A particle of mass `4 m` which is at rest explodes into masses `m, m` & `2m`. Two of the fragments of masses m and `2m` are found to move with equal speeds v each in opposite directions. The total mechanical energy released in the process of explosion is :-

Options

A. `mv^(2)`
B. `2mv^(2)`
C. `1//2mv^(2)`
D. `4mv^(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
A particle of mass `4m` is initially at rest. When it explodes, it breaks into three fragments with masses `m`, `m`, and `2m`.

**Hint:** Remember that the initial momentum of the system is zero since the particle is at rest.

### Step 2: Analyze the Final Conditions
After the explosion, two fragments of masses `m` and `2m` move with equal speeds `v` in opposite directions. Let’s denote the speed of the third fragment (mass `m`) as `v1`.

**Hint:** The direction of the velocities is important; assign a positive direction for clarity.

### Step 3: Apply Conservation of Momentum
Since there are no external forces acting on the system, the total momentum before the explosion must equal the total momentum after the explosion.

- Initial momentum = 0 (since the particle was at rest)
- Final momentum = (mass of 2m) * (velocity of 2m) + (mass of m) * (velocity of m) + (mass of m) * (velocity of m)

Setting up the equation:
\[ 0 = 2m(-v) + mv1 + mv \]

This simplifies to:
\[ 0 = -2mv + mv1 + mv \]

Rearranging gives:
\[ mv1 = 2mv - mv \]
\[ mv1 = mv \]
\[ v1 = v \]

**Hint:** Use the conservation of momentum to relate the velocities of the fragments.

### Step 4: Calculate the Kinetic Energy of Each Fragment
Now, we can calculate the kinetic energy of each fragment after the explosion.

- Kinetic energy of mass `2m` moving with speed `v`:
\[ KE_{2m} = \frac{1}{2} \cdot 2m \cdot v^2 = mv^2 \]

- Kinetic energy of one mass `m` moving with speed `-v`:
\[ KE_{m} = \frac{1}{2} \cdot m \cdot v^2 = \frac{1}{2} mv^2 \]

- Kinetic energy of the other mass `m` moving with speed `v1` (which we found to be equal to `v`):
\[ KE_{m} = \frac{1}{2} \cdot m \cdot v^2 = \frac{1}{2} mv^2 \]

### Step 5: Total Kinetic Energy After Explosion
Now, we sum the kinetic energies of all fragments:
\[ KE_{total} = KE_{2m} + KE_{m} + KE_{m} \]
\[ KE_{total} = mv^2 + \frac{1}{2} mv^2 + \frac{1}{2} mv^2 \]
\[ KE_{total} = mv^2 + mv^2 = 2mv^2 \]

### Step 6: Determine the Total Mechanical Energy Released
Since the initial kinetic energy was zero (the particle was at rest), the total mechanical energy released in the explosion is equal to the total kinetic energy after the explosion:
\[ \text{Total Mechanical Energy Released} = KE_{total} = 2mv^2 \]

### Conclusion
Thus, the total mechanical energy released in the process of explosion is:
\[ \boxed{2mv^2} \]

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