JPJC 2026 Motion and Forces Assignment Solutions

Kinematics Assignment soln 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS/8867 H1 PHYSICS MOTION AND FORCES ASSIGNMENT SOLUTIONS Qn. Solution Mark Allocation 1(a) Displacement-time graph Due to energy loss on hitting the floor, the maximum height to which the ball rises decreases with each bounce. Velocity-time graph The velocity of the ball is negative on its downward motion, positive on its upward motion, and is zero momentarily at maximum height after each bounce. [1] for correct shape of graph [1] for correct shape of graph (i.e. saw-tooth with decreasing height) [1] for same area for adjacent triangles in the same bounce [1] for correct shape of graph [1] for horizontal straight line at −9.81 s / m same area t / s t / s v / m s−1 t3 t2 t1 a/ ms-2 t / s t1 t2 t3

Kinematics Assignment soln 2 Acceleration-time graph The graph is mainly a horizontal straight line for most parts of the motion where the ball is in free-fall. This acceleration of free fall is directed downwards, hence it is negative. However, upon impact with the floor, the change in direction of motion of ball within a short time interval causes the ball to experience a large acceleration directed upwards and is hence positive, which appear as the spikes (i.e. assumed that rebound happened immediately upon impact, and impact time → 0). 1(b) The initial height of the ball, which is the magnitude of the displacement prior to the first bounce, can be found from the area under the velocity -time grap h (refer to the shaded region). [1] for correct description 2 (a) Momentum of the ball, 1 0.16 35 5.6 kg m s (to the left) p mv − = = = [1] (b) (i) Change in momentum p = −1(280)(0.12 0.04)2 [1] −= 111.2 kg m s [1] (ii) Average force 11.2 140 N0.12 0.04 pF t = = =− [1] (iii) The ball approaches the racket from the right. Take rightward as positive: ( )f11.2 0.16 35 p m v v  =  = − − [1] 1 f 35 m sv −= [1]