• 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)                            \[\frac{9}{25}\]

    Correct Answer: B

    Solution :

                    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                 \[\therefore \frac{\Delta v}{{{v}_{1}}}=1-\sqrt{\frac{1.8}{5}}\]                 \[=1-\sqrt{0.36}=1-0.6\]                 \[=0.4=\frac{2}{5}\]

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