A cart of mass M is tied to one end of a massless rope of length 10m. The other end of the rope is in the hands of a man of mass `(M)/(2)`. The entire ststem is on a smooth horizontal surface. If the man pulls the cart by the rope then find the distance by which the man will slip on the horizontal surface before the man and the cart will meet each other.

Options

A. `(20)/(3)m`
B. `(10)/(3)m`
C. 0
D. None

Answer

To solve the problem, we can follow these steps:

### Step 1: Understand the System
We have a cart of mass \( M \) tied to one end of a massless rope of length 10 m. The other end of the rope is held by a man of mass \( \frac{M}{2} \). The entire system is on a smooth horizontal surface, meaning there is no friction.

### Step 2: Identify the Displacements
Let’s denote:
- The distance the man slips as \( x \) (to the left).
- The distance the cart moves as \( 10 - x \) (to the right).

### Step 3: Center of Mass Calculation
Since there are no external forces acting on the system, the center of mass (CM) of the system will not move. The formula for the center of mass \( x_{CM} \) of the system is given by:

\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]

Where:
- \( m_1 = \frac{M}{2} \) (mass of the man)
- \( x_1 = -x \) (displacement of the man)
- \( m_2 = M \) (mass of the cart)
- \( x_2 = 10 - x \) (displacement of the cart)

### Step 4: Set Up the Equation
Since the center of mass does not move, we set the center of mass equation to zero:

\[ \frac{\left(\frac{M}{2} \cdot (-x)\right) + \left(M \cdot (10 - x)\right)}{\frac{M}{2} + M} = 0 \]

### Step 5: Simplify the Equation
The total mass of the system is:

\[ \frac{M}{2} + M = \frac{3M}{2} \]

Substituting this into the equation gives:

\[ \frac{-\frac{Mx}{2} + 10M - Mx}{\frac{3M}{2}} = 0 \]

### Step 6: Clear the Denominator
Multiplying through by \( \frac{3M}{2} \) to eliminate the denominator:

\[ -\frac{Mx}{2} + 10M - Mx = 0 \]

### Step 7: Combine Like Terms
Combine the terms involving \( x \):

\[ 10M - \frac{3Mx}{2} = 0 \]

### Step 8: Solve for \( x \)
Rearranging gives:

\[ 10M = \frac{3Mx}{2} \]

Dividing both sides by \( M \) (assuming \( M \neq 0 \)):

\[ 10 = \frac{3x}{2} \]

Multiplying both sides by 2:

\[ 20 = 3x \]

Finally, divide by 3:

\[ x = \frac{20}{3} \text{ meters} \]

### Conclusion
The distance by which the man will slip before he and the cart meet is \( \frac{20}{3} \) meters.

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