Question

A bullet of mass $m$ moving vertically upwards instantaneously with a velocity $u$, hits the hanging block of mass $m$ and gets embedded in it, as shown in the figure. The height through which the block rises after the collision, is, (Assume that there is sufficient space above the block for the block to rise.)

• A. $u^2/2g$
• B. $u^2/g$
• C. $u^2/8g$
• D. $u^2/4g$

Answer 1

The correct option is C $u^2/8g$

Let the velocity of block and bullet system be $v$, after bullet embedded into block.
Applying conservation of momentum in vertical direction.

$mu+m\times 0=2mv$
$\Rightarrow v=\frac{u}{2}$
since string slackens after collision, the combined mass will move upward under the influence of gravity.

$\Rightarrow h_{max}=\frac{v^2}{2g}=\frac{(u/2{)}^2}{2g}=\frac{u^2}{8g}$

Hence, $\left(C\right)$ is the correct answer.