Question

Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the process?

Answer 1

The boat will travel 0.13 m to the right.

Explanation:

Step 1:

m₁ = $60 \mathrm{kg}$    

m₂ = $40 \mathrm{kg}$      

m₃ = $50 \mathrm{kg}$

Let A be the system's source.

So, we can tell Mr. Verma and Mr. Mathur are in the boat's extreme position

Step 2:

Centre of mass  $=\frac{60 \times 0+40 \times 2+50 \times 4}{150}$

$=\frac{0+80+200}{150}$

$=\frac{280}{150}$

$=1.86$

The center of mass is at a distance of 1.86 from point A, which is the system's origin.

So, when both arrive at the mid-point of the boat, the center of mass is 2 meters from A.

Center move $=2-1.86=0.13 \mathrm{m}$ to the right.

Therefore, since there is no external force in the longitudinal direction, the center of mass will not change.

The boat will therefore travel $0.13 \mathrm{m}$ to the right.