KCET · Physics · Work Power Energy
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
- B \(14 \mathrm{~ms}^{-1}\)
- C \(196 \mathrm{~ms}^{-1}\)
- D \(20 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(14 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
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, \(\frac{\frac{1}{2} m u^2+m g h}{2}=m g h\)
\(\begin{aligned} & \Rightarrow \frac{1}{2} m u^2+m g h=2 m g h \\ & \Rightarrow \begin{aligned} u^2 & =\sqrt{2 g h} \\ & =\sqrt{2 \times 10 \times 10}=14 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}\)
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, \(\frac{\frac{1}{2} m u^2+m g h}{2}=m g h\)
\(\begin{aligned} & \Rightarrow \frac{1}{2} m u^2+m g h=2 m g h \\ & \Rightarrow \begin{aligned} u^2 & =\sqrt{2 g h} \\ & =\sqrt{2 \times 10 \times 10}=14 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}\)
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