A ball of mass 2 kg dropped from a height H above a horizontal surface rebounds to a height h after one bounce. The graph that relates H to h is shown in figure. If the ball was dropped from an initial height of 81 m and made ten bounces, the kinetic energy of the ball immediately after the second impact with the surface was

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Can’t be determined

640 J

480 J

320 J

answer is A.

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Detailed Solution

$e = \sqrt{\frac{h}{H}} = \sqrt{\frac{40}{90}} = \frac{2}{3};$

$V_{n} = {ue}^{n}$

Here n = 2
$V_{2} = \sqrt{2 gh} {(\frac{2}{3})}^{2} \Rightarrow V_{2}^{2} = 2 gh {(\frac{2}{3})}^{4}$

$∴ {KE}_{2} = \frac{1}{2} {mV}_{2}^{2} = \frac{1}{2} m (2 gh) {(\frac{2}{3})}^{4}$

$= \frac{1}{2} (2) (2 \times 10 \times 81) \frac{16}{81} = 320 J$

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