A particle of mass 5 m at rest suddenly breaks on its own into three fragments . Two fragments of mass m of each move along mutully perpendicular direction with speed v each . The energy released suring the process is

Options

A. `4/3mv^(2)`
B. `3/5mv^(2)`
C. `5/3mv^(2)`
D. `3/2mv^(2)`

Explanation

To solve the problem, we need to determine the energy released when a particle of mass \(5m\) breaks into three fragments, two of mass \(m\) moving at speed \(v\) in mutually perpendicular directions, and one fragment of mass \(3m\).

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial mass of the particle is \(5m\).
- The initial velocity of the particle is \(0\) (it is at rest).
- Therefore, the initial kinetic energy (\(KE_i\)) is:
\[ KE_i = \frac{1}{2} \times 5m \times 0^2 = 0 \]

2. **Final Conditions After Breaking**:
- Two fragments of mass \(m\) each move with speed \(v\) in mutually perpendicular directions.
- Let’s assume:
- Fragment 1 (mass \(m\)) moves in the positive x-direction with speed \(v\).
- Fragment 2 (mass \(m\)) moves in the positive y-direction with speed \(v\).
- Fragment 3 (mass \(3m\)) has an unknown velocity \(v'\).

3. **Conservation of Momentum**:
- Since there are no external forces, the momentum before and after the break must be conserved.
- Initial momentum (\(p_i\)) is \(0\) (since the particle is at rest).
- Final momentum (\(p_f\)) can be expressed as:
\[ p_f = m \cdot v \hat{i} + m \cdot v \hat{j} + 3m \cdot v' \]
- Setting initial momentum equal to final momentum gives:
\[ 0 = mv + mv + 3m \cdot v' \]
- This simplifies to:
\[ 0 = mv + mv + 3m \cdot v' \implies 0 = 2mv + 3mv' \implies v' = -\frac{2v}{3} \]

4. **Calculate Final Kinetic Energy**:
- The kinetic energy of each fragment after the break:
- For Fragment 1:
\[ KE_1 = \frac{1}{2} m v^2 \]
- For Fragment 2:
\[ KE_2 = \frac{1}{2} m v^2 \]
- For Fragment 3:
\[ KE_3 = \frac{1}{2} (3m) \left(-\frac{2v}{3}\right)^2 = \frac{1}{2} (3m) \cdot \frac{4v^2}{9} = \frac{2mv^2}{3} \]

5. **Total Final Kinetic Energy**:
- The total final kinetic energy (\(KE_f\)) is:
\[ KE_f = KE_1 + KE_2 + KE_3 = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 + \frac{2mv^2}{3} \]
- Combining the terms:
\[ KE_f = mv^2 + \frac{2mv^2}{3} = \frac{3mv^2}{3} + \frac{2mv^2}{3} = \frac{5mv^2}{3} \]

6. **Calculate Energy Released**:
- The energy released during the process is the difference between the final and initial kinetic energy:
\[ \text{Energy Released} = KE_f - KE_i = \frac{5mv^2}{3} - 0 = \frac{5mv^2}{3} \]

### Final Answer:
The energy released during the process is:
\[ \frac{5mv^2}{3} \]