Work done by Diff type of forces

Question

A ball is thrown vertically downwards from a height of 20 m with an initial velocity ${\nu}_{0}$. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity ${\mathrm{\nu}}_{0}$ is (Take g = $10{\mathrm{ms}}^{-2}$ )

Moderate

Solution

The situation is shown in the figure

Let $\nu $ be the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy,

Gain in kinetic energy = loss in potential energy

$\text{i.e.}\frac{1}{2}m{v}^{2}-\frac{1}{2}m{v}_{0}^{2}=mgh$

$\text{(where}m\text{is the mass of the ball)}$

$\text{or}{v}^{2}-{v}_{0}^{2}=2gh....\left(i\right)$

Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height h.

$\therefore \frac{50}{100}\left(\frac{1}{2}m{v}^{2}\right)=mgh$

$\frac{1}{4}{v}^{2}=gh\text{or}{v}^{2}=4gh$

$\text{Substituting this value of}{v}^{2}\text{in eqn. (i), we get}$

$4gh-{v}_{0}^{2}=2gh$

$\text{or}{v}_{0}^{2}=4gh-2gh=2gh\text{or}{v}_{0}=\sqrt{2gh}$

$\text{Here,}g=10{\mathrm{ms}}^{-2}\text{and}h=20\mathrm{m}$

$\therefore {v}_{0}=\sqrt{2\left(10{\mathrm{ms}}^{-2}\right)(20\mathrm{m})}=20{\mathrm{ms}}^{-1}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Download Now