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:
- \( \frac{M_1}{M_1 + M_2} \)
- \( \frac{M_2}{M_1} \)
- \( \frac{M_2}{M_1 + M_2} \)
- \( \frac{M_1}{M_2} \)
The Correct Option is B
Solution and Explanation
Given that momentum is conserved:
\[ |p_1| = |p_2| \]
Let \( p \) denote the magnitude of the momentum. The kinetic energy (KE) of an object is given by:
\[ KE = \frac{p^2}{2M} \]
Since the momenta of the artillery and the shell are equal, the kinetic energy is inversely proportional to the mass:
\[ KE \propto \frac{1}{M} \]
Thus, the ratio of kinetic energies of the artillery (\( KE_1 \)) and the shell (\( KE_2 \)) is:
\[ \frac{KE_1}{KE_2} = \frac{\frac{p^2}{2M_1}}{\frac{p^2}{2M_2}} = \frac{M_2}{M_1} \]
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