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. will have a magnitude `pi sqrt( 3)`
B. will have a magnitude `pi sqrt( 5)`
C. will have a magnitude `pi`
D. will have a magnitude `2pi`

Answer

To solve the problem, we will use the principle of conservation of momentum. The total momentum before the explosion must equal the total momentum after the explosion. Since the bomb is at rest before the explosion, the total initial momentum is zero.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The bomb is at rest, so the initial momentum \( \vec{P}_{initial} = 0 \).
- The momenta of the first two parts are given as:
- \( \vec{P}_1 = -2p \hat{i} \)
- \( \vec{P}_2 = p \hat{j} \)

2. **Write the Momentum Conservation Equation**:
- According to the conservation of momentum:
\[ \vec{P}_{initial} = \vec{P}_1 + \vec{P}_2 + \vec{P}_3 \]
- Since \( \vec{P}_{initial} = 0 \), we have:
\[ 0 = \vec{P}_1 + \vec{P}_2 + \vec{P}_3 \]
- Rearranging gives:
\[ \vec{P}_3 = -(\vec{P}_1 + \vec{P}_2) \]

3. **Calculate the Resultant Momentum of the First Two Parts**:
- Substitute the values of \( \vec{P}_1 \) and \( \vec{P}_2 \):
\[ \vec{P}_3 = -((-2p \hat{i}) + (p \hat{j})) \]
- This simplifies to:
\[ \vec{P}_3 = -(-2p \hat{i} + p \hat{j}) = 2p \hat{i} - p \hat{j} \]

4. **Find the Magnitude of the Third Part's Momentum**:
- The magnitude of \( \vec{P}_3 \) can be calculated using the Pythagorean theorem:
\[ |\vec{P}_3| = \sqrt{(2p)^2 + (-p)^2} \]
- Simplifying this gives:
\[ |\vec{P}_3| = \sqrt{4p^2 + p^2} = \sqrt{5p^2} = \sqrt{5}p \]

5. **Final Result**:
- The magnitude of the momentum of the third part is:
\[ |\vec{P}_3| = \sqrt{5}p \]