A boy of mass m is standing at one end of a boat of mass M and length l. The boy walks to the other end, the displacement of the

Options

A. CM is zero
B. boat is `(ml)/(M+m)`
C. boy is `(ml)/(M+m)`
D. boat is `(m)/(M)l`

Answer

To solve the problem of the boy walking from one end of the boat to the other, we will analyze the system using the concept of the center of mass.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a boy of mass \( m \) standing at one end of a boat of mass \( M \) and length \( L \).
- The boy walks to the other end of the boat.

2. **Initial Position of the Center of Mass**:
- Let’s denote the initial position of the boy as \( x_1 = 0 \) (one end of the boat).
- The center of mass of the boat is at \( x_2 = \frac{L}{2} \) (the midpoint of the boat).
- The total mass of the system is \( M + m \).

3. **Calculating Initial Center of Mass**:
\[ x_{cm, initial} = \frac{m \cdot x_1 + M \cdot x_2}{m + M} = \frac{m \cdot 0 + M \cdot \frac{L}{2}}{m + M} = \frac{M \cdot \frac{L}{2}}{m + M} \]

4. **Final Position of the Boy and Boat**:
- When the boy walks to the other end, his new position is \( x_1' = L \).
- Let’s assume the boat shifts left by a distance \( z \), so the center of mass of the boat shifts to \( x_2' = \frac{L}{2} - z \).

5. **Calculating Final Center of Mass**:
\[ x_{cm, final} = \frac{m \cdot x_1' + M \cdot x_2'}{m + M} = \frac{m \cdot L + M \cdot \left(\frac{L}{2} - z\right)}{m + M} \]

6. **Setting Initial and Final Center of Mass Equal**:
- Since there are no external forces acting on the system, the center of mass remains constant:
\[ x_{cm, initial} = x_{cm, final} \]
\[ \frac{M \cdot \frac{L}{2}}{m + M} = \frac{m \cdot L + M \cdot \left(\frac{L}{2} - z\right)}{m + M} \]

7. **Simplifying the Equation**:
- Cross-multiplying gives:
\[ M \cdot \frac{L}{2} = mL + M \cdot \frac{L}{2} - Mz \]
- Rearranging leads to:
\[ Mz = mL \implies z = \frac{mL}{M + m} \]

8. **Displacement of the Boat**:
- The displacement of the boat is \( z = \frac{mL}{M + m} \).

9. **Displacement of the Boy**:
- The boy’s displacement is calculated as:
\[ \text{Displacement of the boy} = L - z = L - \frac{mL}{M + m} = \frac{ML}{M + m} \]

### Final Answers:
- The displacement of the boat is \( \frac{mL}{M + m} \).
- The displacement of the boy is \( \frac{ML}{M + m} \).
- The center of mass of the system does not shift.