Two blocks $A$ and $B$ each of mass $m$, are connected by a massless spring of natural length $L$and spring constant $k$. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in fig. A third identical block $C,$ also of mass$m,$moves on the floor with a speed $v$ along the line joining $A$ and $B$, and collides elastically with$\mathrm{A}$. Then : 

After collision between $C$ and $A$, $C$ stops while $A$ moves with speed of $C$ i.e.,$v$ [in head on elastic collision, two equal masses exchange their velocities]. At maximum compression, $A$ and $B$ will move with same speed $\frac{v}{2}$(From conservation of linear momentum).

Let$x$be the maximum compression in this position. $KE$ of $A – B$system at maximum compression

$=\frac{1}{2}\left(2m\right){\left(\frac{v}{2}\right)}^2$ or $K_{max}=\frac{m{v}^2}{4}$

From conservation of mechanical energy in two positions shown in figure.

or $\frac{1}{2}m{v}^2=\frac{1}{4}m{v}^2+\frac{1}{2}K{x}^2$

$\frac{1}{2}K{x}^2=\frac{1}{4}m{v}^2; x = v\sqrt{\frac{m}{2K}}$