Direction (Q. Nos. 22) A solid ball of mass M and radius R is sliding on a smooth horizontal surface with velocity of shown in figure solely and smoothly it comes on the inclined plane of inclination \[30{}^\circ \].  Based on the above information answer the following questions: |
A) If incline is smooth, then the maximum height attained by the ball is \[\frac{v_{0}^{2}}{2g}\].
B) If initially ball is performing pure rolling motion on horizontal surface and Incline is rough enough to prevent any slipping, then the maximum height attained by the ball is \[\frac{7v_{0}^{2}}{10g}\].
C) If initially ball is performing pure rolling motion on horizontal surface and the incline is smooth, then maximum height attained by the ball is \[\frac{v_{0}^{2}}{2g}\].
D) All of the above
Correct Answer: D
Solution :
If incline is smooth, then from energy conservation, \[0-\frac{mv_{0}^{2}}{2}=-mgh\] \[\Rightarrow \] \[h=\frac{v_{0}^{2}}{2g}\] For option \[0-\,\left( \frac{mv_{0}^{2}}{2}+\frac{I\omega _{0}^{2}}{2} \right)=-mgh\] where, \[I=\frac{2}{5}m{{R}^{2}}\] and \[{{\omega }_{0}}=\frac{{{v}_{0}}}{R}\] \[\Rightarrow \] \[h=\frac{7v_{0}^{2}}{10g}\] For option [c] In this case the rotation kinetic energy of ball is not changing as ball moves up the incline as there is no friction and hence from work-energy theorem, \[\frac{I\omega _{0}^{2}}{2}-\,\left( \frac{mv_{0}^{2}}{2}+\frac{I\omega _{0}^{2}}{2} \right)=-mgh\] \[\Rightarrow \] \[h=\frac{v_{0}^{2}}{2g}\]
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