Question

A body is dropped from a height of $4 \,m$ on a surface. If in collision $25\%$ of energy is lost, then the height upto which it will rise after collision is

• A. $3\,m$
• B. $6\,m$
• C. $9\,m$
• D. $12\,m$

Answer 1

Answer: (A)

Let $\upsilon$ be the velocity of the body just before collision with the surface and $\upsilon_{2}$ be the velocity of the body after collision
$\therefore \, \frac{1}{2} m\upsilon_{1}^{2}=mgh_{1} \ldots\left(i\right)$
and $\frac{1}{2}m\upsilon_{2}^{2}=mgh_{2} \ldots\left(ii\right)$
Where $h_{1}$ is the height from where the body is dropped and $h_{2}$ is the height upto which body will rise after collision
Dividing $\left(ii\right)$ by $\left(i\right)$, we get
$\frac{\upsilon_{2}^{2}}{\upsilon_{1}^{2}}=\frac{h_{2}}{h_{1}} \ldots\left(iii\right)$
According to the question,
loss of energy $= 25\%$
$\therefore \frac{1}{2}m\upsilon_{2}^{2}=\left(\frac{75}{100}\right)\frac{1}{2} m\upsilon_{1}^{2}$
$\frac{\upsilon_{2}^{2}}{\upsilon_{1}^{2}}=\frac{75}{100} \ldots\left(iv\right)$
From $\left(iii\right)$ and $\left(iv\right)$, we get
$\frac{h_{2}}{h_{1}}=\frac{75}{100}$
$h_{2}=\frac{3}{4}\times h_{1}$
$\frac{3}{4}\times4=3\,m$
$(\therefore h_{1}=4\,m$ (Given))