Question

A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10 m/s. It then rolls up a smooth inclined plane of inclination ${30}^{\circ }$. Height attained by sphere in meter before stopping is: (take $g=10{\text{ms}}^{-2}$)

Answer 1

As no friction is acting, sphere continues to roll with same angular velocity
$:(T.M.E{)}_f=(T.M.E{)}_i$
$mgh+K.E{.}_{rot}=\frac{1}{2}\times m\times v^2+K.E{.}_{rot}$
$2gh=\frac{1}{2}\times 2\times 100$ (as no rolling occurs on smooth inclined plane)
$h=\frac{100}{2\times 10}=5m$