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

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2022 Page 3 of 6 ( ) ( ) ( ) ( ) 22 22 2 2 2 2 2 22 2 3 2 11of 2 stars 3 22 11 3322 93 22 6 G6 16 3G 8 KE M d M D M D M D M D M D MD MMD D M D =+ =+ =+ = =  =     Wrong answers: (B) ( ) 2G3Total 2(3 ) 2 MM GMKE DD== (C) ( ) 2G3 3Total 22 MM GMKE DD== (D) ( ) ( ) 2G 3 G 3 2GTotal 2(3 ) 2 M M M M MKE D D D= + = 12 D Same temperature implies thermal equilibrium. Hence all 3 objects are in thermal equilibrium and have the same temperature. However the heat capacity of each material is different, so the amount of internal energy is different. Objects in thermal equilib rium can exchange thermal energy, but there will be no net exchange of thermal energy. 13 B As n and T are constant, 1 1 2 2 12 21 30 10 3 pV nRT p V p V VV VV = = = = 14 C 2 2 2 2 2 -1250 300 400 100 500 340 m s (2 s.f.)5 + + + + = 15 C ( ) ( ) 2 0 0 222 2 0 9.81 9.81 6.2cm (2 s.f.) 2 4 2.0 ma mg N ma mg if N m x mg gx f    =− == = = = = = 16 A Graph shows U vs r of object undergoing simple harmonic motion. Hence F = –kr. Alternatively, dUF dr=− , magnitude and direction of F can be obtained from negative gradient of the graph.

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2022 Page 4 of 6 17 D 00 2 2 2 0 2 2 2 R R 0 2 2 T T 0 2 sin where amplitude of wave at transmitter sin sin sin A A A AA P A A APA     =− =   =   = I I 18 B ( ) 0.49 0.152 0.68 330 0.68 490 Hz (2 s.f.) m vf f f    =− = = = = 19 D The glass block has higher refractive index . Hence, the speed of wave and the wavelength is smaller. The number of wavelengths within the thickness of the glass block is more than the same thickness in air. Hence, central maximum position where the 2 waves have no path difference is above the original position. 20 B Let y be the distance from Q, and x be the distance from 4Q ( )0 4 4 =  = = =   22 2 QQ y xy xy y x y y yx 04 + + 4 +2 21 A The particle can be either positively or negatively charged. The particle will be deflected at the point of entry. Electric field lines (and the equipotential lines) are closer together, hence has a stronger electric field strength. 22 C 22 Since and are the same, 21 0.67 (2 s.f.)13 a a a c c c mmdA AL dL L L L d L dRm mA m R dL LR md md       = → = = = = → = =  =  =

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2022 Page 5 of 6 23 A Option A is correct since resistance of A is 15 Ω while resistance of B is 10 Ω. Option B is incorrect since power of A = 0.40 W while power of B = 0.60 W. So power dissipated in B is only 1.5 times of that in A. Option C is incorrect since total current = 0.30 + 0.20 = 0.50 A Option D is incorrect since at 0.20 A, A will have 2.0 V across it while B will have 3.0 V across it. 24 C Assuming Bsolenoid is to the right, 55 tan tan 2.0 10 tan68 4.95 10 solenoid Earth solenoid Earth B B BB   −− = = =   =  57 2 204 95 10 4 10 15 10 0 30 A (2 s.f.) solenoidBn . ( )( ) .    −− − =  =   = 25 B Sliding the rod right or left will induce an e.m.f. across the rod. Since the rod is in the centre of the metal frame, it will produce currents in opposite directions in the left and right sections of the frame. When the magnitude of the magnetic flux den sity increases, the frame experiences an increase in magnetic flux linkage out of the plane of the paper. By Lenz’s law, an induced current will flow in the frame to create a magnetic field into the plane of the paper to oppose this change. Hence, this leads to a clockwise current flowing through the frame. 26 C -1 11 0.02 s 2 div50 0.02time-base 10 ms div2 T f= = = = == -1 2.8 V 1.4 div 2.8Y-gain 2.0 V div1.4 V == == 27 A The cut-off wavelength corresponds to the most energetic photon released. 28 D The uncertainty principle is independent of the experiment equipment. These uncertainties would remain because they originate in the wave like nature of matter. 29 C 28 2 11 [(235.04393 1.00866) (140.91440 91.92617 3 1. 00866)] 3.088 10 () 2.78 10 J (3 s.f.) mu m kg E m c − −  = + − + +   =  = = BEarth Bsoleniod 68

Anglo-Chinese Junior College 2022 J2 Preliminary Exam