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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.

7 years ago


Answers : (1)


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

7 years ago

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