Two identical blocks $$A$$ and $$B$$ each of mass $$m$$ resting on the smooth horizontal floor are connected by a light spring of natural length $$L$$ and spring constant $$K$$. A third block $$C$$ of mass $$m$$ moving with a speed $$v$$ along the line joining $$A$$ and $$B$$ collides with $$A$$. The maximum compression in the spring is:

We need to determine the maximum compression in the spring after block $$C$$ collides with block $$A$$.

1. Analyze Phase 1: The Collision

From the layout shown on the, block $$C$$ of mass $$m$$ moving with velocity $$v$$ makes a head-on collision with an identical stationary block $$A$$ (mass $$m$$).

Assuming the collision is perfectly elastic (standard for such idealized physics problems):

• When two identical masses collide elastically, they exchange their velocities.
• Therefore, immediately after the collision, block $$C$$ comes to a complete rest ($$v_C = 0$$).
• Block $$A$$ instantly acquires the entire initial velocity of block $$C$$, starting its motion with velocity $$v_A = v$$. Block $$B$$ remains stationary ($$v_B = 0$$) at this exact moment.

2. Analyze Phase 2: Maximum Compression of the Spring

As block $$A$$ moves to the right, it begins to compress the spring connecting it to block $$B$$. This compression exerts a forward force on block $$B$$ (speeding it up) and a backward force on block $$A$$ (slowing it down).

The spring reaches its maximum compression ($$x_{\text{max}}$$) at the exact instant when both blocks $$A$$ and $$B$$ are moving with the exact same common velocity ($$V_c$$). At this point, relative motion between them temporarily stops.

• Conservation of Linear Momentum:
  Since the floor is smooth and no external horizontal forces act on the $$A-B$$ spring system, momentum is conserved:

  $$m \cdot v = (m + m) \cdot V_c$$

  $$m v = 2m V_c \implies V_c = \frac{v}{2}$$

• Conservation of Mechanical Energy:
  The total kinetic energy of block $$A$$ immediately after the collision is converted into the combined kinetic energy of both blocks plus the elastic potential energy stored in the compressed spring:

  $$\frac{1}{2} m v^2 = \frac{1}{2} (2m) V_c^2 + \frac{1}{2} K x_{\text{max}}^2$$

3. Solve for Maximum Compression ($$x_{\text{max}}$$)

Substitute $$V_c = \frac{v}{2}$$ into the energy conservation equation:

$$\frac{1}{2} m v^2 = \frac{1}{2} (2m) \left(\frac{v}{2}\right)^2 + \frac{1}{2} K x_{\text{max}}^2$$

Multiply the entire equation by 2 to clear the fractions:

$$m v^2 = 2m \left(\frac{v^2}{4}\right) + K x_{\text{max}}^2$$

$$m v^2 = \frac{1}{2} m v^2 + K x_{\text{max}}^2$$

Isolate the spring term:

$$K x_{\text{max}}^2 = m v^2 - \frac{1}{2} m v^2$$

$$K x_{\text{max}}^2 = \frac{1}{2} m v^2$$

$$x_{\text{max}}^2 = \frac{m v^2}{2K}$$

Taking the square root gives us the maximum compression:

$$x_{\text{max}} = v \sqrt{\frac{m}{2K}}$$

Conclusion

The maximum compression in the spring is $$v \sqrt{\frac{m}{2K}}$$, which corresponds to Option A.