Mass `2m` is kept on a smooth circular track of mass `m` which is kept on a smooth horizontal surface. The circular track is given a horizontal velocity `sqrt(2gR)` towards left. Find the maximum height reached by `2m.`

Answer

`R//3`

Explanation

Let `v` be the final speed o block when it is at maximum height `h`. At that instant the speed of circular track shall also be `v`.

From conservation of momentum
`msqrt(2gR)=(m+2m)v ...(1)`
From conservation of energy
`(1)/(2)m(2gR)=(1)/(2)(m+2m)v^(2)+2mgh .....(2)`
solving `(1)` and `(2)`, we get
`2h=(2)/(3)R Ans. R//3`