What would happen if you had to guns exactly opposite one another and fired them at exactly the same time? Would the bullets shatter each other or deflect each other? Would they both hit each other and render each other useless, thus falling straight down?
Couple questions: Are both the same caliber? What distance are they at from each other?
Exactly the same in every respect. Let’s say 10 feet apart.
Any minute deviation in angle from kick, an unbalanced bullet, wind, etc will cause a deflection. I dunno how much wind or gravity could affect a bullet at 5 feet, but… there it is.
Let us also assume that wind and recoil were of no concern.
what a strange question. so ill defined its impossible to answer.
Most Civil War battlefield museums have lots of bullets culled from the soil. Many of those collections have at least one or two examples of two bullets that ran into each other and fused together.
Of course, the bullets were large (over 1/2 in diameter), made of soft lead, and fired by (relatively) slow black powder rifles at distances well over 10 feet.
Copper-sheathed .30 caliber rounds from a Garand and a Mauser would likely be different story and I have no idea what would happen with the .223 caliber (5.56 mm) rounds from a pair of M-16s.
Taking this as a theoetical physics question…
Assuming that the bullets have exactly the same mass and velocity and that they are moving exactly along a straight line and that they are somehow shatterproof…
Their momenta would be equal but in opposite directions and so would cancel each other out. The bullets would come to a stop and then fall straight down.
The energy of the system would go toward deforming the bullets, producing heat, noise and perhaps a flash of light.
I should add one thing…if there is too much energy to be completely absorbed as above, then each bullet would rebound, each moving straight back at identical speeds which would be slower than the initial speed. Of course, gravity would then take over and both would fall separately.
In realworld, modern-day terms, the bullets would meet in the middle and turn into a cloud of lead vapor and copper fragments.
The Civil War bullets, as noted, were soft, largely unalloyed lead (tin and antimony are added these days to make them harder) and moving relatively slow- under 700 fps, I’d say, and often considerably slower than that when they met halfway across the battlefield.
Even if you had two pointed bullets- say, the thin pointy shape as you see in a modern rifle cartridge- and the point of contact were offset slightly (say, half a diameter or so) the energy imparted by the 1,200 to 3,200 fps velocity would still all but destroy each projectile.
The energy in a modern bullet is enormous- some very “hot” experimental loads, say a .223 bullet leaving the barrel at 4,000 to 5,000 fps (which is actually pretty difficult, and usually causes measurable wear to the gun with each shot) can spontaneously “explode” in midflight simply from the gyroscopic forces from it’s own rifled spin.
Even a lower-power round, say a couple of .45 ACP- 230 grains moving around 700 to 900 fps- if striking each other head-on would still essentially destroy each other in a small burst of lead fragments.
Doc Nickel, do you have a cite for that? I’m not challenging you, I just couldn’t find one when I went looking earlier. I’ve heard of a .220 Swift round doing this, but couldn’t find anything online.
I haven’t looked for one online, but there’s several references I’ve read over the years in either reloading manuals or articles about high-end “wildcat” cartridges.
What it is, is a factor of the spin imparted on the projectile by the rifling. It varies widely, but the rate of twist is usually expressed as inches per turn- a 1:16" is one full turn in sixteen inches of travel down the bore. Rates can be as “fast” as 1:10 or as “slow” as 1:50 (with less-common examples of both faster and slower sometimes found.)
Anyway, a cartridge like a .223 Remington (5.56mm NATO military) typically leaves the barrel at anywhere from 2,200 fps (feet per second) to 3,000, depending on the load and bullet weight. Most .223s have a 1:12 to 1:16" twist to help stabilize the little bullet.
So at 2,200 fps, the bullet is going 1,650 rps (revolutions per second) or a staggering 99,000 RPM. Bump the muzzle velocity to 3,000 fps and you get 135,000 rpm. If you can keep it at 3K fps, and use a faster 1:12 twist barrel, you get a theoretical 180,000 rpm.
Considering that a car engine would be very hard pressed to exceed 5,000 rpm, a hot racebike engine might hit 14,000, and a small, powerful RC car electric motor might see 17,000 with no load, you can see there’s some serious speed there (energy going up with the square of the velocity, isn’t it?)
At those speeds, simple centripetal force tries to sling the jacket (the copper shell of the bullet) outward radially- if it fails (and also keep in mind it’s been “grooved” deeply by being forced down the rifling lands) then the jacket slings away and the core- being softer lead- can’t hold itself together.
Now, this doen’t usually happen by any means. It’s often a symptom of driving a bullet far past it’s design limits via custom-loading handbuilt cartridges (called “wildcatting”.) Someone might, for example, take a .340 Weatherby, and “neck it down” to take a smaller bullet- like a .223. This provides a huge case for powder for a relatively small bullet. The .340 having probably five times the case volume of the usual .223 case.
I only brought it up as an extreme example of the energy involved with a rifle bullet.
Minor nitpick, Doc. 2200 fps is pretty cheesy performance for a .223, unless it’s fired from a target pistol with a short barrel. Remington advertises 3240 fps muzzle velocity for its Express .223 with a 55-gr. soft-point bullet.
Once a buddy of mine and I were firing a .45 caliber gun out in the woods for fun. We took turns and each fired one shot into a section of uprooted tree trunk. On a lark we decided to dig the bullets out since we had a large axe handy. To our enormous surprise we found that even though each bullet entered via a separate path, they came to rest in contact with each other. The second bullet, which was almost unscathed, put a good sized dent in the first. These were copper-jacketed lead bullets.
As far as the OP is concerned, when posed as a thought experiment under ideal conditions, the bullets will stop when they collide and fall straight down to the ground only if it is a perfectly inelastic collision. Lead would be pretty close to inelastic. If it was a perfectly elastic collision, each bullet would rebound and head back the way it came at the same speed, but opposite direction.
That is going to be an inelastic collision, so if there’s too much energy, the bullets will first liquidfy then vapourise. Rebound, no.