c
(c)When block of mass \(M\)  collides with the spring its kinetic energy gets converted into elastic potential energy of the spring.

From the law of conservation of energy \(\frac{1}{2}M{v^2} = \frac{1}{2}K{L^2}\)

\(\therefore \) \(v = \sqrt {\frac{K}{M}} L\)

Where \( v\) is the velocity of block by which it collides with spring.

So, its maximum momentum \(P = Mv = M\sqrt {\frac{K}{M}} \,L\) = \(\sqrt {MK} \,L\)

After collision the block will rebound with same linear momentum.