Answer
$4$ $m/s$
Work Step by Step
The momentum of the system stays constant at $16$ $kg*m/s$. We can write the equation of motion as $0.5*4(2+x)^2+0.5*4(2-x)^2-0.5*8*2^2 = 16$. When this is simplified, it yields $2(8x)-16 = 16$, so $x=2$.
Therefore, the forward part moves at $4$ $m/s$
