Q) Two skaters A and B of weights 40 kg wt and 60 kg wt, respectively, stand facing each other 5 m apart. They then pull a light rope stretched between them. Where do they meet?

A) Use the fact that center of mass remains stationary in absence of any external forces.

So the length of the rope can be divided in inverse ratio of the masses.

M1 / M2 = L2 / L1

L2 / L1 = 2/3

And the sum of lengths is 5.

Therefore, they meet at 3m from initial position of A.

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## Tuesday, September 16, 2008

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