Two persons of masses `55 kg` and `65 kg` respectively are at the opposite ends of a boat. The length of the boat is `3.0 m` and weights `100 kg`. The `55 kg` man walks up to the `65 kg` man and sits with him. If the boat is in still water the centre of mass of the system shifts by.

Options

A. 0.75 m
B. 3.0 m
C. 2.3 m
D. zero

Explanation

To solve the problem, we need to analyze the situation involving the two persons and the boat. We will calculate the center of mass before and after the 55 kg person moves to sit with the 65 kg person.

### Step-by-Step Solution:

1. **Identify the system components**:
The system consists of two persons (mass 55 kg and 65 kg) and the boat (mass 100 kg).

2. **Determine the initial positions**:
- Let’s assume the boat is positioned along a straight line.
- The 55 kg person is at one end of the boat (position 0 m), and the 65 kg person is at the other end (position 3.0 m).

3. **Calculate the initial center of mass (CM)**:
The center of mass of the system can be calculated using the formula:
\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2 + m_b x_b}{m_1 + m_2 + m_b} \]
where:
- \( m_1 = 55 \, \text{kg} \) (mass of the first person)
- \( m_2 = 65 \, \text{kg} \) (mass of the second person)
- \( m_b = 100 \, \text{kg} \) (mass of the boat)
- \( x_1 = 0 \, \text{m} \) (position of the first person)
- \( x_2 = 3.0 \, \text{m} \) (position of the second person)
- \( x_b = 1.5 \, \text{m} \) (position of the boat’s center, which is at the midpoint)

Plugging in the values:
\[ x_{CM} = \frac{(55 \times 0) + (65 \times 3.0) + (100 \times 1.5)}{55 + 65 + 100} \]
\[ = \frac{0 + 195 + 150}{220} = \frac{345}{220} \approx 1.568 \, \text{m} \]

4. **Determine the new positions after the 55 kg person moves**:
After the 55 kg person walks to the position of the 65 kg person, both will be at the same position (3.0 m).

5. **Calculate the new center of mass (CM)**:
Now, the positions of the persons are:
- The 55 kg person is now at 3.0 m.
- The 65 kg person is still at 3.0 m.
- The center of the boat remains at 1.5 m.

The new center of mass is calculated as:
\[ x'_{CM} = \frac{(55 \times 3.0) + (65 \times 3.0) + (100 \times 1.5)}{55 + 65 + 100} \]
\[ = \frac{(165 + 195 + 150)}{220} = \frac{510}{220} \approx 2.318 \, \text{m} \]

6. **Determine the shift in center of mass**:
The shift in the center of mass is given by:
\[ \text{Shift} = x'_{CM} - x_{CM} = 2.318 - 1.568 \approx 0.75 \, \text{m} \]

7. **Conclusion**:
Since the boat is floating in still water and there are no external forces acting on the system, the center of mass of the entire system does not shift in the reference frame of the water. Thus, the center of mass of the system remains stationary.

### Final Answer:
The center of mass of the system does not shift at all.