Question

A bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a $S$ height the initial velocity of the block will be

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

Answer 1

Answer: (D)

Given that, Bullet of $\mathrm{mass}=m$
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,
$(m+M)V=mv+Mu$
$V=\frac{mv}{(m+M)}\dots (I)$
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 $\frac{m+M}{m}\sqrt{2gh}$