Question

A bullet of mass $m$ moving horizontally with a velocity $ v $ strikes a block of wood of mass $ M $ and gets embedded in the block. The block is suspended from the ceiling by a massless string. The height to which block rises is

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

Answer 1

Answer: (A)

The situation is as shown in the figure.

Let $ V $ be velocity of the block-bullet system just after collision. Then by the law of conservation of linear momentum, we get
$ mv = (m + M)V $
$ V = \frac{mv}{m+M} $
Let the block rises to a height $ h $ .
According to law of conservation of mechanical energy, we get
$ \frac{1}{2}\left(m+M\right)V^{2} = \left(m+M\right)gh $
$ h = \frac{V^{2}}{2g} = \frac{v^{2}}{2g}\left(\frac{m}{m+M}\right)^{2} $