EJC 2021 J1 H2 Promo P1 MS

©EJC 2021 9749/J1H2Promo/2021 EUNOIA JUNIOR COLLEGE JC1 Promotional Examination 2021 General Certificate of Education Advanced Level Higher 2 PHYSICS MARK SCHEME 9749 Sep/Oct 2021 Paper 1 Multiple Choice Question Key Question Key Question Key 1 B 6 A 11 C 2 B 7 A 12 B 3 D 8 A 13 D 4 C 9 D 14 D 5 C 10 A 15 B 16 B 21 A 26 C 17 A 22 A 27 C 18 D 23 C 28 A 19 D 24 C 29 B 20 C 25 C 30 D 1 Rubik’s cube is approximately (5 cm)3. 6 18 9 5 4 0 05 0 0 of cu 5 be 00 s 10 5 # . ...                    2 2 2 a b   have same units as v2    22 2 2 2 2 2 2 2 2 2 13 2 units of units of m s m s m sunits of units of m s m s m sunits of 1 kg m munits of kg s 2 2 b b a a m                  3 3 3 34volume 4 6 3 32 mmm dr m d             3 3 3 3 3 8006 g cm1 85 3 3 013 800 6 12 6 0063 12 6 1 85 0 1 g cm 3 8 0 1 g 1 (round UPcm )                     .. md md md md .. .. . .. . . 4 displacement increases with time: j velocity increases until vterminal : h acceleration decrease to zero: k

2 ©EJC 2021 9749/J1H2Promo/2021 5 disregarding wall, at max height, zero vertical velocity:       22 2 max 2 max 2 0 sin 2 10 si mn 30 2 9 81 5 1 v u as u gs s . .       time to reach 5.1 m vertical displacement: s 0 sin 8 0 51 30 91 yy v u at vu ut g. .        time to reach wall: s 35 10 cos 3 0 0 40 xx x x t t sv s . . v    so ball will hit wall before reaching max possible displacement:   2 2 1 2 1 2 m sin 12 yy xx xx s aut ssu vv . t g               6 Rod horizontal so same tension in stings P and Q displace same volume of fluid V U Vg so denser liquid gives more upthrust for either objects: U g W g Tm   object in less-dense liquid has less mass 7 first 3 s no drag so constant a = 9.81 m s-2 eliminate options B and C parachute causes a sudden and large upward force (different direction) so net force/acceleration changes direction 8 total force downramp = 2 sinmg  max possible friction = mg 2 sin sin 3 05 0 mg mg .       9 Inelastic collision so KE not conserved. eliminate B Isolated system so total system momentum conserved eliminate A and C verify option D: 2 2 2 by PCLM 1 2 KE of either mass after collision 11 22 11 42 0 2 4 2mu mv v u u mv m Emu          10      heli lift air heli air heli air heli 2500 15 1680 kg 12 5 9 81 m a F m a g m a g . . W m v m gt m vt m vt                11 gradient of speed-time graph should give constant acceleration thus force Flift W a = 12.5 m s-2 W U T mg 2mg sin 

3 ©EJC 2021 9749/J1H2Promo/2021 12 by PCLM,   cannon recoil ball ball cannon recoil ball ball cannon ball recoil ball 1 m 0 1000 5 s 10 500 vv vv mm mm mv m v       by N3L, Non man =  Non floor 13 so eliminate C A, B ) not true, from free body diagram magnitude is less than weight 14 A) work is done by gravity on the girl as she falls B) tyres are rolling (since no skid), hence work done is done by friction on tyres. C) work is done by the electric field of the plate against on the electron 15 efficiency is useful power / total power expenditure 16 so vvr r  17 vector sum of normal force N and friction f provides centripetal force   2 2 2 horizontally: cos c ___ os sin sin 1 cos xN N mvf r mvf r mv frN         vertically ___ cos 2 sinN W f             2 2 2 so tan tan 30 cos1 2 sin cos sin 900001000 360 0 cos 30100 1000 9.81 s 508 N in 30 mv fr mv Wf mg r f f f f f                    18 gravitational force provides centripetal force GM 2 S R M 2 2 3 23 2 2 4 GS R T RT GS R               19 negative gradient of energy-distance graph gives force ON object; hence gradient of energy-distance graph gives force ON planet. Non car W  Nx Ny  f  Non man W a

4 ©EJC 2021 9749/J1H2Promo/2021 20       31 P QP final final P Q P P 3 P 6 1 10 J kg 1 2 1 2 10 10 J kg 400 2 200 400 22 40 m GM r GM r m m m m                               21 gravitational force provides centripetal force EsGM m s2 mr        2 E 2 3 3 3 3 E 2 EE 11 24 1 3 2 6 10 6 10 60 67 2 624 3080 0 m s GMr GMvr GM GM r T .                 22     0 sin 20 12 sin 8 08 m05 xx T. T . t       23 period is 0.50 s 1 22 05 12 6 rad s 0 . T.     24 X and Y separated by distance of 1.5     2 2 1 5 3 3 1802               s s . 25 Each double slit generates a single slit diffraction pattern 26 350 7 0 m50 v .f    (compression to compression: 1  ) RS 4 28 m 27 All particles in an inter -nodal segment are in-phase with each other, and are in anti - phase with particles in adjacent inter-nodal segment. 28 flight path of UAV is like a screen to two- source interference pattern. Screen distance is much larger than source separation so    23 0 10 2000 4 0 m15x D a . .    maxima’s are every 4.0 m apart. since UAV flies are constant speed: 40 0 044 s90 x T x . v vT.   29    39 3 min distance so shortest wavelength 10 400 10 2 0 m 10 2 100 s r r b . b s s.          30 sindn     sin 30 0 50 0 50 d d a .d .Dx         500 D d   m500 0 50 1 5 0 0015 .. . retina  θ 2.0 mm s =? r = 10 km eye