Question

A man of mass $M=50\text{kg}$ stands at one end of a boat (of length $L$) which is floating in still water. The man walks to the other end of the boat. Length of the boat is $L=10\text{m}$ and mass $M=20\text{kg}$. Then the distance moved by the boat relative to the ground is

• A. $\frac{50}{7}\text{m}$
• B. $7.5\text{m}$
• C. $\frac{25}{7}\text{m}$
• D. $\frac{12.5}{7}\text{m}$

Answer 1

The correct option is A $\frac{50}{7}\text{m}$

Here, no external force is acting on the system. Hence $X_{com}$ will not change its position.



Assume that when the man reaches the other end, boat moves $x$ distance left, relative to the ground.

Hence, distance travelled by the man with respect to the ground, $x_1=(L-x)=(10-x)\text{m}$
Distance travelled by the boat with respect to the ground $\left(x_2\right)=-x\text{m}$

As we know, $X_{com}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
Here $X_{com}=0$
$\Rightarrow 0=\frac{50\times (10-x)+\left(20\right)(-x)}{50+20}$
$\Rightarrow 500-50x-20x=0$
$\Rightarrow 70x=500$
$\therefore x=\frac{50}{7}\text{m}$ towards left