Work Power Energy Question 8
Question 8 - 2024 (31 Jan Shift 1)
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 :
(1) $M_{1} /\left(M_{1}+M_{2}\right)$
(2) $\frac{\mathrm{M}{2}}{\mathrm{M}{1}}$
(3) $M_{2} /\left(M_{1}+M_{2}\right)$
(4) $\frac{M_{1}}{M_{2}}$
Show Answer
Answer: (2)
Solution:
$\left|\overrightarrow{\mathrm{p}{1}}\right|=\left|\overrightarrow{\mathrm{p}{2}}\right|$
$\mathrm{KE}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}} ; \mathrm{p}$ same
$\mathrm{KE} \propto \frac{1}{\mathrm{~m}}$
$\frac{\mathrm{KE}{1}}{\mathrm{KE}{2}}=\frac{\mathrm{p}^{2} / 2 \mathrm{M}{1}}{\mathrm{p}^{2} / 2 \mathrm{M}{2}}=\frac{\mathrm{M}{2}}{\mathrm{M}{1}}$
