Stationary bomb explodesinto three pieces. One piece of 2kg mass moves with a velocity of `8 ms ^(-1)` at right angles to the other piece of mass 1 kg moving. With a velocity of `12 ms^(-1)` . If the mass of the third piece is `0.5kg`, then its velocity is

Options

A. `10 ms^(-1)`
B. `20 ms^(-1)`
C. `30 ms^(-1)`
D. `40ms^(-1)`

Explanation

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

### Step-by-Step Solution:

1. **Identify the masses and velocities of the pieces:**
- Mass of piece 1, \( m_1 = 2 \, \text{kg} \) with velocity \( v_1 = 8 \, \text{m/s} \) (moving in the x-direction).
- Mass of piece 2, \( m_2 = 1 \, \text{kg} \) with velocity \( v_2 = 12 \, \text{m/s} \) (moving in the y-direction).
- Mass of piece 3, \( m_3 = 0.5 \, \text{kg} \) (velocity \( v_3 \) is unknown).

2. **Calculate the momentum of the first two pieces:**
- Momentum of piece 1:
\[ p_1 = m_1 \cdot v_1 = 2 \, \text{kg} \cdot 8 \, \text{m/s} = 16 \, \text{kg m/s} \]
- Momentum of piece 2:
\[ p_2 = m_2 \cdot v_2 = 1 \, \text{kg} \cdot 12 \, \text{m/s} = 12 \, \text{kg m/s} \]

3. **Determine the resultant momentum vector:**
- Since the two pieces are moving at right angles to each other, we can calculate the resultant momentum \( p_3 \) using the Pythagorean theorem:
\[ p_3 = \sqrt{p_1^2 + p_2^2} = \sqrt{(16)^2 + (12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \, \text{kg m/s} \]

4. **Apply conservation of momentum:**
- The total momentum after the explosion must equal the total momentum before the explosion (which is zero):
\[ p_1 + p_2 + p_3 = 0 \]
- Therefore, the momentum of piece 3 must be equal in magnitude and opposite in direction to the vector sum of the momenta of pieces 1 and 2:
\[ p_3 = - (p_1 + p_2) = -20 \, \text{kg m/s} \]

5. **Calculate the velocity of the third piece:**
- Using the momentum of the third piece:
\[ p_3 = m_3 \cdot v_3 \]
- Rearranging gives:
\[ v_3 = \frac{p_3}{m_3} = \frac{20 \, \text{kg m/s}}{0.5 \, \text{kg}} = 40 \, \text{m/s} \]

### Final Answer:
The velocity of the third piece is \( 40 \, \text{m/s} \).

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