Two balls of same mass are dropped from the same height h, on to the floor. The first ball bounces t

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Two balls of same mass are dropped from the same height h, on to the floor. The first ball bounces to a height h/4, after the collision & the second ball to a height h/16. The impulse applied by the first & second ball on the floor are I1 and I2 respectively. Then

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Sol.$O^{2} = V_{2}^{2} - 2g.h/4$⇒ $V_{2} = \sqrt{gh/2}$$I_{1} = {\bar{P}}_{f} - {\bar{P}}_{i} = m.\sqrt{\frac{gh}{2}} + m\sqrt{2gh}$$I_{1} = \frac{3m\sqrt{gh}}{\sqrt{2}}$$0^{2} = v_{3}^{2} - 2g\frac{h}{16}$⇒ $v_{3} = \frac{\sqrt{gh}}{2\sqrt{2}}$∴ $I_{2} = m.\frac{\sqrt{2gh}}{4} + m\sqrt{2gh}$I2 = $\frac{5m\sqrt{2gh}}{4}$Now $\frac{I_{1}}{i_{2}} = \frac{\frac{3}{\sqrt{2}}}{\frac{5}{4}\sqrt{2}} = \frac{6}{5}$⇒ 5I1 = 6I2

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