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A 45 kg woman stands up on a 60 kg platform of length 5m, that sits on a frictionless surface. She walks from a point 1 meter from one end of the platform to a point 1 meter from the other end of the platform. Assume the walking velocity vwalk of the woman is variable!! Calculate how far the canoe (platform) slides during the walk.

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we know that in the absense of a net external force the velocity of an object remains constant or remains zero if it was initially in rest.since the platform was initially at rest,it will not move and as a result centre of mass will not move

now equation of centre of mass is $(\sum m_{i}x_{i})\setminus \sum m_{i}$

now i am taking origin at centre of platform .SO

earlier centre of mass was at (60 x0 + 45 x 1.5)/105=0.643m from centre of platform towards first end (say right side) of platform

now let movement of platform be x m from its centre towards right side when lady moves towards left side.

therefore 0.643=(60* x +45 * (-1.5+x))/105

therefore x =1.2857 m towards the first end(right side) of platform

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