Two skaters A and B of mass 50 kg and 70 kg respectively stand facing each other 6 metres apart. They then pull on a light rope stretched between them. How far each moved when they meet?

Options

A. Both have moved 3 metres.
B. A moves 2.5 metres and B moves 3.5 metres.
C. A moves 3.5 metrs and B moves 2.5 metres.
D. A moves 2 metres and B moves 4 metres.

Answer

To solve the problem of how far each skater moves when they meet, we can use the concept of the center of mass. Here's a step-by-step solution:

### Step 1: Identify the masses and initial positions
- Mass of skater A (m_A) = 50 kg
- Mass of skater B (m_B) = 70 kg
- Initial distance between them = 6 meters

### Step 2: Calculate the center of mass
The center of mass (CM) of the system can be calculated using the formula:
\[ x_{CM} = \frac{m_A \cdot x_A + m_B \cdot x_B}{m_A + m_B} \]
Assuming skater A is at position 0 (x_A = 0) and skater B is at position 6 (x_B = 6):
\[ x_{CM} = \frac{(50 \cdot 0) + (70 \cdot 6)}{50 + 70} \]
\[ x_{CM} = \frac{0 + 420}{120} = \frac{420}{120} = 3.5 \text{ meters} \]

### Step 3: Determine how far each skater moves
Since the skaters pull on each other and there are no external forces acting on the system, they will meet at the center of mass.

- Skater A moves from 0 to 3.5 meters:
\[ \text{Distance moved by A} = 3.5 \text{ meters} \]

- Skater B starts at 6 meters and moves to the center of mass:
\[ \text{Distance moved by B} = 6 - 3.5 = 2.5 \text{ meters} \]

### Step 4: Conclusion
- Skater A moves 3.5 meters.
- Skater B moves 2.5 meters.

Thus, the final answer is:
- A moves 3.5 meters, and B moves 2.5 meters.