A body kept at rest explodes in four identical fragments and it is found that three of them move along mutually perpendicular directions with same kinetic energy K . Total energy released in the explosion is

Options

A. 3k
B. 4K
C. 6K
D. 8K

Explanation

To solve the problem, we need to analyze the explosion of a body at rest that breaks into four identical fragments. We will calculate the total energy released during the explosion based on the kinetic energy of the fragments.

### Step-by-Step Solution:

1. **Initial Setup**:
- Let the mass of the original body be \(4m\).
- After the explosion, the body breaks into four identical fragments, each with mass \(m\).

2. **Kinetic Energy of Fragments**:
- It is given that three of the fragments move along mutually perpendicular directions (x, y, and z axes) with the same kinetic energy \(K\).
- Since the kinetic energy \(K\) of each of these fragments is given by the formula:
\[ K = \frac{1}{2} m v^2 \]
where \(v\) is the velocity of the fragment. Therefore, we can express the velocity of each of the three fragments in terms of \(K\):
\[ v = \sqrt{\frac{2K}{m}} \]

3. **Conservation of Momentum**:
- Since the body was initially at rest, the total initial momentum is zero. By the conservation of momentum, the total final momentum must also equal zero.
- Let the velocity of the fourth fragment be \(V\). The momentum conservation equation can be written as:
\[ m v_i + m v_j + m v_k + m V = 0 \]
- Substituting the velocities of the first three fragments:
\[ m v \hat{i} + m v \hat{j} + m v \hat{k} + m V = 0 \]
- This simplifies to:
\[ v \hat{i} + v \hat{j} + v \hat{k} + V = 0 \]
- Rearranging gives:
\[ V = - (v \hat{i} + v \hat{j} + v \hat{k}) \]

4. **Magnitude of the Fourth Fragment's Velocity**:
- The magnitude of \(V\) can be calculated as:
\[ |V| = \sqrt{(-v)^2 + (-v)^2 + (-v)^2} = \sqrt{3v^2} = v\sqrt{3} \]

5. **Kinetic Energy of the Fourth Fragment**:
- The kinetic energy of the fourth fragment is:
\[ KE_{4} = \frac{1}{2} m (v\sqrt{3})^2 = \frac{1}{2} m (3v^2) = \frac{3}{2} mv^2 \]

6. **Total Kinetic Energy After Explosion**:
- The total kinetic energy of the system after the explosion is the sum of the kinetic energies of all four fragments:
\[ KE_{total} = 3K + \frac{3}{2} mv^2 \]
- Since \(K = \frac{1}{2} mv^2\), we can substitute:
\[ KE_{total} = 3K + 3K = 6K \]

7. **Total Energy Released**:
- The initial kinetic energy of the body before the explosion was 0 (since it was at rest). Therefore, the total energy released during the explosion is:
\[ \text{Energy Released} = KE_{total} - KE_{initial} = 6K - 0 = 6K \]

### Final Answer:
The total energy released in the explosion is \(6K\).