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 the molecules in the air for an electrical current to flow. No molecules in vacuum, hence no current/breakdown.

Hwa Chong Institution (College) 5 5 (a) (i) In the time taken for light to make the round trip, the cogwheel must advance by 0.5/720 of a revolution. 2 0.5 720 2 0.5 2 720 1440 L Tc L c Lc     = = = (ii) 53 0 1 0 1440 (8860)(3.15 10 10 ) 77.6 rad s    −   = = (iii) 000,3 ,5 COMMENT: The wheel must advance by 0.5/720 revolution, 1.5/720 revolutions, 2.5/720 revolutions, and so on (iv) By averaging or otherwise, random errors can be reduced and a more accurate value for c can be obtained. COMMENT: There may also be merit in going for the 2nd or 3rd blocking of light if the teeth and notches are not of equal length. (b) 8 1 6 2 0.5 18 2(88.6) 0.5 2 183.00 10 295,485 rad s 295,485 2 60 2.82 10 rpm L Tc     − = = = =   = (c) (i) 2 2 2 2 2 2211( ) ( ) ( )4 2 4 4 2 4 4 8 d D d D d Dd     + − = + = + (ii) mass 22(2.7)[ (6.0 5.5 )](0.2) 14 g8V = = + = (iii) Smaller value. This approximation moved some mass nearer to the C.G. than they actually are. The nearer the mass is to the C.G., the smaller its contribution to the moment of inertia.

Hwa Chong Institution (College) 6 (iv) Moment of inertia, 2 2 6 21 1 0.0575(0.014)( ) 5.786 10 kg m2 2 2I MR −= = =  Rotational KE, 2 6 21 1 2 (5.786 10 )(15000 ) 7.138 J2 2 60EI  −= =   = rate of increase of rotational KE 7.138(9.0) 4.2 0.189 A VI I I = = = (v) Some power may be lost dues to friction in the axle of the wheel. OR The calculation assumes the rotational KE was increased at a constant rate.

Hwa Chong Institution (College) 7 6 (a) (i) Surface charge density  = charge per unit area Hence Q =  (R2) 24 Q R = (ii) Assume the circular thin disc to be an infinitely large uniformly charged circular sheet and by symmetry we say that the field at 'z' is a uniform outgoing field Ez and solve it by taking a gaussian cylinder with the top and bottom above and below the disc with z  R. Applying Gauss’s Law, let electric field at position (0,0,z) be Ez 9 3 2= where . 2 12 ( )24 2 1( ) 9.0 10 kg m C4 z o z o z o z oo o k QE dA AE dA AAE E k           − = = = == = =   Comment: All necessary steps must be shown since this is a “show” question. It is necessary to include a diagram with the Gaussian surface drawn with the E field lines drawn (iii) 0 since let V be the electric potential at (0,0,z),z z z dVE dr V E dr Ez =− =− =− Comment: The minus sign should not be omitted as it signifies the relationship between V and E.

Hwa Chong Institution (College) 8 (b) (i) Consider infinitesimal charge element dq = r2 and dEz = dE cos 2 2 2 2 2 2 2 2 3 22 2 300 22 2 3300 2 2 2 2 22 22 let be the electric field at (0,0,z), 1cos 4() () () () (2 ) 2 ( ) ( ) ,2 ( z z o z z RR z RR z z E kdqdE where k zr kdq zdE z r z r kzdqdE zr kzdqdE zr kz rdr r drE k z z r z r let u z r du r dr duE k z u      == + = ++ = + = + == ++ = + = =   22 22 2 2 1 2 3 2 2 2 2 2 1) 2 112 1 (shown) 2 zR zR z z o kz u zkz z z R z R    + + −  =  −    = − = −    ++      √z2 + r2 cos  = 𝑧 √𝑧2+𝑟2 dq

Hwa Chong Institution (College) 9 (ii) When z  R, 22zR+  2R   2 2 2 1122 1 1 02 2 2 z oo o o o zzE z R R z R            = − = −    +    = −  − = Comment: The simplification of the expression and the approximation to be made must be clearly shown in the working. (iii) When z  R, 1R z  22 2 2 2 22 2 2 2 1122 1 11122 1 1 111122 1 1 z oo oo oo zzE zR Rz z z R Rz z z R R z z          = − = − +   +    = − = −       + +              −  −     +  +     1 2 1 2 2 2 111122 111 2 11021 oo o RR zz                 = − = −    +  +     = − = Comment: The binomial expansion must be used in the simplification of the expression.

Hwa Chong Institution (College) 10 (iv) When z  R, 2 z o E  = The electric field at (0.0.z) is independent of z, equivalent to the case for an infinite charged sheet, the electric field lines are parallel lines emerging from the surface and constant everywhere (R is ). When z  R, 0zE = . The electric field at (0,0,z) of the disc is equivalent to the electric field of a point charge when R = 0. Comment: Comments on the physical significance and recognition of the common scenarios that led to the expressions are expected.