View Full Version : The 9 didn't do the job tonight-may switch to .40...

bogey3737

12-30-2009, 20:54

...for steel plates that is :) Shot my 2nd match ever tonight...combination of cardboard silhouettes and steel targets. Ran it with my G19 and 115 gr FMJs. Had to engage a plate twice after it failed to drop when I hit it the first time. The G19 / G26 still work for me for defense...but I'm thinking a G22 might be in my future for competitive shooting.

Still finished in the top three, though...so not that bad of a night...

Clem Eastwood

12-30-2009, 20:56

ive never had a 9 fail to put down a plate. .22lr yeah, 9 no.

mesteve2

12-30-2009, 21:02

Just because you know it will :faint:

Glock19Fan

12-30-2009, 21:08

Try 147 grainers if you can find them.

bogey3737

12-30-2009, 21:09

Try 147 grainers if you can find them.

Makes sense...but that doesn't give me an excuse to buy another gun! :whistling:

Glock19Fan

12-30-2009, 21:14

Makes sense...but that doesn't give me an excuse to buy another gun! :whistling:

Well in that case, you need a mans gun. Get something that starts with a .4.

:rofl:

fredj338

12-30-2009, 21:35

...for steel plates that is :) Shot my 2nd match ever tonight...combination of cardboard silhouettes and steel targets. Ran it with my G19 and 115 gr FMJs. Had to engage a plate twice after it failed to drop when I hit it the first time. The G19 / G26 still work for me for defense...but I'm thinking a G22 might be in my future for competitive shooting.

Still finished in the top three, though...so not that bad of a night...

Just switch to heavier bullets. I've had & seen several poppers fail to drop w/ low 115gr or even 124gr hits. Go to a 147gr & they always go down.

...for steel plates that is :) Shot my 2nd match ever tonight...combination of cardboard silhouettes and steel targets. Ran it with my G19 and 115 gr FMJs. Had to engage a plate twice after it failed to drop when I hit it the first time. The G19 / G26 still work for me for defense...but I'm thinking a G22 might be in my future for competitive shooting.

Still finished in the top three, though...so not that bad of a night...

How were the plates "set"? Were the plates calibrated for a specific minimum load or power factor?

With simple momentum (M x V) transfer where the bullet strikes (and splatters) the plate you can increase either M or V by buying and using a different load after confirming its performance over a chronograph or handload using your preferred bullet weigt over a charge that will knock over the plates.

I've always had the best luck (and controlability) running 147s @ 950 fps (PF: 139.65) since it is a gently recoiling load that provides the same momentum that could be obtained by running 115s @ 1215 fps.

Just switch to heavier bullets. I've had & seen several poppers fail to drop w/ low 115gr or even 124gr hits. Go to a 147gr & they always go down.

What he said.

msu_grad_121

12-31-2009, 02:40

I always preferred light and fast in my 9mm when I carried that (got to carry the 40 now, darn it all), but for YOUR particular problem, i'd prescribe a Glock 21 with a conversion to 50 GI! If that don't fix your problem, nothin will!

Then again, I'm all about overkill...

RMTactical

12-31-2009, 04:43

My G19 has never had a problem with steel plates.

Just switch to heavier bullets. I've had & seen several poppers fail to drop w/ low 115gr or even 124gr hits. Go to a 147gr & they always go down.

This.

Well in that case, you need a mans gun. Get something that starts with a .4.

:rofl:

:laughabove:

ComeAndGetThem

12-31-2009, 07:12

Winchester White Box Muzzle Energy

147 grain - 320 lbs/ft

124 grain - 358 lbs/ft

115 grain - 362 lbs/ft

Maybe the OP's plates were heavier than some of yours. Maybe the pivots needed some WD40. But using a lesser powered round would most definitely NOT have caused the plates to fall.

bogey3737

12-31-2009, 07:46

How were the plates "set"? Were the plates calibrated for a specific minimum load or power factor?

8" plates, hinged on a base. I don't think they were "tuned" for specific loads.

winchester white box muzzle energy

147 grain - 320 lbs/ft

124 grain - 352 lbs/ft

115 grain - 362 lbs/ft

maybe the op's plates were heavier than some of yours. Maybe the pivots needed some wd40. But using a lesser powered round would most definitely not have caused the plates to fall.

Incorrect!

I think in this case momentum-oriented, hence, heavier bullet prevails.

And a blunt meplat yields better results than a round nose one in the case of angled plate.

bogey3737

12-31-2009, 11:04

Incorrect!

Yeah...I'm pretty sure that I throw a baseball with less energy than a 115gr 9mm...but I'm willing to bet my fastball will knock down the plate every time.

ComeAndGetThem

12-31-2009, 11:15

Those numbers were taken off the Winchester website. That makes you incorrect.

ComeAndGetThem

12-31-2009, 12:09

Yeah...I'm pretty sure that I throw a baseball with less energy than a 115gr 9mm...but I'm willing to bet my fastball will knock down the plate every time.

You're partly correct. A Major League baseball weighs 5 ounces or 2,187.5 grains. If you could throw a baseball at 90 miles per hour which is MLB class, that would be 132 fps. The energy of the ball leaving your hand would only be 84.61 lbs/ft which is only about 1/4 the energy of a 115 grain 9mm bullet leaving the muzzle.

But if your fast ball knocked over the plate as you say when his 9mm didn't, that of course would be defying Newton's law of inertia and that would be like proving that the world is flat.

There's nothing else at work here.

...for steel plates that is :) Shot my 2nd match ever tonight...combination of cardboard silhouettes and steel targets. Ran it with my G19 and 115 gr FMJs. Had to engage a plate twice after it failed to drop when I hit it the first time. The G19 / G26 still work for me for defense...but I'm thinking a G22 might be in my future for competitive shooting.

Still finished in the top three, though...so not that bad of a night...

If you're looking for a competition gun, perhaps look into a G35.

If you're looking for a competition gun, perhaps look into a G35.

That's just what I was thinking. I have a 35 and it's a great platform for the 40 cal. Very sweet shooting gun..

Winchester White Box Muzzle Energy

147 grain - 320 lbs/ft

124 grain - 358 lbs/ft

115 grain - 362 lbs/ft

Maybe the OP's plates were heavier than some of yours. Maybe the pivots needed some WD40. But using a lesser powered round would most definitely NOT have caused the plates to fall.

fredj338 and coal have it right. It is simple physics.

Higher momentum (ρ), not more kinetic energy, is the solution to the OP's problem.

The mechanism at work when knocking plates over is the simple transfer of momentum (mv = mv) and momentum (ρ = mv), and although it is not completely "conserved" (since the bullets typically shatter on impact with the steel plates), it is still the mechanism that determines the authority with which the plates fall.

Hence:

∂v ½mv² = mv

∫ mv ∂v = ½mv²

While the kinetic energy numbers (taken from Winchester's 'site) are indeed correct; the numbers that actually matter are those that specify momentum (ρ):

147 gr. @ 990 fps = 320 fpe and 0.6462 ft.lb./sec.

124 gr. @ 1140 fps = 358 fpe and 0.6277 ft.lb./sec.

115 gr. @ 1190 fps = 362 fpe and 0.6076 ft.lb./sec.

As you can clearly see, the lower energy bullet (the 147 gr. @ 990fps) possesses 6.35% more momentum than the higher energy 115 gr. bullet and 2.95% more momentum than the 124 gr. bullet as launched at the factory spec'd velocities.

I know that it sounds counter-intuitive but, the lower energy, higher momentum, 147 gr. bullet is the better choice of the three given for the OP's needs.

ETA: The baseball (2187.5 gr.) thrown at 132 fps (~90 mph) for 84.619 fpe would be the most preferable, not to mention the most effective, of all the options considered since it provides 1.2821 ft.lb./sec. of momentum for use against the plate which is slightly over twice that of the highest momentum bullet (the 147 gr.) listed above.

You're partly correct. A Major League baseball weighs 5 ounces or 2,187.5 grains. If you could throw a baseball at 90 miles per hour which is MLB class, that would be 132 fps. The energy of the ball leaving your hand would only be 84.61 lbs/ft which is only about 1/4 the energy of a 115 grain 9mm bullet leaving the muzzle.

But if your fast ball knocked over the plate as you say when his 9mm didn't, that of course would be defying Newton's law of inertia and that would be like proving that the world is flat.

There's nothing else at work here.

The premise of your argument that kinetic energy (KE) is the mechanism by which the plate is made to fall suggests that:

½mv² = ½mv² (which is wholly incorrect)

This violates Newton's Law of (Conservation of) Momentum (for every action there is an equal and opposite reaction) since kinetic energy is expressed as ½mv² and there is no such thing as Newton's Law of Conservation of Energy.

Newton's Laws (relating to Momentum and to Inertia) states that mv = mv (momentum is conserved), whereas ½mv² ≠ ½mv² because energy is not conserved. This should not be mistaken for, nor confused with, the concept that energy can be neither created nor destroyed.

ComeAndGetThem

12-31-2009, 14:43

I'm not sure you're right but I'm in no position to argue.

I'm not sure you're right but I'm in no position to argue.

Well, you could always visit one of the Physics forums available on the 'net. Couldn't hurt. :dunno:

There you might find a better phrased, more understandable explanation than mine. Lots of nice folks on those 'sites with better communication and instructional skills than I'll ever have. Just a thought if you want to become more familiar with the fascinating science that is physics.

:)

I'm not sure you're right but I'm in no position to argue.

Unless physics changed 481 is correct. Of course it's been 30 years since I took physics. LOL!!

Unless physics changed 481 is correct. Of course it's been 30 years since I took physics. LOL!!

Been almost that long for me, too. :whistling:

Hard to believe. Tempus fugit.

hurley842002

12-31-2009, 16:48

Unless physics changed 481 is correct. Of course it's been 30 years since I took physics. LOL!!

X2 But I never took physics, it does however seem quite simple. Think of it this way: You are a 300+ lbs lineman on a football team, who is going to have a better chance knocking you down, a 100lbs person moving at 10mph or a 200lbs person moving at 7mph? Seems to me weight and momentum would have the upper hand.

disclaimer: the weight and speeds in my example do not have any relation to the bullet weight or velocity we are talking about, just numbers thrown out.

blackbirdzach

12-31-2009, 17:02

Just switch to heavier bullets. I've had & seen several poppers fail to drop w/ low 115gr or even 124gr hits. Go to a 147gr & they always go down.

Yup. 147 grain for plates and bowling pins. :cool:

blackbirdzach

12-31-2009, 17:03

Those numbers were taken off the Winchester website. That makes you incorrect.

:rofl::rofl::rofl:

Those numbers were taken off the Winchester website. That makes you incorrect.

The Winchester site was not Incorrect, I meant Your answer was Incorrect.

I'm not sure you're right but I'm in no position to argue.

481 is correct. Physics formulas aside, since all factory service calibers travel at a similar (enough) velocity, you just need to remember that momentum favors mass. And, in plates, it's pretty much all about momemtum. Here's some, I hope, simple/valuable reading: Caliber Talk (http://glocktalk.com/forums/showthread.php?p=11129015)

ComeAndGetThem

12-31-2009, 18:49

Well, you could always visit one of the Physics forums available on the 'net. Couldn't hurt. :dunno:

There you might find a better phrased, more understandable explanation than mine. Lots of nice folks on those 'sites with better communication and instructional skills than I'll ever have. Just a thought if you want to become more familiar with the fascinating science that is physics.

:)

Thanks for the friendly invite.

X2 But I never took physics, it does however seem quite simple. Think of it this way: You are a 300+ lbs lineman on a football team, who is going to have a better chance knocking you down, a 100lbs person moving at 10mph or a 200lbs person moving at 7mph? Seems to me weight and momentum would have the upper hand.

disclaimer: the weight and speeds in my example do not have any relation to the bullet weight or velocity we are talking about, just numbers thrown out.

With miniscule momentum differences between the rounds being discussed, the answer isn't quite as obvious as the analogy of the football players but I get your point. However, the heavier slower moving bullet isn't ALWAYS going to possess more momentum. Could be going slow enough that a slightly lighter, faster bullet could win out.

CTSixshot

12-31-2009, 20:22

I'm hardly the expert on this matter, but I have used the little 95gr JRN in a .38 Special and have knocked down pepper-poppers and plates just fine. Granted, a 6 o'clock hit on a binding plate doesn't always work, but I'd suggest your 115 nine is just fine. Sometimes you get a sticking plate or you hit too low, but given functioning targets, you shouldn't discount your current caliber so quickly.

...

With miniscule momentum differences between the rounds being discussed, the answer isn't quite as obvious as the analogy of the football players but I get your point. However, the heavier slower moving bullet isn't ALWAYS going to possess more momentum. Could be going slow enough that a slightly lighter, faster bullet could win out.

Unfortunately, the statement you make will require the physics "math" 481 supplied earlier. The question is: "How much faster must the 115gr travel to have more momentum than 147gr?". This may exceed the typical velocity of the 115gr load.

However, keep in mind that more mass will RETAIN momentum (i.e. energy) better. So, it's not quite as straightforward as the math because you need to consider "energy retention".

From what I know, heavy bullets rule competitions with steel. I presume, math aside, that's because they have worked best based off the opinion of those who know.

M&P Shooter

12-31-2009, 20:47

http://www.youtube.com/watch?v=4beoMWwtXyw&feature=related

Brass Nazi

12-31-2009, 20:55

Just switch to heavier bullets. I've had & seen several poppers fail to drop w/ low 115gr or even 124gr hits. Go to a 147gr & they always go down.

Best advice in this thread. As usual Fred comes through a thread full of **** smelling like a rose.

Unfortunately, the statement you make will require the physics "math" 481 supplied earlier. The question is: "How much faster must the 115gr travel to have more momentum than 147gr?". This may exceed the typical velocity of the 115gr load.

However, keep in mind that more mass will RETAIN momentum (i.e. energy) better. So, it's not quite as straightforward as the math because you need to consider "energy retention".

From what I know, heavy bullets rule competitions with steel. I presume, math aside, that's because they have worked best based off the opinion of those who know.

coal,

Fortunately, the math for making such a determination is very, very easy. (Yeah, it's my kinda math :supergrin:)

In order to solve the problem, simply solve algebraically for V147 in the equation M147V147 = M115V115 so that you can then determine at what velocity a 115 gr. bullet must be launched in order to match the momentum of a 147 gr. bullet at whatever velocity, say 990 fps, you choose:

So;

M147 = 147 gr.

V147 = 990 fps.

M115 = 115 gr.

V115 = X fps.

M147 x V147 / M115 = V115

then substitute the correct values;

147 gr. x 990 fps / 115 gr. = V115 = 1265.478 fps

So, in order to have a 115 gr. bullet achieve the same momentum of a 147 gr. bullet with a velocity of 990 fps, it needs to be moving no slower than 1265.5 fps.

While the calculated velocity obtained above rests squarely in 9mm +P "territory", it is certainly not unachieveable with the 115 gr. bullet weight. However, it does push at the upper limit of reasonable 9mm performance with the 115 gr. bullet weight which begs the question, "Why operate using a higher pressure load (the 115 gr. +P) when the same result (momentum) can be achieved with a standard pressure load?"

Since a 9mm 147 gr. bullet retains its downrange velocity (and therefore its momentum) better than a 9mm 115 gr. due to its greater sectional density (0.1671 vs. 0.1307), which makes it a more efficient penetrator regardless of the media (air, water, ballistic gelatine, tissue) that it traverses, the 147 gr. bullet is the right choice for putting the maximum retained momentum downrange, and in this case, on steel.

See? Simple.

coal,

Fortunately, the math for making such a determination is very, very easy. (Yeah, it's my kinda math :supergrin:)

In order to solve the problem, simply solve algebraically for V147 in the equation M147V147 = M115V115 so that you can then determine at what velocity a 115 gr. bullet must be launched in order to match the momentum of a 147 gr. bullet at whatever velocity, say 990 fps, you choose:

So;

M147 = 147 gr.

V147 = 990 fps.

M115 = 115 gr.

V115 = X fps.

M147 x V147 / M115 = V115

then substitute the correct values;

147 gr. x 990 fps / 115 gr. = V115 = 1265.478 fps

So, in order to have a 115 gr. bullet achieve the same momentum of a 147 gr. bullet with a velocity of 990 fps, it needs to be moving no slower than 1265.5 fps.

While the calculated velocity obtained above rests squarely in 9mm +P "territory", it is certainly not unachieveable with the 115 gr. bullet weight. However, it does push at the upper limit of reasonable 9mm performance with the 115 gr. bullet weight which begs the question, "Why operate using a higher pressure load (the 115 gr. +P) when the same result (momentum) can be achieved with a standard pressure load?"

Since a 9mm 147 gr. bullet retains its downrange velocity (and therefore its momentum) better than a 9mm 115 gr. due to its greater sectional density (0.1671 vs. 0.1307), which makes it a more efficient penetrator regardless of the media (air, water, ballistic gelatine, tissue) that it traverses, the 147 gr. bullet is the right choice for putting the maximum retained momentum downrange, and in this case, on steel.

See? Simple.

The math is indeed simple and nicely laid out. It is the conclusion, and in particular the question, "Why operate using a higher pressure load (the 115 gr. +P) when the same result (momentum) can be achieved with a standard pressure load?", which is problematical.

The simple answer to the question is that you get 27.7% more kinetic energy for approximately the same recoil. Transfering momentum from the shooter to the shootee does no more damage than shooting the pistol does to the shooter in the first place. What does the damage is the work done by the energy delivered to the shootee's body. Is the damage linear, that is, would twice as much energy produce twice as much incapacitating effect? Almost certainly not but more energy will produce more damage, and that is a major part of what we want to achieve.

English

The math is indeed simple and nicely laid out. It is the conclusion, and in particular the question, "Why operate using a higher pressure load (the 115 gr. +P) when the same result (momentum) can be achieved with a standard pressure load?", which is problematical.

The simple answer to the question is that you get 27.7% more kinetic energy for approximately the same recoil. Transfering momentum from the shooter to the shootee does no more damage than shooting the pistol does to the shooter in the first place. What does the damage is the work done by the energy delivered to the shootee's body. Is the damage linear, that is, would twice as much energy produce twice as much incapacitating effect? Almost certainly not but more energy will produce more damage, and that is a major part of what we want to achieve.

English

No cigar.

The analysis, question and conclusion drawn above is correct.

What you have failed to recognise is that the discussion above relates only to the "bullet on steel" transfer of momentum.

No "shooter's body". No "shootee's body". No "incapacitating effect". No one getting hit by a bullet.

Lead on steel, neither of which are human bodies.

Didn't read the thread, did you?

Phew! What a relief! I know it's been a while, but I couldn't figure out how in the world mv was not equal to mv. Your'e talking PROPORTIONS! Jeeze!

:wow:

Phew! What a relief! I know it's been a while, but I couldn't figure out how in the world mv was not equal to mv. Your'e talking PROPORTIONS! Jeeze!

:wow:

hogfish,

Oh, I am gonna disturb your reality again. Sorry.

If you'll please refer to page 1, post #23 of this thread, I state in the last paragraph of that same post that ½mv² = ½mv² is not a valid expression of Newton's Third Law of Motion since energy is not conserved.

Only momentum is conserved, hence the expression mv = mv is the only valid mathematical expression of Newton's Third Law of Motion.

Energy is not conserved and can be neither created nor destroyed.

The mathematics behind the inability to successfully state a valid expression for the conservation of energy is predicate upon the fact that the exponentially qualified velocity variable, V² , does not permit valid results after algebraic rearrangement and subsequent substitution of the appropriate numerical values occurs.

Here is an example to demonstrate what I attempting to convey:

As above (in post #23) the expression mv = mv yields:

M147 = 147 gr.

V147 = 990 fps.

M115 = 115 gr.

V115 = X fps.

M147 x V147 / M115 = V115

then we "substitute" the correct values;

147 gr. x 990 fps / 115 gr. = V115 = 1265.478 fps

So, in order to have a 115 gr. bullet achieve the same momentum of a 147 gr. bullet with a velocity of 990 fps, it needs to be moving no slower than 1265.5 fps and momentum is conserved.

With me so far?

OK...

However, if we attempt a similar conversion with the expression ½MV² = ½MV² we get:

½MV² / ½M = V² which becomes

(MV² / M) = V² after we reduce the "½" co-effecients from the expression

If we then solve for V, we then get:

√(M147 V147 ² / M115 ) = V115

"Substituting" the previous numerical values for the variables we then get:

√(147 x 990² / 115) = V115 = 1119.295974 fps

which cannot be since we've already proven under mv = mv that

147 x 990 = 115 x 1265.4873 fps is 145,530 = 145,530

and therefore, the correct velocity of the 115 gr. bullet must be 1265.4873 fps in order to agree with mv = mv , as opposed to 1119.295974 fps which would require that:

145,530 = 128,719.037

which is not true.

So momentum cannot be reconciled against kinetic energy without disagreement between the two expressions and therefore, Newton's Third Law of Motion which says, "To every action, an equal and opposite reaction" cannot be construed to apply in a similar manner to the kinetic energy relationship that might exist between two objects.

Sorry this ran so long. Hope that it clears things up for you.

:)

hogfish,

I got into this thread by a kind of accident without reading earlier posts or the title but 481 has given me a summary and so I hope I can contribute.

481's post just above is essentially correct though I would quibble about his "energy is not conserved". What he means is correct. Energy is not conserved in the motion of the objects, but it is conserved in other forms provided we don't get into Einstein. I wouldn't even say his post was too long in spite of his appology in the last sentence.

With every energetic transaction, some energy is transformed into unuseable heat. If we consider two idealised pool balls hitting each other without spin we can imagine two extreme possibilities that conserve momentum. In one, the first ball comes to a complete stop and the other ball continues in the same direction at the same speed. Momentum is obviously conserved since where there was one ball travelling at a certain speed in a certain direction, there is another ball travelling at the same speed in the same direction. Momentum has been transfered perfectly from one to the other.

In the second possibility, the two balls "stick" together and both travel on at half the speed. At twice the mass and half the speed, mv remains the same.

So what is the difference? The answer is their kinetic energy. In the first case the KE is mvv/2. In the second it is (2mvv/4)/2 = mvv/4. So, half the KE has been lost. Where did it go? Into heat. The two balls stuck together will be a little warmer than they were to start with but in the first case their temperature will be unchanged.

In both cases, momentum remains the same but the KE of the system is reduced in the second case relative to the first.

If ever you are hit by a bullet bouncing back from a steel target, it will be hot. The heat will not have come just from the heat of the propellant and barrel friction but from the transformation of KE into heat.

Because of this loss of KE to what is called hysteresis, you can't use conservation of energy in momentum calculations.

English

hogfish,

I got into this thread by a kind of accident without reading earlier posts or the title but 481 has given me a summary and so I hope I can contribute.

481's post just above is essentially correct though I would quibble about his "energy is not conserved". What he means is correct. Energy is not conserved in the motion of the objects, but it is conserved in other forms provided we don't get into Einstein. I wouldn't even say his post was too long in spite of his appology in the last sentence.

With every energetic transaction, some energy is transformed into unuseable heat. If we consider two idealised pool balls hitting each other without spin we can imagine two extreme possibilities that conserve momentum. In one, the first ball comes to a complete stop and the other ball continues in the same direction at the same speed. Momentum is obviously conserved since where there was one ball travelling at a certain speed in a certain direction, there is another ball travelling at the same speed in the same direction. Momentum has been transfered perfectly from one to the other.

In the second possibility, the two balls "stick" together and both travel on at half the speed. At twice the mass and half the speed, mv remains the same.

So what is the difference? The answer is their kinetic energy. In the first case the KE is mvv/2. In the second it is (2mvv/4)/2 = mvv/4. So, half the KE has been lost. Where did it go? Into heat. The two balls stuck together will be a little warmer than they were to start with but in the first case their temperature will be unchanged.

In both cases, momentum remains the same but the KE of the system is reduced in the second case relative to the first.

If ever you are hit by a bullet bouncing back from a steel target, it will be hot. The heat will not have come just from the heat of the propellant and barrel friction but from the transformation of KE into heat.

Because of this loss of KE to what is called hysteresis, you can't use conservation of energy in momentum calculations.

English

Yeah, and then you "hit one outta the park" like this. Good post!

NMGlocker

01-02-2010, 16:34

I don't need a physics lesson to know .357sig slams the plates down.

I can see it happen at the range where math meets reality.

481's post just above is essentially correct though I would quibble about his "energy is not conserved". What he means is correct. Energy is not conserved in the motion of the objects, but it is conserved in other forms provided we don't get into Einstein.

Just a difference in semantics between you and I in this case.

I am using the term "conservation" (as it relates in my post about kinetic energy realtionships) in the sense that there can be no mathematical reconciliation between the two equations. (for momentum and KE)

I make no argument against your second statement, since I agree that energy can be neither created nor destroyed; its form and existence being transitory and exchangeable, yet unextinguishable.

We are in total agreement.

:)

hogfish,

I got into this thread by a kind of accident without reading earlier posts or the title but 481 has given me a summary and so I hope I can contribute.

481's post just above is essentially correct though I would quibble about his "energy is not conserved". What he means is correct. Energy is not conserved in the motion of the objects, but it is conserved in other forms provided we don't get into Einstein. I wouldn't even say his post was too long in spite of his appology in the last sentence.

With every energetic transaction, some energy is transformed into unuseable heat. If we consider two idealised pool balls hitting each other without spin we can imagine two extreme possibilities that conserve momentum. In one, the first ball comes to a complete stop and the other ball continues in the same direction at the same speed. Momentum is obviously conserved since where there was one ball travelling at a certain speed in a certain direction, there is another ball travelling at the same speed in the same direction. Momentum has been transfered perfectly from one to the other.

In the second possibility, the two balls "stick" together and both travel on at half the speed. At twice the mass and half the speed, mv remains the same.

So what is the difference? The answer is their kinetic energy. In the first case the KE is mvv/2. In the second it is (2mvv/4)/2 = mvv/4. So, half the KE has been lost. Where did it go? Into heat. The two balls stuck together will be a little warmer than they were to start with but in the first case their temperature will be unchanged.

In both cases, momentum remains the same but the KE of the system is reduced in the second case relative to the first.

If ever you are hit by a bullet bouncing back from a steel target, it will be hot. The heat will not have come just from the heat of the propellant and barrel friction but from the transformation of KE into heat.

Because of this loss of KE to what is called hysteresis, you can't use conservation of energy in momentum calculations.

English

Great post there English. I was watching one of the Smithsonian channels (or something like that) the other day and they were talking about exactly what you posted. They showed a few slow motion shots were the bullet and the metal turned to liquid. Pretty cool stuff, well actually it was hot stuff.. LOL!!!

481 & English,

Thanks for the explanations. I was kin of speed reading, so it might have been clear from an earlier statement, but what got me (LOL) was that m & v weren't expressed as being different (bullet or time of flight). Only when I saw them later with the identifiers (sub 1 & 2) did it register. Sorry about that...my brain works in a silly way sometimes. ALL I could see was mv (simbol for not equal to) mv.

bogey3737

01-02-2010, 19:53

Thanks for the education guys...been a while since my last physics class. In all honesty, though...I just want another gun :)

DWARREN123

01-02-2010, 22:07

Wait for the GEN 4 guns then get a new new gun, G22.:supergrin:

Just a difference in semantics between you and I in this case.

I am using the term "conservation" (as it relates in my post about kinetic energy realtionships) in the sense that there can be no mathematical reconciliation between the two equations. (for momentum and KE)

I make no argument against your second statement, since I agree that energy can be neither created nor destroyed; its form and existence being transitory and exchangeable, yet unextinguishable.

We are in total agreement.

:)

481,

It was not a serious quibble but an interesting one that led to an interesting aspect of energy and momentum. I could as easily have said, "I entirely agree ...." and I am sure everyone reading your words understood their meaning.

Regards,

English

BleedNOrange

01-05-2010, 14:47

Maybe ya just "winged" it. Sounds like it wasnt a direct it.

Thanks for the education guys...been a while since my last physics class. In all honesty, though...I just want another gun :)

Then I would look at a G35 before you buy a G22. I have the 35 and it's a very sweet shooter plus you get a little more velocity from the longer barrel.

More better to knock them plates over with. Bang!!! :faint:

PAPACHUCK

01-06-2010, 04:50

Now my brain hurts.....too early in the morning for this kind of learning!!

Now my brain hurts.....too early in the morning for this kind of learning!!

Excedrin.

What pain?

:supergrin:

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