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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.```
8 years ago

8 Points
```										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=(Δab)*(5sec.)=1.5 m/s
```
8 years ago
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