Question

A rope is stretched between two boats at rest. A sailor in the first boat pulls the rope with a constant force of 100 N. First boat with the sailor has a mass of 250 kg whereas the mass of second boat is double of this mass. If the initial distance between the boats was 30 m, the time taken for two boats to meet each other is (neglect water resistance between boats and water)

Answer 1

Given, force acts on both boat ,f = 100
M * a1 = f
So, 250 * a1 = 100

a1 = 100/250 = 0.4 m/s²

Given, mass of second boat is double.(i.e 250 * 2 = 500)
So, 500 * a1 = 100
a1 = 100/500 = 0.2 m/s²

Relative acceleration = 0.4 + 0.2 = 0.6

We know that S = ut + 1/2 at²
ie,
The distance between the boats , S=30 m
30 = 0 + 1/2(0.6)* t²
30 = (0.3) * t²
t² = 30/0.3
t² = 100
t = 10 seconds.