A girl is sitting with her dog at the left end of a boat of length 10.0 m. The mass of the girl, her dog and the boat are 60.0 kg, 30.0 kg and 100.0 kg respectively. The boat is at rest in the middle of the lake. Calculate the centre of mass of the system. If the dog moves to the other end of the boat, the girl staying at the same place, how far and in what direction does the boat move?

Introduction

This question is based on the principle of conservation of momentum and center of mass in a system with no external horizontal forces. Since the boat is on a lake, the system is isolated, and the center of mass of the system must remain unchanged in the absence of external forces.

Given Data

Mass of girl = 60.0 kg
Mass of dog = 30.0 kg
Mass of boat = 100.0 kg
Length of boat = 10.0 m
Boat is initially at rest, in calm water.

Step 1: Define the Coordinate System

Let us take the left end of the boat as the origin (0 m). The positions of the components of the system are as follows:

Girl: x₁ = 0 m
Dog: x₂ = 0 m (initially beside the girl)
Boat’s center: x₃ = 5.0 m (since the boat is 10 m long)

Step 2: Calculate Initial Center of Mass (COM)

Total mass of system: M = 60 + 30 + 100 = 190 kg

Initial COM (X_initial):

X_initial = (60×0 + 30×0 + 100×5) / 190 = (0 + 0 + 500)/190 ≈ 2.63 m

Step 3: Dog Moves to Right End of Boat

The girl stays at 0 m relative to the boat. The dog moves to 10.0 m (right end of boat). To keep COM same in the water frame, the boat must shift slightly in the opposite direction.

Let the boat move a distance x to the left. Then all objects inside it move the same amount to the left.

New positions relative to original coordinate:

Girl: 0 – x = -x
Dog: 10 – x
Boat center: 5 – x

Step 4: Use Conservation of Center of Mass

New COM = Old COM = 2.63 m

So,

(60×(-x) + 30×(10 – x) + 100×(5 – x)) / 190 = 2.63

Simplifying:

[-60x + 300 – 30x + 500 – 100x] / 190 = 2.63

(800 – 190x) / 190 = 2.63

800 – 190x = 190 × 2.63 = 499.7

190x = 800 – 499.7 = 300.3

x = 300.3 / 190 ≈ 1.58 m

Final Answers

Initial center of mass: 2.63 m from left end of boat
Boat moves: 1.58 m to the left

Conclusion

This is a classic problem demonstrating conservation of the center of mass. Since no external force is acting on the system, the center of mass remains fixed, and the boat moves in the opposite direction of the dog’s motion to maintain balance. The boat shifts approximately 1.58 meters to the left.