A bullet of mass $‘m^{\prime }$, moving with a speed $‘v^{\prime }$ strikes a wooden block of mass $M$ kept at rest and gets embedded in it. The speed of this embedded block will be:

The correct option is C $\left(\frac{m}{M+m}\right)v$

Let the velocity of the embedded block be $V$.
Given: the mass of bullet is $m$, and mass of the wooden block is $M$, initial velocity of the bullet be $v$.
Using law conservation of momentum:
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
Initially the block is at rest. So $u_1=0m{s}^{-1}$
$\therefore$ Total momentum before impact $=mv$
Finally the bullet is embedded in the block
$\therefore$Total momentum after impact $=(M+m)V$
Now, applying conservation of momentum,
$mv=(M+m)V$
$\therefore V=\left(\frac{m}{M+m}\right)v$