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

At B, total kinetic energy =m g h

Here, m= mass of ball

The ratio of rotational to translational kinetic energy would be,

$\frac{{\mathrm{K}}_{\mathrm{R}}}{{\mathrm{K}}_{\mathrm{T}}}=\frac{2}{5}\Rightarrow \therefore {\mathrm{K}}_{\mathrm{R}}=\frac{2}{7}\mathrm{mgh}\text{ and }{\mathrm{K}}_{\mathrm{T}}=\frac{5}{7}\mathrm{mgh}$

In portion B C, 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 B C, then

$\mathrm{mgH}={\mathrm{K}}_{\mathrm{T}}$ or $\mathrm{mgH}=\frac{5}{7}\mathrm{mgh}$ or $\mathrm{H}=\frac{5}{7}\mathrm{h}$