Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a fri

Q: Wait, why doesn't the spring force change the velocity of the center of mass?

The spring force is an *internal* force. It pulls the $10\text{ kg}$ block backward and the $4\text{ kg}$ block forward with equal and opposite force. Because these forces cancel out, they cannot change the total momentum of the system, and therefore, the center of mass velocity remains constant.

Q: Why is the velocity of the $4\text{ kg}$ block zero right after the impulse?

An impulse acts over an infinitesimally small time interval. During this tiny moment, the spring hasn't compressed yet. Without compression, there is no spring force to accelerate the $4\text{ kg}$ block, so its velocity remains zero at that exact instant.

Q: What happens to the center of mass velocity later, when the blocks are oscillating?

It stays exactly the same! As long as there is no external friction acting on the blocks, the total momentum is conserved. The blocks will perform a complex dance of speeding up and slowing down, but their center of mass will steadily cruise at $10\text{ m/s}$.

Q: Could we solve this using conservation of energy?

No, not easily. The impulse adds an unknown amount of kinetic energy to the system. While energy is conserved *after* the impulse, we don't need it to find the center of mass velocity. Momentum is the direct and correct tool here.

Q: What if the surface wasn't frictionless?

If there were friction, it would act as an *external* force on the system. This external force would cause the total momentum to decrease over time, meaning the velocity of the center of mass would gradually slow down until the entire system came to a halt.

Question

Two blocks of masses $10	ext{ kg}$ and $4	ext{ kg}$ are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of $14	ext{ m/s}$ to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

Options:

(A) $30	ext{ m/s}$

(B) $20	ext{ m/s}$

(C) $10	ext{ m/s}$

(D) $5	ext{ m/s}$

Sigma Percentile · Accepted Answer

### The Beauty of the Center of Mass

Imagine a chaotic system: blocks sliding, springs compressing and expanding, forces acting in multiple directions. Tracking the motion of every single particle can quickly become a mathematical nightmare. But physics offers us a beautiful, elegant shortcut: **The Center of Mass**.

The center of mass is a unique point that represents the average position of all the mass in a system. The magic of this point is that it moves as if all the mass of the system were concentrated there, and all external forces were applied directly to it. If there are no net external forces acting on the system, the velocity of the center of mass remains perfectly constant, regardless of the internal chaos!

### Setting the Stage: The Impulse

In our problem, we have two blocks, $m_1 = 10	ext{ kg}$ and $m_2 = 4	ext{ kg}$, connected by a spring on a frictionless surface. Initially, everything is peaceful and at rest.

Suddenly, an impulse is delivered to the heavier block. An impulse is a large force applied over a very short time, and it instantly changes the momentum of the object it acts upon. This impulse gives the $10	ext{ kg}$ block a velocity of $v_1 = 14	ext{ m/s}$.

Crucially, right after this impulse, the spring hasn't had time to compress. Because the spring hasn't compressed, it isn't exerting any force on the lighter $4	ext{ kg}$ block yet. Therefore, the initial velocity of the lighter block is exactly zero ($v_2 = 0$).

### The Master Equation

We are asked to find the velocity of the center of mass, $v_{cm}$. The formula for the velocity of the center of mass is a weighted average of the individual velocities:

$$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$

Notice the numerator: $m_1 v_1 + m_2 v_2$. This is simply the **total linear momentum** of the system. The denominator, $m_1 + m_2$, is the **total mass**. So, the velocity of the center of mass is just the total momentum divided by the total mass.

### The Final Calculation

Let's plug in our values. We know the heavier block is moving, so its momentum is:

$$p_1 = m_1 v_1 = 10 	imes 14 = 140	ext{ kg m/s}$$

The lighter block is at rest, so its momentum is zero:

$$p_2 = m_2 v_2 = 4 	imes 0 = 0	ext{ kg m/s}$$

The total momentum of our system is $140 + 0 = 140	ext{ kg m/s}$.

Now, we divide this total momentum by the total mass of the system, which is $10 + 4 = 14	ext{ kg}$:

$$v_{cm} = \frac{140}{14} = 10	ext{ m/s}$$

And there we have it! The center of mass of this system will glide along at a constant $10	ext{ m/s}$. Even as the spring compresses and expands, and the two blocks speed up and slow down relative to each other, that imaginary point—the center of mass—will continue moving steadily at $10	ext{ m/s}$ forever, thanks to the absence of external friction.

Select Answer:

Visualized Solution

The Physical Setup

The Impulse

Center of Mass Velocity Formula

Substituting the Values

Calculating the Numerator

Calculating the Denominator

Final Calculation

The Sigma Insight: Motion of Centre of Mass

The Beauty of the Center of Mass

Setting the Stage

The Master Equation

The Final Calculation

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