Physics 101 OMG

[DaveBeal]

Don't believe me? Get a medicine ball, put on roller skates and "throw" the ball without letting go. Note you will wobble but have no net "recoil'

True. And if you fire a round in a barrel with a solidly blocked bore, the same thing will happen. When the bullet starts to move, you are immediately pushed backward. When the bullet hits the blockage, you're pulled forward. No net movement, just like with the medicine ball.

When firing through an unblocked barrel, again you are immediately pushed backward by the recoil. The only difference is that this time there's nothing to stop the bullet's escape and reverse your motion. Whenever anything moves forward, something must move backward. Immediately. Physicists call it conservation of momentum.

Recoil basically doesn't happen until the bullet exits the muzzle because the gun/bullet is a closed system until the bullet exits.

You can certainly consider the bullet and the gun to be a closed system. But they're the same system before and after the bullet leaves the barrel. Before you pulled the trigger, the system had zero momentum. After you've pulled the trigger, no outside force has acted on the system, so it still must have zero total momentum. That is, the bullet's forward motion must be compensated for by the gun's backward motion, both before and after the bullet exits the barrel.

So if two bullets are exactly the same except for density the more dense bullet would actually hit the ground slightly before the less dense one.

That's true. As lions said, when you drop a bullet, not only does the bullet accelerate downward toward the earth, the earth also accelerates upward toward the bullet. The earth's acceleration is extremely small, but it's real. If you double the bullet's mass and drop it from the same height, the force felt by both the bullet and the earth doubles. The heavier bullet will drop at the same acceleration as before, because its doubled mass - and therefore doubled inertia - exactly cancels the increased force it feels. But the earth has no increase in mass to counteract the doubled force, so it accelerates twice as quickly toward the bullet. They will meet sooner. Very, very slightly sooner.

 

[everallm]

Only real quibble is that in these case we should not talk of WEIGHT but MASS.

We did the demonstration at my old school with 3 clear sealed, thick walled perspex tubes with an object in each and a tap valve in the base.

One had a large feather, one had a ping-pong ball, one had a small lead weight.

When at normal atmospheric pressure turn the tubes, watch the objects and time the drops.

Next put a hard vacuum on the tubes and repeat.

As a first year (11 years old) it was illuminating

 

[SN13]

The distance traveled in free-fall is s=g*t*t/2 so the time to fall a given distance is t=SQRT(2*distance/g) or the square root of 2 (~1.41 seconds) to fall 9.8 meters. Not the one second you claim. But you are correct the downward component of velocity will be 9.8 m/s after one second of free-fall.

As Einstein said: "Things should be as simple as possible, but no simpler." You've oversimplified to the point of not being correct.

--wally.

Yes, you are correct, I was using an improper formula... I was calculating alot of things, including the acceleration of the .308 150gr bullet inside a 24" barrel... so when I got to the end, i used the wrong equation but the point I was making is still valid. It's not the weight of the object, it's the time it takes to get to the target that causes the increased drop of the bullet.

http://www.remington.com/products/a..._ballistics_results.aspx?data=R223R6*PRA7MMRB

Go there, look closely... why does the smaller, lighter .223 drop so much faster than the heavier 7mm?

You're going to say "BC" but the BC of the .223 just means it has more drag and more friction and slows down faster giving gravity more time to work on it and making it hit lower at the same distance, but in a vacuum, BC wouldn't matter and they'd hit almost at the same distance away from the origin. (well 3110 vs 3025fps so within 100 feet if the flight time was 1second or less.)

 

[dmazur]

Whether he's using it on purpose or not, "Thus" was a more or less constant presence in physics problems...

I find it more nostalgic than annoying, but then I don't have to fight my way through physics problems anymore. Yay!

 

[Tarvis]

Your example of bullet drop was correct, just like if you had the different bullets sitting on a table and they all dropped off simultaneously. Although, wouldn't their individual aerodynamics (not ballistic coefficient) change how fast they fell due to air resistance as they accelerated towards the ground? Think of 2 200 gr objects, one being larger than the other.

Edit: my point was touched on earlier. As long as the bullets being fired/dropped have the same shape and are fired exactly parallel to the ground (considering the ground to be a flat surface that at each point is exactly the same distance to the center of the earth), they will hit the ground at the same time.

In the end, the only thing we need be concerned about is the fact that bullets do not "rise," that a bullet that "shoots flatter" is traveling faster and loses speed at a slower rate (read higher BC), causing the bullet to strike the target sooner giving gravity less time to pull it down; and lastly that true knowledge is knowing that we know nothing.

Had to throw that last bit in .

 

[MCMXI]

BC and "a rule of thumb"

I like useful "rule of thumb" stuff ...

BC (for a given bullet) * 100 is approximately equal to the distance at which the bullet has half of its initial (muzzle) energy.

Interesting stuff but as the saying goes "In theory there's no difference between theory and practice, but in practice there is".

 

[Sean S. Sulminers]

While I realize that the disparity between the two masses involved is so drastic as to render my point scientifically insignificant, it doesn't change the fact that...
Quote:
DUE TO GRAVITY AND TIME... Not WEIGHT.
...is an incorrect statement.

The fact that it can be neglected does not mean that it is not true, I merely intended to demonstrate that weight does have something to do with it.

Hi guys. I'm new to this forum and this thread caught my eye. It is interesting to see how things like this go back and forth due to definitions of language and our various perspectives. I think lions will agree that for all practical purposes, saying the bullets fall at the same rate is accurate but, I do have a couple of points that might shed some light on the subject.

1) We really should talk in terms of mass rather than weight. They are too different things and gravity is best associated with mass only. This point is not the focus of the discussion, but it's important in terms of definition.

2) In order to allow for the movement of the Earth toward the bullet, the observer has to be standing somewhere other than the Earth. Someone might have mentioned the concept of reference in the thread and that is the answer. When you work out the acceleration due to gravitational force and the mass of one of the bodies drops out, you have the complete answer. There is no need to work it out the other way and use some sort of superposition to account for the motion an observer would see if he was standing on the bullet.

So, what we are calculating is the force felt by the bullet and the acceleration that results from that force. Since our reference is the Earth and we should assume a constant gravitational force as if the experiments were being conducted in the same geographical location at the same altitude, everything is equal except the mass of the bullets. And, since the mass of the bullets drop out of the acceleration equation, the bullets do, in fact, fall at the same rate.

With the Earth as our reference, we are unable to observe it move, so the acceleration of the Earth toward the bullet does not play a role in this calculation. Neglecting wind resistance and such -- putting our experiment in a vacuum -- the bullets fall toward the Earth at exactly the same rate.

 

[RedLion]

Sean is correct ( And Welcome to THR), The gravitational pull of a bullet onto the earth is so small you could not measure it. The mass of the earth is 5.97*10^24 kg... a bullet is about 10 to 20 grams so if Force of gravity = [(gravitational constant of 6.67*10^-11)*(Mass of earth)*(mass of bullet)]/(radius between bullet and center of earth squared). you get an extremely small number. That is the force. If you take that force and apply it to a 10 to 20 gram bullet you get a very noticeable acceleration, but if you apply it to the earth you would get such a small acceleration you could not measure it.

Mass and Weight are different. Mass is not dependent on gravity. An object with Mass 'X' will have mass 'X' anywhere in the known universe, except maybe black holes or something, whereas an object with Weight 'A' will not have weight 'A' on mars or jupiter or in space because the gravitational acceleration is different.

Also Momentum and Kinetic energy are different. Momentum is calculated with M*V because it takes into account thing like sound and light being emitted by collisions or explosions hence a loss in total energy. Kinetic energy is .5*m*(v^2) where the collision happens inelastically with no energy lost to heat, light, sound ect. So a recoiling rifle should only be calculated using M*V and not .5*M*(V^2).

 

[Art Eatman]

Harvster, there is NO time lag on this "equal and opposite reaction" deal. When the bullet begins to travel forward, the rifle begins to travel back the other direction. Instantaneously. Right then. It does not sit there, mumbling to itself, "Er, uh, what do I do now?"

Go to the school books. Newton's laws of motion. Conservation of momentum. Or ask any physics prof, or sumdood in an engineering college who teaches statics and dynamics.

, Art

 

[Art Eatman]

Harvster, there is NO time lag on this "equal and opposite reaction" deal. When the bullet begins to travel forward, the rifle begins to travel back the other direction. Instantaneously. Right then. It does not sit there, mumbling to itself, "Er, uh, what do I do now?"

Go to the school books. Newton's laws of motion. Conservation of momentum. Or ask any physics prof, or sumdood in an engineering college who teaches statics and dynamics.

, Art

 

[GarandOwner]

This thread makes me want to pull my hair out

WOW :banghead:.....where to even start, it looks like this thread was attacked by a bunch of wiki-nazi's many things used here are incorrect. At first I wasnt even going to touch this post, but the engineer in me wouldnt let me walk away. Ok from the beginning..... I will use metric units since that is what the original poster used.

First and foremost, I just want to point out a few errors in prevoius posts. I am not "attacking" anyone, just making corrections. Some problems require higher level physics using differential equations to be accurate. Being an engineer neener: John) I have a background dealing with some of the forces that are ignored in lower level physics

I dont know about you guys, but I dont shoot in a vacuum. because of this ALL forces need to be accounted for, they can not be neglected. They are neglected in simple high school physics problems so you can learn the concepts, not because they are so small that they are insignificant.

In a Vacuum, we can calculate gravity's acceleration by a = Dv/Dt where a = acceleration, D(delta) = the change, v = velocity and t = time. In other words, if an object is moving at 9.8m/s after 1 second and it started at 0m/s at 0 seconds then the calculation a = (9.8m/s - 0m/s) / (1s - 0s) or 9.8m/s^2.

You can not calculate gravity the way you described because gravity is not a constant acceleration, using the change in velocity over the change in time will only give you average acceleration. Gravity varies with distance from the centriod of the earth. (oh and somewhere it was posted that this difference isn't measurable, yes it is )

First problem that I saw in multiple posts: lots of people keep posting the equation for the FORCE of gravity not the ACCELERATION due to gravity. When most people talk of gravity, they mean the acceleration due to gravity. This is:

g = (Gm)/r^2

where G is the gravitational constant 6.67 x 10^-11 N*m^2/kg^2
m is the mass of the earth
r the distance from the centroid of the earth

here is where the math starts to get complicated, because g is a function of r (r is a function of t because the bullet changes over time...... so we have a function g(r(t)) but we wont concern ourselves with that. Close to the earth ,as SN13 assumed correctly, gravity can be assumed to be constant (here in florida gravity is: 9.79 m/s^2 though the generally accepted "constant" value for gravity is 9.81 m/s^2 (this changes by quite a bit depending on the elevation above sea level of where you live)

However Gravity is not the only acceleration acting on the bullet. Air resistance is as well.

Air resistance does SLOW the acceleration towards the earth and there is a point at which air resistance and gravity are equal so that there is no more acceleration (Terminal Velocity) which is as fast as the object can fall, but we won't concern ourselves with that for now.

Sorry, but air resistance is SIGNIFICANT. If you want to talk about bullets in a vacuum, go shoot in a vacuum, but the results from those assumptions dont mean squat in the real world. As I will show in a moment...Warning: this will involve differential equations!

IF YOU SHOOT A BULLET WITH A TRAJECTORY THAT PARALLELS THE EARTH, IT WILL NOT RISE. Not for the first 25yards, 100yards or EVER. When you THINK a bullet is rising, it's because you shot it at an upward angle. It's beginning to accelerate DOWN instantly.

This is correct.

RECOIL BEGINS THE MOMENT THE BULLET MOVES.

Correct, while some of your math is wrong due to the fact that the acceleration is exponential not constant while the bullet is traveling down the barrel, the concept is correct. Most people think of perceived recoil. The Rifle has a much larger mass, there for a much larger inertia than the bullet, so it doesn't accelerate as fast

aerodynamics play a massive role. Shape and drag coefficient of jacket material both play a role in how fast something drops. If the jacket is of the same material, and they are the exact same shape, then yes they fall at the same speed.

incorrect, wind resistance IS dependent on mass as well as shape and material

#3 - A 65 gr bullet and a 500gr bullet DROP AT THE SAME RATE!

If you fire a 65gr bullet at 100fps and a 500gr bullet at 100fps, they'll have the SAME drop rate.

So, if you fire a 150gr bullet with the same BC as a 180gr bullet, and they travel at the SAME velocity when exiting the barrel, which one will have more drop at 200yrds for a 100yrd zero?

If you said 180gr you're WRONG.

Gravity pulls objects towards the earth. If you set up 5 guns, 9.8 meters above the ground, with all barrels parallel to the earth, one shooting a 500gr bullet at 100fps, one shooting a 200gr bullet at 100fps, one shooting a 500gr bullet at 2000fps, one more shooting a 200gr bullet at 2000fps, and ONE last one shooting a 50gr bullet at 4000fps, and you fire ALL OF THEM AT THE SAME TIME, the bullets will contact earth at the EXACT same time... ONE second after firing.

Well why is there a difference in drop between 150gr .308 and a 168gr .308?

Simple really.... one's traveling at 3000fps and the other is traveling at 2700fps. The heavier bullet takes a bit longer to reach the target, thus it will fall a bit more due to gravity and gravity ALONE. A bullet moving at 3000fps (in a vacuum to eliminate drag and air friction) will travel 3000 feet in one second and fall 32.15 (9.8m) due to gravity. For a bullet traveling 2700fps to move 3000 feet, requires 1.11 seconds which results in 35.68ft of drop DUE TO GRAVITY AND TIME... Not WEIGHT.

INCORRECT!!!!!!!

Reason: AIR RESISTANCE!!!!

Air resistance IS dependant on mass as I will show:

Lets start with Newton's second Law: F = ma

a is the first derivative of velocity (dv/dt) air resistance is defined as bv where b is a positive constant that depends on the density of air and the shape of the object, v is the vertical velocity. So we have:

m(dv/dt) = mg - bv

using separation of variables this First Order Differential equation can be solved:

dv/(mg - bv) = dt/m

integrating we get:

-(1/b)ln|mg - bv| = t/m +c where c is a possible constant from integrating

so we get:

mg - bv = e^(-bc)*e^(-bt/m)

or mg - bv = Ce^(-bt/m) because e^(-bc) is a constant

solving for v we get:

v = mg/b - A/be^(-bt/m) it can be seen that velocity IS dependent on mass

I could go further but I doubt anyone would want to follow it, but the end result is that position is the function:

h(t) = (gm/b)t+(m/b)((gm/b)-v_initial)(e^(-(b/t)t)-1)


so in the real world, due to air resistance, a heavier bullet DOES drop faster than a lighter one of the same shape in the same air density.

 

[cracked butt]

cracked butt, in air you're correct when you're talking about Ma Bell distances. In vacuum, however, BC doesn't have anything to do with anything. Irrelevant. In vacuum, the bullet could be a flat-end cyclinder.

If we lived in a vacuum, this would be useful, but in the real world, air drag has more effect on bullet flight than gravity, at least past spitting distances.

 

[Jim Watson]

Yup.

Guess we are going to have to sneak into the physics department stockroom at night and do some tests in their vacuum chambers with their frictionless surfaces.

Operating in air with real materials changes things a lot.

 

[ctdonath]

Recoil basically doesn't happen until the bullet exits the muzzle because the gun/bullet is a closed system until the bullet exits. The force from momentum backward as the bullet moves forward is basically balanced by the friction force of the bullet dragging the gun forward with it.

Don't believe me? Get a medicine ball, put on roller skates and "throw" the ball without letting go. Note you will wobble but have no net "recoil'

ARGH! So wrong I can't not respond.

Recoil ONLY happens while the bullet is in the barrel! (OK, and for that miniscule time while the gasses are still pushing after.) Get a medicine ball, put on roller skates and "throw" the ball without letting go. Note you will have gone from holding the ball close, to the "closed system" being expanded with both you and the ball farther from where you started from, connected with your stretched-out arms still connecting the two. The moment you let go of that ball, you're coasting from that "recoil" movement. Someone standing behind & contacting you will most certainly feel "recoil" even though you didn't let go of the ball.

The gasses are pushing the bullet one way, and the rest of the gun the other way. Once the bullet exits and the gasses are released, there is no more energy being introduced into the system; subsequent movement is of the "an object in motion stays in motion" kind.

 

[JohnBT]

"First and foremost, I will just correct things due to the fact that I have a higher understanding of physics than most since I have an engineering degree."

You know, that's the same attitude that bothered me when I was a physics major and they hired grad students out of the engineering department to teach calculus. Or as the old timers called it, The Calculus. Why do engineers think they know it all? Does anybody know why that is?

Heck, I learned all of this in high school. My high school physics teacher had worked with von Braun's group and even got me an all-day behind-the-scenes guided tour of NASA's Goddard Space Flight Center in Greenbelt Maryland. Now that was a fun day. If only I'd had the money to tour all of the NASA facilities around the country. Nothing like having a connection to a bunch of rocket scientists.

Imagine a high school teacher using an air cannon to shoot a 2-ounce brass bullet at a coffee can dropped from an electromagnet suspended from the ceiling. No matter how often you do it, the falling bullet always hits the falling can. I wonder if Mongomery County MD schools would allow him to use a big air gun in the classroom these days?

John

 

[Harvster]

Harvster, there is NO time lag on this "equal and opposite reaction" deal. When the bullet begins to travel forward, the rifle begins to travel back the other direction. Instantaneously. Right then.

Well I would agree that there is some recoil from the bullet starting its journey (yes I do in fact believe in conservation of momentum) most of what you feel in terms of recoil however, is from the gas escaping the barrel after the bullet leaves.

 

[EHCRain10]

wow this stuff is too deep for me (its summer time dont wanta think too hard) and im halfway through a physics degree
ive read it and what i spotted as being wrong was soon pointed out and corrected

 

[Tarvis]

most of what you feel in terms of recoil however, is from the gas escaping the barrel after the bullet leaves

Did Al Gore tell you that? Equal and opposite reaction. What you are saying is the gas has thrust that far outweighs the energy from the bullet? If this was true, a muzzle brake would in fact make a rifle jump forward when it is fired. I don't have any cool formulas to prove you wrong, but if you use the same amount of powder in a large bore rifle as you do in, say, a 22-250 firing a 40 gr bullet with everything being the same except the bullet weight, the heavier bullet will have more recoil.

 

[Shear_stress]

most of what you feel in terms of recoil however, is from the gas escaping the barrel after the bullet leaves.

Sorta true. The escaping gas does generate momentum and this momentum can contribute a little to the felt recoil. In reality, the small weight of the propellant limits this "rocket effect", even though the gas may travel much faster than the bullet. Definitely not "most" of the recoil, especially in pistols, which have a relatively small amount of propellant.

 

[lions]

As far as the medicine ball example goes, forget about throwing it but not letting go, that isn't how a gun works so it really doesn't prove anything. Put on your roller skates and throw the ball, take note of when you start to move backwards. I bet you will start moving as soon as you start throwing, not when you let go.

 

[hso]

In the real world, where scientists still can't tell how a bumble bee is able to fly

This is a common myth. The story goes that a 19thC dinner party "back of the napkin" calculation was done "showing" that the bumblebee couldn't fly. In reality the calculations for bumblebee flight have been knows for some time and this myth is addressed in almost any college physics course as an example of how simple linear equations may not be adequate to describe a complex process requiring advance modeling. Bumblebees, and other bees and insects, fly because their wings are flexible instead of rigid airfoils. That flex allows vortexes to form on the up and down stroke providing lift for the little critters. Recent modeling coupled with advanced imagery gives a much better detailed explanation of the flight of bees than ever before.

 

[GarandOwner]

You know, that's the same attitude that bothered me when I was a physics major and they hired grad students out of the engineering department to teach calculus. Or as the old timers called it, The Calculus. Why do engineers think they know it all? Does anybody know why that is?

Sorry if it came out sounding arogant John, I didnt mean for it to sound like "im better than you" or "I know it all" just pointing out that most of the debate here deals with theory rather than practical application, and many posts look like they just cut and pasted something from wikipedia :banghead: Until you get to calculus and differential equations most physics classes leave out alot of important variables, which makes it easier to learn the concept, but the equations from lower level physics cannot generally be applied accuratly to the real world.....Oh and I dont think that I know everything, and dont claim to. But from what I have seen, you have it wrong, mathematicians are the ones who think they know everything. we engineers are happy to approximate when we can... in many ways we are the lazy part of the scientific community

By the way I rephrased my previous post to a more "humble" introduction

 

[jackdanson]

You can see .45 ACP bullets in flight, and I have done it many times when I was younger

Yeah, I'm 25 and I see .45's all the time. I also can see .410 shot. Never seen a .22 though, maybe just haven't caught the light right yet.

 

[Husker1911]

Speaking of physics and Newtonian laws, my favorite western movie has a scene that still pisses me off. It's in Open Range, where Duvall shoots BG with shotgun, and the blast propels BG across the room, slamming into the opposite wall. Yet Duvall isn't similarly affected by the recoil. A sour note in an otherwise marvelous movie.

 

[cracked butt]

Speaking of physics and Newtonian laws, my favorite western movie has a scene that still pisses me off. It's in Open Range, where Duvall shoots BG with shotgun, and the blast propels BG across the room, slamming into the opposite wall. Yet Duvall isn't similarly affected by the recoil.

The Hollyweird writers probably get their physics understanding from the same sources or leaps of logic I've often seen on Hunting forums: If a x magnum has a muzzle energy 3000ft-lbs, it will have the same effect on y animal as getting hit by a Buick.
I've long since quit arguing with those types because stupid is impossible to fix.