Question 68: Centre of Mass and Rotational Motion

Question

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:

Accepted Answer

$\begin{aligned}
& \left|\overrightarrow{\mathrm{p}_1}ight|=\left|\overrightarrow{\mathrm{p}_2}ight| \
& \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} 	ext { same } \
& \mathrm{KE} \propto \frac{1}{\mathrm{~m}} \
& \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1}
\end{aligned}$

Answer

$M_1 /\left(M_1+M_2\right)$

Answer

$\frac{M_2}{M_1}$

Answer

$\frac{M_1}{M_2}$

Answer

$M_2 /\left(M_1+M_2\right)$

Centre of Mass and Rotational Motion

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:

Choose the correct answer: