A cart of mass 4m holds a block of mass m which is attached to the former by means for spring of spring constant K, as shown in the figure. All surfaces are frictionless. Now a sharp impulse J is given to block m as shown. The maximum compression of spring will be  $\frac{2J}{\sqrt{xmk}}$. The  value of $x$ is 

Impulse =change in momentum=J
Initially mass m is at rest . 
$\therefore$ it will gain the momentum J.
At maximum compression the block and car will move with same velocity and momentum J. 
$\therefore$ KE of system at this instant
$=\frac{J^2}{2\left(5m\right)}$
Initial KE=  $\frac{J^2}{2m}$
$\therefore$ From conservation of energy 
$\begin{array}{l}\frac{J^2}{2m}=\frac{1}{2}k{x}^2+\frac{J^2}{2\left(5m\right)}\Rightarrow \frac{1}{2}k{x}^2=\frac{4J^2}{2\left(5m\right)}\\ orx=\sqrt{\frac{4J^2}{5mk}}=\frac{2J}{\sqrt{5mk}}\end{array}$