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. Then from the principle of conservation of momentum, we have
Mv = (M + m)V
or $V=\frac{mv}{(M+m)}$ (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
$\frac{1}{2}(M+m)V^2=(M+m)gh$
or $V=\sqrt{2gh}$ (ii)
Using (ii) in (i), we get
$\sqrt{2gh}=\frac{mv}{(M+m)}$
Squaring this equation, we find that h is given correctly by choice (a).