Two identical blocks A and B, each of mass m resting on smooth floor are connected by a light spring of natural length L and spring constant K, with the spring at its natural length. A third identical block ‘C’ (mass m) moving with a speed v along the line joining A and B collides with A elastically. the maximum compression in the spring is

Initial momentum of the system = mv. After colliding with ‘A’, the block ‘C’ comes to rest and now both block A and ‘B’ moves with a velocity ‘V’. When compression in spring is maximum. By law of conservation of linear momentum

$mv=(m+m)V\Rightarrow V=\frac{v}{2}$

By using law of conservation of energy

K.E of block ‘c’ = K.E of system + P.E of system

$\frac{1}{2}m{v}^2=\frac{1}{2}\left(2m\right)V^2+\frac{1}{2}K{x}^2\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}\left(2m\right){\left(\frac{V}{2}\right)}^2+\frac{1}{2}K{x}^2\Rightarrow K{x}^2=\frac{1}{2}m{v}^2$

$x=v\sqrt{\frac{m}{2K}}$