Problem source

A small bullet of mass $m$ is fired into a block of wood of mass $M$ which is
at rest. 
The speed
of the bullet on entering the block is $u$.  Its trajectory 
within the block is a horizontal
straight line and  the resistance to the 
bullet's motion is $R$, which is constant.

\begin{questionparts}
\item The block is fixed. The bullet travels a distance
$a$ inside the block before coming to rest.
Find an expression for $a$ in terms of $m$, $u$ and $R$.

\item Instead, the block is free to move on a smooth horizontal 
table. The bullet travels a distance $b$ inside the block
before coming to rest relative to the block, at which time the 
block has moved a distance $c$ on the table. 
Find expressions for $b$ and $c$ in terms of $M$, $m$ and $a$.
 \end{questionparts}

Solution source

\begin{questionparts}
\item Since $R$ is constant, $F=ma \Rightarrow \text{acc} = \frac{R}{m}$ and $v^2 = u^2 + 2as$ so $0 = u^2 - 2 \frac{R}{m}a$, ie $a = \frac{m u^2}{2R}$

\item By conservation of momentum, the bullet/block combination will eventually be travelling at $v = \frac{m}{m+M}u$.

The bullet will slow down to this speed in a time of $\frac{m}{m+M}u = u - \frac{R}{m} T \Rightarrow T = \frac{Mm}{R(m+M)}u$ and will travel $b+c = \frac{\left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)u^2m}{2R}= \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a$.

The block will travell $c = \frac12 \frac{R}{M} \frac{M^2m^2u^2}{R^2(m+M)^2} = \frac{Mm}{(m+M)^2}a$

Therefore the $b = \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a -  \frac{Mm}{(m+M)^2}a = \frac{M}{M+m}a$

\end{questionparts}

Work done by friction is the relative gain in KE for the block. ie $R \cdot c = \frac12 M \left (  \frac{m}{m+M}u\right)^2 \Rightarrow c = \frac{Mm}{(M+m)^2}a$.