HCI 2026 H3 Physics Prelim (Ans)

1 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 HCI H3 Prelim 2026 Suggested Solutions 1(a) Velocity of centre of mass (CM) with respect to Earth (E) P P Q Q CM PQ m v m vv mm += + The horizontal component of VCM (along East-West direction) ( ) ( ) horizontal -1 -15.00 5.00 10.0 10.00cos 45 3.047 m s 3.05 m s (3 s.f.) 5.0 10.0 CMv  +  − = = − = − + The vertical component VCM (along North-South direction) ( )vertical -1 -10 10.0 10.00sin45 4.714 ms 4.71ms (3 s.f.) 5.0 10.0 CMv +  = = = + A1 A1 1(b) CM P CM E E P E P E CMv v v v v− − − − −= + = − The horizontal component of velocity of P ( ) ( )horizontal -15.00 3.047 8.05 msCM Pv − = − − = The vertical component of velocity of P ( ) ( )vertical -10.00 4.714 4.71msCM Pv − = − =− Complete Galilean transformation calculations correctly Hence, ( ) ( ) 22 -1 8.047 4.714 9.3261 9.33 m s (3 s.f.)CM Pv − = + − = = A1 A1 1(c)(i) In the centre of mass frame, P must move due South after the collision with the same speed of 9.33 m s-1, because • total momentum is zero in the centre of mass frame (zero -momentum frame): if Q moves due North with a certain momentum, P must move due South with the same momentum • total kinetic energy remains the same for elastic collisions: Using 2 2 k pE m= , for the same magnitude of momentum of P and Q, the speed of P must remain the same after the collision. B1 B1

2 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 1(c)(ii) The horizontal component of velocity of P in the Earth’s frame is ( ) ( ) ( )horizontal horizontal horizontal -13.05 0.00 3.05 m sE P E CM CM Pv v v− − − = + =− + =− The vertical component of velocity of P in the Earth’s frame is ( ) ( ) ( ) ( )vertical vertical vertical -14.71 9.33 4.62 m sE P E CM CM Pv v v− − − = + = + − =− Complete Galilean transformation calculations correctly Speed of P after the collision in Earth frame ( ) ( ) 22 -13.05 4.62 5.54 m s= − + − = Direction: 1 4.62tan 56.6 south of west3.05 − = =   A1 A1 A1

3 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 2(a)(i) 0i R= A1 2(a)(ii) qC = A1 2(b)(i) Differentiating the equation with respect to t, 2 2 2 2 2 2 0 1 0 1 d di qLdt dt C d i dq dL dt C dtdt d i dqL dt Cdt d i i LCdt   − − + = − − + = =− =− where 0ddt = and dqi dt= C1 2(b)(ii) Differentiating 0 sin( )i i t =+ with respect to t twice gives ( ) 2 22 02 sindi i t idt    =− + =− where 2 11 LC LC= → = Hence, 0 sin( )i i t =+ satisfies the differential equation. A1 A1 2(b)(iii)1 At t = 0, the charge on the capacitor is ( )0q t C== ε 0 0 di qL dt C di C dt L LC L L − − + = −= + = − =−− ε ε ε ε ε M1

4 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 2(b)(iii)2 ( )0 cos 0di itdt = + =ω at t = 0. Hence, ( )0 cos 0 0 cos 0 2 i    += = = ω M1 2(b)(iv)1 The current through the inductor will decrease from i0 to zero. B1 2(b)(iv)2 As the current decreases, t he induced e.m.f. across the inductor maintains a current through the diode to the capacitor. Additional charge accumulates on the capacitor plates. Since /CV q C= , the potential difference across the capacitor increases above ε . B1 B1 2(b)(iv)3 When the current reaches zero, the diode becomes reverse -biased. It prevents the capacitor from discharging back through the circuit. B1 2(b)(v)1 Correct shape for 0 < t < 0.25 T Correct shape for 0.25 T < t < 1.00 T B1 B1 t / s i / A 0 1.00 T 0.75 T 0.50 T 0.25 T

5 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 2(b)(v)2 The accumulated charge, q ( ) ( ) ( ) ( ) ( ) 0 00 /4 /4 /400 0 0 00 sin sin / 2 cos cos sin TT T i i t i i t i i t iiq idt i t dt t q LCR        =+ = + = =  = = = = =  ( )total total at 0q q t q q C LC R = = + =+ B1 B1 B1 A0

6 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 3(a) ( ) ( ) ( )( )0 sin sin sin 2PE E t t t     = + − + − Using the hint provided, , A t B  = − = ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 1 3 0 0 1 3 2 0 0 0 sin sin 2 2 sin cos 2 sin cos sin 1 2cos sinP E E E t t E t E E E E t E t E E t               + = + − = − + + = − + − = + − M1 B1 B1 A0 3(b)(i) Using the principle of superposition, the phase difference at P for the constructive interference which leads to the maximum intensity is ( )00 1 2cos 3 cos 1 2 where 0, 1, 2, ... EE mm    += = == M1 A0 3(b)(ii) Using the principle of superposition, the phase difference at P for the destructive interference which leads to zero intensity is ( )0 1 2cos 0 1cos 2 2 2 where 0, 1, 2, ...3 E mm    += =− = + = and 4 2 where 0, 1, 2, ...3 mm= + = M1 A1 A1 3(c) The resultant intensity is proportional to the square of the resultant amplitude ( ) ( ) 22 0 2 0 1 2cos 1 2cos E   + + I I=I when 60 = ( )( ) 2 00 1 2cos 60 4+  =I = I I M1 A1

7 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 4(a) Terminal velocity is reached when the net force acting on the ball bearing is zero. Weight – (buoyant force + drag force) = 0 ( )60ball ball oil ball TV g V g rv   − + = ( ) 6 ball oil ball T Vgv r −= ( ) 34 3 6 ball oil rg r     − = ( ) 22 9 ball oil rg=− M1 A0 4(b)(i) Net force, F = Weight – (buoyant force + drag force) ( )6ball ball oil ballF V g V g rv  = − + ( ) 6ball oil ball V g rv  = − − ( ) 34 63ball oil r g rv   = − −  ( ) ( ) ( ) 3342700 800 2.50 10 9.813 −= −   ( )( )( ) 236 1.102 10 2.50 10 2.92 −−−   42.96 10 N−=−  Mass of ball bearing, ( ) 33 3 444 2700 2.5 10 1.77 1033ballm r kg   −−= =   =  Acceleration, 4 -2 4 2.96 10 1.67 ms1.77 10Fa m −−−= = =−  , upwards direction M1 A1

8 © Hwa Chong Institution 9814 H3 Physics / Preliminary Exam 2026 4(b)(ii) B1 – shape of velocity (demonstrate terminal velocity) B1 – shape of acceleration B1 – velocity and acceleration are in opposite directions velocity/ m s -1 time

9 © Hwa Chong Institution