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A cannon ball is fired with a velocity $200 \mathrm{~m} / \mathrm{sec}$ at an angle of $60^{\circ}$ with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity $100 \mathrm{~m} / \mathrm{sec}$, the second one falling vertically downwards with a velocity $100 \mathrm{~m} / \mathrm{sec}$. The third fragment will be moving with a velocity
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$300 \mathrm{~m} / \mathrm{s}$ in the horizontal direction
Momentum of ball (mass $m$) before explosion at the highest point $=m \hat{i}=m u \cos 60^{\circ} \hat{i}$
$=m \times 200 \times \frac{1}{2} \hat{i}=100 \mathrm{mi}_{\mathrm{i}} \mathrm{kms}^{-1}$
Let the velocity of third part after explosion is
After explosion momentum of system =
$\begin{aligned}
& \dot{j}_2+\bar{P}_3 \\
&=\frac{m}{3} \times 100 \hat{j}-\frac{m}{3} \times 100 \hat{j}+\frac{m}{3} \times \hat{w}
\end{aligned}$
By comparing momentum of system before and after the explosion
$\frac{m}{3} \times 100 \hat{j}-\frac{m}{3} \times 100 \hat{j}+\frac{m}{3} \hat{W}=100 m \hat{V}=300 \mathrm{~m} / \mathrm{s}$
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