Set up the physical situation

• Blocks $A$ and $B$ are identical, each of mass $m$, connected by a light spring of spring constant $K$.
• Block $C$ of mass $m$ moves with speed $v$ and collides with $A$.
• The floor is smooth, so there is no external horizontal force.

What happens just after collision?

Since the collision is between identical masses and is along the line of motion, we take it as an elastic head-on collision of equal masses.

Therefore, immediately after collision:

• Block $C$ comes to rest.
• Block $A$ moves with speed $v$.
• Block $B$ is still at rest.

So now we only need to study the two-block spring system $A$-$B$ with initial velocities: $u_A=v, \qquad u_B=0$

Velocity of center of mass of $A$ and $B$

Total momentum of $A+B$ just after collision is $p = mv + 0 = mv$

Total mass of $A+B$ is $2m$, so center of mass speed is $V_{cm} = \frac{mv}{2m} = \frac{v}{2}$

Condition for maximum compression

Maximum compression occurs when the relative velocity of $A$ and $B$ becomes zero, i.e. both blocks move with the same velocity. At that instant both move with the center of mass velocity: $u = \frac{v}{2}$

Use conservation of energy

Just after collision, spring is uncompressed, so total energy of system $A+B$ is only kinetic: $E_i = \frac12 mv^2$

At maximum compression $x$, both blocks move with speed $v/2$, so kinetic energy is $E_k = 2\left(\frac12 m\left(\frac v2\right)^2\right)= m\frac{v^2}{4}$

Hence energy stored in spring at maximum compression is $\frac12 Kx^2 = E_i - E_k$ $\frac12 Kx^2 = \frac12 mv^2 - \frac14 mv^2 = \frac14 mv^2$

Solve for $x$

$Kx^2 = \frac12 mv^2$ $x^2 = \frac{mv^2}{2K}$ $x = v\sqrt{\frac{m}{2K}}$

Match with options

This corresponds to: $\boxed{\text{D: } v\sqrt{\frac{m}{2K}}}$