Ans.

The masses of the bullet and the block of wood are \( m \) and \( M \) respectively. The velocity of the bullet before collision is \( v \). Since the wood block is at rest, so the velocity of the wood block before collision is Zero.

After collision the bullet becomes embedded in the wood block.

(i) Subsequent velocity of the system:

Let \( V \) be the velocity of the wood block with bullet after collision.

According to the conservation of linear momentum,

\( mv=(M+m)V \)

or, \( \displaystyle{V=\frac{m}{m+M}v} \)

(ii) Loss in the kinetic energy:

The loss in the kinetic energy is

Total kinetic energy before collision – Total kinetic energy after collision

= \( \displaystyle{\frac{1}{2}mv^2-\frac{1}{2}(m+M)V^2} \)

= \( \displaystyle{\frac{1}{2}mv^2-\frac{1}{2}(m+M)\frac{m^2}{{(m+M)}^2}v^2} \), [putting the value of V ]

= \( \displaystyle{\frac{1}{2}mv^2\left(1-\frac{m}{m+M}\right)} \)

= \( \displaystyle{\frac{1}{2}\frac{mM}{m+M}v^2} \)