Assuming m to be the coefficient to kinetic friction between the block and the floor

Question

A bullet of mass ‘m’ travel at a very high velocity ‘V’ (as shown in the figure) and gets embedded inside the block of mass ‘M’ initially at rest on a rough horizontal floor. The block with the bullet is seen to move a distance ‘S’ along the floor. Assuming $\mu$ to be the coefficient to kinetic friction between the block and the floor and ‘g’ the acceleration due to gravity, what is the velocity ‘V’ of the bullet?  

 

      (a) $\frac{M+m}{m}\sqrt{2\mu gs}$                                                  (b) $\frac{M-m}{m}\sqrt{2\mu gs}$ 

 

      (c) $\frac{\mu \left ( M+m \right )}{m}\sqrt{2gs}$                                                (d) $\frac{M}{m}\sqrt{2\mu gs}$

Answer 1

So above is the derivation for the velocity of the bullet according to the problem