Question

A boat of length $10\text{m}$ and mass $450\text{kg}$ is floating without motion in still water. A man of mass $50\text{kg}$ standing at one end of it walks to the other end of it and stops. The magnitude of the displacement of the boat (in $\text{metre}$) relative to ground is

• A. $0$
• B. $1$
• C. $2$
• D. $5$

Answer 1

The correct option is B $1$

Given:
$m_1=450\text{kg};m_2=50\text{kg}$

If the net external force on a system of the particle is zero, the centre of mass of the system will not accelerate.

So, we can also say that if a man moves from one side of the boat to the other, then there will be no change in centre of mass since there is no horizontal force on the boat.

Thus,
$\because |{\overrightarrow{F}}_{net}{|}_{ext}=0\Rightarrow {\overrightarrow{a}}_{com}=0$

$\Rightarrow m_1{x}_1+m_2{x}_2=0$

Where, $x_1$ is the displacement of the boat with respect to ground.


As distance covered by man relative to boat is $10\text{m}$, so if

$x_1=-x$, then $x_2=(10-x)$

$\Rightarrow 450(-x)+50(10-x)=0$

$\Rightarrow 450x-500+50x=0$

$\therefore x=1\text{m}$

Hence, $\left(B\right)$ is the correct answer.