1. Given

• Mass of block $= M$
• Spring constant $= k$
• Maximum compression of spring $= L$
• Surface is frictionless

We need the maximum momentum of the block after collision.

2. Use conservation of mechanical energy

At maximum compression, the block momentarily comes to rest, so all its kinetic energy gets stored in the spring.

If the speed of the block just before touching the spring (and hence just after leaving it) is $v$, then

$\frac{1}{2}Mv^2 = \frac{1}{2}kL^2$

So,

$Mv^2 = kL^2$

$v = L\sqrt{\frac{k}{M}}$

3. Find momentum

Momentum of the block is

$p = Mv$

Substituting $v$:

$p = M \cdot L\sqrt{\frac{k}{M}}$

$p = L\sqrt{Mk}$

4. Why is this the maximum momentum after collision?

After the block leaves the spring, the spring force becomes zero and the block moves with constant speed on the frictionless surface. Its largest momentum magnitude is therefore the same as its momentum when it separates from the spring:

$p_{\max} = L\sqrt{Mk}$

5. Check options

• A: $\dfrac{kL^2}{2M}$ → this has units of energy per mass, not momentum
• B: $\sqrt{Mk}\,L$ → correct
• C: $\dfrac{ML^2}{k}$ → wrong dimensions
• D: Zero → incorrect

6. Final answer

$\boxed{\sqrt{Mk}\,L}$

So the correct option is B.