A rubber ball is dropped from a height of 5 m on a plane. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of

3/5

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2/5

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16/25

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9/25

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Solution

The correct option is B 2/5
By conservation of energy, all the potential energy at $5 m$ height is converted into kinetic energy.
$m g h = \frac{1}{2} m v^{2}$
$v^{2} = 2 g h$

Let $h_{1} = 5 m, h_{2} = 1.8 m, v_{1} :$ velocity just before collision, $v_{2} :$ velocity just after collision
$v_{1}^{2} = 2 g h_{1}$
$v_{2}^{2} = 2 g h_{2}$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{5}{1.8}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$
$v_{2} = \frac{3}{5} v_{1}$
Velocity loses by the factor of $= v_{1} - v_{2} = v_{1} - \frac{3}{5} v_{1} = \frac{2}{5} v_{1}$ .
Therefore, reduction factor of velocity = $\frac{2}{5}$

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