What happens to the 400 ft-lbs of energy from a typical handgun bullet when it hits a target?
470082
You must not think about forces but impulses p=mv! The sum of all impulses must always be the same (0) and therefore m1v1=m2v2. If you observe the velocity of the gun you get v1=m2v2/m1. Since the mass of the gun is so much bigger than the mass of the bullet you get a smaller velocity of the gun (and lets not forget the man holding the gun and is adding to the impulse)!
892312
Hmm-m. Interesting question...\n\nI thought about this for a while and see no\nreason why this math would not be correct.\nAssuming that the bullet did explode when\nstruck by the firing pin, and the bullet left \nthe end of the barrel of the gun at say \n2000 feet per second velocity...\n\n2000 fps = 1800 Mph\n\nThe Moon is 239,000 miles from Earth.\nLet's say you were 10 miles up in the sky\nwhen you fired the gun.  Distance = 238,990 Miles.\n\nSo, 238,990 / 1800 = 132.7 hours \n\nSince there is no atmosphere in space to slow down the \nbullet as it traveled, we can estimate that the bullet would \ncontinue traveling at 1800 mph for the entire trip. The \ngravitational pull of the Moon would influence the bullet's\nvelocity somewhat towards the end of its flight, but my\nfigures show that with 1/6 the gravity of Earth, the Moon\nwould not significantly change the time of flight.\n\nAnswer: 132 Hours.   or   5.5 Days.\n\nNow, I have two questions for you:\n\n#1.)  If the rifle did fire off in space, would you see any smoke?\n\n#2.)  If you were 10 miles up in the atmosphere - say 50,000 ft.,\nhow fast must you be going, to be able to remain at that altitude,\nand in which direction?\n\nYour question makes reference to the Space Station as the point of firing the gun. In my answer above I did not use that as the platform from which the gun was fired because I do not have at my finger tips the approximate altitude of the Space Station though I know it is much more than 50,000 Ft. I also did not take into account the thin atmosphere which exists at 50,000 feet, or the pull backwards on the bullet by the gravity of Earth. These omissions were made simply to run the numbers and see what I came up with since I thought this was an interesting question. Now, having dispensed with the mundane aspects of my quick math, let's discuss the real problem with this question...\n\nThe Earth is spinning. The Moon is orbiting the Earth at a certain velocity, hence it is a moving target. The moving target is being shot at from a moving platform. If you have any hope of hitting the Moon with your bullet, you would have to fire at the place where the Moon will be 5.5 days from the time of firing your gun, not at where the Moon is now. To do that requires more math than a few simple divisions of distance and velocity. For example, the moon swings completely around the Earth at a radius of 238,900 miles in 27.3 Earth days. The circumfrence of that circle is Pi x D = 1,500,292 miles (assuming the orbit is a circle, which it is not). So, the velocity that the Moon is moving away from its present spot right this minute is 2,290 miles per hour.  Since the Moon rises and sets on our horizon each night, you can see that in our calculation, the Moon would have passed overhead at least 5 times in 5.5 days. So your rifle would have to be pointed at the place you calculate the Moon should be at in 5.5 days.        \nRegards,\nZah
526447
I actually had my class conduct an experiment on this very thing.  The problem was: does the percentage of energy lost from a ball bouncing increase, decrease, or stay the same as the height increases.  This is akin to your question, because the height at which it bounces back to depends on how much energy is left after the bounce.  \n\nWe found that the percantage of energy lost stays a constant, which means the the answer to your question is yes.\n\nFor example:\nIf you start with a height of 10 meters and drop a 1.0 kg ball, it will have 98 Joules of potential energy (PE = mgh).  If it loses 10% of its energy on the bounce, it will have 88.2 Joules of kinetic energy, which will translate to a height of 9.0 meters.  \n\nIf you double the original height, you will have 196 Joules of energy.  After losing 10% from the bounce, you are left with 176.4 Joules, which translates to a height of 18 meters after the bounce.
256108
depends on how big it is. here is a link that gives you how much damage that would be caused depending on its size, velocity, ect:\n\nhttp://www.lpl.arizona.edu/impacteffects/
862116
KE = 0.5*m*(v)^2 = 0.5*0.023*(848)^2 = 8269.696 J\n\nThis must = work to accelerate the bullet\n\nW = F*d = 8269.696\n\nF= 8269.696/0.803 = 10298.5 N
516788
Rollo's bullet hits the ground first, with or without taking air resistance into account. \nThe dropped bullet starts at zero speed and increases it's speed as it falls. The fired bullet starts at high speed and (neglecting air resistance) increases it's speed even more as it falls. Since the fired bullet starts out faster and keeps going faster than the dropped bullet all the way to the ground, it gets there first.\n\nNow let's include air resistance. The fired bullet will still start at high speed but it will start slowing down right away due to friction with the air. At some point it would have slowed to terminal velocity, the velocity at which air resistance equals the force of gravity, and then it will then fall the rest of the way at that speed. The dropped bullet will start from zero speed and speed up until it reaches terminal velocity and will then fall that that speed for the rest of the way. So, although both bullets will hit the ground at the same speed, the fired one traveled part of the way faster than the dropped bullet and never slower than the dropped bullet and still hits the ground first, although the difference is not as great as in the no air resistance case.\n\n(EDIT) Oh, I agree with dei48399. You probably meant he fired the bullet directly to the side. Nobody would doubt the outcome it it was fired down, would they?
1058405
I think it would land behind you because as it rises, the rotational speed of the bullet around the earth would have to increase in order to keep up with you because it's going around a larger circle.  Of course, if you're going around a sphere, that could also be considered to be in front of you.  Also, I doubt that the bullet would fall at the same speed that it was shot.  I don't think that gravity would get it going as fast as the gun did.
939061
The base of the bullet contains a small bit of phosphorous which is ignited by the burning gases from the ignition of the cartridge. This burns for as long as it takes the bullet to travel @400 meters. The tracer effect can be used under low light conditions to show you where your bullets are impacting (and the enemy to see where your bullets are coming from), allow you to 'designate' targets visually, and (to a limited extent) act as an incendiary device on inflammable materiel.
407036
The bullet will eventually fall back to Earth, however, it will not be going nearly as fast as it was once shot out of the gun.\n\nThe muzzle velocity of a bullet out of a gun is much higher than the terminal velocity of that same bullet falling down through our atmosphere.  As soon as that bullet leaves the barrel, it begins to slow down due to air resistance.  As the bullet climbs higher and higher, gravity too is working to slow the bullet down.  The bullet will reach its highest point and begin to fall back down.  Gravity works to accelerate the bullet downward, but air resistance works to slow it down. The bullet will reach some maximum downward velocity when the force of friction (air resistance) equals the bullet's weight (force of gravity) and will not accelerate any more.  The bullet will hit the ground traveling much slower than it did when it was fired from the gun.\nThat said, I still wouldnt want to be the person it falls on before I do any terminal velocity calculations for the particular bullet.
710427
You've got some good answers here, espeically the guy with the links, but my two cents.\nSolid, non expanding bullets, such as most military and police ammunition tend to go "through and through" the target when they hit it.  Since the bullet goes out the other side basically undamaged, it transmits relatively little of its kinetic energy to the target.\nSoft-point bullets do a little better, flattening on impact to present a broader wound track and transmitting more energy to the target.\nHollow-point bullets take this a step farther, the thin walls of the bullet around the hollow (in theory) peel back, expanding the bullet faster than a soft-point bullet does - with a corresponding increase in energy transferred to the target.  In addition, many hollow point bullets fragment, resulting in multiple wound tracks and more rapid blood loss.\nRemember, when you're hunting the critter you've shot, unless it is very small, is not likely to die instantly of the shock of being shot.  Unless you make a head shot (which hunters generally do not do due to high miss probability), the animal is going to run until it bleeds out.  The larger the wound, and the more energy transferred from the bullet in flight to the animal, the faster and more humanely this happens.
744405
a) (speed is actually closer to 3.97...not a big deal)\n\nb) Conservation of energy:\nThe sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:\nU(bottom) + K(bottom) = U(top) + K(top)\nNow, the speed at the top is zero, so the ewuation becomes:\nU(bottom) + K(bottom) = U(top)\nSolve for the change in potential energy:\n[U(top) - U(bottom)] = K(bottom)\nThe change in potential energy due to gravity is: (m1+m2)gh\nThe kinetic energy at the bottom is: (1/2)(m1+m2)v²\nSo the equation becomes:\n(m1+m2)gh = (1/2)(m1+m2)v²\nThe (m1+m2) terms cancel:\ngh = (1/2)v²\nSolve for h:\nh = v² / (2g) = 0.8 m
649550
It is the reaction of shooting force on gun's body. let m=bullet mass and M=gun mass. Then if bullet is shot from gun by velocity of v, the recoil velocity would be V=m*v/(m+M) and its shock is M*V=m*M*v/(m+M)\n\nIt does not depend of gun/bullet type.\n\nIf you want to measure it, you must first measure the mass of bullet and empty gun, then measure the bullet shooting speed. This measurement could be done using a common method of shooting to a tough wooden block, which is hung by four equal size lines from a support. You must measure the maximum height (vertically) of displacement of block after shooting (name it H). Then if the mass of block be MB, then the speed of bullet when going out of gun can be calculated as follows:\n\n(MB+m)*g*H=1/2*m*v^2 then v=sqrt[2*(MB+m)*g*H/m]\nwhere m=bullet mass and g=9.81N/m^2 (gravity factor)\n\nAll units are metric (mass=grams; H=meters; v=meters/second)\n\nGood luck!
686704
The increase of the gravitational energy (m1+m2)gh is the loss of kinetic enery m(v^2)/2.\nThe bullet gets the same speed as the bob.\nUse conservation of momentum.
1024308
Friction aside, think of it like this:\nwhen the bullet is fired it has a certain amount of speed.  every second that speed is being reduced by gravity by 9.8 m/s.  So when it reaches as far up as it will go it stops because gravity has subtracted 9.8 from from the original speed every second until the speed is 0.  Like taking out money from an account.  So now you can spend all the money you have-it's got potential.  Then it starts moving toward earth.  gravity is still working at the same rate. so it picks up speed at the same rate that it lost it last time. And it's the same distance, so it'll take the same amount of time.  So now you're depositing the money into a different account.   You only have the money you took out of the first account, so that's how much the second account will have.\nI hope that explains it for you.
234424
Depends on the gun and the bullet.  Different kinds of bullets have aerodynamic designs and powder charges that vary their speed when leaving the barrel.  For most handguns (say a Colt .38), that speed is around 300 meters per second.  On other kinds of wapons (say a Desert Eagle Cal. .50) The speed can go to around 400 meters per second with a heavier projectile, and so on.
832794
It depends drastically on the power of the charge, the weight of the bullet, and the length of the barrel.  It also depends on other factors such as: the action of the firearm (rifle, revolver, semi-auto), the rifling in the barrel, the fit of the bullet to the barrel, and a few other minor details.  \n\nThe answer you are looking for is about 1200 to 1500 feet per second.
294744
How big of a bullet?\n\nIf we're talking standard handgun round, It would definitely hurt, but there's not 100% certainty that it would kill them. The terminal velocity of a bullet free-falling is not nearly as fast as a bullet shot from a gun, due to the lack of spiral flight pattern and the wind resistance involved in such a fall.\n\nIt would basically be about the same as dropping a penny off the empire state building, which has been shown by Mythbusters that it won't kill you, but will certainly sting.\n\nNow, if we're talking larger ammo, i think we're definitely getting into the range of being lethal.