I'm interested in the underlying physics of the recoil-operated loading mechanism in a general rifle system (e.g. M107A1 rifle). A technical of the rifle can be found on wikipedia https://en.wikipedia.org/wiki/Barrett_M82, where it states that the loading mechanism is of the type "short-recoil". A general description of the short-recoil principle is also provided by wikipedia https://en.wikipedia.org/wiki/Recoil_operation#Short_recoil_operation.

In short the cycle consists of four steps

  1. Ready to fire position. Bolt is locked to barrel, both are fully forward.
  2. Upon firing, bolt and barrel recoil backwards (left in figure) a short distance (usually 25 mm) while locked together. Near the end of the barrel travel, the bolt and barrel unlock.
  3. The barrel stops, but the unlocked bolt continues to move to the rear, ejecting the empty shell and compressing the recoil spring.
  4. The bolt returns forward under spring force, loading a new round into the barrel.
  5. Bolt locks into barrel, and forces barrel to return to battery.

The process is sketched in the diagrams below:

Key Short-recoil principle


From the description it seems that the entire mechanism is driven by the recoil of the bullet. In that case step 3-4 can be described as a single mass $m$ first compressing a spring and then being pushed by a spring, according to the equation $$m\ddot{x} = - kx$$ subjected to the boundary conditions $x(t=t_2) = x_2 =25\, \textrm{mm}$ and $v(t=t_2) = v_2$, where $t_2$, $x_2$, and $v_2$ are the time, position, and velocity after step 2 is finished. There might be a unknown velocity dependent frictional term missing from the equation above.

In step 4-5 the mass includes both the bolt $(m)$ and barrel $(M)$, and assuming a inelastic collision momentum conservation can be used to determine the initial condition on the velocity.

However, how to model step 1-2 in the process is a bit more mysterious to me as there is a gas pressure in the barrel pushing the bullet forward, and perhaps the barrel backwards simultaneously. In addition, the initial velocity of the bolt and barrel should be related to the recoil of the bullet in some way. I hope someone can shed light on the issue related to modeling step 1-2 from a physics point of view. I suspect part of the problem is that it is not clear to me whether or not the bullet is still inside the bore when the barrel begins to move.


2 Answers 2


We need to find out the mass of the bullet, $m_b$ and its exit velocity, $V_E$ and the mass of the rifle $m_R$

Then based on conservation of momentum:

$$m_r*V_r= m_b *V_b \rightarrow \ V_R=\frac{m_b *V_b}{m_r}$$

This kinetic energy of the rifle, $E_R$ is

$$E_R= 1/2 m_R*V_R^2=1/2kx^2=1/2k(25mm)^2$$

By plugging the known constants in the above equations we can get the rifle's recoil speed and momentum.

  • $\begingroup$ To address the question specifically, it would be the mass of the barrel and bolt, not the whole rifle, but otherwise I believe you are correct. $\endgroup$
    – Drew
    Mar 18, 2022 at 4:19
  • $\begingroup$ @Drew, you are probably right. I have never even seen a rifle outside of the movies! $\endgroup$
    – kamran
    Mar 18, 2022 at 4:57
  • $\begingroup$ Thanks for the responses! I think indeed momentum conservation is the key, albeit a bit more complex than described above. I think the point is that momentum has to be conserved at all times $t$. Since the bullet is accelerated by the pressure $p(t)$ inside the barrel the bullets time-dependent momentum will be $P(t) = \int_0^t p(t) A dt$. This momentum has to be balanced at all times by the bolt and barrel system. Therefore the Eom should be something like $(m+M)\ddot{x} = - kx +p(t) A $ Do you agree? $\endgroup$
    – MOOSE
    Mar 18, 2022 at 8:41

Calculating the recoil physics is actually fairly easy. It's conservation of momentum. When the bullet leaves, it will have some momentum, and the barrel and bolt (which are rigidly joined at this point) will have equal and opposite momentum.

Use conservation of momentum to calculate the velocity of the barrel-bolt assembly.

The gas blowing out of the barrel after the bullet adds additional momentum but that's not easy to calculate so we'll ignore it.

The barrel and bolt will recoil a short distance, and the the barrel will hit a hard stop, maybe with some kind of buffer to absorb the impact? In any case the barrel will stop, while the bolt continues with the same velocity it had prior to the separation. From here you can figure the recoil operation just using the bolt velocity and inertia,and forget about the barrel.


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