Question

A rope is stretched between two boats at rest. A sailor in the first boat pulls the rope with a constant force of $100N$. First boat with the sailor has a mass of $250kg$ whereas the mass of second boat is double of that mass. If the initial distance between the boats was $100m$, the time taken for two boats to meet each other is:

• A. $13.8s$
• B. $3.18s$
• C. $18.3s$
• D. $31.8s$

Answer 1

Answer: (C)

Here $100N$ force acts on both boat.

Thus, $250a_1=100\Rightarrow a_1=0.4m/s^2$
and $500a_1=100\Rightarrow a_1=0.2m/s^2$
The relative acceleration ,$a=a_1-(-a_2)=0.4+0.2=0.6m/s^2$
Using $S=ut+\frac{1}{2}a{t}^2,$ we get
$100=\frac{1}{2}\left(0.6\right)t^2$ as$(u=0)$
$\therefore t=18.3s$