A bomb at rest explodes into three parts of the same mass. The momentum of the two parts are `xhat(i) and -2 x hat(j)` . The momentum of the third part will have a magnitude of

Options

A. `x`
B. `sqrt(5x)`
C. `xsqrt(5)`
D. Zero

Answer

To solve the problem, we need to find the momentum of the third part after the bomb explodes. The total momentum before the explosion is zero since the bomb is at rest. According to the law of conservation of momentum, the total momentum after the explosion must also be zero.

### Step-by-step Solution:

1. **Identify the Momentum of the Parts**:
- Let the momentum of the first part be \( \vec{p_1} = x \hat{i} \).
- Let the momentum of the second part be \( \vec{p_2} = -2x \hat{j} \).
- Let the momentum of the third part be \( \vec{p_3} \).

2. **Apply Conservation of Momentum**:
- The total momentum before the explosion is zero:
\[ \vec{p_{total}} = \vec{p_1} + \vec{p_2} + \vec{p_3} = 0 \]
- Rearranging gives:
\[ \vec{p_3} = -(\vec{p_1} + \vec{p_2}) \]

3. **Calculate the Sum of the First Two Momenta**:
- Substitute the values of \( \vec{p_1} \) and \( \vec{p_2} \):
\[ \vec{p_1} + \vec{p_2} = x \hat{i} + (-2x \hat{j}) = x \hat{i} - 2x \hat{j} \]

4. **Find the Momentum of the Third Part**:
- Now substitute this back into the equation for \( \vec{p_3} \):
\[ \vec{p_3} = - (x \hat{i} - 2x \hat{j}) = -x \hat{i} + 2x \hat{j} \]

5. **Calculate the Magnitude of the Third Momentum**:
- The magnitude of \( \vec{p_3} \) is given by:
\[ |\vec{p_3}| = \sqrt{(-x)^2 + (2x)^2} = \sqrt{x^2 + 4x^2} = \sqrt{5x^2} = \sqrt{5} |x| \]

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