ACJC 2025 J2 H2 Physics Prelim P1 Guide

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 1 of 6 Qn Ans Discussion 1 C The object travels at the same speed. Hence, fivv = . ( ) fi fi v v v vv  = − = + − 2 A For the upward motion, air resistance and gravitational force are both acting downwards, acceleration is greater than the acceleration of free fall and hence it decelerates faster to reach the maximum height. As for the downward motion, air resistance and gravitational force are in opposite direction s and the resultant force equals to gravitational force minus the air resistance. Acceleration is smaller than the acceleration of free fall and hence it accelerates slower downwards than it decelerates upwards. 3 D Consider cart 3 and cart 4 as single object of mass 800 kg. Only T is acting on this object. F = ma T = 800 (2) = 1600 N 4 A Assuming the mass of each ball is m. Since total initial momentum = mv By Conservation of linear momentum, total final momentum should be mv A) Total final momentum = 0 + mv = mv B) Total final momentum = 22 mv mv mv+= C) Total final momentum = 31 4 4 2 mv mv mv−= D) Total final momentum = mv – mv = 0 Only A and B’s momentum are conserved. Elastic means relative velocity of approach = relative velocity of separation By relative velocity of approach and separation, KE not conserved for option B. 5 A By conservation of energy, elastic potential energy is converted to kinetic energy and work done against friction. As height is the same at initial and final positions, gravitational potential energy is unchanged. ( ) ( ) friction 2 2 1 2 1 since 02 loss gain final initial final initial EPE KE WD kx KE KE fd KE kx fd KE =+ = − + = − =

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 2 of 6 6 D 2 2 29.30.52 9.811.3 39.7 40 N mvN mg r mvN mg r −= =+ =+  =  7 A 11 2 2 2 11 1 (6.67 10 )(30)Resultant 2 cos 45 ( 10 10 ) 1.4 10 N kg upwards g − −− =   + = 11 22 10 1 (6.67 10 )(30)Resultant 2 ( 10 10 ) 2.8 10 J kg  − −− = − + =−  8 C k p p at 3.2 MJ2 at 6.4 MJ at 2 3.2 MJ2 GMmEr R GMmEr R GMmEr R == = − = − = − = − 9 B Since they have the same volume and pressure initially, they must have the same T. Therefore when the partition is removed, the P and T remains the same. 10 B Since temperature is the same, the average kinetic energy is the same. 21 2 kE m c= 2 , , 1 1 2 0.71 rms X Y rms Y X c m c m cm  = = = 11 B U Q W = + For path 1, ( )10 4 6 JU = + − =+ For path 2, U is the same. ( )62 8 J Q Q = + − =+

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 3 of 6 12 A Characteristic of heavy damping. 13 C ( ) 2 00 2 1 6.0 0.30 4.47 4.5 rad s ax    − = = =  14 B P and S are at the centre of regions of compression. T is at the centre of a region of rarefaction. The centre of regions of compression has a phase difference of  from the centre of regions of rarefaction. 15 D ( ) 544 4.0 10 1.6 10 sT −−=  =  4 11 6250 Hz1.6 10f T −= = =  330 6250 5.3 cm v f = = = 16 B ax D Dx a   = = x is directly proportional to  . However, the graph does not start from the origin. 17 B applying small angle approximation, b   ( )( ) 6 25 18 0.40 10 2.8 10 5.1 2.2 10 m x Db Dx b   − = =  = = 18 D 33.5no. of wavelengths 2.7912== Since X is an intensity maxima (antinode), an intensity minima (node) will be detected at 0.25 , 0.75 , 1.25 , 1.75 , 2.25 and 2.75       away from X.

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 4 of 6 19 C Between parallel charged plates, the electric field E is uniform. Hence, the force F qE= experienced by the α-particle is uniform at any position in between the two plates. 20 D Option A is incorrect as potential is a work done per unit charge / work done per unit mass. Option B is incorrect as potential is inversely proportional to r and not r2. Option C is incorrect as electric potential can be positive if the object is positively charged. Option D is correct since moving along a field line implies moving from one equipotential line to another so the potential changes. 21 A I = nevA v = I /neA since I , n, e are constants v  1/A  1/d2 2 2 2 2 () 0.25(2 ) = == y x xy y x v d vd v d vd 22 A As V increases, I/V ratio increases, V/I ratio decreases, R decreases. 23 B Effective resistance of right portion increases. Using potential divider principle, S is brighter than before, while P and R are dimmer than before. 24 A Using the right-hand grip rule at each wire position, the resultant magnetic field due to the contributions from the four wires are as follows: - Downwards at P - Rightwards at Q - Upwards at R - Leftwards at S 25 D Electron experiences a magnetic force out of the page. B sinF Bev = For it to remain undeflected, the electric force must have the same magnitude. Hence, E sinFE Bve == The electric force must point into the page. Since an electron is negatively charged, the electric field points out of the page.

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 5 of 6 26 B for one rotation BA= ( ) ( )( ) ( ) 2 2 2 magnitude of 2 1 2 1 0.23 0.65 1202 5.8 V d dt BA T Br Br      = = =   = = = Alternatively, ( )( ) ( ) 2 2 2 2 magnitude of 1 2 1 2 1 2 1 0.23 0.65 1202 5.8 V d dt dBrdt dBr dt Br     = =    =      = = = 27 B For a rectified square wave, 2 0 rms 0 2 2 TV V T V  = = 2 rms ave 2 0 2 0 1 2 2 VP R V R V R = = = 28 D ( )( ) ( ) 34 8 9 energy of photon 6.63 10 3.0 10 650 10 1.9 eV hc  − − =  =  = Transition from E3 to E2 gives the same energy. 29 B The p.d. through which the electrons are accelerated will increase the kinetic energy of the electrons. This will increase the maximum energy of X -rays emitted, reducing the cut -off wavelength of the spectrum. The characteristic wavelengths remain unchanged as the target is not changed.

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 6 of 6 30 A The correct answer describes a piece of evidence that uniquely identifies α-radiation. - Option A is correct because the count rate only reduces significantly 1 cm away from the source in air if it is α-radiation. β and γ radiation have greater penetrating power. - Option B is incorrect because a lead block can shield α, β and γ radiation. - Option C is incorrect because both α and β particl