Suppose a suspended a rifle from 2 strings so it is free to swing. How much of the energy would go back into the gun when it was fired as opposed to someone holding the gun up against their shoulder. And, is their a difference in velocity from a bullet depending on who and how the gun is being held?
I’m no physicist but since every action has an equal and opposite reaction, half the power goes into the bullet and half into the firearm whether it’s being held or is suspended by strings. I think the velocity difference in the bullet would be almost immeasurable but since the string-supported gun is accelerated to a higher velocity opposite the direction of the bullet, I would expect the bullet to lose a corresponding amount of velocity going forward. I’m guessing this is a few feet per second for a bullet that is travelling at high hundreds or a few thousand feet per second. Someone else can probably do the math.
Both energy and momentum are conserved. With this problem, it’s easier too look at momentum than energy.
I was going to point out that Newton’s third law would cover the first instance, but in the second, the mass of the shooter would surely make a difference. To say half the power goes into the bullet and half into the firearm is to ignore the relative mass of firearm and bullet. It would only be true if both were the same.
Caveat: I don’t know the actual masses of typical guns and bullets, so adjust these numbers as appropriate.
Let’s say that the gun has a mass of 1 kg, and the bullet has a mass of 1 g (that is, a mass ratio of 1000 to 1). If the bullet goes forward with a velocity v, then the gun goes backwards with a velocity of v/1000, in order to conserve momentum (p = mv).
Now we look at energy. KE = 1/2 m*v^2. So the gun has more mass than the bullet, but less velocity, and it’s the velocity that matters more, because it’s squared. Specifically, if we take the ratio of the two and cancel, we find that the bullet has 1000 times the energy of the gun.
And this works just as well for any other mass ratio. If the bullet is instead 1/2000 times the mass of the gun, then it will get 2000 times as much energy, and so on.
Corollary: It’s momentum that knocks things over, but roughly speaking energy that does damage. So a bullet fired by a man-portable gun won’t knock down a human target, because if it did, then it’d knock down the shooter, too. But it’ll do a lot more damage to the target than to the shooter.
Indeed. The key is that energy can be in all sorts of forms, such as chemical energy of the gunpowder or heat energy when the bullet gets stopped, and most of those are harder to account for than the mechanical energy. But there’s only one form of momentum, so it can’t hide.
Oh, and to the question of whether the gun is held firmly or suspended: The held gun will transfer a bit more to the bullet, but not very much. Shoulders and wrists are less rigid than steel, and will flex a fair bit. If the bullet has already left the barrel by the time the flesh stops squishing, then the shooter’s mass will be almost irrelevant. And even if the shooter had shoulders of steel, it’d still be basically irrelevant from the bullet’s perspective: It’d mean getting 99.999% of the energy, instead of 99.9% of it, or somewhere thereabouts (assuming the same 1 kg gun and 1 g bullet, and 100 kg shooter).
The bullet leaves the barrel before the gun even moves in an automatic rifle. The gun will not move until the bolt moves backwards and compresses the recoil spring. It cannot do this until the bolt unlocks. To exploit this, there have been rifles designed to fire two quick rounds and they both leave the barrel before it starts to rise thus giving two shots to the point of aim.
In this ultra slow motion you can see the bullet hits the target before the slide is back and the barrel starts upwards:
DennisI would have never guessed, amazing!
Well, yes so the guns recoil acceleration is less than the bullets/projectiles acceleration. F=mA. But the original question is about energy, and the energy is going to into accelerating the gun and recoil and compressing springs and all that.
A more simple thing to consider…
work = force * distance.
But for every force there is an equal and opposite force.
So since the bullet has the same force on it, but its moving greater distance… its getting more work done on it… more energy.
Strictly speaking what you say is true but there is an implicit assumption that this is a case of rigid body dynamics i.e. the gun is a ball of iron. The gun is far from a rigid body (by design) and although the center of mass has to follow the equations/expressions you write, the overall movement is remarkably different.
Back to the OP : I am using a Glock just because the numbers are easy. The Glock 17 uses 9x19 mm cartridges and it weighs 625g (22 oz) when unloaded. Glock 17 vs 19 - Difference and Comparison | Diffen
The bullet mass is around 8g and the muzzle velocity is about 400 m/s. 9×19mm Parabellum - Wikipedia
So for your hang by string case, the gun will go back at the speed of about (8/625) x 400 ~ 5 m/s or about 15 ft/s or about 10 miles per hour.
Now visualize a tube with one end closed and a bullet in the middle of the tube. Say there is a connection to this tube from a high pressure gas source (sort of like a captive bolt piston used to stun cattle). You’ll quickly see that the final speed of the bullet is a strong function of the inimitial pressure and how long it is accelerated through the tube. It doesn’t make much difference how you hold the tube.
How do they chamber the second round without moving the bolt?