An artillery piece of mass $M_1$ fires a shell of mass $M_2$ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:
Solution
<p>$$\begin{aligned}
& \left|\overrightarrow{\mathrm{p}_1}\right|=\left|\overrightarrow{\mathrm{p}_2}\right| \\
& \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text { same } \\
& \mathrm{KE} \propto \frac{1}{\mathrm{~m}} \\
& \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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