question_answer

A one kg ball rolling on a smooth horizontal surface at\[20m{{s}^{-1}}\] comes to the bottom of an inclined plane making an angle of\[{{30}^{o}}\] with the horizontal. Calculate K.E. of the ball when it is at the bottom of incline. How far up the incline will the ball roll? Neglect friction.

Answer:

 As is known from theory,                 Total KE of the ball \[=\frac{7}{10}m{{\upsilon }^{2}}=\frac{7}{10}\times 1\times {{\left( 20 \right)}^{2}}=280J\] As work done in stopping the ball = K.E. of ball \[\therefore \]                                  \[(mg\,\sin \theta )\times s=280\] \[s\,=\frac{280}{mg\,\sin \,\theta }=\frac{280}{1\times 9.8\times \frac{1}{2}}=57.14\,m\]