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:

1. the initial potential energy of the ball.
2. the speed of the ball at which it hits the ground.
3. the loss in kinetic energy on striking the ground.

[g = 10 m/s²]

Answer 1

Given: Mass of the ball (m) = 20 g = 0.02 kg

Initial height (h₁) = 45 m

Final height (h₂) = 40 m

g = 10 m/s²

a. Initial potential energy of the ball is given by,

PE = mgh₁

= 0.02 × 10 × 45

= 9 J

So, the initial potential energy is 9 J.

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

Let the speed of the ball be v.

So,

Kinetic energy = 9

⇒ `1/2 mv^2` = 9

⇒ `1/2 xx 0.02 xx v^2` = 9

⇒ 0.01 × v² = 9

⇒ v² = `9/0.01`

⇒ v² = 900

⇒ v = `sqrt 900`

⇒ v = 30 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 × 10 × 40

= 8 J

Loss in KE = Initial kinetic energy − Final kinetic energy

= 9 − 8

= 1 J

So, the loss in kinetic energy is 1 J.