A MAN OF MASS M STANDS AT ONE END OF A PLANK OF LENGTH L WHICH LIES AT REST ON A FRICTIONLESS HORIZONTAL SURFACE. THE MAN WALKS TO THE OTHER END OF THE PLANK. IF THE MASS OF THE PLANK IS (M/3), THE DISTANCE THAT THE MAN MOVES RELATIVE TO THE GROUND WILL BE _________ A) L. B) L/4. C) 3L/4. D) L/3

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2014-10-18T12:54:39+05:30

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Mass of man = M
mass of plank = M/3
let man is standing at the starting of plank(end A), at x=0
end of plank(end B) is at x=L
CM of man is at x=0
CM of plank is at x=L/2

position of CM of system =  \frac{M*0+(M/3)*(L/2)}{M+(M/3)} = \frac{ML/6}{4M/3} = \frac{L}{8}

Let after walking to end A from end B, he travels a distance z.
CM of man is at x=z
CM of plank will shift (L-z) towards left,
CM of plank is at = [L/2 - (L-z)] = (z-L/2)
Since there is no external force, CM of system won't change its position.

 \frac{L}{8}= \frac{(M*z)+(M/3)*(z-L/2)}{M+M/3}

 \frac{L}{8}= \frac{Mz+Mz/3-ML/6}{4M/3}

 \frac{L}{8}= \frac{4Mz/3-ML/6}{4M/3}

 \frac{L}{8}=  \frac{4Mz/3}{4M/3} -  \frac{ML/6}{4M/3}

 \frac{L}{8} = z- \frac{L}{8}

z= \frac{L}{8} + \frac{L}{8} = \frac{L}{4}

So answer is L/4

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2014-10-20T19:55:27+05:30

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See  diagram.  When man is at one end of the plank the center of mass is at a distance from the end of the plank near he stands:
                (M/3 * L/2 + M 0 ) / (M/3 + M ) = L/8

Center of mass of the system is L/8 distance away from the man, when he stands at one end of the plank.

Now the man walks to the other end of the plank. The center of mass of the system remains at the same position as there is no net external force on the system.  The man is on the other side of center of mass now, and he is at distance L/8 from the center of mass.

Distance moved by the man =  L/8 + L/8 = L/4 

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