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I have looked through various questions about guns and bullets but I haven't found this one.

If someone fires a handgun there is a recoil. Shooters learn to control this, the gun does not fly out of their hand.

Surely if a bullet is fired accurately and precisely into a gun barrel that someone is holding, they should be able to catch it without the gun being knocked out of their hand. The 'recoil' should be less or equal to that operating on the shooter.

Question (Edited to take account of a comment)

Is there a flaw in my argument? Apart from the difficulty of making such a shot, what else can we say about the interaction? For example, how does the heat generated in the receiving barrel (due to compression of the air) relate to the heat produced by the explosion in the shooting barrel? Must the (rate of) deceleration* be the same as the initial acceleration?

*I say 'rate of deceleration' rather 'deceleration' because presumably the mean value must be the same.

P.S. It occurs to me that presumably the bullet will rebound back out of the receiving barrel. That does not have an obvious parallel in the shooting barrel. How does a bounce versus no bounce affect the 'recoil' in the receiving gun?

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  • $\begingroup$ I don't think there's any flaw. When the gun is fired the momentum on the bullet is the same but opposite to the momentum on the gun. Catching a bullet in a gun is just the opposite. $\endgroup$ – John Duffield Sep 14 '15 at 12:52
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    $\begingroup$ Note that this type of question (Am I right?) Is not conducive to this Q&A site as the answer (yes or no) is too short to be a valid answer. Try modifying your question to be slightly more open (but still respecting site policies as set on Physics Meta.) $\endgroup$ – Kyle Kanos Sep 14 '15 at 12:55
  • $\begingroup$ @KyleKanos - Is my new question more suitable? $\endgroup$ – chasly from UK Sep 14 '15 at 13:04
  • $\begingroup$ To me, it's not immediately obvious that the acceleration of the bullet as it leaves the first gun (which ostensibly happens over the entire length of the barrel) would be equal to the deceleration of the bullet when it strikes the chamber of the second gun; so forces won't necessarily be the same, even though impulses should be (through conservation of momentum) $\endgroup$ – Sean Sep 14 '15 at 13:39
  • $\begingroup$ @Sean: recoil experienced by shooter and catcher will be the same if both guns are of the same mass. $\endgroup$ – Gert Sep 14 '15 at 14:01
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Assuming the two weapons have the same caliber, the incoming bullet will fit in the target barrel, but the process of stopping the bullet will not be graceful.

In a well-engineered gun, the burn rate of the propellant has two goals: first, to evolve gas at a rate which is approximately proportional to the volume of the barrel behind the exiting bullet, and second to stop burning just before the bullet exits the barrel. The first goal produces a (more or less) constant pressure in the barrel, and a constant acceleration of the bullet. By keeping the barrel pressure, and bullet acceleration, constant, the muzzle velocity of the bullet can be maximized for a given barrel strength. If the propellant continues to burn after the bullet exits the muzzle, you get increased muzzle flash, louder report, and waste propellant in the process.

For a bullet being intercepted by a gun barrel, there is no propellant gas to slow the bullet. Ignoring friction, the force decelerating the bullet is provided by the atmospheric gas trapped in the barrel and compressed by the bullet. As the bullet moves down the barrel, the resisting pressure will rise exponentially, but will only be appreciable at the very end of the bullet travel. As a result, either the pressure in the end of the chamber will rise to extraordinary levels as the bullet approaches the end of the chamber, or the bullet will contact the back of the chamber and stop as a result of collision with the frame of the gun.

The first possibility is only going to occur with what are called wadcutters: that is, bullets which are essentially perfect cylinders. For any "normal" bullet, with a pointed or rounded nose, the tip of the nose will contact the rear of the chamber.

Here is a site which lists maximum chamber pressures for firing a wide range of rounds, and let's take a .45 Auto as an example. Maximum chamber pressure for normal loads is 21,000 psi. Assuming the target gun is a classic M1911, the barrel is 5 inches long. As the bullet enters the barrel, the air pressure inside the barrel is 15 psi, which will not slow the bullet appreciably. To calculate the pressure in the barrel, keep in mind that $$P\times V = constant$$ assuming that the entire process will be over quickly enough that there will be no time for the heat of compression to be conducted away from the air in the barrel - which certainly seems reasonable. Since the interior of the barrel is a cylinder, the volume of compressed air can, to a first approximation, be considered proportional to the distance between the nose of the bullet and the back of the chamber, which we'll call x. Then $$P = 15 psi \times \frac{5}{x} = \frac{75}{x}$$ or alternatively, $$x = \frac{75}{P}$$ So in this case, normal peak pressure will not be reached until $$x = \frac{75}{21,000} = .0036 inches$$

It should also be clear that, for any reasonable nose shape, the tip of the bullet will contact the rear of the chamber before the displaced gas will get anywhere near normal firing pressure. In other words, the bullet will be stopped by hitting the frame of the gun. In a weapon which uses a bottleneck cartridge, such as virtually any modern rifle, the chamber has a larger diameter than the barrel, and pressure will not rise nearly as much, since there is room for expansion in the chamber and the bullet will certainly hit the rear. This is bad enough for a revolver. In a semi-automatic handgun, this impact will occur on the rear of the slide, and will probably cause catastrophic damage such as unseating the slide and sending it who-knows-where. While the two situations (normal firing and down-the-barrel-impact) will have identical total momentum transfers, the impact scenario will have much greater peak forces, and there is no reason to believe the two situations will behave identically.

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Yes you can I believe it had happened once. A perp had fired on a police officer and by a huge stroke of luck for the officer the perps bullet went directly down the barrel. Not sure of he dropped the gun or not he may have just out of surprise. The bullet does not carry any more energy then you feel in the recoil so catching a bullet in a gun would feel very similar to firing it

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    $\begingroup$ Do you have a link to a news article or anything similar? $\endgroup$ – LogicianWithAHat Sep 14 '15 at 13:07
  • $\begingroup$ Can't find a link I had seen it may years ago in a documentary $\endgroup$ – newguy Sep 14 '15 at 13:21
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    $\begingroup$ Sounds very much like an urban legend. What does Snopes have to say? (not available to me at the moment). Alt: Mythbusters had a devil of a time, on a range, firing a bullet such that it flew thru the scope sight on a rifle. $\endgroup$ – Carl Witthoft Sep 14 '15 at 13:43
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    $\begingroup$ @CarlWitthoft have found the opposite case seattlepi.com/local/article/… .I think he dropped the gun though :) $\endgroup$ – Matas Vaitkevicius Sep 14 '15 at 14:06
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    $\begingroup$ @newguy The bullet is carrying the same energy but the duration of the compression/collision is much smaller. That means higher deceleration in order to stop it. Another example of a bullet decelerating is when it hits a wall. Same energy.. extreme deceleration. In other words, having the same energy is not enough to determine whether it will stop gently, or break the gun into pieces. $\endgroup$ – Fermi paradox Sep 14 '15 at 17:34

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