A bomb at rest explodes into three parts of the same mass the momenta of the two parts are `- 2 p hati` and `p hat j` The momentum of the third part will have a magnitude of :

Options

A. `p`
B. `sqrt3p`
C. `psqrt5`
D. zero

Explanation

To solve the problem, we will use the principle of conservation of momentum. Since the bomb is initially at rest, the total momentum before the explosion is zero. Therefore, the total momentum after the explosion must also be zero.

### Step-by-Step Solution:

1. **Identify the given momenta**:
- The momentum of the first part is \( \vec{p_1} = -2p \hat{i} \).
- The momentum of the second part is \( \vec{p_2} = p \hat{j} \).

2. **Write the conservation of momentum equation**:
- Let the momentum of the third part be \( \vec{p_3} \).
- According to the conservation of momentum:
\[ \vec{p_1} + \vec{p_2} + \vec{p_3} = 0 \]
- This can be rearranged to find \( \vec{p_3} \):
\[ \vec{p_3} = -(\vec{p_1} + \vec{p_2}) \]

3. **Substitute the known momenta into the equation**:
- Substitute \( \vec{p_1} \) and \( \vec{p_2} \):
\[ \vec{p_3} = -(-2p \hat{i} + p \hat{j}) = 2p \hat{i} - p \hat{j} \]

4. **Calculate the magnitude of \( \vec{p_3} \)**:
- The magnitude of a vector \( \vec{p_3} = a \hat{i} + b \hat{j} \) is given by:
\[ |\vec{p_3}| = \sqrt{a^2 + b^2} \]
- Here, \( a = 2p \) and \( b = -p \):
\[ |\vec{p_3}| = \sqrt{(2p)^2 + (-p)^2} = \sqrt{4p^2 + p^2} = \sqrt{5p^2} \]

5. **Simplify the expression**:
- Thus, the magnitude of the momentum of the third part is:
\[ |\vec{p_3}| = \sqrt{5}p \]

### Final Answer:
The magnitude of the momentum of the third part is \( \sqrt{5}p \).