ACJC 2022 Prelim P1 Guide

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2022 Page 1 of 6 Qn Ans Discussion 1 B 11 2 mass of car 2000 kg speed 100 km h 28 m s 1KE 0 784 MJ2 mv . −−   == 80 gsm (grams per square metre) of A4 paper A4 area 210 mm x 297 mm = 0.062 m2 Mass of A4 paper = 5.0 g 2 C ( ) 2 top 2 2initial 1 KE 2 1KE 2 m u cos cos mu  == 3 A At steady speed of 1.4 m s–1, air resistance = weight component along slope 1.4 k = (80.0)(9.81)sin5° k = 48.857 At steady speed of 5.5 m s–1, air resistance = (5.5)(48.857) = 268.71 N Additional force required = 268.71 – (80.0)(9.81)sin5°= 200 N (2 s.f.) 4 B 1 1 1 2 12 12 1 2 10 0 7 5 2 5 cm At constant speed: (1) When slowing down: (2) Sub(1) into (2) (3) (3) (1) 1 99 2 0 cm new length 10 0 2 0 8 0 cm x . . . mg kx mg kx mg kx ma kx kx ma xxa gx x . . . . . = − = = =− − = − − = − −→= = = − = 5 B Applying cosine rule, ( ) ( ) ( ) ( )( ) 2 2 2 1 20 15 2 20 15 cos 45 14.2 ms 0.140(14.2) 2.0 N s (2 s.f.) v v p m v    − = + −  = = = = 6 C Considering X, Y and Z as a system, F = 6 ma Considering X as a system, F – FYX = ma FYX = 5 ma = 5/6 F

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2022 Page 2 of 6 7 A Taking moments about the contact of the beam with the wall, Sum of clockwise moments = sum of anti-clockwise moments (Tcos30)(Lsin60) = (Tsin30)(Lcos60) + (1/2W)(Lsin60) T = 85 N (2 s.f.) 8 B Ep = ET mgr = ½ m (42) gr = 8 If the mass possesses 75% of E, it means E has decreased by 25%. Let x be the point where GPE has decreased by E/4, Ep = Ex mgr = mg(3r/4) + ½ m v2 v2 = ½ gr v = 2.0 m s–1 9 C As v = r ω, Since ω is constant and r is decreasing at a steady rate with time, v  r  t Graph is a straight line, negative gradient but with a non-zero speed (as r  0) 10 D All geostationary satellites must have the same radius, and angular velocity 2 T  = . Hence, they will have the same linear speed. Why C is incorrect ( ) ( )( ) 2 2 2 23 22 3 -11 24 7 7 G where - radius of orbit 4 G 424 60 60 6.67 10 6.0 10 4.231 10 m height from surface, 4.231 10 Mmmr r r Tr M r r hR = =   =  = =  − 11 A ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 3 3 (1) 33 (2) Compare (1) & (2), 3 (3) 3 1From (2) 33 (4)16 GM MMd dD GM MMD dD dD GM M MDDD GM D     = + = + = = +  = 3r / 4

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