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)

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

Answer 2

Given that, Bullet of $\operatorname{mass}= m$ 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, $( m + M ) V = mv + Mu$ $V =\frac{ mv }{( m + M )} \ldots ( I )$ As block will rise to a height $h$ Potential energy of combination = Kinetic energy of the combination $( m + M ) gh =\frac{1}{2}( m + M ) V ^{2}$ $2 gh =\left(\frac{ mv }{ m + M }\right)^{2}$ $v =\frac{ m + M }{ m } \sqrt{2 gh }$ Hence, the initial velocity $v$ is $\frac{ m + M }{ m } \sqrt{2 gh }$