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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2023
  • Hard
  • Numerical

A body is dropped on ground from a height \(h_1\) and after hitting the ground, it rebounds to a height \(h _2\) If the ratio of velocities of the body just before and after hitting ground is \(4\), then percentage loss in kinetic energy of the body is \(\frac{x}{4}\). The value of \(x\) is \(.......\)

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Answer & Solution

Correct Answer

(C) \(375\)

Step-by-step Solution

Detailed explanation

Let \(V_1\) and \(V_2\) are velocity just before and just after hitting the floor.

\(\frac{V_1}{V_2}=4 \Rightarrow V_1=4 V_2\)

\(K E_{\text {bygove }}=\frac{1}{2} m V_1^2\)

\(K E_{\text {affor }}=\frac{1}{2} m V_2^2=\frac{1}{2} \frac{m \cdot V_1^2}{16}\)

\(\Delta K E=\frac{1}{2} m V_1^2\left(\frac{1}{16}-1\right)=\frac{-15}{32} m V_1^2\)

\(\% \text { change }=\frac{\Delta K E}{K E_{\text {before }}} \times 100 \%\)

\(=\frac{-15}{16} \times 100=\frac{-375}{4} \%\)

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