# The figure shows two boats A and B with B connected by a string on the surface of a lake. A person in boat A pulls the string by a constant force of 50N. The combined mass of boat A and the person in it is equals to 250 kg and that of boat B is 500 kg. Show that the velocity of boat A relative to the boat B , 5 sec after the person in the first boat begins to pull the rope is 1.5 m/s.

Grade:12

## 1 Answers

askIITianexpert IITDelhi
8 Points
14 years ago

Considering both boats A & B as a system will render pull of the person as an internal force for the system.

So their "Center of Mass" will not shift.Taking it as origin,

-MaXa + MbXb = 0  where  Ma & Mb are respectively 250kg. & 500kg.

Xa=distance of boat A from origin(Center of Mass of the system here)

Xb=distance of boat B from origin in (+ve) X-direction.

or,            MbXb = MaXa

Differentiating w.r.t time 't' twice,

MbΔb = MaΔa               where  Δa & Δb are respectively accelerations towards each other.

Hence relative acceleration Δab = Δb + Δa = [ (Ma+ Mb)/Ma]*Δb = 3Δb = 3*(50N/500kg)=0.3 mtrs./sec.2

Vab=ab)*(5sec.)=1.5 m/s

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