An explosion blows a rock into three parts. Two parts go off at right angles to each other . These two are `1 kg` first part moving with a velocity of `12 ms^(-1) and 2 kg` second part moving with a velocity of `8 ms^(-1)`. If the third part flies off with a velocity of `4 ms^(-1)`. Its mass would be

Options

A. 5Kg
B. 7Kg
C. 17Kg
D. 3Kg

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.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have three parts of a rock after an explosion.
- The first part has a mass of \( m_1 = 1 \, \text{kg} \) and moves with a velocity of \( v_1 = 12 \, \text{m/s} \).
- The second part has a mass of \( m_2 = 2 \, \text{kg} \) and moves with a velocity of \( v_2 = 8 \, \text{m/s} \).
- The third part has an unknown mass \( m_3 \) and moves with a velocity of \( v_3 = 4 \, \text{m/s} \).
- The first two parts move at right angles to each other.

2. **Set Up the Momentum Conservation Equation**:
- Since the rock was initially at rest, the initial momentum is \( 0 \).
- Therefore, the total final momentum must also equal \( 0 \).
- We can express the momentum of each part in vector form:
- The momentum of the first part:
\[ \vec{p_1} = m_1 \cdot v_1 \hat{i} = 1 \cdot 12 \hat{i} = 12 \hat{i} \, \text{kg m/s} \]
- The momentum of the second part:
\[ \vec{p_2} = m_2 \cdot v_2 \hat{j} = 2 \cdot 8 \hat{j} = 16 \hat{j} \, \text{kg m/s} \]
- The momentum of the third part:
\[ \vec{p_3} = m_3 \cdot v_3 \hat{r} \]
where \( \hat{r} \) is the direction of the third part.

3. **Calculate the Resultant Momentum**:
- The total momentum in the x-direction:
\[ p_x = 12 \, \text{kg m/s} \]
- The total momentum in the y-direction:
\[ p_y = 16 \, \text{kg m/s} \]
- The resultant momentum \( P \) can be calculated using the Pythagorean theorem:
\[ P = \sqrt{p_x^2 + p_y^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, \text{kg m/s} \]

4. **Equate the Momentum of the Third Part**:
- The momentum of the third part can be expressed as:
\[ \vec{p_3} = m_3 \cdot v_3 \]
- Setting the magnitudes equal, since the total momentum must equal zero:
\[ m_3 \cdot 4 = 20 \]

5. **Solve for the Mass of the Third Part**:
- Rearranging gives:
\[ m_3 = \frac{20}{4} = 5 \, \text{kg} \]

### Final Answer:
The mass of the third part is \( 5 \, \text{kg} \).