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A cannon shell fired breaks into two equal parts at its highest point. One part retraces the path to the cannon with kinetic energy $E_1$ and kinetic energy of the second part is $E_2$, relation between $E_1$ and $E_2$ is
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The correct answer is:
$E_2=9 E_1$
At highest point
$m u \cos \theta=-\frac{m}{2} u \cos \theta+\frac{m}{2} v$
$m u \cos \theta+\frac{m}{2} u \cos \theta=\frac{m}{2} v$
$\frac{2 m u \cos \theta+m u \cos \theta}{2}=\frac{m}{2} v$
$$
\begin{aligned}
m v & =3 m u \cos \theta \\
v & =3 u \cos \theta
\end{aligned}
$$
The kinetic energy
$$
\begin{aligned}
& E_1=\frac{1}{2} \times \frac{m}{2} u^2 \cos ^2 \theta \\
& E_1=\frac{1}{4} m u^2 \cos ^2 \theta
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
& E_2=\frac{1}{2} \times \frac{m}{2} \times 9 u^2 \cos ^2 \theta \\
& E_2=\frac{9}{4} m u^2 \cos ^2 \theta
\end{aligned}
$$
The relation between the $E_1$ and $E_2$
$$
\begin{aligned}
& E_2=9 \times E_1 \\
& E_2=9 E_1
\end{aligned}
$$
(from the Eq. (ii))
$m u \cos \theta=-\frac{m}{2} u \cos \theta+\frac{m}{2} v$
$m u \cos \theta+\frac{m}{2} u \cos \theta=\frac{m}{2} v$
$\frac{2 m u \cos \theta+m u \cos \theta}{2}=\frac{m}{2} v$
$$
\begin{aligned}
m v & =3 m u \cos \theta \\
v & =3 u \cos \theta
\end{aligned}
$$
The kinetic energy
$$
\begin{aligned}
& E_1=\frac{1}{2} \times \frac{m}{2} u^2 \cos ^2 \theta \\
& E_1=\frac{1}{4} m u^2 \cos ^2 \theta
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
& E_2=\frac{1}{2} \times \frac{m}{2} \times 9 u^2 \cos ^2 \theta \\
& E_2=\frac{9}{4} m u^2 \cos ^2 \theta
\end{aligned}
$$
The relation between the $E_1$ and $E_2$
$$
\begin{aligned}
& E_2=9 \times E_1 \\
& E_2=9 E_1
\end{aligned}
$$
(from the Eq. (ii))
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