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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 . (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$100 \mathrm{~N}(+\hat{\mathbf{j}})$
Here, mass of ball $m=0.1 \mathrm{~kg}$
Velocity attained by the ball before hitting the
$\begin{aligned}
\text { ground }(\mathrm{v}) & =\sqrt{2 \mathrm{gh}}(-\hat{\mathrm{j}}) \\
& =\sqrt{2 \times 10 \times 20}(-\hat{\mathrm{j}})=-20 \hat{\mathrm{j}} / \mathrm{s}
\end{aligned}$
Velocity of ball when bounce back to the same height after hitting the ground,
$V^{\prime}=-v=-(-20 \hat{j})=20 \hat{j} \mathrm{~m} / \mathrm{s}$
$\therefore$ Change in velocity
$\Delta V=V^{\prime}-V=20 \hat{j}-(-20 \hat{j})=40 \hat{j} \mathrm{~m} / \mathrm{s}$
$\therefore$ Force exerted on the ball
$\begin{aligned}
\mathrm{f} & =\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\mathrm{m} \Delta \mathrm{V}}{0.04}[\text { given } \Delta \mathrm{t}=0.04] \\
& =\frac{0.1 \times 40 \hat{\mathrm{j}}}{0.04}=100 \mathrm{~N}(\hat{\mathrm{j}})
\end{aligned}$
Velocity attained by the ball before hitting the
$\begin{aligned}
\text { ground }(\mathrm{v}) & =\sqrt{2 \mathrm{gh}}(-\hat{\mathrm{j}}) \\
& =\sqrt{2 \times 10 \times 20}(-\hat{\mathrm{j}})=-20 \hat{\mathrm{j}} / \mathrm{s}
\end{aligned}$
Velocity of ball when bounce back to the same height after hitting the ground,
$V^{\prime}=-v=-(-20 \hat{j})=20 \hat{j} \mathrm{~m} / \mathrm{s}$
$\therefore$ Change in velocity
$\Delta V=V^{\prime}-V=20 \hat{j}-(-20 \hat{j})=40 \hat{j} \mathrm{~m} / \mathrm{s}$
$\therefore$ Force exerted on the ball
$\begin{aligned}
\mathrm{f} & =\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\mathrm{m} \Delta \mathrm{V}}{0.04}[\text { given } \Delta \mathrm{t}=0.04] \\
& =\frac{0.1 \times 40 \hat{\mathrm{j}}}{0.04}=100 \mathrm{~N}(\hat{\mathrm{j}})
\end{aligned}$
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