Influence of slide mass to recoil of a pistol

Question

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).

Answer 1

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.

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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...

Answer 2

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.

Answer 3

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).