A small bucket of mass `M` is attached to a long inextensible cord of length `L`. The bucket is released from rest when the cord is in a horizontal position. In its lowest position the bucket scoops up m of water, what is the height of the swing above the lowest position?

Answer

`l`

Explanation

Let the bucket of the mass `M` reach the lowest position with a speed `v`. As it scoops up a water of mass `m`, at the lowest position, its peed decreases to `v`,. The speed of the bucket just after scooping of water is given by conserving momentum of the system just before and after scooping of water we obtain
`Mv=implies (M+m)v'`
`implies v'=(Mv)/(M+m)`.....i
Conservation of energy of the bucket between `A` and `B` (just before scooping) yields, `(1/2)mv^(2)-mgl=0`
`implies v=sqrt(2gl)`......ii

Conservation of energy of the bucket `+` water between `B` and `C` after the scooping) of water, yields
`1/2mv^('2)-mgh=0impliesh=v^('2)/(2g)`.............iii
By using i, ii and iii we obtain `h=(M/(M+m))^(2)l`