Q127 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Physics

A particle is released freely from a height $H$. At a certain height, its kinetic energy is two times of its potential energy. Then, the height and the speed of the particle at that instant are respectively ( $g=$ acceleration due to gravity)

  • A
    $\frac{H}{3}, \sqrt{\frac{2 g H}{3}}$
  • Correct answer
    B
    $\frac{H}{3}, 2 \sqrt{\frac{g H}{3}}$
  • C
    $\frac{2 H}{3}, \sqrt{\frac{2 g H}{3}}$
  • D
    $\frac{H}{3}, \sqrt{2 g H}$

Solution

If particle falls by a distance $x$, then $$ \begin{aligned} \mathrm{KE}=\frac{1}{2} m v^2 & =\frac{1}{2} m(2 g x)=m g x \\ \mathrm{PE} & =m g(H-x) \end{aligned} $$ As, $\mathrm{KE}=2(\mathrm{PE}) \Rightarrow m g x=2 m g(H-x) \Rightarrow x=\frac{2 H}{3}$ So, height of particle is $H-x=\frac{H}{3}$. Speed of particle at $\frac{H}{3}$ distance $=\sqrt{2 g\left(\frac{2 H}{3}\right)}=2 \sqrt{\frac{g H}{3}}$

Hence, the correct answer is option: 'B'.

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