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) 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]

(a) Given,

• Mass of the ball (m) = 20 g = 0.02 kg
• Initial height (h₁) = 45 m
• Final height (h₂) = 40 m
• g = 10 m s^-2

(a) Initial potential energy of the ball is given by,

PE = mgh₁
= 0.02 x 10 x 45
= 9 J

So, the initial potential energy is 9 J.

(b) Due to law of conservation of mechanical energy, just before hitting the ground, the initial potential energy converts into kinetic energy.

Let speed of the ball be v.

So,

$\text {Kinetic energy} = 9 \\[1em] \Rightarrow \dfrac{1}{2} \text {mv}^2 = 9 \\[1em] \Rightarrow \dfrac{1}{2} \times 0.02\times \text {v}^2 = 9 \\[1em] \Rightarrow 0.01\times \text {v}^2 = 9 \\[1em] \Rightarrow \text {v}^2 = \dfrac{9}{0.01} \\[1em] \Rightarrow \text {v}^2 = 900 \\[1em] \Rightarrow \text {v} = \sqrt {900} \\[1em] \Rightarrow \text {v} = 30 \text { m s}^{-1}$

So, the speed of the ball on hitting the ground = 30 m s^-1.

(c) Final kinetic energy after rebound is equal to potential energy at 40 m which is given by,

Final kinetic energy = mgh₂
= 0.02 x 10 x 40
= 8 J

Loss in KE = Initial kinetic energy - Final kinetic energy = 9 − 8 = 1 J

So, the loss in kinetic energy = 1 J.