In figure, an inextensible string that connects two bodies of mass M and `m`, passing over a fixed smooth pulley. The body M slides along a smooth vertical rigid bar. If the body M is released from the given position, find the maximum distance raised by body `m`.

Explanation

Here we can use `DeltaK+DeltaU=0`.
Since initially, the system is at rest and finally it comes to rest (instaneously), the change of KE of the system is zero.
`impliesDeltaK=0`
Also `DeltaU=0`.
`implies(DeltaU)_M+(DeltaU)_m=0`
Since the block m rises through a height `h` (say),
`(DeltaU)_m=mgh`
Similarly, the change in PE of the mass M is `Mgh^'`, as it falls through a height `h^'`, say
`implies mgh+(-Mgh^')=0impliesh^'=m/Mh` (i)
When the mass `m` ascends through a distance `h`, the length of the string between the pulley and the bar will be equal to `(a+h)`.
Therefore, `h^'=sqrt((a+h)^2-a^2)impliesh^'=sqrt(h(2a+h))` (ii)
Eliminating `h^'` from (i) and (ii), we obtain
`sqrt(h(2a+h))=m/Mh`
`implies2a+h=m^2/M^2himpliesh=(2a)/((m^2/M^2-1))`