Two blocks A and B each of mass m, are connected by a massless spring of natural length L and spring

Q: Wait, why does block C come to a complete stop?

In a perfectly elastic head-on collision between two objects of equal mass, they exchange their velocities. Since block A was initially at rest, block C takes on A's initial velocity (zero), and A takes on C's initial velocity ($v$). It's a perfect transfer of momentum and kinetic energy!

Q: Why do blocks A and B have the same velocity at maximum compression?

Imagine if A was moving faster than B; the spring would still be compressing. If B was moving faster than A, the spring would be expanding. Therefore, the point of maximum compression (the turning point between compressing and expanding) must be the exact instant their velocities are equal.

Q: Why can't we just say the kinetic energy is zero at maximum compression?

That would only be true if the system was attached to a rigid wall! Here, both blocks are free to move on a frictionless surface. Because the system had an initial momentum ($mv$), the law of conservation of momentum dictates that the system must continue moving forward. It cannot simply stop.

Q: Why do we use $2m$ when calculating the final kinetic energy?

At the instant of maximum compression, blocks A and B are moving together at the exact same speed $v'$. For that specific moment, they act like a single combined mass of $m + m = 2m$ moving with velocity $v'$.

Q: Is the mechanical energy of the A-B system conserved throughout?

Yes! After the initial collision (which transfers energy to A), there are no non-conservative forces (like friction) acting on the A-B system. The total mechanical energy constantly shifts back and forth between the kinetic energy of the blocks and the elastic potential energy of the spring, but their sum remains perfectly constant.

Question

Two blocks $A$ and $B$ each of mass $m$, are connected by a massless spring of natural length $L$ and spring constant $k$. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in figure. A third identical block $C$, also of mass $m$, moves on the floor with a speed $v$ along the line joining $A$ and $B$, and collides elastically with $A$. Then

Options:

(A) the kinetic energy of the $A$-$B$ system, at maximum compression of the spring, is zero

(B) the kinetic energy of the $A$-$B$ system, at maximum compression of the spring, is $\frac{mv^2}{4}$

(C) the maximum compression of the spring is $v\sqrt{\frac{m}{k}}$

(D) the maximum compression of the spring is $v\sqrt{\frac{m}{2k}}$

Sigma Percentile · Accepted Answer

The beauty of physics often lies in its ability to break down complex, chaotic events into a series of elegant, predictable steps. This problem is a classic example of such elegance. We are presented with a system of three identical blocks and a spring, and we need to determine the state of the system at the moment of maximum spring compression.

At first glance, it might seem like a tangled mess of moving parts, but if we apply our fundamental conservation laws step-by-step, the solution unfolds beautifully. Let's embark on this journey!

### The Initial Spark: The Elastic Collision

The problem begins with block $C$ hurtling towards block $A$ with a velocity $v$. Block $A$ is peacefully resting, connected to block $B$ via a relaxed spring. The collision between $C$ and $A$ is described as perfectly elastic.

Here is where we deploy our first powerful physics principle. In a perfectly elastic, head-on collision between two objects of **identical mass**, the objects simply exchange their velocities. It is as if block $C$ transfers its very essence—its momentum and kinetic energy—entirely to block $A$.

Immediately after the collision, block $C$ comes to a dead stop. Its job is done. Block $A$, on the other hand, instantly acquires the velocity $v$. From this moment forward, block $C$ is irrelevant to the dynamics of the spring. We now shift our entire focus to the $A$-$B$ system.

### The Dance of the Spring: Seeking Maximum Compression

With block $A$ now moving at velocity $v$ and block $B$ still at rest, the spring between them begins to compress. As the spring compresses, it exerts a restoring force. It pushes backward on $A$, slowing it down, and pushes forward on $B$, speeding it up.

When does the spring reach its maximum compression? This is a crucial conceptual checkpoint. If $A$ is moving faster than $B$, it is still catching up to $B$, and the spring is still compressing. If $B$ is moving faster than $A$, $B$ is pulling away, and the spring is expanding. Therefore, the maximum compression occurs at the exact instant when block $A$ and block $B$ are moving at the **exact same velocity**. Let's call this common velocity $v'$.

### The Master Equation: Conservation of Momentum

To find this common velocity $v'$, we look at the $A$-$B$ system as a whole. In the horizontal direction, there are no external forces acting on this two-block system. The spring force is an internal force. Therefore, the total linear momentum of the $A$-$B$ system must be conserved.

The initial momentum of the system (just after the collision) is simply the momentum of block $A$:
$$p_{	ext{initial}} = mv$$

At the moment of maximum compression, both blocks are moving together with velocity $v'$. The final momentum is:
$$p_{	ext{final}} = (m + m)v' = 2mv'$$

Equating the initial and final momentum:
$$mv = 2mv'$$

Solving for $v'$, we find:
$$v' = \frac{v}{2}$$

At maximum compression, the entire system is drifting forward at half the original speed!

### Evaluating the Kinetic Energy

Now that we know the state of the system at maximum compression, we can easily calculate its kinetic energy. The system consists of two masses, each of mass $m$, moving together at velocity $\frac{v}{2}$.

$$K_{\max} = \frac{1}{2} (2m) \left(\frac{v}{2}ight)^2$$

$$K_{\max} = m \left(\frac{v^2}{4}ight) = \frac{mv^2}{4}$$

This elegant result confirms that option (b) is correct. Notice that the kinetic energy is not zero! The system must keep moving to conserve its initial momentum.

### The Final Calculation: Conservation of Mechanical Energy

Our final task is to find the actual physical compression of the spring, let's call it $x$. For this, we turn to the conservation of mechanical energy. Since there is no friction, the total mechanical energy of the $A$-$B$ system remains constant after the initial collision.

The initial mechanical energy is purely the kinetic energy of block $A$:
$$E_{	ext{initial}} = \frac{1}{2}mv^2$$

At maximum compression, this energy has been partitioned into two forms: the kinetic energy of the moving blocks (which we just calculated) and the elastic potential energy stored in the compressed spring.
$$E_{	ext{final}} = K_{\max} + U_{	ext{spring}}$$
$$E_{	ext{final}} = \frac{mv^2}{4} + \frac{1}{2}kx^2$$

Equating the initial and final energies:
$$\frac{1}{2}mv^2 = \frac{mv^2}{4} + \frac{1}{2}kx^2$$

Subtracting the kinetic energy term from both sides:
$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2 - \frac{1}{4}mv^2$$
$$\frac{1}{2}kx^2 = \frac{1}{4}mv^2$$

Now, we simply solve for $x^2$:
$$x^2 = \frac{mv^2}{2k}$$

Taking the square root of both sides yields the maximum compression:
$$x = v\sqrt{\frac{m}{2k}}$$

This confirms that option (d) is also correct.

### Conclusion

By methodically applying the principles of elastic collisions, momentum conservation, and energy conservation, we have completely unraveled the dynamics of this system. It is a profound reminder that even when objects are interacting through complex internal forces like a spring, the overarching conservation laws provide a clear and unwavering path to the solution.

Select Answer:

Visualized Solution

Initial Setup

The Collision

Velocity Exchange

System Dynamics

Maximum Compression Condition

Conservation of Momentum

Common Velocity

Kinetic Energy at Max Compression

Calculating $K_{m a x}$

Conservation of Mechanical Energy

Energy Equation

Solving for Spring Energy

Final Compression $x$

The Sigma Insight: Head-on Collision

The Initial Spark

The Dance of the Spring

The Master Equation

Evaluating the Kinetic Energy

The Final Calculation

Conclusion

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