A body of mass (4m) is laying in xy-plane at rest. It suddenly explodes into three pieces. Two pieces each mass (m) move perpedicular to each other with equal speeds (v). Total kinetic energy generated due to explosion is

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
The body of mass \(4m\) is at rest in the xy-plane. Therefore, the initial momentum of the system is zero.

### Step 2: Analyze the explosion
After the explosion, the body splits into three pieces:
- Two pieces of mass \(m\) each, moving with speed \(v\) and perpendicular to each other.
- One piece of mass \(2m\) that moves in an unknown direction with speed \(V_1\).

### 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 (before explosion):
\[ \text{Initial momentum} = 0 \]

Final momentum (after explosion):
\[ \text{Final momentum} = m \cdot v \hat{i} + m \cdot v \hat{j} + 2m \cdot V_1 \]
Where \( \hat{i} \) and \( \hat{j} \) are unit vectors in the x and y directions, respectively.

Setting the initial momentum equal to the final momentum:
\[ 0 = mv \hat{i} + mv \hat{j} + 2m \cdot V_1 \]

### Step 4: Solve for \(V_1\)
Rearranging the momentum equation gives:
\[ 2m \cdot V_1 = -mv \hat{i} - mv \hat{j} \]
Dividing through by \(2m\):
\[ V_1 = -\frac{v}{2} \hat{i} - \frac{v}{2} \hat{j} \]

### Step 5: Calculate the magnitude of \(V_1\)
The magnitude of \(V_1\) can be found using:
\[ |V_1| = \sqrt{\left(-\frac{v}{2}\right)^2 + \left(-\frac{v}{2}\right)^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{2v^2}{4}} = \frac{v}{\sqrt{2}} \]

### Step 6: Calculate the total kinetic energy after the explosion
The kinetic energy \(KE\) for each piece is given by:
- For the first piece:
\[ KE_1 = \frac{1}{2} m v^2 \]
- For the second piece:
\[ KE_2 = \frac{1}{2} m v^2 \]
- For the third piece:
\[ KE_3 = \frac{1}{2} (2m) V_1^2 = \frac{1}{2} (2m) \left(\frac{v}{\sqrt{2}}\right)^2 = \frac{1}{2} (2m) \cdot \frac{v^2}{2} = mv^2 \]

### Step 7: Sum the kinetic energies
Total kinetic energy after the explosion:
\[ KE_{\text{total}} = KE_1 + KE_2 + KE_3 = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 + mv^2 = mv^2 + mv^2 = \frac{3}{2} mv^2 \]

### Final Answer
The total kinetic energy generated due to the explosion is:
\[ \text{Total Kinetic Energy} = \frac{3}{2} mv^2 \]

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