Question

A ball is dropped vertically from rest at a height of 12m. After striiking the ground , it bounces back to a height of 9m. What fraction of kinetic energy does it lose on striking the ground?

Answer 1

Here,
Let v = velocity of the ball just before it strikes the ground
We know
v² = 2gh
Multiplying both sides by m
mv² = 2mgh
=> ½mv² = mgh
Now loss in KE = ½mv₁² – ½mv₂²
=> ½mv₁² – ½mv₂² = mgh₁ – mgh₂
Loss in KE = Loss in PE
Potential energy of the ball at height 12 m is
PE₁ = mg12
Potential energy of the ball at height 9 m is
PE₂ = mg9
Loss in energy = PE₁ – PE₂ = mg(12 - 9)
= 3mg
Fraction of KE lost = 3mg / 12mg
= ¼ or 25%