Question
A batsman deflects a ball by an angle of $45^{\circ}$ without changing its initial speed which is equal to $54 km / h$. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg .)
✓
Answer
The ball struck by the bat is deflected back such that the total angle is $45^{\circ}$.
Now, initial momentum of ball $= mu \cos \theta$
$
\begin{aligned}
& =\frac{0.15 \times 54 \times 1000 \times \cos 22.5}{3600} \\
& =0.15 \times 15 \times 0.9239 \text { along ON }
\end{aligned}
$
Final momentum of ball $=$ mucos $\theta$ along ON
Impulse $=$ change in momentum
$=\operatorname{mucos} \theta-(-m u c o s \theta)$
$=2 mucos \theta$
$=2 \times 0.15 \times 15 \times 0.9239$
i.e., Impulse $=4.16 kg ms ^{-1}$
Now, initial momentum of ball $= mu \cos \theta$
$
\begin{aligned}
& =\frac{0.15 \times 54 \times 1000 \times \cos 22.5}{3600} \\
& =0.15 \times 15 \times 0.9239 \text { along ON }
\end{aligned}
$
Final momentum of ball $=$ mucos $\theta$ along ON
Impulse $=$ change in momentum
$=\operatorname{mucos} \theta-(-m u c o s \theta)$
$=2 mucos \theta$
$=2 \times 0.15 \times 15 \times 0.9239$
i.e., Impulse $=4.16 kg ms ^{-1}$
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