# Influence of slide mass to recoil of a pistol

All other things being equal, how does the mass of a pistol's slide effect the recoil imparted to the shooter?

Background: firearm enthusiasts often assert that the larger the "reciprocating mass" the more recoil. I am no physicist, but this seems completely wrong to me. Here is my argument. I would appreciate anyone pointing out what I have right and wrong here.

Suppose we have two pistols $$A$$ and $$B$$ having equal mass and each firing bullets that impart identical forces. Suppose also that despite being equal in mass on the whole, the slide of pistol $$B$$ is more massive than $$A$$'s (obviously the frame of $$B$$ is proportionally less massive).

Now, since momentum is a conserved force the momentum imparted to the shooter is identical between $$A$$ and $$B$$: the additional mass of $$B$$'s slide is canceled, for lack of a better word, by the proportional decrease in its velocity. Kinetic energy, however, is proportional to the square of the velocity. So the more massive slide of $$B$$ in fact means less energy is imparted. So all things being equal, $$B$$ imparts less felt recoil--at least to the degree that a shooter feels both the momentum and kinetic energy of recoil.

Of course, there's more to the story in the real world admittedly. Having a lighter slide, $$A$$ would probably have a "heavier" recoil spring which would slow the velocity and acceleration of its slide I assume. On the other hand, a heavier spring means a greater force is imparted when the spring begins to expand as the slide begins moving in the opposite direction during recoil.

Finally, what can be said objectively about what one feels when a pistol recoils and the nature of momentum vs energy vs force (i.e. mass times acceleration)?

Is there any sense or truth in saying something like, "Momentum is what feels like the 'shove' you receive, while energy is what causes that feeling of the impact radiating into the bones of your firearm"?

(Again, I have no training in physics).

• It looks to me you have an assumption that you can actually corroborate with real world data: You say that for a heavier slide a lesser velocity will be imparted to the bullet -- are you sure that's true, and if so, how much less? You got the general idea correctly that the momenta of the slides if written as $m_1v_1$ and $m_2v_2$ with $m_1,m_2$ their masses and $v_1,v_2$ their maximal velocities it may happen that $m_1v_1=m_2v_2$ if you assume $m_1=k\times m_2$ and $v_1=v_2/k$ for some constant $k$. But that's a very special case, and requires data to confirm something even close happens
– Amit
May 14 at 23:01
• "You say that for a heavier slide a lesser velocity will be imparted to the bullet..." No, I'm saying, assuming the force imparted by the bullet is the same in each case, then the heavier slide will reciprocate with less velocity.
– RTF
May 14 at 23:04
• That's right, I also now realize I got confused by a linguistic error on my part, I confused the slide with what is apparently called the hammer/striker :) Clearly I'm no firearm expert :) But anyway, force is not quite the right term -- the momentum that the bullet obtains as it leaves the muzzle (I suppose can be calculated by muzzle speed times bullet mass) is equal and opposite to the momentum that the gun will obtain. Indeed, a more massive gun will recoil less! I see no reason that the individual recoil of the slide will affect the overall recoil experienced by the holder
– Amit
May 14 at 23:10
• consider also, an explanation that may justify the subjective experience of the firearm enthusiasts that you mention: even for a lesser recoil, we must consider that as a more massive object becomes unstable in our hand, we will experience more effort in stabilizing it. In other words, for a gun in particular, the shooter will make an effort to keep it stable and straight at all times. So even for a smaller overall recoil, the act of restabilizing the gun and keeping it pointing straight may actually require more effort done by smaller muscles such as wrist, forearm, etc.
– Amit
May 14 at 23:21
• That's exactly related to what I just wrote :) a material will react differently according to different impulses. Impulse is force applied for a duration of time.
– Amit
May 15 at 1:03

Actually, this is a question requiring a multi-part answer.

First of all, we should talk about the momentum. The basic version, $$p=mv$$, is sufficient for us. If the bullet flies out with momentum $$p$$, then the whole gun will recoil with momentum $$-p$$ too.

Your arm does not get a hole in it because the kinetic energy of the bullet will be much greater than the kinetic energy of the gun. This is because $$KE=\frac12mv^2=\frac{p^2}{2m}$$ and since the bullet has a much smaller mass than the gun, it would have much bigger energy.

The above does not care about the slide. As long as the pistol has a bigger mass, the above property will work.

What the slide does, is to lengthen the time with which the recoil acts on your arm. Instead of all the recoil from the bullet acting on your arm at once, the slide takes away some of the recoil, and delivers it on your arm after it reaches the end. This means that instead of one big jolt, you get two smaller ones. This means that, not only is there considerations of material maximum stress tolerance (you do not want the slide to break away from the pistol after just a few firings, so no to hugely massive slides v.s. the rest of the pistol), really, the best situation would be that the slide and the rest of the pistol should be equal mass, so that this distribution is overall minimal.

Needless to say, longer time of recoil means lower average force and lower power, as in, lower rate of delivery of the energy, all good.

It would be fun to have a slide on the slide...

• "Needless to say, longer time of recoil means lower average force and lower power, as in, lower rate of delivery of the energy, all good." Right, so if I'm following, a more massive slide would at least lower the average force by extending the time of part of the recoil since the more massive slide is moving slower. At any rate, a gun where half of the mass is in the slide would be a very massive slide indeed! Thanks for your answer.
– RTF
May 15 at 14:27
• "since the more massive slide is moving slower" is not the thing, though. What happens is that the base of the pistol is transferring momentum and energy into your arm quickly after the shot, and the slide, by using time to move backwards, delivers momentum and energy a few moments later, when it hits the back catch of the pistol. That is what is causing the time extension. May 15 at 14:41
• Not sure I follow then. Wouldn't a slide that is moving more slowly naturally extend the amount of time between the first part of recoil which you described and the second ( i.e. when the back catch of the pistol is hit by said slide)?
– RTF
May 15 at 14:50
• I never denied that if it is more massive, it would move slower, and thereby increase the time taken. I am saying that this is not the reason. If you wanted that mass based slowdown, you could just make a rigid pistol with no slide, and just increase the mass. Having the slide would make a much longer time taken than just having more mass on a rigid pistol. May 15 at 15:09
• This answer doesn't reflect the reality of the design, which uses a stiff spring to spread the impulse out continuously (with a small jolt at the end of the traverse). Maybe this whole question should be in Engineering SE instead, since it is a question about the physics principles of a specific kind of machine.
– g s
May 15 at 15:28

The momentum imparted to the shooter plus weapon plus planet system is identical regardless of the mass of the weapon or any part thereof. The time in which that momentum is transferred from the bullet plus gun part of the system to the shooter plus planet part can be increased, thus proportionately decreasing the force applied to the shooter (or for artillery, the mounting by which the weapon is secured to the earth), by use of a spring to apply a smaller force over a longer time. Only that part of the weapon’s mass which is suspended on the spring can have its momentum transfer to the shooter plus planet part of the system spread out in time this way, hence, for a given total weapon mass, the larger the mass of the reciprocating part, the smaller the time averaged force. Shock absorbers on a vehicle are analogous.

Increasing the total mass of the weapon will have results as described in Joseph H's answer.

• Thanks. You mention "the larger the mass of the reciprocating part, the smaller the time averaged force." Just so I'm clear, does "smaller time averaged force" in any way mean less total force? Or is the force equal but delivered over a shorter time frame? It's difficult to shake the intuition that a reciprocating slide moving at a tremendous speed would deliver much more force when it slams back than a much more massive slide moving slowly but with equal momentum.
– RTF
May 15 at 0:51
• @RTF physics doesn't have a "total force" in the way you mean. Impulse (momentum transfer), the integral of force with respect to time, is unchanged. The work done on the shooter, the integral of force with distance, is slightly increased, but still very small.
– g s
May 15 at 1:13
• @RTF specifics of specific weapons might get an interesting answer on Engineering SE, but... the linked weapon predates the ballpoint pen. Side by side comparison with a modern firearm probably involves many different factors, including many design decisions made on the basis of experimental results, not theoretical predictions.
– g s
May 15 at 4:27
• Just a note: No answer by Dale here
– Amit
May 15 at 12:53
• @Amit Oops! Somehow my brain decided that Dale had written the answer by Joseph H. I'll edit.
– g s
May 15 at 15:21

Assuming ideal conditions, and as you have stated, all things being equal, a heavier slide should mean a decreased recoil.

When a gun is fired, the cartridge recoils pushing the slide and gun backward, while the projectile (bullet) is propelled forward. Because momentum is conserved (and looking at the gun and bullet system only) $$m_gv_g+m_bv_b=0 \\ \frac{m_g}{m_b}={\large\mid}\frac{v_b}{v_g}{\large\mid}$$ with subscripts b for the bullet and g for gun. We know that in practice the bullet leaves the muzzle with a far greater velocity than the gun moves backward, intuitively because the gun is heavier$$^1$$, and we can see this confirmed in the above equation, since $$m_g\gg m_b\to v_b\gg v_g$$

This would suggest that the greater the ratio $$\frac{m_g}{m_b}$$ then the the greater $$v_b$$ and the smaller ($$v_g$$) the recoil would be. The conclusion should be that the heavier you make a gun, the smaller will be the recoil.

Therefore, since making the slide heavier makes the gun heavier, this indicates a smaller recoil, all other things being equal. But I'd guess that the difference would not be very noticeable.

$$^1$$ Again, for simplicity, we have ignored the human and earth part, looking at the gun and bullet system only. If a gun floating in midair was to discharge, the gun should move further backwards than if held by a human, but still at a far smaller speed than the bullet. In reality, of course momentum is transferred into the human. Also, the slide moves backward absorbing momentum, transferring it to the whole gun structure. Again, it should be emphasized that this is an idealization of reality, but is nevertheless true if momentum conservation is true (and it is always true).

• Thank you. "Therefore, since making the slide heavier makes the gun heavier, this indicates a smaller recoil," So a heavier slide only dissipates recoil insofar as it increases the total mass of the gun? In your explanation aren't you only looking at momentum? What about the energy of a light vs heavy slide (total mass of the weapon aside for the moment). Don't we have to take into consideration the square of the velocity?
– RTF
May 15 at 1:05
• OK, I guess I was partially wrong then. I was under the impression that a more massive slide would exert less energy when it collides with the rear of the frame (since it is moving more slowly) even if the total mass of the weapon is held constant (Imagine a fantasy scenario where we can transfer some of the mass in the frame to the slide). Anyway, thanks for the details.
– RTF
May 15 at 1:30
• @RTF You're correct that if two slides have the same momentum, the less massive one will have more energy. (Due to $\text{KE}=p^2/2m$). But talking about "exertion" of energy is not very useful in the analysis of the recoil, the bullet carries off kinetic energy that was stored as various types of potential energy (such as chemical energy), and hence the total amount of energy is conserved, but momentum conservation is more useful here since it is conserved as a vector. The bullet's and the gun's momentum are equal and opposite, but we can't talk about "opposite" KE, KE is always positive.
– Amit
May 15 at 13:40
• @Amit Very interesting. I was unaware of the vector aspect. Thanks for the insight.
– RTF
May 15 at 14:20