A few things to keep in mind when talking about ballistics:
The momentum imparted upon the shooter is the same (or very close) to the momentum imparted upon the bad person being hit by the bullet, assuming the bullet gets lodged. If you don’t move much after shooting the gun, then the bad person won’t move much either. In fact, there’s no reason for a person to even fall down when being hit by a bullet, except for the fact that they’re (obvious) hurting and don’t feel like standing anymore.
There’s no law that says kinetic energy must be conserved.
The kinetic energy of the gun is much less (typically < 0.5%) than the kinetic energy of the bullet.
Most experts agree “energy dumping” and “hydrostatic shock” are myths.
QED has it correct, and I think the whole “bracing” argument is mostly a red herring. It seems obvious what is meant in the OP, and that’s that the shotgun blasts shown on TV and movies that lift a person completely off their feet or knock them flat are pure bullshit.
I’ve fired a 12-gauge unbraced before (off-balance, while hunting), and I’ve never come close to be being knocked back or over (about 120 pounds at the time). I’ve also fired a double-barreled 10-gauge too, and it did not make me take even one step back (but did wreck my shoulder).
I disagree. Bracing might matter if we’re talking about only enough force to make someone lose their balance (e.g., a shove). That’s what “bracing” is – standing in a more stable way so that you’re not as easily unbalanced. But if we’re talking about enough force to send someone flying off their feet, then the only relevant question seems to be “Can the ground exert enough friction on their feet to hold them in place?”
So to avoid the issue of balance and falling over, could we just have the shooter lying flat on his stomach shooting the victim in the top of his head while also laying flat? I think they’d both jiggle a little, nothing more. Neither would go sliding backwards 10 feet.
Don’t forget that there’s a big difference between accelerating a mass down a 30" barrel, and stopping it with your bones. Hold a lightweight but bulletproof plate in front of you, and have someone shoot it with a 12 ga slug. That plate is going to do you some damage, because the energy is being absorbed by your body instantly.
That’s also why it hurts a lot more if the shotgun isn’t tight into your shoulder. When the shotgun is locked into your body, the acceleration is transferred slowly as the bullet accelerates, with the heavy shotgun acting as a damper. But hold it away from your body, and the shotgun will accelerate before it hits you, and you’ll take an instant shock from the gun. Hence the bruise. If the shotgun were lighter, it would hurt a lot more.
Another example: A gun with a very lightweight, space-age frame will hurt a lot more from recoil than will a heavy gun like a Colt Python. People have made .44 magnum derringers, but they’ve been known to break people’s wrists wtih the recoil.
None of this means a person will go flying from a gunshot - in fact, the hard impact makes it more likely that the bullet will either penetrate and carry the energy with it, or expend its energy shattering bones and deforming/tearing flesh. A human body is a mass that will resist a fast push more than a slow one.
I saw a sequence of photos in a gun magazine many years ago, they had an article on 2ga and 4ga shotguns. That’s “two” and “four”, not “twelve” and “fourteen”. Not surprisingly they pack quite a whallop.
Anyhow, one of the testers sat on a cheap plastic sled in the snow and fired the 2ga. It did actually scoot him back a bit (we’re talking a foot or two).
So if that scaled-down bazooka will only budge a man who is on a relatively slippery surface, I don’t think that any firearm found in normal experience is going to send anyone flying through the air by dint of the actual impact.
Una, you are too tuff for me, You know how big I am and I have fired off a 10 gauge also and it hurt me !!! I can really relate to your shoulder hurting… I am impressed you could do it, that is a lot of gun even for big folks…
But if the mass of the projectile(s) is more or less constant, equal forces implies equal accelelerations. If the shotgun has a 24" barrel, whereas the projectiles come to a stop in the target over a distance of a few inches, it seems implausible that the accelerations are equal on each end.
That’s true. But acceleration is not the same as force, though they are related. The shot pellets or bullet stops in a much shorter distance (and therefore experiences greater acceleration) in teh body of the target because it’s more dense than the air in the barrel of the weapon.
Strictly speaking, some energy is lost to atmospheric drag between the barrel and the target, so saying the forces are the same isn’t completely accurate. I’ll give you that.
Momentum is conserved, and is defined by mass times velocity. In this case, mass x velocity is 100 lb*ft/sec. (1/16 x 1600)
Take that exact momentum and convert it in a 100lb person, and you get 1ft/sec velocity, that’s slow, think three seconds per 3ft stride. Convert it in a 200lb person and it’s 0.5ft/sec, even slower. Motion of 1ft per second doesn’t make anyone fly anywhere, at best you lose your balance and fall over.
Right - which is why I feel that the forces must not be equal. F = ma says that if masses are the same, accelerations must be substantially equal for the forces to be so, too.
I think for the purposes of this discussion we can stipulate that the target is very close to the end of the barrel, and therefore the loss of energy to drag is minimal.
F = ma is completely ignoring drag. Drag is defined a force acting in opposition to a body moving through a medium. Force is also defined as 1/2 mv[sup]2[/sup], and since the mass, m, of the shot doesn’t change measurably between the etime it leaves the barrel to the time it hits the target, and we’ve stipulated that the velocity hasn’t changed either, then the force exerted must remain constant as well.
Well, drag is (as you indicate) another force, so it can be accounted for by F = ma. Specifically, it produces a deceleration inversely proportional to the mass. But if we stipulate that the target is, say, 10 cm from the end of the barrel, it will be safe to pretty much ignore the drag encountered over that distance and concentrate on comparing the force on the shooter with the force on the target.
I believe (mv[sup]2[/sup])/2 is kinetic energy, not force.
