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 the muzzle velocity in m/s. 1J = 1 (kg m²) /s²

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 the muzzle velocity in m/s. 1J = 1 (kg m²) /s²

Quizplus · Accepted Answer

First, convert the mass of the propellant from grains to grams: $5.7 \text{ grains} \times \frac{1 \text{ gram}}{15.43 \text{ grains}} \approx 0.369 \text{ grams}$. Then, calculate the energy produced by the propellant: $0.369 \text{ grams} \times 4545 \text{ J/g} \approx 1677.105 \text{ J}$. Considering the efficiency, the usable energy is $1677.105 \text{ J} \times 0.59 \approx 989.492 \text{ J}$. Convert the bullet weight from grains to kilograms: $115 \text{ grains} \times \frac{1 \text{ gram}}{15.43 \text{ grains}} \times \frac{1 \text{ kg}}{1000 \text{ grams}} \approx 0.00745 \text{ kg}$. Use the kinetic energy formula $KE = \frac{1}{2}mv^2$ to solve for velocity: $989.492 = \frac{1}{2} \times 0.00745 \times v^2$. Solving for $v$, we get $v \approx 516 \text{ m/s}$.

Given the Following Information Regarding a Typical Cartridge of 9mm