Question

A wooden plank of mass 90 kg and length 3.3 m is floating on still water. A girl of mass 20 kg walks from one end to the other end of the plank. The distance through which the plank moves is

• A. \(30\) cm
• B. \(40\) cm
• C. \(80\) cm
• D. \(60\) cm

Hint:

In the absence of external horizontal forces, the center of mass of a system remains unchanged. When a person walks on a floating plank, the plank moves in the opposite direction to conserve the center of mass.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept 
Since there is no external force acting on the system  in the horizontal direction, the center of mass of the system must remain unchanged.
Step 2: Initial and Final Position of the Center of Mass 
Let \( x \) be the distance moved by the plank in the opposite direction when the girl moves from one end to the other. Initially, the center of mass of the system is: \[ X_{\text{initial}} = \frac{(M x_{\text{plank}} + m x_{\text{girl}})}{M + m} \] where: - \( M = 90 \) kg (mass of plank), - \( m = 20 \) kg (mass of girl), - \( x_{\text{plank}} = \frac{3.3}{2} \) m (center of the plank initially at its midpoint), - \( x_{\text{girl}} = 0 \) m (girl starts at one end). Step 3: Final Center of Mass Position 
After the girl moves to the other end: \[ X_{\text{final}} = \frac{(M (x_{\text{plank}} + x) + m (3.3 - x))}{M + m} \] Since the center of mass remains unchanged: \[ \frac{90 \times \frac{3.3}{2} + 20 \times 0}{90 + 20} = \frac{90 \times (\frac{3.3}{2} + x) + 20 \times (3.3 - x)}{110} \] Solving for \( x \): \[ x = \frac{20 \times 3.3}{90 + 20} = \frac{66}{110} = 0.6 \text{ m} = 60 \text{ cm} \] 
Step 4: Conclusion 
Thus, the plank moves a distance of: \[ \boxed{60 \text{ cm}} \]