A body of mass 1 kg moving in the x-direction, suddenly explodes into two fragments of mass 1/8 kg and 7/8 kg. An instant later, the smaller fragment is 0.14 m above the x-axis. The position of the heavier fragment is:

Options

A. 1/510 m above x-axis
B. 1/ 50 m below x-axis
C. 7/50 m below x-axis
D. 7/50 m above x-axis

Answer

To solve the problem, we need to apply the principle of conservation of momentum and the concept of the center of mass.

### Step-by-Step Solution:

1. **Understanding the System**:
- A body of mass \( M = 1 \, \text{kg} \) is moving in the x-direction.
- It explodes into two fragments: one of mass \( m_1 = \frac{1}{8} \, \text{kg} \) and the other of mass \( m_2 = \frac{7}{8} \, \text{kg} \).

2. **Initial Conditions**:
- Before the explosion, the entire mass is moving in the x-direction with some velocity \( v \).
- The center of mass of the system is initially at the origin (0,0) in the y-direction since there is no vertical motion.

3. **After the Explosion**:
- The smaller fragment \( m_1 \) is observed to be \( 0.14 \, \text{m} \) above the x-axis (let's denote its position as \( y_1 = 0.14 \, \text{m} \)).
- We need to find the position of the heavier fragment \( m_2 \) in the y-direction, denoted as \( y_2 \).

4. **Using the Center of Mass Formula**:
- The center of mass (COM) in the y-direction can be expressed as:
\[ y_{COM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} \]
- Since the system was initially at rest in the y-direction, the y-coordinate of the center of mass remains at 0 after the explosion:
\[ 0 = \frac{\left(\frac{1}{8}\right)(0.14) + \left(\frac{7}{8}\right)y_2}{1} \]

5. **Setting Up the Equation**:
- Rearranging the equation gives:
\[ 0 = \frac{0.0175 + \frac{7}{8}y_2}{1} \]
- This simplifies to:
\[ 0.0175 + \frac{7}{8}y_2 = 0 \]

6. **Solving for \( y_2 \)**:
- Rearranging gives:
\[ \frac{7}{8}y_2 = -0.0175 \]
- Multiplying both sides by \( \frac{8}{7} \):
\[ y_2 = -0.0175 \cdot \frac{8}{7} \]
- Calculating \( y_2 \):
\[ y_2 = -0.02 \, \text{m} \]

7. **Conclusion**:
- The position of the heavier fragment \( m_2 \) is \( 0.02 \, \text{m} \) below the x-axis.

### Final Answer:
The position of the heavier fragment is \( 0.02 \, \text{m} \) below the x-axis.

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