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