A cart of mass `M` is tied to one end of a massless rope of length `10 m`. The other end of the rope is in the hands of a man of mass `M`. The entire system is on a smooth horizontal surface. The man is at `x=0` and the cart at `x=10 m`. If the man pulls the cart by the rope, the man and the cart will meet at the point

Options

A. they will never meet
B. `x=10 m`
C. `x=5 m`
D. `x=0`

Answer

To solve the problem, we need to determine the point where the man and the cart will meet when the man pulls the cart towards him. We can use the concept of the center of mass (COM) of the system to find this point.

### Step-by-Step Solution:

1. **Identify the masses and positions**:
- Let the mass of the man be \( M \) and the mass of the cart also be \( M \).
- The initial position of the man is \( x_m = 0 \) m.
- The initial position of the cart is \( x_c = 10 \) m.

2. **Determine the center of mass (COM)**:
The center of mass \( x_{com} \) of a two-particle system can be calculated using the formula:
\[ x_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Here, \( m_1 = M \) (mass of the man), \( x_1 = 0 \) m (position of the man), \( m_2 = M \) (mass of the cart), and \( x_2 = 10 \) m (position of the cart).

3. **Substituting values into the COM formula**:
\[ x_{com} = \frac{M \cdot 0 + M \cdot 10}{M + M} = \frac{0 + 10M}{2M} = \frac{10}{2} = 5 \text{ m} \]

4. **Conclusion**:
The center of mass of the system is located at \( x_{com} = 5 \) m. Since there are no external forces acting on the system, the center of mass will remain stationary. Therefore, the man and the cart will meet at the point \( x = 5 \) m.

### Final Answer:
The man and the cart will meet at the point \( x = 5 \) m.