Work Energy Power

JEE Main level, numerical answer, medium

72 of 183

Question 72 of 183

A body is dropped on ground from a height h1{h}_{1} and after hitting the ground, it rebounds to a height h2{h}_{2}. If the ratio of velocities of the body just before and after hitting ground is 44, then percentage loss in kinetic energy of the body is x4\dfrac{x}{4}. The value of xx is _____.

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Answer: 375

The given ratio of the velocities is
v1v2=4\dfrac{{v}_{1}}{{v}_{2}}=4
The percentage loss in kinetic energy is
mv212−mv222mv212×100=x4⇒v21−v22v21=x400⇒1−v22v21=x400⇒1−116=x400⇒x=375\dfrac{\dfrac{m{{v}^{2}}_{1}}{2}-\dfrac{m{{v}^{2}}_{2}}{2}}{\dfrac{m{{v}^{2}}_{1}}{2}}\times 100=\dfrac{x}{4} \Rightarrow \dfrac{{{v}^{2}}_{1}-{{v}^{2}}_{2}}{{{v}^{2}}_{1}}=\dfrac{x}{400} \Rightarrow 1-\dfrac{{{v}^{2}}_{2}}{{{v}^{2}}_{1}}=\dfrac{x}{400} \Rightarrow 1-\dfrac{1}{16}=\dfrac{x}{400} \Rightarrow x=375

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