• # 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)                                                                                                                                                                                            $\frac{16}{25}$                               B)                 $\frac{2}{5}$    C)                            $\frac{3}{5}$    D)                            $\frac{9}{25}$

Key Idea: According to conservation of energy, potential energy at height h = kinetic energy at ground                 Potential energy = Kinetic energy                 i.e.,        $mgh=\frac{1}{2}m{{v}^{2}}$                 $\Rightarrow$               $v=\sqrt{2gh}$                 If ${{h}_{1}}$ and ${{h}_{2}}$ are initial and final heights, then                 ${{v}_{1}}=\sqrt{2g{{h}_{1}}},\,{{v}_{2}}=\sqrt{2g{{h}_{2}}}$                 Loss in velocity                 $\Delta v={{v}_{1}}-{{v}_{2}}=\sqrt{2g{{h}_{1}}}-\sqrt{2g{{h}_{2}}}$                 $\therefore \,\,\,$        Fractional loss in velocity = $\frac{\Delta v}{{{v}_{1}}}$                 $=\frac{\sqrt{2g{{h}_{1}}}-\sqrt{2g{{h}_{2}}}}{\sqrt{2g{{h}_{1}}}}$                 $=1-\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}$                 Substituting the values, we have                 $\therefore \frac{\Delta v}{{{v}_{1}}}=1-\sqrt{\frac{1.8}{5}}$                 $=1-\sqrt{0.36}=1-0.6$                 $=0.4=\frac{2}{5}$