2024 HCI H3 Prelim P1 SS with comments

H3 Prelim 2024 SS 1(a)(i) 1 -1 80 rev 2 rad 1 min 8.4 rad s1 min rev 60 sec (8.4)(0.60) 5.0 m svr   −  ==      = = = B1 A1 1(a)(ii) -20 8.4 0.21 rad s40 fi tt  −−= = = =− A1 Comments: The angular acceleration must be negative. 1(a)(iii) ( ) 2211 (8.4)(40) 0.21 (40) 168 rad22 itt   = + = + − = 1 rev168 rad 26.8 rev2 rad  = Accept answers with 2 s.f. B1 A1 1(b) Using the principle of conservation of angular momentum, totalL mvd == I where L is the angular momentum of the blu-tac and 2 2 total 2 MR mR=+I ( ) 2 2 2 2 2 2 MRmvd mR mvd M m R   =+ = + B1 B1 A1 Comments: About half of students had difficulties in expressing the distance from the blu -tac to the centre of rotational axis correctly. They cited d rather than R. Some could not conceptualize the conservation of angular momentum, hence they either used “conservation” of kinetic energy or linear momentum. However, since the collision is an inelastic one, the “conservation” of kinetic energy of the system will not work.

2 © Hwa Chong Institution 9814 H3 Physics / Prelim 2024 2(a) In the Earth frame, after the collision, 12 2 1 2 1 11 12 0 and mu mv mv u v v v u v u v u v v v u =+ = − → = + = + + → = = Finding velocities in the zero-momentum frame before the collision, 22 cm mu uv m== 1 , 1 , , 2, 2, , 22 0 22 cm E E cm cm E E cm uuu u u u uuu u u = + = − = = + = − = − After the elastic collision, their velocities change sign in the zero -momentum frame (CM frame), 1 , 2, and 22 cm cm uuvv = − = Hence, the kinetic energy of the system after the collision in zero-momentum frame is 22 2 , 11 2 2 2 2 4 after cm u u muKE m m    = − + =       M1 M1 A0 Comments: Everyone could obtain the velocity of CM in Earth frame, well done. However, many did not figure out the velocities in the CM frame after the collision. Some used the velocities before the collision to calculate the kinetic energy after the collision. Only a handful of them mentioned that since the collision was elastic, hence the kinetic energy would remain the same after the collision. 2(b)(i) By PCOLM, , -1 -1 cos30 cos 45 sin45sin30 sin45 0 2 sin30 40.0 2 cos30 cos 45 20.7 m s 29.3 m s A x A B B A B A B BB B A mu mv mv vmv mv v v vv v v = +  −  = → = =  = +  = = B1 A2 Comments: Many successfully applied the PCOLM in 2D. A handful of students need to brush up their concept of the conservation of momentum. 2(b)(ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 222 2 1/ 2 40.0 1/ 2 29.3 1/ 2 20.7fraction 1/ 2 40.0 before after before m m mKE KE KE m −−−== 19.6% of initial kinetic energy is dissipated. B1 A1 Comments: Unfortunately, many did not interpret the term fraction correctly and went on to find the ratio of KE of asteroid A before and after the collision.

3 © Hwa Chong Institution 9814 H3 Physics / Prelim 2024 3(a)(i) y2 = A2 cos (t + ) A1 Comments: This question was well done. 3(a)(ii) Using graphical method, 2 2 2 1 2 1 2 22 1 2 1 2 22 1 2 1 2 + - 2 cos ( ) = + - 2 (cos cos + sin sin ) = + + 2 cos rA A A A A A A A A A A A A       =− C1 B1 Comments: Majority have difficulty handling the vector addition. It would be helpful to draw the vector diagram involving the vector addition and obtain the required angle. 3(b)(i) Since intensity I is proportional to A2, for the combined intensity of the waves arriving at D: 2 2 2 1 1 2 2 , , rrI A I A I A   2 2 2 12 12 + 2 cos = 0 cos = 0 = + 2 n 2     =  → rA A A AA B1 A1 Comments: Many students were lost in the mathematics and failed to arrive at a logical deduction of the phase difference. 3(b)(ii) Path difference between the two waves arriving at Q1 and Q2: 2 21 11LQ LQ d− = + − ( ) 21 2 2 2 2 2 2 2 211 24 11 4 1 1 2 4 16 16 8 LQ LQ d d d d        − = + − = = + = + + = + + =+ B1 B1 B1 A0 Comments: Those who were able to apply the answer from (b)(i) to this part were most likely to succeed in answering this question and they also presented their workings clearly. Students were also more successful in handling the mathematics for this question as they persevered to arrive at the given relationship.

4 © Hwa Chong Institution 9814 H3 Physics / Prelim 2024 4(a) From Faraday’s Law = = ( cos ) cos = = - sin dd BAN tdt dt dtBNA BNA tdt     2 , 4 ==LNA 2 1 = - ( )( )(2 )sin = - sin42      LB f t BL f t Magnitude of maximum induced e.m.f. = 1 2 BL f B1 B1 B1 Comments: Note that in derivation as in this case-should start with an expression for the magnetic flux linkage and then relate it to the induced e.m.f. A few students who did not start off with the correct expression for the magnetic flux linkage being a cosine expression rather than sine probably did not take note of the initial position of the coil relative to the magnetic field. A few also forgot to express all terms in the final expression in terms of the stated physical quantities. 4(b) As the coil rotates through T/4, the magnetic flux linkage through the coil decreases to zero (induced e.m.f reaches maximum) and remains zero till 3T/4. After 3T/4, the flux linkage through the coil increases to a maximum (induced e.m.f decreases to zero) till T. Comments: A common mistake: did not label the time for which the induced e.m.f. changes.

5 © Hwa Chong Institution 9814 H3 Physics / Prelim 2024 5(a)(i) 32 2 0dU A B dr r r= − + = Only one value of r for circular orbit, 0dU dr = ( ) ( ) 22 15 7 2 8.39 102 3.98 10 4.22 10 m c c Ar B r  ==  = B1 B1 A0 Comments: Most students can do it correctly. 5(a)(ii) ( ) ( ) 2 2 2 2 215 22 7 4 2 4 3.98 10 4 8.39 10 4.72 10 J cc A B B B BU r r A A A U U = − = − = −  =−  = −  TE = Ueff = - 4.72  107 J, because in circular orbits the radial component of velocity is zero, hence the radial kinetic energy is zero. B1 A1 B1 Comments: Most students can do it correctly. 5(a)(iii) 2 22 2 22 2 7 4 5 -1 22 2 2(10)(8.39 102 10(4.22 10 ) 7.27 10 rad s cc cc c c c AL L mAr mr L mvr m r mA mA mr    − = → = = = = ==  = B1 B1 A0 Comments: Most students can do it correctly. 5(b)(i) 77 2 7 2 1.00 10 4.72 10 3.72 10 AB rr AB rr  −  = − −  = − At perihelion (closest distance) and aphelion (farthest distance), the radial component of velocity is zero. 7 2 22 15 7 2 15 22 7 7 3.72 10 8.39 10 3.98 10 3.72 10 3.98 10 8.39 10 0 2.89 10 m 7.81 10 m closest farthest rr rr r r −  =  −  −  +  −  = = = B1 B1 A1 Comments: Some students could not get the answers because they forgot about using the given equation.

6 © Hwa Chong Institution 9814 H3 Physics / Prelim 2024 5(b)(ii) 2 far nearrra += By Kepler’s third law, 23 23 32 2 2 3 4