RI Y5 Remedial Dynamics.2 Soln

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 5 Chapter 3: Dynamics (Momentum and Collisions) Solutions Worked Example 1 B       -1 -1 3 -1 -1 change in momentum, area under - graph 11 84 1 2 2 1 222 60 kg m s Change in velocity, 60 200 m s300 10 Since 0 m s , 200 200 m s pF t pv m u vu v             ANS: B 1 A Subtract initial speed instead of adding it to the change in speed B Forget to add the initial speed C       -1 -1 3 -1 -1 change in momentum, area under - graph 11 84 1 2 2 1 222 60 kg m s Change in velocity, 60 300 m s200 10 Since 50 m s , 300 350 m s pF t pv m u vu v             D Add both areas under graph ANS: C 2 A Did not account for change in direction and just take final speed minus initial speed as the change in velocity B    taking direction away from wall as positive, Impulse on ball by the wall, 0.080 18 23 3.3 N s pm v   

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 6 C Did not account for change in direction and just take final speed minus initial speed as the change in velocity and wrong direction D Wrong direction ANS: B Worked Example 2 A Take total final momentum as the initial momentum of Y. B Add both final momentum without considering the direction and equate it as the initial momentum of Y. C Take rightward direction as positive. By Principle of Conservation of Momentum, Total initial momentum = Total final momentum 22 + py = 2 + 10 py = 14 N s Magnitude of momentum of Y = 14 N s D Take initial momentum of X as the initial momentum of Y. ANS: C 3 C By Principle of Conservation of momentum Momentum of system before collision Momentum of system after collision Momentum of system after collision P and Q are identical (having same mass) momentum p       2 2 2 22 2 2 of Q after collision 2 11Intial kinetic energy, 22 2 11Kinetic energy of system after collision 22 2 2 2 2 11Kinetic energy of Q after collision 24 2 4 p mv pEm v mm pp Emm p p Emm           ANS: C p R (Before) (After) m, E m Q p 2m

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 7 4 (a) Impulse exerted on the heavier truck by the lighter truck = change in momentum of the heavier truck = M (V – U) = (66 000) (2.00 – 1.00) = + 66 000 N s (towards the right) M1 A1 (b) Total KE before collision = ½ (22 000)(3.00)2 + ½ (66 000) (1.00)2 = 132 000 J Total KE after collision = 0 + ½ (66 000) (2.00)2 = 132 000 J Since Total KE before and after collision are the same, the collision is elastic. OR Since relative speed of approach = relative speed of separation = 2.00 m s1, the collision is elastic. M1 A1 (c) At that instant, some of the kinetic energy has been converted into elastic potential energy in the springs. B1 5 (a) By Principle of Conservation of momentum, 3( ) ( 3) 0.5 Speed 0.5 C C mu m u m m v vu u      M1 A1 (b) Assume after the interaction: 12 12 21 21 1 2 1 Conservation of momentum: 3( ) 3 2 3 ----- (1) Elastic collision: () 2 ----- (2) Solving (1) and (2) 2 (2 travelling to the left), 0 Speeds are mu m u mv mv uv v uu v v uv v vu u v v              22 and 0.uv  M1 M1 A2 (c) (d) 1 mark for shape of curve 1 mark for labelling v c correctly 1 mark of labelling T correctly B3 + + v1 v2 u u velocity time vc T 2u Tritium