 10 sin 3 6 1 6 nN N dn ..             27 for L1, waves formed are integers of multiples of 2  Let 1 2Lp  , p is integer for L2, waves formed are odd integers of 4  Let  2 21 4Lq  , q is integer   2 1 21 214 2 2 qL q Lp p    

5 ©EJC 2021 9749/J1H2Promo/2019 28 P and Q in series so same amount of current flowing through both of them p.d. across Q is 5 V, total pd is 15 29 8 8 4 2 Q Q Q / Q tT     I 30 Re-drawing:    YZ XY 2 5 25 10 52 2 1 9 5 7 5 10 RR R R R. . . ..         Y Z 5 2R X R R + 5 Y