JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 6)
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:
$$M_1 /\left(M_1+M_2\right)$$
$$\frac{M_2}{M_1}$$
$$\frac{M_1}{M_2}$$
$$M_2 /\left(M_1+M_2\right)$$
Explanation
$$\begin{aligned}
& \left|\overrightarrow{\mathrm{p}_1}\right|=\left|\overrightarrow{\mathrm{p}_2}\right| \\
& \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text { same } \\
& \mathrm{KE} \propto \frac{1}{\mathrm{~m}} \\
& \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1}
\end{aligned}$$
Comments (0)
