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                          [CBSE PMT 1998]

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 \[{{v}_{1}}\] and \[{{v}_{2}}\] respectively then \[e=\frac{{{v}_{2}}}{{{v}_{1}}}\] So \[\frac{{{v}_{2}}}{{{v}_{1}}}=\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}=\sqrt{\frac{1.8}{5}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\] i.e. fractional loss in velocity \[=1-\frac{{{v}_{2}}}{{{v}_{1}}}=1-\frac{3}{5}=\frac{2}{5}\]