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, we will follow these steps:

### Step 1: Understand the Problem
We have an object of mass 3 kg at rest that explodes into three equal parts, each with a mass of 1 kg. The momenta of the first two parts are given as \(4 \hat{i}\) and \(2 \hat{j}\). We need to find the energy released in the explosion.

### Step 2: Conservation of Momentum
Since the object was initially at rest, the total initial momentum is zero. According to the conservation of momentum, the total final momentum must also be zero.

Let the momentum of the third part be \(\vec{p_3}\). Therefore, we can write:
\[ \vec{p_1} + \vec{p_2} + \vec{p_3} = 0 \]
Substituting the known momenta:
\[ 4 \hat{i} + 2 \hat{j} + \vec{p_3} = 0 \]
This implies:
\[ \vec{p_3} = - (4 \hat{i} + 2 \hat{j}) = -4 \hat{i} - 2 \hat{j} \]

### Step 3: Calculate the Magnitude of the Momenta
Now we can calculate the magnitudes of the momenta:
- For the first part:
\[ |\vec{p_1}| = |4 \hat{i}| = 4 \, \text{kg m/s} \]
- For the second part:
\[ |\vec{p_2}| = |2 \hat{j}| = 2 \, \text{kg m/s} \]
- For the third part:
\[ |\vec{p_3}| = |-4 \hat{i} - 2 \hat{j}| = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \, \text{kg m/s} \]

### Step 4: Calculate the Kinetic Energy of Each Part
The kinetic energy (KE) for each part can be calculated using the formula:
\[ KE = \frac{p^2}{2m} \]
where \(p\) is the momentum and \(m\) is the mass of the part (1 kg).

1. For the first part:
\[ KE_1 = \frac{(4)^2}{2 \times 1} = \frac{16}{2} = 8 \, \text{J} \]
2. For the second part:
\[ KE_2 = \frac{(2)^2}{2 \times 1} = \frac{4}{2} = 2 \, \text{J} \]
3. For the third part:
\[ KE_3 = \frac{(2\sqrt{5})^2}{2 \times 1} = \frac{20}{2} = 10 \, \text{J} \]

### Step 5: Calculate Total Final Kinetic Energy
Now we can find the total final kinetic energy:
\[ KE_{total} = KE_1 + KE_2 + KE_3 = 8 + 2 + 10 = 20 \, \text{J} \]

### Step 6: Calculate Energy Released
The initial kinetic energy was zero (since the object was at rest), so the energy released in the explosion is equal to the total final kinetic energy:
\[ \text{Energy Released} = KE_{total} - KE_{initial} = 20 \, \text{J} - 0 \, \text{J} = 20 \, \text{J} \]

### Final Answer
The energy released in the explosion is **20 Joules**.

---