2023 GCE A Level 9814 H3 P1 SS [HCI]

Hwa Chong Institution (College) 2023 H3 A level Suggested Solutions 1(a)(i) The restoring force (tangential component of weight) is the net force. By Newton’s 2nd law, sinmg ma−= sin r = l where r is the amplitude of the oscillation. For small angles, sin , so the amplitude r becomes l . The motion of pendulum is approximated to be a simple harmonic motion, hence, 22ar   =− =− l Rewriting Newton’s 2nd law, ( ) 2 2 2 2 mg m g T g       − = − = == l l l 1(a)(ii) 11 2 1 9.81 0.910 Hz (3 s.f.)2 0.300 gf T f   == == l 1(b)(i) 2 2 2 2 0 0 0 1 0 LCVV di qL dt C d q qL dt C dq qdt LC += += += += The differential equation for q is analogous to the simple harmonic motion. θ mg Tension mgsinθ m l r

Hwa Chong Institution (College) ( ) 22 1 2 1 2 fLC f LC   == = 1(b)(ii) 36 11 2.81 Hz 2 2 82 10 (39000 10 ) 2.81 3.090.910 3 LC LC pendulum LC pendulum f LC f f ff   −− = = =  ==  1(c) The natural frequency of LC circuit must be 3 times smaller. So, the effective capacitance of all the capacitors should be 9 times greater. To achieve this, 9 capacitors should be connected in parallel. However, this does not exactly match, 0.938 Hz and 0.910 HzLC pendulumff == . To exactly match the frequencies , connect 10 capacitors in series this gives 0.890 HzLCf = and increase the length of the pendulum to 31.4 cm, which gives 0.890 Hzpendulumf = .

Hwa Chong Institution (College) 2(a) The magnitude of the electric dipole moment is the product of charge on an atom and distance between the centres of positive and negative charges. 2(b) 30 20 12 20 19 1.50 3.34 10 6.67 10 C(0.784)(95.8 10 ) 6.67 10 C 1.60 10 C 0.417 pq d e qe − − − − − = = =     = 2(c) By applying cosine rule, 22(1.50D) (1.50D) 2(1.50D)(1.50D)cos104 1.85 net net p pD = + −  = 2(d)(i) The centre of mass should be closer to the oxygen atom as it is heavier. 2(d)(ii) electric sin sin sinF r qEr pE   = = = 2(d)(iii) Initially, the dipole is stationary. It has zero rotational kinetic energy. Its potential energy is cos(63 )pE  . 104° 1.50D 1.50D Anti-clockwise Felectric Felectric

Hwa Chong Institution (College) The dipole will have maximum kinetic energy when it has minimum potential energy (the dipole vector is parallel with the electric field, hence the angle is zero). The potential energy at this moment is cos(0 )pE  . The maximum rotational kinetic energy would be the change in potential energy between its initial state and when the dipole moment vector is parallel with the electric field. 34 26 cos(0 ) cos(63 ) 1.85 3.34 10 9500 (1 cos(63 )) 3.21 10 J pE pE − − =  −  =     −  =

Hwa Chong Institution (College) 3(a) 3(b) The net force in vertical direction: cos sin 0mg− − =NF (1) The net force in horizontal direction: 2 maxcos sin mv r+=FN (2) The frictional force is N . Substituting this into (1): cos sin cos sin mg mg       =+ = − NN N The equation (2) then becomes ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 max 2 max 2 max 2 max max cos sin cos sincos sin cos tan cos 1 tan tan 1 tan tan 1 tan mv r mvmg r mg mv r rgv rgv                   += +=− + =− += − += − NN 3(c)(i) 1 tan− should be the smallest possible so that the speed would be the largest. This can be achieved when the angle of 45°, not 12°. 20.0 0.90 cos45 20.637 20.6 m (3 s.f.)r = +  = = N F center of circle

Hwa Chong Institution (College) 3(c)(ii) ( ) ( ) -1 max 20.6(9.81) 0.70 tan45 33.9 m s1 0.70tan45v +== −

Hwa Chong Institution (College) 4(a) By PCOLM, 0 1 2 2 0 1 2 cos cos 2 cos mv mv mv mv v v v   = + + =+ From the diagram, 2 2 cos 16 x Lx  = + 0 1 2 2 2 2 16 xv v v Lx =+ + Therefore, 2 2 2() 16 xfx Lx = + 4(b) 1 1 2 2 L xLv t x t v + = + → = 22 22 2 2 2 2 1 16 16 16 2 LLxxLv t x v v Lt x ++ = + → = = + θ L/4

Hwa Chong Institution (College) 2 2 0 1 1 2 2 1 01 2 16 216 41 2 6 2 Lxxv v v LL xx xv Lx xLvv xL  + =+  ++  =+  + += + 4(c) Since the collision is elastic, the kinetic energy would remain the same after the collision. 2 2 2 0 1 2 2 2 2 0 1 2 1 1 1 22 2 2 2 mv mv mv v v v =+  =+ From earlier part, express this in terms of 1v only, ( ) ( ) 2 22 22 1 1 1 2 22 22 2 22 6 / 16 22 /2 6 2 / 16 222 2 / 4 24 8 / 2 0 x L x Lv v vxL xL x L x L x L x L x L xL x xL L ++ =+ + + + + +    −=    ++    + + − = Solving this quadratic equation for x, 12 0.0538 and 0.378x L x L= =− Substituting this into ( )212 Lxb=− x1 produces 77bb= → = and x2 produces 2.644b =− . Hence, x1 is accepted and x2 is rejected. Therefore, the value of b is 7. 4(d) The friction between the table and the balls produces same deceleration effect on the target balls and cue ball. They will all be moving at slower speed and the trick shot may still be achieved. However, the time for the cue ball hitting the right-hand cushion and the time for target balls being potted would be different.

Hwa Chong Institution (College) 5(a)(i) By parallel-axis theorem, single square plateI through the central axle at a distance l away from the centre of square plate is ( ) 22 22 single square plate 2 22 2 12 12 0.120(8360)(0.12) (0.003) 0.25012 0.023 kg m Ma a MM = + = +  =+  = I l l 5(a)(ii) ( ) 2 22 2 cylindrical rod 2 total single square plate cylindrical rod 2 total 8360( (0.003) 0.250(0.250) 0.00123 kg m33 8 0194 kg m 019 kg m (2 s.f.) rodm = = = = + = = lI I I I I 5(a)(iii) By PCOE, 22 total 2 2 2 total drum 2 22 total drum -1 112 0 0 0 2 22 1122 22 4 4(1.81)(9.81)(11.0) 2 0.194 2(1.81)(0.043) 62.4 rad s mgh mv mgh m r mgh mr     + + = + + =+ == ++ = I I I 5(a)(iv) 2 2 0 2 12 where is constant2 220 2 2(11.0) 8.2 s(62.4)(0.086 / 2) drum drum drum drum hh at a a t a h t htt r t r tr ht r     = → = → = + = + = = = = = 5(a)(v) 2 3 (Area of square plate)(distance travelled) ( ) ( )( ) because is constant (0.12) (62.4 / 2)(0.25)(8.2) 0.92 m air air avg avg air air V A vt V A t V V   == = → = = l 5(b)(i) Assume that, the loss in gravitational potential energy is fully converted to increase the temperature of water only.

Hwa Chong Institution (College) water 3 water 3 2 2 2(1.81)(9.81)(11.0) (999 93.2 10 )(4190) 1.0 10 K 1.0 mK m c T mgh mghT mc T − − =  = =   =  = 5(b)(ii) Total supply of energy is the lost in GPE of system, 2 2(1.81)(9.81)(11.0) 390.6 Jmgh== 1. Translational kinetic energy of moving masses, 2212 1.81 (0.305) 0.1684 J2 mv == 2. Rotational kinetic energy of rotating system, 2 21 0.3050.194 4.880 J2 0.043 ==  I 3. Ignoring other forms of energies conversion (due to friction, radiation etc), the thermal energy converted, 390.6 0.1684 4.880 385.6 J− − = Hence, the energy associated with paddles & insulated copper cylinder would then be 385.6 / 13 29.66 J= Since 29.66494 > (0.1684 + 4.880), item (2) has a higher impact on temperature rise. 5(b)(iii) The resolution of the thermometer is the smallest change in the marking of the thermometer. In the older days where instruments are mainly analogue, so from one marking (say a line) to the adjacent marking is a temperature difference of 2.8 mK. Using a magnifier etc, it is possible that Joule could have read the interval in between the two successive markings to tell a temperature difference of 1.0 mK or even less as it is in the same order of magnitude. This can even be enhanced if Joules allowed the masses to fall through an even longer distance of 11 m (there is no change in apparatus used) and a higher temperature change will result.