Question

A block of wood of mass M is suspended by means of a thread. A bullet of mass m is fired horizontally into the block with a velocity v. As a result of the impact, the bullet is embedded in the block. The block will rise to vertical height given by

• A. $\frac{1}{2g}{\left(\frac{mv}{M+m}\right)}^2$
• B. $\frac{1}{2g}{\left(\frac{mv}{M-m}\right)}^2$
• C. $\frac{1}{2g}\frac{m{v}^2}{(M+m)}$
• D. $\frac{1}{2g}\frac{m{v}^2}{(M-m)}$

Answer 1

The correct option is A

$\frac{1}{2g}{\left(\frac{mv}{M+m}\right)}^2$

Let V be the velocity of the block with the bullet embedded in it at the time of impact. From the principle of conservation of momentum, we have
$mv=(M+m)V$
$\Rightarrow V=\frac{mv}{(M+m)}$ .... (i)
If the block, with the bullet embedded in it, rises to a vertical height h, then from the principle of conservation of energy, we have
$\frac{1}{2}(M+m)V^2=(M+m)gh$
$\Rightarrow V=\sqrt{2gh}$
Using (i) and (ii), we get
$\sqrt{2gh}=\frac{mv}{(M+m)}$
$\Rightarrow h=\frac{1}{2g}{\left(\frac{mv}{M+m}\right)}^2$
Hence, the correct choice is (a).