Given the following information regarding a typical cartridge of 9mm ammunition:
$$\begin{array} { | l | l | l | }
\hline 1 \text { gram: } & 15.43 & \text { grains } \
\hline 1 \mathrm { oz } & 28.35 & \mathrm {~g} \
\hline \text { Propellant energy: } & 4545 & \mathrm {~J} / \mathrm { g } \
\hline \text { Mass of propellant } & 5.7 & \text { grains } \
\hline \text { Efficiency } & 59 \% & \
\hline \text { Bullet weight } & 115 & \text { grains } \
\hline
\end{array}$$
Calculate how much chemical energy from the propellant will be available to impart to the kinetic energy of the bullet. Report in J

Question

Given the following information regarding a typical cartridge of 9mm ammunition:
$$\begin{array} { | l | l | l | } 
\hline 1 \text { gram: } & 15.43 & \text { grains } \
\hline 1 \mathrm { oz } & 28.35 & \mathrm {~g} \
\hline \text { Propellant energy: } & 4545 & \mathrm {~J} / \mathrm { g } \
\hline \text { Mass of propellant } & 5.7 & \text { grains } \
\hline \text { Efficiency } & 59 \% & \
\hline \text { Bullet weight } & 115 & \text { grains } \
\hline
\end{array}$$
Calculate how much chemical energy from the propellant will be available to impart to the kinetic energy of the bullet. Report in J

Quizplus · Accepted Answer

The mass of the propellant in grams is $ \frac{5.7}{15.43} \approx 0.369 $ grams. The total chemical energy is $ 0.369 \times 4545 \approx 1677 $ J. With 59% efficiency, the energy imparted is $ 0.59 \times 1677 \approx 991 $ J.

Given the Following Information Regarding a Typical Cartridge of 9mm