Question

A bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to height $h$, then the initial velocity $v$ the bullet must have been

• A. $\sqrt{2gh}$
• B. $\frac{M+m}{2m}\sqrt{2gh}$
• C. $\frac{m}{M+m}\sqrt{2gh}$
• D. $\frac{M+m}{m}\sqrt{2gh}$

Answer 1

The correct option is D $\frac{M+m}{m}\sqrt{2gh}$

Given that,

Bullet of mass $=m$

Block of mass $=M$

Velocity $=v$

As the bullet comes to rest with respect to the block, the two behaves as one body.

Let V be the velocity of the combination

Applying the conservation linear momentum,

$(m+M)V=mv+Mu$

$V=\frac{mv}{(m+M)}.....\left(I\right)$

As block will rise to a height h

Potential energy of combination = Kinetic energy of the combination

$(m+M)gh=\frac{1}{2}(m+M)V^2$

$2gh={\left(\frac{mv}{m+M}\right)}^2$

$v=\frac{m+M}{m}\sqrt{2gh}$

Hence, the initial velocity $v$ is $\frac{m+M}{m}\sqrt{2gh}$