Question

A body falling from a height of 10m rebounds from hard floor . if it losses 20%energy in the imapct, then coefficient of restitutiion is

Answer 1

coefficient of restitution is 0.89
As 20% energy is lost during the collision,
mgh'/mgh = 80/100
h'/h = 0.8
coeff. of restitution e =√(h'/h) = √0.8 = 0.894

Answer 2

Answer:

Explanation:

Height of the body = h = 10m, 

PE = mgh = 10mg

KE = 0, and the total energy (TE) = 10mg

efore impact,

PE = 0

KE = TE - PE = 10mg

Energy lost after impact = 20%

Thus, energy lost = 20% of 10mg = 2mg

Total Energy Left = 10 - 2 = 8mg

Let the maximum height after impact = H

At height H, KE = 0 (as velocity is 0 at maximum height)

PE = mgH, but since PE = TE = 8mg

Therefore,  mgH = 8 mg

As 20% energy lost in collision therefore, coefficient -

e = √h2/ h1 = √0.8

= 0.89