Posted: 3/29/2012 7:36:51 PM EDT
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Mouse over no more and come on in!!
ARFCOM physics threads always go so well.... so let's try this one... In this episode, they are discussing Newton's 3rd law in regards to a pistol being shot out of someone's hand. Jamie proposes that due to Newton's 3rd law, to test the amount of force of a bullet hitting the gun, he could design a gun that fires a bullet at the same angle as the proposed incoming bullet, and the recoil from shooting the gun would be equal to the amount of force generated by the bullet's impact on the pistol. effectively, the force of the recoil would equal the force of the bullet impact. I don't think that's right. am I wrong? 
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you are basically wrong. The force on the impacted gun would be VERY slightly less due to the loss of air friction on the bullet. But they are pretty much the same. i understand that, but i would argue his premise is flawed on the fact that not all of the opposite reaction is absorbed by the shooter's hand. a lot of that energy follows the bullet out the end of the barrel, right? |
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you are basically wrong. The force on the impacted gun would be VERY slightly less due to the loss of air friction on the bullet. But they are pretty much the same. i understand that, but i would argue his premise is flawed on the fact that not all of the opposite reaction is absorbed by the shooter's hand. a lot of that energy follows the bullet out the end of the barrel, right? Third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear The force following bullet - e.g. hot explosive gasses are exerting a force exactly opposite their escape route. |
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If the gun is free in space, and the bullet trap is 100% effective (no flying shards for example) then the impulse delivered should be equal and opposite so yes, the net result should be a stationary gun/bullet/trap combo.
This assumes that the fired and captured bullet have the same mass and speed but exactly opposite velocity vectors and neglects air friction and such. It also assumes that the center of mass of the gun is where the two impulses are applied. I will be surprised if mythbusters gets it right without at least four tries if not more. ETA: along with air friction, this disregards the effect of the expelled propellant gases. In theory, theory and practice are the same. In practice, they are different. 
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you are basically wrong. The force on the impacted gun would be VERY slightly less due to the loss of air friction on the bullet. But they are pretty much the same. i understand that, but i would argue his premise is flawed on the fact that not all of the opposite reaction is absorbed by the shooter's hand. a lot of that energy follows the bullet out the end of the barrel, right? Third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear The force following bullet - e.g. hot explosive gasses are exerting a force exactly opposite their escape route. The bullet imparts a forward momentum to the barrel. Not much for sure so I think for the purpose of this myth, it is technically equal. |
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you are basically wrong. The force on the impacted gun would be VERY slightly less due to the loss of air friction on the bullet. But they are pretty much the same. i understand that, but i would argue his premise is flawed on the fact that not all of the opposite reaction is absorbed by the shooter's hand. a lot of that energy follows the bullet out the end of the barrel, right? This is the notion that makes no sense, Energy in these physics cases is something moving, ie kinetic energy, what is moving other than the bullet? |
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The big word is Impulse.
Impulse = force x time The time it takes to accelerate the bullet down the barrel is less than the time it takes the bullet to stop against a solid surface, therefore the force will be higher, but the duration will be shorter. Basically, it will feel "sharper" but it won't throw the gun across the room... |
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Actually, the recoil will be larger in total impulse on the firing end because the firing weapon also experiences recoil due to the acceleration and ejection of the spent gases from the powder. Mass does not disappear, just changes form.
ETA: Damnit. Less than a minute. |
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I would think that recoil would actually have more energy than the bullet, reason is that the weapon would also be expelling gasses form the power which would add to recoil but not to the bullets energy after the bullet left the barrel.
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you are basically wrong. The force on the impacted gun would be VERY slightly less due to the loss of air friction on the bullet. But they are pretty much the same. i understand that, but i would argue his premise is flawed on the fact that not all of the opposite reaction is absorbed by the shooter's hand. a lot of that energy follows the bullet out the end of the barrel, right? Third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear The force following bullet - e.g. hot explosive gasses are exerting a force exactly opposite their escape route. The bullet imparts a forward momentum to the barrel. Not much for sure so I think for the purpose of this myth, it is technically equal. |
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That was a .308 with a frangible projectile. |
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so, since i'm not a physicsologist.... can someone double check the forces we're talking about here?
the action of firing a bullet out of a barrel is: expansion of gases caused by burning of propellant (action) + bullet being propelled forward with equal amount of energy (reaction) right? |
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I would think that recoil would actually have more energy than the bullet, reason is that the weapon would also be expelling gasses form the power which would add to recoil but not to the bullets energy after the bullet left the barrel. I agree. The forward momentum is more than overcome by the recoil, it is just a force that makes a little bit of the recoil must overcome. Thus reducing the recoil by a small amount. At least that is my way of thinking. |
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That was a .308 with a frangible projectile. Source? |
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Quoted: Not the same. The gun firing also has the muzzle gas blast, which is a considerable portion of the recoil. For most pistol rounds, it isn't. Higher muzzle pressure and greater powder mass, yes. Like .38 Super where compensators are quite effective use this energy to counteract recoil. |
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That was a .308 with a frangible projectile. Source? Watched a special on police sniper shootings, they showed that video. They went into detail on the frangible ammo used and what they showed was a .308. |
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so, what still confuses me is that if you take something like a shotgun slug, and look at the amount of energy distributed upon impact, it SEEMS greatly more than the amount of energy your shoulder absorbs, even accounting for surface area differences, and yes, i even thought about the recoil pad, but even if you fire a shotgun slug with a hard stock, you don't take near the amount of damage you would being impacted by a slug (again surface area taken into consideration)
just having trouble wrapping my brain around this one. is anyone good enough with physics to help with some of the math involved? |
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so, what still confuses me is that if you take something like a shotgun slug, and look at the amount of energy distributed upon impact, it SEEMS greatly more than the amount of energy your shoulder absorbs, even accounting for surface area differences, and yes, i even thought about the recoil pad, but even if you fire a shotgun slug with a hard stock, you don't take near the amount of damage you would being impacted by a slug (again surface area taken into consideration) just having trouble wrapping my brain around this one. is anyone good enough with physics to help with some of the math involved? Force = mass x acceleration. Force on the shotgun is approximately equal to the force on the slug. The shotgun has many times the mass of the slug, so it's acceleration during firing is much less, therefore it's velocity into your shoulder is fairly low. |
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so, what still confuses me is that if you take something like a shotgun slug, and look at the amount of energy distributed upon impact, it SEEMS greatly more than the amount of energy your shoulder absorbs, even accounting for surface area differences, and yes, i even thought about the recoil pad, but even if you fire a shotgun slug with a hard stock, you don't take near the amount of damage you would being impacted by a slug (again surface area taken into consideration) just having trouble wrapping my brain around this one. is anyone good enough with physics to help with some of the math involved? Force = mass x acceleration. Force on the shotgun is approximately equal to the force on the slug. The shotgun has many times the mass of the slug, so it's acceleration during firing is much less, therefore it's velocity into your shoulder is fairly low. that makes sense. that's probably the part i was missing. thanks! |