Question

A man of mass $M=50\text{kg}$ stands at one end of the boat which is floating on still water. The man walks to the other end of the boat. Length of the boat is $L=10\text{m}$ and mass $M^{\prime }=100\text{kg}$. Then find the distance moved by the boat relative to the ground

• A. $\frac{10}{3}\text{m}$
• B. $\frac{5}{3}\text{m}$
• C. $\frac{2}{3}\text{m}$
• D. $0\text{m}$

Answer 1

The correct option is A $\frac{10}{3}\text{m}$

Here no external force is acting on the system. Hence $x_{com}=0$.
COM will not change its position.

Assume


When man reaches the other end boat moved $x$ distance relative to the ground.

Hence,
Distance travelled by the man w.r.t ground,
$x_1=(L-x)=(10-x)\text{m}$
Distance travelled by the boat w.r.t ground is $x\text{m}$
As we know,
$\Delta x_{com}=\frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}$
$\Delta x_{com}=0$
$0=\frac{50\times (10-x)+\left(100\right)(-x)}{50+100}$
($-x$ because we assumed rightward movement as positive)
$\Rightarrow 500-50x-100x=0$
$\Rightarrow x=\frac{500}{150}$
$\therefore x=\frac{10}{3}\text{m}$