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 f R  = =  ABff= B1 A0 (c) Let Nrod,A and Frod,A be the normal contact force and frictional force of the hoop on the stick. Let Nrod,C and Frod,C be the normal contact force and frictional force of the ground on the stick.

4 By Newton’s third law, ,, and A rod A A rod AN N f f== . For the stick, taking moments about C, , ,A ( cos ) ( cos )22 1 cos2 rod A rod A LLW mg N L N N mg = =  ==   B1 M1 A0 (d) Resolving horizontally for forces on the hoop, cos sinA B Af f N += Since ABff = 1cos cos sin 2 1 cos sin2 1 cos BB B f f mg mg f +=   = +      Since tan 2 R L  = and using the identity sintan 2 (1 cos )  = + sin (1 cos ) sin(1 cos ) R L L R    =+ += Thus, 1 cos sin 12 cossin 2 B mg mgRf L L R  ==   M1 A1 3 (a) Since the ring is smooth, the tensions in the left (of ring) and right (of ring) sections of the string are the same. Resolving forces horizontally, 2 00sin73 sin51 mvTT r−= −−(1) Resolving forces vertically, 00cos73 cos51T T mg += −−(2) M1 M1

5 0 0 2 00 1 (1) sin73 sin51 (2) 250 9.81cos73 cos51 21.8 m s v v − − = + = A1 (b) 22 221.83 1.91 m s250 c va r −= = = B1 (c) Although the speed of the car is constant, its velocity is not as its direction is constantly changing. Hence the car is accelerating and there is a resultant force acting on it. The resultant force is directed towards the centre of the circle, hence it is always perpendicular to the velocity of the car. As a result, no work is done by this force and hence there is no change in the kinetic energy of the car. B1 B1 4 (a) When the platform is at displacement y from the equilibrium position, resultant force on the platform is ( )12 12 2 2 k y k y ma kkay m − + = +=−  Since 122kk m + is constant, acceleration ay− . This satisfies the definition for simple harmonic motion where the angular frequency 122kk m += . Hence the platform and the ball oscillate in simple harmonic motion. M1 M1 (b) (i) At the equilibrium position of the three -springs system, extension of each spring is e. 12 12 2 2 k e k e mg mge kk =+ = − After bottom spring breaks, at the equilibrium position of the two -springs system, extension of each spring above the platform is e’. 1 1 2' ' 2 k e mg mge k = = The amplitude of oscillation of the two-springs system in SHM, M1 M1

6 0'y = distance from top support where spring breaks – distance from top support to equilibrium of two-springs system ( ) ( )00 0 1 2 1 0 1 2 1 '' 22 11 22 y L e y L e mg mg yk k k ymg k k k mg = + + − + = + −− = − + − A0 (ii) Maximum speed of the platform is at the equilibrium position. max 0 01 1 2 1 0 1 1 2 1 '' 2 11 22 112 22 vy yk mgm k k k mg yg mk k k k mg =  = − +   − = − +  − A1 (iii) ( ) 2 max 0 01 1 2 1 0 1 1 2 1 '' 2 11 22 112 22 ay yk mgm k k k mg ykg k k k mg =  = − +  − = − + − Ball will lose contact with the platform if max 0 1 1 2 1 1011 1 2 1 101 12 101 12 112 22 222 122 22 22 1 (shown)2 ag yk g g k k k mg kykk k k k mg kyk k k mg kyk k k mg   − +  − − + − +− +− M1 A0

7 5 (a) The radius of the cone decreases linearly with the length of the conductor. Consider a section of the cone of radius y, and thickness dx, located a distance x from the left end of the cone as shown in the diagram above. The radius y can be expressed as 222 r r ry r x r x LL −= − = −  The resistance across opposite sides of this small section is 22 2 dx dxdR y rrx L == −     Therefore, the total resistance across the truncated cone is 0 2 0 2 2 1 2 11 2 2 L L dxR rrx L L rr rx L L r r r L r =  − = − =− =                  C1 C1 M1 2r r L dx y x

8 (b) ( ) ( ) ( ) ( ) 2 22 2 22 22 2 222 1 1 7Original volume of cone is 2 2 3 3 3 Volume of hollow cylinder is ' 2 ' ' 3 ' Since ' 7 3' 3 7' 9 ' ' 7' 3 272 V r L r L r L V r L r L r L VV r L r L LL L L LR rrrr = − = = − = = = = = = = −            2 2 272 1.937' 14 27 L R r LR r = = =     C1 B1 A1 6 (a) To accelerate the particles, adjacent tubes must have opposite polarities. Their polarity must change continuously at a constant frequency to synchronize with the movement of particles from one tube to the next. This means that the period T of the alternating voltage supply