First slide

Work done by Diff type of forces

Question

A ball is thrown vertically downwards from a height of 20 m with an initial velocity $ν_{0}$ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $ν_{0}$ is (Take g = $10 {ms}^{- 2}$ )

Moderate

A

$28 {ms}^{- 1}$
B

$10 {ms}^{- 1}$
C

$14 {ms}^{- 1}$
D

$20 {ms}^{- 1}$

Solution

The situation is shown in the figure
Let $ν$ be the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy,
Gain in kinetic energy = loss in potential energy
$i.e. \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = m g h$
$(where m is the mass of the ball)$
$or v^{2} - v_{0}^{2} = 2 g h . . . . (i)$
Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height h.
$∴ \frac{50}{100} (\frac{1}{2} m v^{2}) = m g h$
$\frac{1}{4} v^{2} = g h or v^{2} = 4 g h$
$Substituting this value of v^{2} in eqn. (i), we get$
$4 g h - v_{0}^{2} = 2 g h$
$or v_{0}^{2} = 4 g h - 2 g h = 2 g h or v_{0} = \sqrt{2 g h}$
$Here, g = 10 {ms}^{- 2} and h = 20 m$
$∴ v_{0} = \sqrt{2 (10 {ms}^{- 2}) (20 m)} = 20 {ms}^{- 1}$

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