Question

A body is dropped on ground from a height ‘h₁’ and after hitting the ground, it rebounds to a height ‘h₂’ If the ratio of velocities of the body just before and after hitting ground is \(4\). then percentage loss in kinetic energy of the body is\(\frac{x}{4}\). The value of x is_____.

Answer 1

Correct Answer: 375

Let \(u\) and \(v\) be the speeds, just before and after the body strikes the ground.
Given \( \frac{u}{v} = 4 \).
We know that the loss in kinetic energy is: \[ \Delta K.E. = \frac{1}{2} m u^2 - \frac{1}{2} m v^2 \] Thus, \[ \Delta K.E. = \frac{1}{2} m \left( u^2 - v^2 \right) \] Substitute the values: \[ \Delta K.E. = \frac{1}{2} m \left( \left( 4v \right)^2 - v^2 \right) \] \[ = \frac{1}{2} m \left( 16v^2 - v^2 \right) = \frac{1}{2} m \times 15v^2 \] Thus, the percentage loss is: \[ \frac{15}{16} \times 100 = 93\% \] Therefore, the percentage loss in kinetic energy is 93\%.