A ball falls freely from a height of $180 \mathrm{~m}$ on to a hard horizontal floor and repeatedly bounces.…

A ball falls freely from a height of $180 \mathrm{~m}$ on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is 0.5 , the average speed and average velocity of the ball before it ceases to rebound are respectively (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
  1. $10 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}$
  2. $50 \mathrm{~ms}^{-1}, \frac{50}{3} \mathrm{~ms}^{-1}$
  3. $\frac{50}{3} \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}$
  4. $\frac{20}{3} \mathrm{~ms}^{-1}, \frac{50}{3} \mathrm{~ms}^{-1}$

Solution


When ball dropped from height $h$, then time taken to reach the ground $ t_0=\sqrt{\frac{2 h}{g}} \text { and speed, } v_0=\sqrt{2 g h} $ After first collision, its speed will become $v_1=e v_0=e \sqrt{2 g h}$ where, $e=$ coefficient of restitution. Now, the ball will go up and will take time $t_1$, when it stops $ \begin{aligned} & v=u+a t \\ & \quad 0=v_1-g t_1 \\ & t_1=\frac{v_1}{g} \end{aligned} $ It will come down and take same time $t_1$ before second collision. So, time taken between first and second collision is $2 t_1$. Similarly, time taken between second and third collision will be $2 t_3=\frac{2 v_2}{g}$ Total time before it ceaser to rebound $ \begin{aligned} & T=t_0+2 t_1+2 t_2+\ldots \ldots \ldots \\ & T=t_0+\frac{2 v_1}{g}+\frac{2 v_2}{g}+\ldots \ldots \ldots \\ & T=t_0+\frac{2 e v_0}{g}+\frac{2 e^2 v_0}{g}+\ldots \ldots \ldots \\ & \quad\left\{\text { as } v_2=e v_1=e^2 v_0\right\} \\ & T=\sqrt{\frac{2 h}{g}}\left[1+2 e\left(1+e+e^2+e^3+\ldots\right)\right] \\ & \left\{\text { as } v_0=\sqrt{\frac{2 h}{g}}\right\} \end{aligned} $ It forms a GP, $ \begin{aligned} T & =\sqrt{\frac{2 h}{g}}\left[1+2 e\left(\frac{1}{1-e}\right)\right] \\ T & =\sqrt{\frac{2 h}{g}}\left(\frac{1+e}{1-e}\right) \end{aligned} $ Given, $h=180 \mathrm{~m}$ and $e=0.5$ $ \begin{aligned} T & =\sqrt{\frac{2 \times 180}{10}}\left(\frac{1+0.5}{1-0.5}\right) \end{aligned} $ $ T=6 \times 3=18 \mathrm{~s} $ Calculation of distance in entire motion initial height $h_0=h$ After first collision $=h_1$ As $ \begin{aligned} & e=\frac{v}{u}=\sqrt{\frac{2 g h_1}{2 g h_0}}=\sqrt{\frac{h_1}{h}} \\ & \quad h_1=e^2 h, \text { similarly } h_2=e^2\left(e^2 h\right) \\ & h_2=e^4 h \end{aligned} $ Total distance, $ \begin{aligned} & H=h_0+2 h_1+2 h_2+\ldots \ldots \\ & H=h+2 e^2 h+2 e^4 h+\ldots \ldots \\ & H=h\left[1+2 e^2\left(1+e^2+e^4+\ldots \ldots\right)\right] \\ & H=h\left[1+2 e^2\left(\frac{1}{1-e^2}\right)\right] \\ & H=h\left(\frac{1+e^2}{1-e^2}\right) \\ & H=180\left(\frac{1+\frac{1}{4}}{1-\frac{1}{4}}\right), e=0.5=\frac{1}{2} \end{aligned} $ $\begin{aligned} & H=180 \times \frac{5}{3} \\ & H=300 \mathrm{~m} \\ & \text { Average speed, } v=\frac{\text { Total distance }}{\text { Time }} \\ & v=\frac{300}{18}=\frac{50}{3} \mathrm{~m} / \mathrm{s} \\ & \text { Average velocity, } v=\frac{\text { Total displacement }}{\text { Time }} \\ & v=\frac{180}{18}=10 \mathrm{~m} / \mathrm{s} \\ & \end{aligned}$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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