Question

If a ball of mass $0.1 kg$ hits the ground from the height of $20 m$ and bounce back to the same height then find out the force exerted on the ball if the time of impact is $0.04 sec$.
$(g = 10\, m/s^2$)

• A. -100N
• B. -10N
• C. -1N
• D. None of these

Answer 1

Answer: (A)

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 v₂= 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]