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:

• A. $\frac{M_1}{M_1 + M_2}$
• B. $\frac{M_2}{M_1}$
• C. $\frac{M_2}{M_1 + M_2}$
• D. $\frac{M_1}{M_2}$

Answer 1

Answer: (B)

$\frac{M_2}{M_1}$

Answer 2

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:

$\frac{KE_1}{KE_2} = \frac{\frac{p^2}{2M_1}}{\frac{p^2}{2M_2}} = \frac{M_2}{M_1}$