Question

A person of mass 60 kg is sitting at one of the extreme ends of a boat mass 40 kg and having a length of 4 m (assume that mass is uniformly distributed across the boat). The boat is floating on still water. The person then moves to the middle of the boat. Neglecting friction with the water, the distance moved by the boat on the water, in meters, is

• A. 1.2

Answer 1

Answer: (A)

1.2 m

Answer 2

The boat moves 1.2 meters.

Solution:

1. Initial Setup:

   • Mass of person, $m_p = 60$ kg
   • Mass of boat, $m_b = 40$ kg
   • Length of boat = 4 m → The boat’s center (geometrical center) is at 2 m from either end.

2. Initial Positions:
   Assume the boat’s center is at position X. The person sits at the extreme end, 2 m from the center; choose the right end (+2 m).

   • Person’s initial position = X + 2
   • Boat’s center (mass center of boat) = X

3. Global Center of Mass (COM):
   Since no external horizontal forces act, the global COM remains fixed.

   $$\text{COM} = \frac{m_p (X+2) + m_b (X)}{m_p + m_b} = \frac{60(X+2) + 40X}{100} = X + \frac{120}{100} = X + 1.2$$

   Choose the water frame such that the COM is at 0 initially:

   $$X + 1.2 = 0 \quad \Rightarrow \quad X = -1.2 \text{ m}$$

   (Thus, the boat’s center initially is at –1.2 m.)

4. Final Position:
   After the person moves to the middle of the boat, they sit at the boat’s center. Let the new boat center be X'.

   • Person’s final position = X'
   • Boat’s center remains the same as the reference for boat mass.

   Now, the global COM is:

   $$\text{COM} = \frac{60X' + 40X'}{100} = X'$$

   Since the COM must remain at 0:

   $$X' = 0 \text{ m}$$

5. Boat’s Displacement:
   The boat’s center moves from –1.2 m to 0 m.
   Hence, the displacement = 1.2 m.