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A canon shell fired breaks into two equal parts at its highest point. If one part retraces the path to the canon with kinetic energy $E_1$ and kinetic energy of the second part is $E_2$, then
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Verified Answer
The correct answer is:
$E_2=9 E_1$
Momentum before explosion
$$
=u \cos \theta \times 2 \mathrm{~m}
$$
Momentum after explosion
$$
=-m u \cos \theta+m v_2
$$
Momentum is conserved, so
$$
\begin{aligned}
& 2 m u \cos \theta=-m u \cos \theta+m v_2 \\
& \Rightarrow \quad v_2=3 u \cos \theta
\end{aligned}
$$
Kinetic energy of first part
$$
\begin{aligned}
& =E_1=\frac{1}{2} m v_1^2=\frac{1}{2} m(-u \cos \theta)^2 \\
\Rightarrow \quad E_1 & =\frac{1}{2} m u^2 \cos ^2 \theta
\end{aligned}
$$
Now, kinetic energy of second part
$$
=E_2=\frac{1}{2} m v_2^2=9 \times \frac{1}{2} m u^2 \cos ^2 \theta=9 E_1
$$
So, $\quad E_2=9 E_1$
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