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)

$3\,m$

Answer 2

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))