A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take ) [2025]

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Solution

Q.
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8 {m/s}^{2}$ ) [2025]

1 0

2 84 NS

3 21 NS

4 7 NS

Ans.
(3)

Mass of ball, $m = 0.5$ kg

$h_{1} = 40 m$ and $h_{2} = 10 m$

So, velocity of ball before hitting the ground

$v_{1} = \sqrt{2 g h_{1}} = \sqrt{2 \times 9.8 \times 40} \Rightarrow v_{1} = 28 m/s$

After hitting the ground, let final velocity of the ball be $v_{2} .$

$∴ v_{2} = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 \times 10} \Rightarrow v_{2} = 14 m/s$

Change in momentum, $Δ p = m [v_{2} - (- v_{1})]$

$= \frac{1}{2} [14 - (- 28)]$

$Δ p = 21 kg \Rightarrow Impulse, Δ p = 21 N s$

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