A spring block system is resting on a frictionless floor as shown in the figure. The spring constant

Q: Why does the first block move to the left after the collision?

When the lighter block ($1.0\text{ kg}$) collides elastically with the heavier stationary block ($2.0\text{ kg}$), the momentum transfer causes the lighter block to rebound. Mathematically, this is reflected in the negative sign of its post-collision velocity ($v_1 = -\frac{2}{3}\text{ ms}^{-1}$).

Q: Why is the time taken for the spring to return to its natural length equal to $T/2$?

After the collision, the second block moves right, compressing the spring to maximum compression (one-quarter of a cycle, $T/4$), and then moves back to the starting position (another $T/4$). Thus, it takes exactly half a time period ($T/2$) to return to the unstretched position for the first time.

Q: Does the spring constant affect the final separation between the blocks?

Yes, the spring constant $k$ determines the angular frequency $\omega = \sqrt{k/m}$ and thus the time period $T$. A stiffer spring (higher $k$) means a shorter time period, giving the first block less time to travel, resulting in a smaller final separation.

Q: What would happen if the collision was inelastic?

If the collision was inelastic ($e < 1$), the post-collision velocity of the first block would be different (less negative or even positive), and the second block would start with a lower velocity. However, the time period of oscillation for the second block would remain unchanged as it only depends on $m_2$ and $k$.

Q: Why does the first block move with constant velocity after the collision?

The floor is completely frictionless, and after the collision, there are no horizontal forces acting on the first block. According to Newton's first law of motion, it will continue to move with a constant velocity.

Question

A spring block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0	ext{ Nm}^{-1}$ and the mass of the block is $2.0	ext{ kg}$. Ignore the mass of the spring. Initially, the spring is in an unstretched condition. Another block of mass $1.0	ext{ kg}$ moving with a speed of $2.0	ext{ ms}^{-1}$ collides elastically with the first block. The collision is such that the $2.0	ext{ kg}$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ............ .

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Blocks

Imagine a perfectly smooth, frictionless horizontal floor. On this floor, we have two blocks. The first block has a mass of $1.0	ext{ kg}$ and is moving with a velocity of $2.0	ext{ ms}^{-1}$ to the right. The second block, with a mass of $2.0	ext{ kg}$, is initially at rest and is connected to a spring with a force constant of $2.0	ext{ Nm}^{-1}$. The other end of the spring is anchored to a rigid wall.

Our goal is to find the distance between these two blocks when the spring returns to its unstretched position for the first time after the collision. This problem beautifully combines the principles of linear momentum, elastic collisions, and simple harmonic motion.

### The Collision: Momentum and Elasticity

When the moving block strikes the stationary block, a perfectly elastic collision occurs. Because the collision is elastic, two key physical quantities are conserved: total linear momentum and total kinetic energy. We can write the equation for the conservation of linear momentum as:

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$

Substituting the given values ($m_1 = 1.0	ext{ kg}$, $m_2 = 2.0	ext{ kg}$, $u_1 = 2.0	ext{ ms}^{-1}$, and $u_2 = 0$):

$$(1.0)(2.0) + (2.0)(0) = (1.0)v_1 + (2.0)v_2 \implies v_1 + 2v_2 = 2.0$$

Since the collision is perfectly elastic, the coefficient of restitution $e$ is equal to $1$. This gives us the relative velocity equation:

$$e = \frac{v_2 - v_1}{u_1 - u_2} = 1 \implies v_2 - v_1 = 2.0$$

Now we have a system of two linear equations:

1) $v_1 + 2v_2 = 2.0$

2) $v_2 - v_1 = 2.0$

Adding these two equations together, we get:

$$3v_2 = 4.0 \implies v_2 = \frac{4}{3}	ext{ ms}^{-1}$$

Substituting $v_2$ back into the second equation yields:

$$v_1 = v_2 - 2.0 = \frac{4}{3} - 2 = -\frac{2}{3}	ext{ ms}^{-1}$$

The negative sign for $v_1$ indicates that the first block rebounds and moves to the left at a speed of $\frac{2}{3}	ext{ ms}^{-1}$.

### The Dance of the Spring: Simple Harmonic Motion

Immediately after the collision, the second block starts moving to the right with a velocity of $\frac{4}{3}	ext{ ms}^{-1}$. As it moves, it compresses the spring. Because the block is attached to the spring on a frictionless floor, it executes Simple Harmonic Motion (SHM).

The angular frequency $\omega$ of this motion is given by:

$$\omega = \sqrt{\frac{k}{m_2}}$$

Substituting $k = 2.0	ext{ Nm}^{-1}$ and $m_2 = 2.0	ext{ kg}$:

$$\omega = \sqrt{\frac{2.0}{2.0}} = 1.0	ext{ rad s}^{-1}$$

The time period $T$ for one complete oscillation is:

$$T = \frac{2\pi}{\omega} = 2\pi	ext{ seconds}$$

The spring will compress to its maximum limit and then expand back. It returns to its unstretched, natural length for the first time when the block completes exactly half of a full oscillation cycle. The time taken for this is:

$$t = \frac{T}{2} = \frac{\pi}{\omega} = \pi	ext{ seconds} \approx 3.1416	ext{ s}$$

### The Final Calculation: Bringing It All Together

During this time interval of $t = \pi	ext{ seconds}$, the first block continues to slide to the left at a constant speed of $\frac{2}{3}	ext{ ms}^{-1}$ because there is no friction to slow it down. The distance $d_1$ traveled by the first block in this time is:

$$d_1 = |v_1| 	imes t = \frac{2}{3} 	imes \pi \approx \frac{2}{3} 	imes 3.1416 \approx 2.094	ext{ m}$$

At the exact moment $t = \pi	ext{ s}$, the second block has returned to its initial starting position (where the spring is unstretched). Therefore, the distance between the two blocks is simply the distance traveled by the first block:

$$d = 2.09	ext{ m}$$

This elegant result shows how different areas of mechanics—collisions and oscillations—intertwine to create a beautiful physical puzzle.

Enter Numerical Value:

Visualized Solution

Understanding the Physical Setup

Applying Collision Mechanics

Substituting the Given Values

Solving for Post-Collision Velocities

Analyzing the Motion of Block 2

Calculating the Half-Period Time

Tracking the Motion of Block 1

Computing the Final Separation

Exploring Variations of the Problem

The Sigma Insight: Simple Harmonic Motion (SHM)

The Setup

The Collision

The Dance of the Spring

The Final Calculation

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