An object of mass 3 kg at rest in space suddenly explodes into three parts of the same mass. The momenta of the two parts are `4hati` and `2hatj` respectively. Then the energy released in the explosion is

Options

A. 60 J
B. 40 J
C. 20 J
D. 10 J

Answer

To solve the problem step by step, we will use the principles of conservation of momentum and kinetic energy.

### Step 1: Understand the problem
We have an object of mass 3 kg at rest that explodes into three parts of equal mass. The momenta of the two parts are given as \( \vec{P_1} = 4\hat{i} \) and \( \vec{P_2} = 2\hat{j} \).

### Step 2: Determine the mass of each part
Since the total mass of the object is 3 kg and it explodes into three equal parts, the mass of each part is:
\[ m_1 = m_2 = m_3 = \frac{3 \text{ kg}}{3} = 1 \text{ kg} \]

### Step 3: Apply conservation of momentum
The total momentum before the explosion is zero (since the object is at rest). According to the conservation of momentum:
\[ \vec{P_1} + \vec{P_2} + \vec{P_3} = 0 \]
Substituting the known momenta:
\[ 4\hat{i} + 2\hat{j} + \vec{P_3} = 0 \]
This implies:
\[ \vec{P_3} = -4\hat{i} - 2\hat{j} \]

### Step 4: Determine the velocity of each part
Since the mass of each part is 1 kg, the momentum is equal to the mass times the velocity:
\[ \vec{P} = m \vec{V} \]
Therefore, the velocities of the parts can be expressed as:
- For part 1:
\[ \vec{V_1} = \frac{\vec{P_1}}{m_1} = \frac{4\hat{i}}{1} = 4\hat{i} \]
- For part 2:
\[ \vec{V_2} = \frac{\vec{P_2}}{m_2} = \frac{2\hat{j}}{1} = 2\hat{j} \]
- For part 3:
\[ \vec{V_3} = \frac{\vec{P_3}}{m_3} = \frac{-4\hat{i} - 2\hat{j}}{1} = -4\hat{i} - 2\hat{j} \]

### Step 5: Calculate the kinetic energy of each part
The kinetic energy (KE) of an object is given by:
\[ KE = \frac{1}{2} m v^2 \]
Calculating the kinetic energy for each part:
- For part 1:
\[ KE_1 = \frac{1}{2} \cdot 1 \cdot (4)^2 = \frac{1}{2} \cdot 1 \cdot 16 = 8 \text{ J} \]
- For part 2:
\[ KE_2 = \frac{1}{2} \cdot 1 \cdot (2)^2 = \frac{1}{2} \cdot 1 \cdot 4 = 2 \text{ J} \]
- For part 3:
\[ KE_3 = \frac{1}{2} \cdot 1 \cdot \left(\sqrt{(-4)^2 + (-2)^2}\right)^2 = \frac{1}{2} \cdot 1 \cdot (4^2 + 2^2) = \frac{1}{2} \cdot 1 \cdot (16 + 4) = \frac{1}{2} \cdot 20 = 10 \text{ J} \]

### Step 6: Calculate the total energy released
The total energy released in the explosion is the sum of the kinetic energies of all three parts:
\[ E_{total} = KE_1 + KE_2 + KE_3 = 8 \text{ J} + 2 \text{ J} + 10 \text{ J} = 20 \text{ J} \]

### Final Answer
The energy released in the explosion is \( \boxed{20 \text{ J}} \).