A man of mass M stands at one end of a plank of length L Which lies at rest on a frictionless surface . The man walks to the other end of the plank. If the mass of the plank is 3M, the distance that the man moves relative to the ground is

Options

A. `L/4`
B. `(3L)/4`
C. `(2L)/3`
D. `L/3`

Answer

To solve the problem, we need to analyze the motion of the man and the plank while considering the conservation of momentum and the center of mass of the system.

### Step-by-Step Solution:

1. **Identify the System**:
- We have a man with mass \( M \) and a plank with mass \( 3M \) lying on a frictionless surface. The total mass of the system is \( 4M \).

2. **Initial Condition**:
- Initially, both the man and the plank are at rest. Therefore, the center of mass of the system is stationary.

3. **Define Variables**:
- Let \( L \) be the length of the plank.
- Let \( x \) be the distance the plank moves to the left when the man walks to the right.
- The distance the man moves relative to the plank is \( L - x \).

4. **Conservation of Center of Mass**:
- Since no external forces are acting on the system, the center of mass of the system must remain stationary. The position of the center of mass before and after the man walks must be the same.

5. **Set Up the Equation**:
- The initial position of the center of mass \( x_{cm, initial} \) is:
\[ x_{cm, initial} = \frac{M \cdot 0 + 3M \cdot \frac{L}{2}}{4M} = \frac{3L}{8} \]
- After the man walks, the new position of the center of mass \( x_{cm, final} \) is:
\[ x_{cm, final} = \frac{M \cdot (L - x) + 3M \cdot \left(\frac{L}{2} - x\right)}{4M} \]
- Setting \( x_{cm, initial} = x_{cm, final} \):
\[ \frac{3L}{8} = \frac{M(L - x) + 3M\left(\frac{L}{2} - x\right)}{4M} \]

6. **Simplify the Equation**:
- Multiply through by \( 4M \):
\[ \frac{3L}{2} = (L - x) + 3\left(\frac{L}{2} - x\right) \]
- Expand and simplify:
\[ \frac{3L}{2} = L - x + \frac{3L}{2} - 3x \]
\[ \frac{3L}{2} = 2L - 4x \]
\[ 4x = 2L - \frac{3L}{2} = \frac{4L - 3L}{2} = \frac{L}{2} \]
\[ x = \frac{L}{8} \]

7. **Find the Distance the Man Moves**:
- The distance the man moves relative to the ground is:
\[ L - x = L - \frac{L}{8} = \frac{8L - L}{8} = \frac{7L}{8} \]

### Final Answer:
The distance that the man moves relative to the ground is \( \frac{7L}{8} \).