Question

A bullet of mass $m$ hits a target of mass $M$ hanging by a string and gets embedded in it. If the block rises to height $h$ as a result of this collision, the velocity of the bullet before collision is :

• A. $v=\sqrt{2gh}$
• B. $v=\sqrt{2gh}[1+\frac{m}{M}]$
• C. $v=\sqrt{2gh}[1+\frac{M}{m}]$
• D. $v=\sqrt{2gh}[1-\frac{m}{M}]$

Answer 1

The correct option is C $v=\sqrt{2gh}[1+\frac{M}{m}]$

Let $v^{{}^{\prime }}$ is the velocity of ($M+m$) after collision. Then from conservation of linear momentum in the horizontal direction,
$mv=(M+m)v^{{}^{\prime }}$.......(1)
Both the bullet and the block have the same velocity after collision since the bullet gets embedded in the block
Since the block+bullet rise to a height $h$, from conservation of energy,

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

$\therefore v^{\prime }=\sqrt{2gh}$.....(2)
$\therefore$ by putting $v^{\prime }$ from eqn (2) in eqn (1), we get:

$v=\left(\frac{M+m}{m}\right)\sqrt{2gh}=\sqrt{2gh}(1+\frac{M}{m})$