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. B