An isolated particle of mass `m` is moving in horizontal plane` xy` 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` cm. The larger fragment at this instant is at

Options

A. `y=-5cm`
B. `y=+20cm`
C. `y=+5cm`
D. `y=-20cm`

Explanation

To solve the problem, we need to find the position of the larger fragment after the explosion of the isolated particle. Let's break down the solution step by step.

### Step 1: Understand the System
We have a particle of mass `m` moving along the x-axis. It explodes into two fragments with masses `m/4` (smaller fragment) and `3m/4` (larger fragment).

### Step 2: Conservation of Momentum
Since the explosion occurs in an isolated system, the center of mass of the system does not change. The initial position of the center of mass can be expressed as:
\[ x_{cm} = \frac{m \cdot x_0}{m} = x_0 \]
where \(x_0\) is the initial position of the mass `m`.

### Step 3: Position of the Center of Mass After Explosion
After the explosion, the center of mass position can be expressed as:
\[ x_{cm} = \frac{m_1 \cdot y_1 + m_2 \cdot y_2}{m_1 + m_2} \]
where \(m_1 = \frac{m}{4}\), \(y_1 = 15 \, \text{cm}\) (position of the smaller fragment), \(m_2 = \frac{3m}{4}\), and \(y_2\) is the position of the larger fragment we need to find.

### Step 4: Set Up the Equation
Since the center of mass does not change, we have:
\[ \frac{\frac{m}{4} \cdot 15 + \frac{3m}{4} \cdot y_2}{\frac{m}{4} + \frac{3m}{4}} = 0 \]
This simplifies to:
\[ \frac{\frac{m}{4} \cdot 15 + \frac{3m}{4} \cdot y_2}{m} = 0 \]
Multiplying through by \(m\) gives:
\[ \frac{m}{4} \cdot 15 + \frac{3m}{4} \cdot y_2 = 0 \]

### Step 5: Solve for \(y_2\)
Now we can simplify and solve for \(y_2\):
\[ \frac{m}{4} \cdot 15 + \frac{3m}{4} \cdot y_2 = 0 \]
Dividing through by \(m/4\):
\[ 15 + 3y_2 = 0 \]
Rearranging gives:
\[ 3y_2 = -15 \]
Thus,
\[ y_2 = -5 \, \text{cm} \]

### Conclusion
The position of the larger fragment is at \(y = -5 \, \text{cm}\).

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