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Physics · System Of Particles

JEE Main 2024 — 4 April, Shift 1 — Question 41

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 :

  1. Option A:

    50%,gh250 \%, \sqrt{\frac{\mathrm{gh}}{2}}

  2. Option B:

    50%, gh 50 \%, \sqrt{\text { gh }}

  3. Option C:

    40%,2gh40 \%, \sqrt{2 \mathrm{gh}}

  4. Option D:

    50%,2gh50 \%, \sqrt{2 \mathrm{gh}}

    Correct

Answer: D

Step-by-step solution

Velocity just before collision =2gh=\sqrt{2 \mathrm{gh}}

Velocity just after collision =2g(h2)=\sqrt{2 g\left(\frac{h}{2}\right)}

∴ΔKE=12 m(2gh)−12mgh\therefore \Delta \mathrm{KE}=\frac{1}{2} \mathrm{~m}(2 \mathrm{gh})-\frac{1}{2} \mathrm{mgh} =12mgh=\frac{1}{2} \mathrm{mgh}

∴%\therefore \% loss in energy =ΔKEKEi×100=12mgh12mg2 h×100=50%=\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 \%

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 1
Subject
Physics
Chapter
System Of Particles
Topic
Impulsive Forces & Impulse-Momentum Equation