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5.Work, Energy, Power and Collision
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An artillery piece of mass $M_1$ fires a shell of mass $\mathrm{M}_2$ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:
A$\mathrm{M}_1 /\left(\mathrm{M}_1+\mathrm{M}_2\right)$
B$\frac{M_2}{M_1}$
C$\mathrm{M}_2 /\left(\mathrm{M}_1+\mathrm{M}_2\right)$
D$\frac{M_1}{M_2}$
(JEE MAIN-2024)
Solution
$\left|\overrightarrow{\mathrm{p}_1}\right|=\left|\overrightarrow{\mathrm{p}_2}\right|$
$\mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text { same }$
$\mathrm{KE} \propto \frac{1}{\mathrm{~m}}$
$\frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1}$
$\mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text { same }$
$\mathrm{KE} \propto \frac{1}{\mathrm{~m}}$
$\frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1}$
Standard 11
Physics
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