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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Aicharya · Accepted Answer

Momentum before explosion $ =u \cos 	heta 	imes 2 \mathrm{~m} $ Momentum after explosion $ =-m u \cos 	heta+m v_2 $ Momentum is conserved, so $ \begin{aligned} & 2 m u \cos 	heta=-m u \cos 	heta+m v_2 \ & \Rightarrow \quad v_2=3 u \cos 	heta \end{aligned} $ Kinetic energy of first part $ \begin{aligned} & =E_1=\frac{1}{2} m v_1^2=\frac{1}{2} m(-u \cos 	heta)^2 \ \Rightarrow \quad E_1 & =\frac{1}{2} m u^2 \cos ^2 	heta \end{aligned} $ Now, kinetic energy of second part $ =E_2=\frac{1}{2} m v_2^2=9 	imes \frac{1}{2} m u^2 \cos ^2 	heta=9 E_1 $ So, $\quad E_2=9 E_1$

A canon shell fired breaks into two equal parts at its highest point. If one part retraces the path to the…

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