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 this mass. If the initial distance between the boats was $100m$, the time taken for two boats to meet each other is (neglect water resistance between boats and water)

• A. $13.8s$
• B. $18.25s$
• C. $3.18s$
• D. $31.8s$

Answer 1

The correct option is B $18.25s$

Due to tension, $2^{nd}$ boat will got a force of $100$N and due to reaction force first will get a force of $100$N.

$a_1=\frac{100}{250}=\frac{2}{5}m/s^2$

$a_2=\frac{100}{500}=\frac{1}{5}m/s^2$

We have, $S=ut+\frac{1}{2}{at}^2$

$\therefore \frac{1}{2}\times a_1\times t^2+\frac{1}{2}{a}_2{t}^2=100$

$\Rightarrow \frac{1}{2}.\frac{2}{5}\times t^2+\frac{1}{2}.\frac{1}{5}{t}^2=100$

${3t}^2=1000$

$\Rightarrow t=18.26s$