A rope is strecthed 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 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)

Options

A. `13.8`
B. `18.3`
C. `3.18`
D. `31.8`

Answer

B

Explanation

`F =100N, m_(1) = 250kg, m_(2) =500kg, S =100m`
Force acts on both the boats is same.
using `F =ma, F =m_(1)a_(1),F =m_(2)a_(2)`
`a_(1)=(F)/(m_(1)) = (100)/(250) , a_(2) = (F)/(m_(2)) = (100)/(500)`
`a_(1) =0.4ms^(-2) a_(2)=0.2ms^(-2)`
Relative acceleration `a_(rel) =a_(1) +a_(2)`
`a_(rel) =0.4 +0.2 =6.0 ms^(-2)`
using ` s = ut + (1)/(2) a_(rel) t^(2) ,u =0`
` t =sqrt((2s)/(a_(rel))) =sqrt((2xx100)/(0.6)) =18.3s` .