Question

Mr. Verma $\left(50kg\right)$ and Mr. Mathur $\left(60kg\right)$ are sitting at the two extremes of a $4m$ long boat $\left(40kg\right)$ 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?

• A. $40/3cm$
• B. $50/3cm$
• C. $110/3cm$
• D. $30/10cm$

Answer 1

Answer: (A)

$40/3cm$

$m_1=60kg{m}_2=40kg{m}_3=50kg$

Let A be the origin of the system

Initially, Mr. Verma and Mr. Mathur are at the extreme position of the boat.

$\therefore$ The center of mass will be at distance,

$=\frac{(60\times 0)+(40\times 2)+(50\times 4)}{150}=\frac{280}{150}=1.87m$ from 'A''

When they came to the mid point of the boat the center of moment lies at 2m from 'A'

$\therefore$ The shift in center of moment$=2-1.87=0.13m=13cm=40/3cmtowardsright$.

But as there is no external force in longitudinal direction their center of moment would not shift. So, the boat moves 0.13m or 13cm or 40/3cm towards right.