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A cricket ball of mass $150 \mathrm{~g}$ moving with a speed of $126 \mathrm{~km} /$ $\mathrm{h}$ hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for $0.001 \mathrm{~s}$, the force that the batsman had to apply to hold the bat firmly at its place would be
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Verified Answer
The correct answer is:
$1.05 \times 10^4 \mathrm{~N}$
As given that,
$$
\begin{aligned}
&m=150 \mathrm{~g}=\frac{150}{1000} \mathrm{~kg}=0.15 \mathrm{~kg} \\
&\Delta t=\text { time of contact }=0.001 \mathrm{~s} \\
&u=126 \mathrm{~km} / \mathrm{h}=\frac{126 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s} \\
&=126 \times \frac{5}{18}=35 \mathrm{~m} / \mathrm{s} \\
&v=-126 \mathrm{~km} / \mathrm{h}=-126 \times \frac{5}{18}=-35 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
So, final velocity is acc. to initial force applied by batsman.
So, change in momentum of the ball
$$
\Delta p=m(v-u)=\frac{3}{20}(-35-35) \mathrm{kg}-\mathrm{m} / \mathrm{s}
$$
$$
=\frac{3}{20}(-70)=-\frac{21}{2}
$$
As we know that, force
$$
F=\frac{\Delta p}{\Delta t}=\frac{-21 / 2}{0.001} \mathrm{~N}=-1.05 \times 10^4 \mathrm{~N}
$$
Hence negative sign shown that direction of force will be opposite to initial velocity which taken positive direction. Hence verify the option (c).
$$
\begin{aligned}
&m=150 \mathrm{~g}=\frac{150}{1000} \mathrm{~kg}=0.15 \mathrm{~kg} \\
&\Delta t=\text { time of contact }=0.001 \mathrm{~s} \\
&u=126 \mathrm{~km} / \mathrm{h}=\frac{126 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s} \\
&=126 \times \frac{5}{18}=35 \mathrm{~m} / \mathrm{s} \\
&v=-126 \mathrm{~km} / \mathrm{h}=-126 \times \frac{5}{18}=-35 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
So, final velocity is acc. to initial force applied by batsman.
So, change in momentum of the ball
$$
\Delta p=m(v-u)=\frac{3}{20}(-35-35) \mathrm{kg}-\mathrm{m} / \mathrm{s}
$$
$$
=\frac{3}{20}(-70)=-\frac{21}{2}
$$
As we know that, force
$$
F=\frac{\Delta p}{\Delta t}=\frac{-21 / 2}{0.001} \mathrm{~N}=-1.05 \times 10^4 \mathrm{~N}
$$
Hence negative sign shown that direction of force will be opposite to initial velocity which taken positive direction. Hence verify the option (c).
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