Two persons are facing each other on two boats floating on still water. They are holding the two ends of a rope. When either of them or both pull the rope, then the boats meet at the same point whatever the pull on the rope may be. Give reasons for this. Is there any difference in the times taken by the boats to meet, for different forces applied on the rope?

Two persons are facing each other on two boats floating on still water

1.	Two persons are facing each other on two boats floating on still water. They are holding the two ends of a rope. When either of them or both pull the rope, then the boats meet at the same point whatever the pull on the rope may be. Give reasons for this. Is there any difference in the times taken by the boats to meet, for different forces applied on the rope?
Answer» Solution :LET the MASSES of the first and the second MEN together with their respective boats be `m_(1) and m_(2)`. Let the force applied by ONE of them on the rope be F. Acceleration of the 1 st boat = `(F)/(m_(1)) = a_(1) ` , and that of the 2nd boat = `(F)/(m_(2)) = a_(2)`. Let the displacements of the two boats before meeting be `s_(1) and s_(2)` respectively. They MEET after a time t. `therefore""s_(1) = (1)/(2)a_(1)t^(2) = (1)/(2) (F)/(m_(1))t^(2)`, ` s_(2) = (1)/(2) a_(2) t^(2)= (1)/(2) (F)/(m_(2)) t^(2)` Hence, `(s_(1))/(s_(2)) = (m_(2))/(m_(1))`. Thus for any value of the force F, the meeting point of the two boats divide the initial distance in the ratio of `m_(2) : m_(1)` . Hence, the contact point is a fixed point. Now,`t^(2) = (2s_(1) m_(1))/(F) = (2 s_(2) m_(2))/(F)`. As ` s_(1) m_(1) = s_(2) m_(2)` , time remains unchanged for a fixed F. But the time taken would be different for different values of F.