Question

A block of m falls vertically on a horizontal floor of coefficent of restitution e = 0.8. The fraction of energy lost during collision of

Answer 1

energy fraction retained = e^2 = .64
hence energy lost = .36 pr 36%

Answer 2

It is interesting and dynamic question , thanks for asking .

A block of mass m falls vertically on a horizontal floor .
it means initial velocity of block , u = 0
Let distance between block and horizontal floor = H
Then, velocity of block before striking , v₁ = -√{2gH} [ negative sign shows body moves downward]
Now, after strikes block moves upward , let it reaches h height .
then, velocity of block is required to reach h height from the floor is v₂ = √{2gh}

∵ Coefficient of restitution = - velocity of separation/velocity of approach
⇒0.8 = -(-√{2gh}/√{2gH}
⇒ 0.8 = √{2gh}/√{2gH}
Squaring both sides,
⇒0.64 = h/H

∵ % loss in energy = $\bold{\frac{mgh_i-mghf}{mgh_i}\times100}$
Here , hi = H and hf = h
∴ % loss in energy = (H - h)/H × 100
= {1 - h/H} × 100
= (1 - 0.64) × 100
= 0.36 × 100
= 36 %