Question

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

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

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
$\therefore \frac{1}{2}m{\upsilon }_1^2=mg{h}_1\dots \left(i\right)$
and $\frac{1}{2}m{\upsilon }_2^2=mg{h}_2\dots \left(ii\right)$
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
$\frac{{\upsilon }_2^2}{{\upsilon }_1^2}=\frac{h_2}{h_1}\dots \left(iii\right)$
According to the question,
loss of energy $=25\%$
$\therefore \frac{1}{2}m{\upsilon }_2^2=\left(\frac{75}{100}\right)\frac{1}{2}m{\upsilon }_1^2$
$\frac{{\upsilon }_2^2}{{\upsilon }_1^2}=\frac{75}{100}\dots \left(iv\right)$
From $\left(iii\right)$ and $\left(iv\right)$, we get
$\frac{h_2}{h_1}=\frac{75}{100}$
$h_2=\frac{3}{4}\times h_1$
$\frac{3}{4}\times 4=3m$
$(\therefore h_1=4m$ (Given))