2020 GCE A Level 9814 H3 P1 SS [HCI]

Hwa Chong Institution (College) 1 H3 A-Level 2020_Suggested Solutions 1 (a) (i) By Principle of Conservation of Linear Momentum,     = = − → = = = 4 u 10 228 u 57 if Ra Ra Ra Ra pp vmm v m v vm (ii)   = = =  2 2 2 1 KE 228 u 1 12 1KE 4 u 57 57 2 Ra Ra Ra mv mv (b) Total KE = 4.08 x 106 x 1.60 x 10-16 J      − + = + = =    6 19 1 58Total KE = KE KE KE KE57 KE 57 57KE 4.08 10 1.60 10 J58 Ra Given that ( )   − −  =  → =  19 71 7 2 1.60 10 4.81 10 C kg kg 4.81 10 q mm ( ) ( )( )       − − −    = → = =   =   =  6 19 2 19 7 6 7 7 1 572 4.08 10 1.60 102 KE1 58KE 2 2 1.60 10 4.81 10 57 4.08 10 4.81 10 1.39 10 m s58 m v v m

Hwa Chong Institution (College) 2 2 (a) Let t be the time that the ball remains on the circular plate. Horizontal displacement = x = 0.50t Vertical displacement = y = ( ) 21 sin302 gt ( ) ( ) ( ) += += 2 2 2 2 22 210.50 sin30 0.202 x y r t g t Solving the quadratic equation yields t2 = 0.06337. As a result, t = 0.252 s. (b) Net force acting on the rolling ball:  = −− = → = sinsin net CM CM CM F ma mg Fmg F ma a m (1) Net torque on the rolling ball: = = = → = = = 22 2 5 2 2 5 CM CM CM CM CM a Fr Fr FFr I I a r I m mr (2) Equating (1) and (2), we obtain −= = → =sin 5 2 sin 27 CM mg F F mgaF mm Now, we substitute F into (2)  −= = = = 25 5 2 sin 5 sin30 3.5 m s2 2 7 7 CM F mg ga mm (c) Moment of inertia for solid cylinder about its central axis is 2 2 MR Rotational Kinetic Energy = ( ) = = =   22 22 2 1 1 0.080 1.84 0.068 J2 2 2 4 CM MR vI R

Hwa Chong Institution (College) 3 3 (a) Take top to bottom. 9.2 cm: 460 V. Therefore Y-gain = 50 V cm-1. 8.0 cm: 2T = 0.040 s. Therefore time-base = 5.0 × 10-3 s cm-1. (b) (i) (ii) 130 0.130 1500 195 VV =  = 0 t RCV V e − = At 0 1950.5 0.01 s, 230 VtT V= = = 0.01 1500195 230 Ce − = 40.4 FC = (iii) Total capacitance will drop, time constant will drop and the discharge is more rapid. Current will drop to under 130 mA in 0.01 s and the component is more likely to malfunction.

Hwa Chong Institution (College) 4 4 (a) (i) 2 -2 2 -1 -3Units for W kg m s kg m A sUnits for q A s== (ii) 0 Cd A = and QC V= ( ) 2 -1 -3 2 4 -1 -3 0 2 2 2 4 -1 -3 A s m A s mkg m A sF m F mUnits for A s kg mm m m kg m A s = = = = = (b) Vernier caliper is used to measure diameter, because the measurement was stated to 0.01 cm. Micrometer screw gauge is used to measure separation, because the measurement was stated to 0.01 mm. (c) (i) If V reaches breakdown Vb, charged accumulated is 0 b b b Aq CV V d == . Since b TqI N= , 0 b ATIVNd = Rearranging, 0 b ITdV NA= (ii) ( )( ) ( ) 63 6 12 20 4.0 10 12 60 14 5.42 10 2.71 10 V 200 8.85 10 0.0654 b ITdV NA  −− −   +  = = =    (iii) It is necessary to ionize th