An object , initially at rest , explodes into three fragments of equal mass . The momenta of two parts are `2phati and phat j` where p is a positive number . The momentum of the third part is

Options

A. Will be of the magnitude 3p inclined at `tan ^(-1) (1/2)` with x - axis
B. Will be of the magnitude `sqrt(5)p` inclined at `tan^(-1) (2)` with x - axis
C. Will be of the magnitude 3p inclined at `pi-tan ^(-1) (1/2)` with x - axis
D. Will be of the magnitude `sqrt(5)p` inclined at `pi+tan^(-1) (1/2)` with x - axis

Answer

To solve the problem, we will use the principle of conservation of momentum. The object is initially at rest, meaning its initial momentum is zero. After the explosion, the total momentum of the three fragments must also equal zero.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The object is at rest initially, so the initial momentum \( \mathbf{P}_{initial} = 0 \).
- The momenta of the first two fragments are given as:
- Fragment 1: \( \mathbf{p_1} = 2p \hat{i} \)
- Fragment 2: \( \mathbf{p_2} = p \hat{j} \)
- We need to find the momentum of the third fragment, \( \mathbf{p_3} \).

2. **Apply the Conservation of Momentum:**
According to the conservation of momentum:
\[ \mathbf{P}_{initial} = \mathbf{p_1} + \mathbf{p_2} + \mathbf{p_3} \]
Since the initial momentum is zero:
\[ 0 = \mathbf{p_1} + \mathbf{p_2} + \mathbf{p_3} \]
Rearranging gives:
\[ \mathbf{p_3} = -(\mathbf{p_1} + \mathbf{p_2}) \]

3. **Substitute the Values of \( \mathbf{p_1} \) and \( \mathbf{p_2} \):**
\[ \mathbf{p_3} = -\left(2p \hat{i} + p \hat{j}\right) \]
This simplifies to:
\[ \mathbf{p_3} = -2p \hat{i} - p \hat{j} \]

4. **Express the Momentum of the Third Fragment:**
Thus, the momentum of the third fragment is:
\[ \mathbf{p_3} = -2p \hat{i} - p \hat{j} \]

5. **Calculate the Magnitude of \( \mathbf{p_3} \):**
The magnitude of \( \mathbf{p_3} \) can be calculated using the formula:
\[ |\mathbf{p_3}| = \sqrt{(-2p)^2 + (-p)^2} = \sqrt{4p^2 + p^2} = \sqrt{5p^2} = \sqrt{5}p \]

6. **Determine the Direction of \( \mathbf{p_3} \):**
The direction can be found using the tangent function:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{p}{2p} = \frac{1}{2} \]
Therefore,
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]
Since \( \mathbf{p_3} \) is in the third quadrant (both components are negative), the angle with respect to the positive x-axis is:
\[ \text{Angle} = 180^\circ + \theta = 180^\circ + \tan^{-1}\left(\frac{1}{2}\right) \]

### Final Answer:
The momentum of the third part is:
\[ \mathbf{p_3} = -2p \hat{i} - p \hat{j} \]
The magnitude is \( \sqrt{5}p \) and the direction is \( 180^\circ + \tan^{-1}\left(\frac{1}{2}\right) \).