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: Understanding the Concept:
Gravitational Potential Energy (P.E.) is the energy possessed by a body due to its position above the ground.
Step 2: Key Formula or Approach:
\(P.E. = mgh\)
Step 3: Detailed Explanation:
Given:
Mass, \(m = 20 \text{ g} = \frac{20}{1000} \text{ kg} = 0.02 \text{ kg}\)
Height, \(h = 45 \text{ m}\)
Acceleration due to gravity, \(g = 10 \text{ m/s}^{2}\)
\[P.E. = 0.02 \times 10 \times 45\]
\[P.E. = 0.2 \times 45 = 9 \text{ J}\]
Step 4: Final Answer:
The initial potential energy of the ball is 9 J.
(b)
Step 1: Understanding the Concept:
According to the law of conservation of energy, the potential energy at the top is converted into kinetic energy at the bottom (neglecting air resistance).
Step 2: Key Formula or Approach:
\(mgh = \frac{1}{2}mv^{2}\) or \(v = \sqrt{2gh}\)
Step 3: Detailed Explanation:
Initial Height, \(h = 45 \text{ m}\)
Gravity, \(g = 10 \text{ m/s}^{2}\)
\[v = \sqrt{2 \times 10 \times 45}\]
\[v = \sqrt{900}\]
\[v = 30 \text{ m/s}\]
Step 4: Final Answer:
The speed of the ball when it hits the ground is 30 m/s.
(c)
Step 1: Understanding the Concept:
The loss in kinetic energy upon collision is the difference between the energy just before impact and the energy just after rebound.
Step 2: Key Formula or Approach:
Loss in Energy = Initial energy - Final energy after rebound.
Step 3: Detailed Explanation:
1. Total initial energy = 9 J (from part a).
2. Potential energy at rebound height (40 m):
\(P.E._{\text{rebound}} = mgh_{2} = 0.02 \times 10 \times 40 = 8 \text{ J}\).
3. Since it reaches 40 m, its kinetic energy just after rebound must have been 8 J.
4. Loss in energy = \(9 \text{ J} - 8 \text{ J} = 1 \text{ J}\).
Step 4: Final Answer:
The loss in kinetic energy on striking the ground is 1 J.