# Does energy of explosion channelise?

Whenever there is an explosion inside a barrel of a gun, the bullet shoots off with the energy of explosion. My question is:

Suppose $E$ amount of energy is liberated from explosion. Then, does the bullet use almost all of this energy or only a fraction of it, as the rest of the energy goes into pushing against the walls of the barrel which is then just wasted as vibrational energy and then eventually into heat. I mean, does this mean that the bullet starts off with only,say, 1/6 of the total energy of explosion (considering the barrel to be cubical for over simplification).

• See the "Firearm energy efficiency" section of en.wikipedia.org/wiki/Physics_of_firearms for an energy breakdown of a typical small firearm. It claims that only 32% of the energy is converted to bullet kinetic energy. @JohnRennie: I thought the percentage would be higher, as you also suggest in your answer. – James Aug 19 '16 at 14:44

When the propellant explodes it creates a hot high pressure gas, and the pressure of this gas pushes the bullet along the barrel.

The energy of the bullet is equal to the work done on it, so the energy at some distance $d$ along the barrel will be:

$$E = \tfrac{1}{2}mv^2 = A \int_0^d P(x)\,dx$$

where $A$ is the area of the bullet, though actually calculating the integral will be hard. The propellant does not burn instantly so the pressure is some complicated function of time as well as distance moved by the bullet. There will also be some frictional term to subtract from that energy, though this is likely to be a small correction.