Editing Physics of firearms (section)

==Kinetic energy==
However, the smaller mass of the bullet, compared to that of the gun-shooter system, allows significantly more [[kinetic energy]] to be imparted to the bullet than to the shooter. The kinetic energy for the two systems are <math>\begin{matrix}\frac{1}{2}\end{matrix}MV^2</math> for the gun-shooter system and <math>\begin{matrix}\frac{1}{2}\end{matrix}mv^2</math> for the bullet. The energy imparted to the shooter can then be written as:

:<math>\frac{1}{2}MV^2 = \frac{1}{2}M\left(\frac{mv}{M}\right)^2 = \frac{m}{M}\frac{1}{2}mv^2</math>

For the ratio of these energies we have:

:<math>\frac{\frac{1}{2}MV^2}{\frac{1}{2}mv^2} = \frac{m}{M} \qquad (2)</math>

The ratio of the kinetic energies is the same as the ratio of the masses (and is independent of velocity). Since the mass of the bullet is much less than that of the shooter there is more kinetic energy transferred to the bullet than to the shooter. Once discharged from the weapon, the bullet's energy decays throughout its flight, until the remainder is dissipated by colliding with a target (e.g. deforming the bullet and target).