A solid ball rolls down a parabolic path ABC from a height h as shown in figure. Portion AB of the path is rough while BC is smooth. How high will the ball climb in BC?

Explanation

At B, total kinetic energy `=mgh`
Here, `m=` mass of ball
The ratio of ratational to translational kinetic energy would be.
`(K_(R))/(K_(T))=(2)/(5)impliesthereforeK_(R)=(2)/(7)mgh` and `K_(T)=(5)/(7)mgh`
In portiona BC, friction is absent. therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy. Hence, if `H` be the height to which ball climbs in BC, then
`mgH=K_(T)` or `mgH=(5)/(7)mgh` or `H=(5)/(7)h`