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A small sphere of radius R is held again...

A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The masses of large and small spheres are 4M and M, respectively , this arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the centre of the larger sphere when the smaller sphere reaches the other extreme position.

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The correct Answer is:
`(L + 2R, 0)`

`x_(cm) = (4 ML + M(L + 5R))/(5M) = (4MX + M(X - 5R))/(5M)`
`rArr x = L + 2R`
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