Work Energy Power

JEE Main level, single correct, medium

56 of 183

Question 56 of 183

If a rubber ball falls from a height hh and rebounds upto the height of h/2h / 2. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :

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Answer: (A) 50%,2gh50 \%, \sqrt{2 \mathrm{gh}}

 Velocity just before collision =2gh Velocity just after collision =2 g(h2)∴ΔKE=12 m(2gh)−12mgh=12mgh∴% loss in energy =ΔKEKEi×100=12mgh12mg2 h×100=50%\begin{aligned} & \text { Velocity just before collision }=\sqrt{2 \mathrm{gh}} \\ & \text { Velocity just after collision }=\sqrt{2 \mathrm{~g}\left(\frac{\mathrm{h}}{2}\right)} \\ & \therefore \Delta \mathrm{KE}=\frac{1}{2} \mathrm{~m}(2 \mathrm{gh})-\frac{1}{2} \mathrm{mgh} \\ & =\frac{1}{2} \mathrm{mgh} \\ & \therefore \% \text { loss in energy } \\ & =\frac{\Delta \mathrm{KE}}{\mathrm{KE}_{\mathrm{i}}} \times 100=\frac{\frac{1}{2} \mathrm{mgh}}{\frac{1}{2} \mathrm{mg} 2 \mathrm{~h}} \times 100=50 \% \end{aligned}

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