A ball is dropped from a height of 10 m.It strikes the ground and rebounds up to a height of 2.5 m. During the collision, the per cent loss in the kinetic energy is:
As per conservation of energy and momentum,
KEinitial = PEinitial=mghinitial
Similarly,
KEfinal = PEfinal = mghfinal
Percentage loss in KE =Â $$\frac{(mgh_{initial}-mgh_{final})}{mgh_{initial}}$$
= $$ \frac{h_{initial}-h_{final}}{h_{initial}}$$
= $$\frac{10-2.5}{ 10 }\times100$$
=75%
Option D is correct.
Create a FREE account and get:
