question_answer

A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing, it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of

A) 16/25               

B) 2/5

C) 3/5                               

D) 9/25

Correct Answer: B

Solution :

If ball falls from height \[{{h}_{1}}\] and bounces back up to height\[{{h}_{2}}\], then \[e=\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}\] Similarly if the velocity of ball before and after collision are \[{{\nu }_{1}}\,\,and\,\,{{\nu }_{2}}\] respectively, then \[e=\frac{{{\nu }_{2}}}{{{\nu }_{1}}}\] So \[\frac{{{\nu }_{2}}}{{{\nu }_{1}}}=\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}=\sqrt{\frac{1.8}{5}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\] That is, fractional loss in velocity \[=\,\,\,1\,-\frac{{{\nu }_{2}}}{{{\nu }_{1}}}=1-\frac{3}{5}=\frac{2}{5}\]