An object initially at rest explodes into three equal fragments `A, B` and `C`.The momentum of `A` is `phati` and that of `B` is `sqrt(3)phatj`
where `p` is `a +ve` number. The momentum of `C` will be

Options

A. `(1+sqrt(3))p` in a direction making angle `120^(@)` clockwise with that of `A`
B. `(1+sqrt(3))p` in a direction making angle `150^(@)` anticlockwise with that of `B`
C. `2p` in as direction making angle `150^(@)` with that `A`
D. `2p` in a direction making angle `150^(@)` with that of `B`.

Explanation

To solve the problem, we need to apply the principle of conservation of momentum. The object is initially at rest, which means its initial momentum is zero. After the explosion, the total momentum of the fragments must still equal zero.

### Step-by-Step Solution:

1. **Identify the given momenta:**
- Momentum of fragment A: \( \vec{p}_A = p \hat{i} \)
- Momentum of fragment B: \( \vec{p}_B = \sqrt{3} p \hat{j} \)

2. **Set up the conservation of momentum equation:**
Since the initial momentum is zero, the total momentum after the explosion must also equal zero:
\[ \vec{p}_A + \vec{p}_B + \vec{p}_C = 0 \]
Rearranging gives:
\[ \vec{p}_C = -(\vec{p}_A + \vec{p}_B) \]

3. **Substitute the known momenta into the equation:**
\[ \vec{p}_C = -\left(p \hat{i} + \sqrt{3} p \hat{j}\right) \]
This simplifies to:
\[ \vec{p}_C = -p \hat{i} - \sqrt{3} p \hat{j} \]

4. **Express the momentum of C in terms of magnitude and direction:**
The momentum of C can be expressed in polar coordinates. The magnitude \( |\vec{p}_C| \) is given by:
\[ |\vec{p}_C| = \sqrt{(-p)^2 + (-\sqrt{3}p)^2} = \sqrt{p^2 + 3p^2} = \sqrt{4p^2} = 2p \]

5. **Determine the direction of momentum C:**
The angle \( \theta \) with respect to the negative x-axis can be found using:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{-\sqrt{3}p}{-p} = \sqrt{3} \]
This gives \( \theta = 60^\circ \) below the negative x-axis. Therefore, the angle with respect to the positive x-axis is \( 180^\circ - 60^\circ = 120^\circ \).

6. **Determine the angle with respect to fragment B:**
Fragment B is moving in the positive y-direction. The angle between the direction of C and the direction of B can be calculated as:
\[ 180^\circ - 60^\circ = 150^\circ \]
This angle is measured in the anticlockwise direction from B.

### Final Result:
The momentum of fragment C is:
\[ \vec{p}_C = 2p \text{ at an angle of } 150^\circ \text{ with respect to B.} \]