A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is ............

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the system
We have a particle of mass \(4m\) that is initially at rest. After the explosion, it breaks into three fragments: two fragments of mass \(m\) each, moving with speed \(v\) in mutually perpendicular directions, and the third fragment of mass \(2m\).

### Step 2: Apply conservation of momentum
Since the initial momentum of the system is zero (the particle is at rest), the total momentum after the explosion must also be zero. We can denote the velocities of the fragments as follows:
- 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 \(2m\)): moves at an angle \(\theta\) with speed \(v'\).

The momentum conservation equations in the x and y directions are:
- In the x-direction:
\[ mv + 0 + 2m v' \cos(\theta) = 0 \]
- In the y-direction:
\[ 0 + mv + 2m v' \sin(\theta) = 0 \]

### Step 3: Solve for \(v'\) and \(\theta\)
From the x-direction equation:
\[ mv = -2m v' \cos(\theta) \implies v' \cos(\theta) = -\frac{v}{2} \]

From the y-direction equation:
\[ mv = -2m v' \sin(\theta) \implies v' \sin(\theta) = -\frac{v}{2} \]

Dividing the two equations gives:
\[ \frac{\sin(\theta)}{\cos(\theta)} = 1 \implies \tan(\theta) = 1 \implies \theta = 45^\circ \]

### Step 4: Find \(v'\)
Using \(\theta = 45^\circ\):
\[ v' \cos(45^\circ) = -\frac{v}{2} \implies v' \frac{1}{\sqrt{2}} = -\frac{v}{2} \implies v' = -\frac{v}{\sqrt{2}} \]

### Step 5: Calculate the kinetic energy of each fragment
Now, we can calculate the kinetic energy of each fragment:
1. Kinetic energy of Fragment 1:
\[ KE_1 = \frac{1}{2} m v^2 \]
2. Kinetic energy of Fragment 2:
\[ KE_2 = \frac{1}{2} m v^2 \]
3. Kinetic energy of Fragment 3:
\[ KE_3 = \frac{1}{2} (2m) \left(-\frac{v}{\sqrt{2}}\right)^2 = \frac{1}{2} (2m) \left(\frac{v^2}{2}\right) = mv^2 \]

### Step 6: Total kinetic energy and energy released
Now, we sum the kinetic energies:
\[ KE_{total} = KE_1 + KE_2 + KE_3 = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 + mv^2 = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 + \frac{2}{2} mv^2 = 3 \cdot \frac{1}{2} mv^2 = \frac{3}{2} mv^2 \]

Thus, the total energy released in the process of explosion is:
\[ \text{Total Energy Released} = \frac{3}{2} mv^2 \]