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. A third identical block C, also of mass m, moving on the floor with a speed v along the line joining A and B, collides with A see Fig. Then 

Since the spring is massless, the momentum imparted to the system A - B by block C is equally shared between A and B. Since A and B have the same mass, they will have the same velocity; let it be u. From the law of conservation of momentum, we have 

mv = mu + mu = 2mu 

which gives u = v/2. Let x be the maximum compression of the spring. From the law of conservation of energy, we have 

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+\frac{1}{2}m{u}^2+\frac{1}{2}k{x}^2 \\ {v}^2=2u^2+\frac{k{x}^2}{m}=2{\left(\frac{v}{2}\right)}^2+\frac{k{x}^2}{m} \\ \frac{k{x}^2}{m}=\frac{v^2}{2} \\ x=v\sqrt{\frac{m}{2k}} \left(i\right) \end{gathered}$$

The kinetic energy is equally shared between masses A and B and the spring is at maximum compression,

i.e.,  $\frac{1}{2}k{x}^2=\frac{1}{2}m{u}^2+\frac{1}{2}m{u}^2=m{u}^2$
Thus, kinetic energy of A-B at maximum compression $m{u}^2=m{\left(\frac{v}{2}\right)}^2=\frac{m{v}^2}{4}.$

Using (i) notice that $\frac{1}{2}k{x}^2=\frac{m{v}^2}{4}$.