An isolated particle of mass m is moving in horizontal plane (x -y), along the X-axis, at a certain

Q: Wait, why doesn't the explosion change the path of the center of mass?

Because an explosion is the result of internal forces between the fragments. According to Newton's laws, only external forces can change the momentum and trajectory of the center of mass of a system.

Q: Do we need to know the X-coordinates of the fragments to solve this?

No! The motion in the X and Y directions are completely independent. We only need to balance the Y-coordinates to ensure the center of mass stays on the X-axis.

Q: What if the particle was initially moving at an angle instead of along the X-axis?

If it had an initial Y-velocity, the center of mass would continue moving in the Y-direction. We would then need to calculate $y_{CM}$ at that specific instant using $y_{CM} = v_{y} t$. But here, $v_y = 0$, making it much simpler.

Q: Why is the final answer negative?

The negative sign indicates direction. Since the smaller fragment flew upwards (positive Y), the larger fragment must move downwards (negative Y) to keep the center of mass exactly on the X-axis.

Q: Could the fragments have different X-velocities after the explosion?

Absolutely. The explosion could push one fragment forward and the other backward. But regardless of their X-positions, their Y-positions must always average out (weighted by mass) to zero.

Question

An isolated particle of mass $m$ is moving in horizontal plane $(x -y)$, along the $X$-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses $m/4$ and $3m/4$. An instant later, the smaller fragment is at $y = + 15	ext{ cm}$. The larger fragment at this instant is at

Options:

(A) $y = - 5	ext{ cm}$

(B) $y = + 20	ext{ cm}$

(C) $y = + 5	ext{ cm}$

(D) $y = - 20	ext{ cm}$

Sigma Percentile · Accepted Answer

### The Peaceful Journey

Imagine a particle of mass $m$ cruising smoothly along the horizontal $X$-axis. It is moving at a constant velocity, completely undisturbed.

Because its motion is strictly restricted to the $X$-axis, it has absolutely no velocity in the $Y$-direction. This means its $Y$-coordinate is exactly zero, and it will stay that way as long as no external forces act upon it.

### The Internal Explosion

Suddenly, the particle explodes into two fragments! One fragment has a mass of $\frac{m}{4}$, and the other has a mass of $\frac{3m}{4}$.

Here is the most crucial physics principle of this problem: an explosion is caused entirely by **internal forces**. According to Newton's laws of motion, internal forces cannot change the momentum or the trajectory of the center of mass of a system.

Therefore, even though the fragments fly off in different directions, their collective center of mass continues to move exactly as it did before the explosion—straight along the $X$-axis.

### Balancing the Fragments

Since the center of mass remains on the $X$-axis, its $Y$-coordinate must still be zero. We can express this using the center of mass formula for the $Y$-axis:

$$y_{CM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$$

We are given that the smaller fragment ($m_1 = \frac{m}{4}$) is located at $y_1 = +15 	ext{ cm}$. We need to find the position $y_2$ of the larger fragment ($m_2 = \frac{3m}{4}$). The total mass is simply $m$.

Substituting these values into our master equation, we get:

$$0 = \frac{(\frac{m}{4})(15) + (\frac{3m}{4})(y)}{m}$$

### The Final Calculation

To simplify this equation, we can multiply both sides by the total mass $m$, which completely eliminates the denominator:

$$0 = \frac{m}{4}(15) + \frac{3m}{4}(y)$$

Notice that $\frac{m}{4}$ is a common factor in both terms on the right side. By dividing the entire equation by $\frac{m}{4}$, we can strip away the mass variables entirely:

$$0 = 15 + 3y$$

Now, it is just a matter of basic algebra. Moving the $15$ to the other side gives:

$$3y = -15$$

Finally, dividing by $3$, we arrive at our answer:

$$y = -5 	ext{ cm}$$

The negative sign is physically profound. It tells us that while the lighter fragment flew upwards, the heavier fragment had to move downwards to perfectly balance the system and keep the center of mass anchored to the $X$-axis.

Select Answer:

Visualized Solution

Initial State

The Explosion

CM Y-Coordinate

CM Formula

Substitution

Simplification

Canceling Mass

Solving for $y$

Final Answer

The Sigma Insight: Motion of Centre of Mass

The Peaceful Journey

The Internal Explosion

Balancing the Fragments

The Final Calculation

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