Question

A dog weighting $5kg$ is standing on a flat boat so that it is $10m$ from the shore. The walks on the boat towards the shore and then halts. The boat weights $20kg$ and one can assume that then no friction between it and the water. How far is the dog from the shore at the end of this time ?

• A. $3.2m$
• B. $m$
• C. $10m$
• D. $6.8m$

Answer 1

The correct option is D $6.8m$

Let $x$ be the distance of the center of mass of boat from the shore and $x_{cm}$ be the distance of the center of mass.

$x_{cm}=\frac{(5\times 10+20\times 1)}{(20+5)}=2+\frac{4}{5}$

Now, new distance from the shore$=x_1+x$

New distance of dog from the shore$=6+x$

Now, $x_{cm}=5(6+x)+\frac{20(x_1+x)}{20+5}=\frac{6}{5}+x+\frac{4}{5}+1$

$x=2-\frac{6}{5}=\frac{4}{5}$

So, distance$=6+\frac{4}{5}=\frac{30+4}{5}m=\frac{34}{5}m=6.8m$