Question:
If a ball of mass 0.1 kg hits the ground from the height of 20m and bounce back to the same height then find out the force exerted on the ball if the time of impact is 0.04 sec.
(g = 10 m/s)
If a ball of mass 0.1 kg hits the ground from the height of 20m and bounce back to the same height then find out the force exerted on the ball if the time of impact is 0.04 sec.
(g = 10 m/s)
Updated On: Aug 12, 2024
- -100N
- -10N
- -1N
- None of these
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The Correct Option is A
Solution and Explanation
The velocity the ball before hitting the ground,
$v^{2}_{1} = u^{2}+2gh$
$u = 0$
so, $v_{1} = \sqrt{2gh}=\sqrt{2\times10\times20}$
= 20m/s in downward direction
After hitting the ground, ball reaches to the same height. So the collision is elastic. So velocity just after hitting the ground v= 20 m/s in upward direction
From impulse - momentum theorem,
Applied force = Rate of change of momentum
Force $= \frac{\Delta P}{\Delta t}=\frac{mv_{2}-mv_{1}}{\Delta t}$
$= \frac{m\left(v_{2}-v_{1}\right)}{\Delta t}=\frac{-0.1\times40}{0.04}= - 100N$
$\quad\quad\quad\quad\quad\quad\quad$[in upward direction]
$v^{2}_{1} = u^{2}+2gh$
$u = 0$
so, $v_{1} = \sqrt{2gh}=\sqrt{2\times10\times20}$
= 20m/s in downward direction
After hitting the ground, ball reaches to the same height. So the collision is elastic. So velocity just after hitting the ground v= 20 m/s in upward direction
From impulse - momentum theorem,
Applied force = Rate of change of momentum
Force $= \frac{\Delta P}{\Delta t}=\frac{mv_{2}-mv_{1}}{\Delta t}$
$= \frac{m\left(v_{2}-v_{1}\right)}{\Delta t}=\frac{-0.1\times40}{0.04}= - 100N$
$\quad\quad\quad\quad\quad\quad\quad$[in upward direction]
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