Question

A boat of mass $M=100\text{kg}$ and length $L=20\text{m}$ is on a frictionless water surface. Two men $A$ of mass $m_1=25\text{kg}$ and $B$ of mass $m_2=50\text{kg}$ are standing at the two opposite ends. Now, both start travelling in opposite directions to each other and they reach the opposite positions of each other. Find the distance travelled by the $A$ w.r.t the ground.

• A. $\frac{160}{7}\text{m}$
• B. $\frac{80}{7}\text{m}$
• C. $\frac{20}{7}\text{m}$
• D. $\frac{40}{7}\text{m}$

Answer 1

The correct option is A $\frac{160}{7}\text{m}$

Initial and final positions are shown in diagram below:


Assume that the distance travelled by the boat when $A$ and $B$ reach the opposite ends is $x\text{m}$ towards left
Then,
Distance covered by man $A$ w.r.t ground $x_1=(L-x)\text{m}$
Distance covered by man $B$ w.r.t ground $x_2=-(L+x)\text{m}$
Distance covered by boat w.r.t ground $x_3=-x\text{m}.$

There is no external force on the system.
Hence, $x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=0$
where
$m_1=$ mass of man $A=25\text{kg}$
$m_2=$ mass of man $B=50\text{kg}$
$m_3=$ mass of boat $=100\text{kg}$

$x_{com}=0=\frac{25\times (20-x)+50[-(20+x\left)\right]+100\times (-x)}{25+50+100}$
$\Rightarrow 500-25x-1000-50x-100x=0$
$\Rightarrow 175x=-500$
$\Rightarrow x=\frac{-20}{7}\text{m}$
(-ve sign means boat will move rightwards)

Hence, the distance travelled by the man $A$ w.r.t ground is $=(L-x)$
$=20-\left(\frac{-20}{7}\right)=\frac{160}{7}\text{m}$