A ball is released from rest above a horizontal surface. It strikes the surface and

Q#20 (Past Exam Paper – November 2015 Paper 12 Q9)

A ball is released from rest above a horizontal surface. It strikes the surface and bounces several times.

The velocity-time graph for the first two bounces is shown.

What is the maximum height of the ball after the first bounce?

A 0.20 m                     B 0.25 m                     C 0.45 m                     D 0.65 m

Solution:

Answer: A.

Initially, the ball is at rest above the horizontal surface. In the graph, this is the point (0, 0).

As the ball is released, it falls down and its velocity increases until its velocity rises to + 3.00 m s^-1 in the graph. In the graph, this motion is shown as positive. So, the downward direction is taken as positive. When the velocity is positive, the ball is moving downwards.

As the ball strikes the ball, its velocity decreases instantly to zero (at time = 0.30 s). The ball then bounces and starts to move upwards with a velocity of – 2.00 m s^-1. The downward motion is taken as positive and so, when the ball rises up, its velocity is negative.

As the ball moves upwards, the force of gravity (which is downwards) acts against its upward motion, decreasing its velocity until it becomes zero. This is the maximum height reached (at time = 0.50 s).

The area under a velocity-time graph gives the displacement. So, to obtain the maximum height reached we calculate the area under the graph between t = 0.30 s and t = 0.50 s.

Maximum height = area under graph

Maximum height = ½ × (0.50 – 0.30) × 2.00 = 0.20 m