Q143 / 180 · KCET 2023 · Physics

A ball of mass $0.2 \mathrm{~kg}$ is thrown vertically down from a height of $10 \mathrm{~m}$. It collides with the floor and loses $50 \%$ of its energy and then rises back to the same height. The value of its initial velocity is

  • A
    Zero
  • Correct answer
    B
    $14 \mathrm{~ms}^{-1}$
  • C
    $196 \mathrm{~ms}^{-1}$
  • D
    $20 \mathrm{~ms}^{-1}$

Solution

Given, mass of the ball, $m=0.2 \mathrm{~kg}$ Height from the surface of floor, $h=10 \mathrm{~m}$ Let initial velocity be $u$. Total energy as initial point $\mathrm{TE}=\frac{1}{2} m v^2+m g h$ After collision, remaining energy $=\frac{\frac{1}{2} m u^2+m g h}{2}$ With this remaining energy the ball bounces upto height $h$ Therefore, $\frac{\frac{1}{2} m u^2+m g h}{2}=m g h$ $\begin{aligned} & \Rightarrow \frac{1}{2} m u^2+m g h=2 m g h \\ & \Rightarrow \begin{aligned} u^2 & =\sqrt{2 g h} \\ & =\sqrt{2 \times 10 \times 10}=14 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$

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

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