Question

A ball of mass 20 g falls from a height of 45 m. It rebounds from the ground to a height of 40 m. Calculate:
(a) the initial potential energy of the ball.
(b) the speed of the ball at which it hits the ground.
(c) the loss in kinetic energy on striking the ground.
[g = 10 m/s\(^2\)]

Hint:

The energy lost during an inelastic collision (like a ball bouncing) is the difference between the potential energies at the initial height and the rebound height: \( \Delta E_{loss} = mg(h_{initial} - h_{rebound}) \).

Answer 1

Step 1: List the given data and convert units:
- Mass (\(m\)) = 20 g = 0.02 kg - Initial height (\(h_1\)) = 45 m - Rebound height (\(h_2\)) = 40 m - Acceleration due to gravity (\(g\)) = 10 m/s\(^2\)
(a) Initial Potential Energy:
The potential energy (PE) of an object is given by the formula \(PE = mgh\). \[ PE_{initial} = m \times g \times h_1 = 0.02 \text{ kg} \times 10 \text{ m/s}^2 \times 45 \text{ m} \] \[ PE_{initial} = 9 \text{ J} \] (b) Speed on hitting the ground:
By the principle of conservation of energy (assuming no air resistance), the initial potential energy is completely converted into kinetic energy (KE) just before the ball hits the ground. \[ KE_{before} = PE_{initial} \] \[ \frac{1}{2}mv^2 = 9 \text{ J} \] \[ v^2 = \frac{2 \times 9}{m} = \frac{18}{0.02} = 900 \] \[ v = \sqrt{900} = 30 \text{ m/s} \] (c) Loss in kinetic energy:
The loss in kinetic energy during the collision is the difference between the kinetic energy just before hitting the ground and the kinetic energy just after rebounding. - Kinetic energy just before impact: \(KE_{before} = PE_{initial} = 9\) J. - The kinetic energy just after rebound is equal to the potential energy it gains on reaching the rebound height \(h_2\). \[ KE_{after} = PE_{rebound} = m \times g \times h_2 = 0.02 \times 10 \times 40 = 8 \text{ J} \] - The loss in kinetic energy is: \[ \Delta KE_{loss} = KE_{before} - KE_{after} = 9 \text{ J} - 8 \text{ J} = 1 \text{ J} \]