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 of a man walking on a plank on a frictionless surface, we can follow these steps:

### Step-by-Step Solution:

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

2. **Initial Conditions**:
- Initially, both the man and the plank are at rest. Therefore, the initial velocity of the center of mass of the system is zero.

3. **Movement of the Man**:
- When the man walks from one end of the plank to the other, he moves a distance \( L \) relative to the plank.

4. **Displacement of the Plank**:
- As the man walks, the plank will also move in the opposite direction due to the conservation of momentum. Let \( x_p \) be the displacement of the plank.

5. **Center of Mass Consideration**:
- The center of mass of the system must remain stationary since there are no external forces acting on it. The equation for the center of mass is given by:
\[ 0 = \frac{M \cdot x_m + 3M \cdot x_p}{M + 3M} \]
- Here, \( x_m \) is the displacement of the man and \( x_p \) is the displacement of the plank.

6. **Expressing Displacements**:
- The displacement of the man relative to the ground when he walks to the other end of the plank is:
\[ x_m = L + x_p \]
- Substituting this into the center of mass equation:
\[ 0 = M(L + x_p) + 3M x_p \]
- Simplifying gives:
\[ 0 = ML + 4Mx_p \]
- Rearranging gives:
\[ 4Mx_p = -ML \implies x_p = -\frac{L}{4} \]

7. **Finding the Man's Displacement**:
- Now substituting \( x_p \) back to find \( x_m \):
\[ x_m = L + x_p = L - \frac{L}{4} = \frac{3L}{4} \]

8. **Final Result**:
- The distance that the man moves relative to the ground is:
\[ \boxed{\frac{3L}{4}} \]