2024 RI H3 Prelim_SS

1 2024 RI H3 Physics Prelims Solutions & Mark Scheme 1 (a) (i) By the conservation of linear momentum 0 ...... (1)A A A A B Bm u m v m v+ = + Since the collision is elastic ..... (2) A B B A A B A u u v v v v u − = − =− M1 Sub (2) into (1) ( ) ( ) 2 ..... (3) A A A B A B B A A A A A B B B AA B AB m u m v u m v m u m u m v m v muv mm = − + + = + = + M1 When Bm reduces to zero, ABmm , ( )A B Am m m+ This means that the term ( ) 2 A AB m mm+ tends towards 2 2A A m m = . Hence Bv cannot exceed 2 Au . A1 (ii) 2 2 2 2 1 2 ..... (4)1 2 BB BB AA AA mv mvf mumu == Sub [3] into [4] ( ) ( ) ( ) 2 22 2 2 2 2 24 4 B A A B A A ABA A A A AB AB AB m m u m m uf mmm u m u mm mm mm  ==  + + = + B1 (iii) 1. Fraction of kinetic energy of A transferred to B: ( ) 2 4 AB AB AB mmf mm → = + Fraction of kinetic energy of B transferred to C: ( ) 2 4 BC BC BC mmf mm → = + Fraction of kinetic energy of A transferred to C: ( )( ) ( ) ( ) ( ) ( ) 22 2 22 44 16 A B B C BCAB A B B C A B C A B B C F f f mmmm m m m m m m m m m m m →→= = ++ = ++ B1

2 2. For the largest F, 0 B dF dm = ( ) ( ) ( )( )( ) ( ) ( ) 22 2 44 32 2 2 16 0 A C B A B B C A B B C A B C B B A B B C m m m m m m m m m m m m m m mdF dm m m m m  + + − + + + += ++ = C2 ( ) ( ) ( )( )( ) ( )( ) ( ) 22 2 22 2 32 32 2 0 20 20 0 B A B B C B A B B C A B C A B B C B A B C A B A C B B C A B B B C A C B B A C m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m m + + − + + + + = + + − + + = + + + − − − = −= = M1 A1 1. (b) (i) 1 2 1 2 1 2 1 2 11 1 2 1 2 22 1 2 1 2 12 2 ' ' 22 ' 22 ' ' ' 0 (shown) 22 CM CM total mu mu u uv mm v v v u u u uuu u u u uuu u u u up p p m m −−== + =− −+= − = −+=− − =− ++   = + = + − =       B1 B1 (ii) 1. In the laboratory frame, in the vertical direction, momentum must be zero after the collision. Hence, 12 12 sin sin (shown) vv vv = = B1 2. For momentum to remain zero in the centre -of-mass frame, the two spheres must be moving in opposite directions in that frame, with the same speed 12 2 uu+ as before. And since ' CMv v v=− and therefore ' CMv v v=+ , by drawing a vector diagram, the velocities of the two spheres in the centre of mass frame must be in the vertical direction. Hence, since the vectors form a right-angled triangle, B1

3 ( ) ( ) ( ) 12 12 12 12 12 12 1 2 1 2 12 1 2 1 2 2tan 2 60 tan60 3 3 3 1 3 1 31 31 1 where 31 uu uu uu uu uu uu u u u u uu u u u u     + +== − − = += = − + = − − = + += − + = = − B1 B1 2 (a) Draw and label the normal contact forces NA and NB, and the frictional forces fA and fB correctly. B1 (b) Since the hoop is in equilibrium, taking moments about its centre O, 0 ABf R