2023 HCI H3 Prelim P1_SS

2023 H3 Prelim Suggested Solutions 1 (a) It is a frame, which moves at a constant velocity and Newton’s laws are applicable in such frames. B1 B1 Comments: Several students forgot to mention about Newton’s laws in their answers. (b) Let the identical mass be m. The kinetic energy of the system in this frame is ( ) ( ) ( ) 22 21 1 12 3 132 2 2K m u m u mu= + = A frame that moves towards cart A with a speed of u. In this frame, the speed of cart A becomes 3u and the speed of cart B becomes 2u. Hence, the same kinetic energy K is obtained in this inertial frame. B1 A1 Comments: Many students managed to express the correct frame of reference. (c) Let the speed of the mass m be u. The collision must be a perfectly inelastic one. By the principle of conservation of linear momentum, 0.5(0) 1.5 0.667 common common mu mv vu += = The initial kinetic energy of the system before the collision is 21 2 mu . The kinetic energy after the collision is ( )( ) 2 21 1.5 0.667 0.3332 m u mu = . The increase in internal energy of the system is ( ) 220.5 0.333 0.167mu mu−= The fraction of the initial KE that is convertible to internal energy of the system would be 2 2 0.167 1 30.5 mu mu = B1 B1 A1 Comments: This part was a distinguishing factor. Weaker students could not properly set their thought process on the inelastic nature of collision, only in which, energy losses take place.

2 © Hwa Chong Institution 9814 H3 Physics / Prelim 2023 2 (a)(i) ( ) 22 2 22 M R d R d   = + A1 (a)(ii) Moment of inertia of a circular disc about an axis through its center is given by 𝐼𝑑𝑖𝑠𝑐 = ∫ ( 𝑚 𝜋𝑅2) (2𝜋𝑟𝑑𝑟) 𝑅 0 𝑟2 = 𝑚𝑅2 2 ( ) ( )( ) ( ) ( ) ( )( )( )( ) ( ) ( )( ) 222 2222 22 2 2 2 2 2 2 2 22 2 2 2 2 22 2 2 3218 2 2 33 11 2 9 3 inner outerinner outer reel reel reel V R V RM M r M R d R R d R R d R d M R d R RRd MR MR     ++==   + +=  + += == I I I B1 B1 B1 Comments: The proof for the moment of inertia of a disc should be shown in the working as the expression is not given in the formula list.

3 © Hwa Chong Institution 9814 H3 Physics / Prelim 2023 (b) Let the tension in the string be T. Consider the box falling with acceleration a, 𝑀𝑔 − 𝑇 = 𝑀𝑎 Consider the translational motion of reel, 𝑇 − 2𝑀𝑔𝑠𝑖𝑛30° − 𝑓 = 2𝑀(2𝑅𝛼) Consider the rotational motion of reel, 𝑇(𝑅) + 𝑓(2𝑅) = (11𝑀𝑅2 3 ) 𝛼 Solving, 3𝑇 2 − 𝑀𝑔 = 4𝑀𝑅𝛼 + 11𝑀𝑅𝛼 6 = 35𝑀𝑅𝛼 6 Further, as the reel rotates, 𝑣𝑀 = 𝑣𝑂𝑌 + 𝑣𝑂 where vM is velocity of box wrt earth VOY is velocity of Y wrt reel center of mass VO is velocity of reel wrt earth differentiat