Question 53

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:

Solution

As per conservation of energy and momentum,

KEinitial = PEinitial=mghinitial

Similarly,

KEfinal = PEfinal = mghfinal

Percentage loss in KE = $$\frac{(mgh_{initial}-mgh_{final})}{mgh_{initial}}$$

= $$ \frac{h_{initial}-h_{final}}{h_{initial}}$$

= $$\frac{10-2.5}{ 10 }\times100$$

=75%

Option D is correct.

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