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JEE Main 2016 · Physics · Work, Energy and Power · medium

A ball is dropped from a height of 5 m onto a horizontal floor and the coefficient of restitution for every impact is 0.6. Taking g=10 m s^-2, the…

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A ball is dropped from a height of 5 m5\ \mathrm{m} onto a horizontal floor and the coefficient of restitution for every impact is 0.60.6. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the total time taken by the ball from the release until it finally comes to rest is
  1. A. 2.5 s2.5\ \mathrm{s}
  2. B. 3.2 s3.2\ \mathrm{s}
  3. C. 5 s5\ \mathrm{s}
  4. D. 4 s4\ \mathrm{s}
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Answer: (D) 4 s4\ \mathrm{s}

The first fall takes t0=2h/g=1 st_0=\sqrt{2h/g}=1\ \mathrm{s} and the speed on first impact is v0=gt0=10 m s−1v_0=gt_0=10\ \mathrm{m\,s^{-1}}. After each bounce the speed is multiplied by ee, so the flight times (up and down) are 2et0, 2e2t0,…2et_0,\ 2e^2t_0,\dots The total time is t=t0+2t0(e+e2+⋯ )=t0(1+2e1−e)=t01+e1−e=1×1.60.4=4 st=t_0+2t_0(e+e^2+\cdots)=t_0\left(1+\dfrac{2e}{1-e}\right)=t_0\dfrac{1+e}{1-e}=1\times\dfrac{1.6}{0.4}=4\ \mathrm{s}. Summing only the downward falls would not include the upward flights.
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