Math Problem

Serendipity

Hi,

I was just wondering if any of you could help me figure out this calculus problem, as I am totally stumped and not sure where to start.

"A solider in a firing range fires an 8-shot burst from a weapon at a rate of 1000 rounds per minute. Each bullet has a mas of 7.45g and a speed of 293 m/s relative to the ground as it leaes the barrel of the weapon. Calculate the average recoil force exerted on the weapon during that burst."

I'm not sure if I'm supposed to use integrals here, since force = the derivative of momentum and the integral of F(t)dt = the integral of (dp(t)/dt)*dt.

And I also think that since I can calculate the difference between the initial and final momenta p0 and p1 and since p = mv, then delta p = p0-p1 = mv0-mv1, am I right in this regard?

Thanks a lot for your help!

ben62670

The average recoil would be the same as one shot. The second shot isn't going to place anymore force on the weapon than the first shot. Once the action is fully cycled it starts all over again. Its a trick question. Now if they are looking for the summed force to the weapon it would be 8x the first shot. Also another thing to be considered is that the bullet doesn't reach max velocity till after it leaves the barrel. I am assuming that this was taken into consideration when the info was given to you.

Serendipity

ben62670 wrote: »

The average recoil would be the same as one shot. The second shot isn't going to place anymore force on the weapon than the first shot. Once the action is fully cycled it starts all over again. Its a trick question. Now if they are looking for the summed force to the weapon it would be 8x the first shot. Also another thing to be considered is that the bullet doesn't reach max velocity till after it leaves the barrel. I am assuming that this was taken into consideration when the info was given to you.

Thanks, this helped me a bit, but I am still unsure how I should tackle this problem. I think the problem is looking for the summed force on the weapon, but the problem (quoted above) isn't really clear if this is true or not.

I am thinking of using p = p0-pf = mv0-mvf = m(v0-vf) = m*delta v to find the momentum, bt there is an impulse applied by the weapon (on the bullet?) I'm not sure if I am correct in that matter. I am thinking the impulse i, which = the integral of the applied force (integral F over time) will help.

Since force = d/dx of momentum, I = the integral of F(t)dt = integral of dp(t)/dt * dt = p0-pf = delta p. Since I also = F*delta t = delta p / delta t, then how do I solve this equation?

I'm still confused.

Thanks a lot for your help!

I-SIG

Actually the effects of recoil are cumulative. Keep in mind that the GAU-8 on an A-10 slows it down like 30 or 40 MPH.

Wes

ben62670

There are to many variables missing. Is the gun is solid mounted? It would increase the overall force applied to said weapon. If it is not solid mounted. You have to take into consideration the weapon moving rearward do to recoil. This also depends on force/weight of holding device. At that firing rate there wouldn't be enough time to fully recoup between shots. Also if the weapon isn't solidly mounted then you also need to take into consideration the weight of the gun to figure velocity*projectile weight to figure force applied to gun. Also not mentioned is the weapon gas operated, or recoil operated? Now you have to figure in the force of the cartridge moving back wards, and the weight/spring rate of recoil mechanism. If it is a gas operate gun then you also lose some force do to escaping gas. Also with gas you have to consider there is a piston/spring mechanism moving rearward, so you would also need to know the mass of the piston, and the spring rate of said mechanism. With all these variables there is no way to calculate total force applied to said weapon. Thats why I say it is a trick question. If this is for a class you could easily disprove any answer given with all these missing variables.

beardog03

appadv wrote: »

Hi,


Each bullet has a mas of 7.45g and a speed of 293 m/s relative to the ground as it leaes the barrel of the weapon. Calculate the average recoil force exerted on the weapon during that burst."
QUOTE]

Not to be a d*ck ben, but they didn`t ask how the weapon operates ...!!

mas , speed, length of burst, # of bursts, length of travel relative to the ground, etc.

I don`t know !
I suck at math !!!


Sorry gents, carry on...

Not sure why I even stepped in !
My Bad.....:rolleyes:

ben62670

Ever fire a semi auto gun compared to a bolt action? Both guns being equal weight same shell the semi auto has considerable lest recoil.

beardog03

Blowback....
gases push slide to the rear and eject cartridge, etc.


I`ve fired one or two in my time....;)

I`m good with weapons, I suck at math !!

mrbigbluelight

Answer to math problem:

Kill 'em all, let God sort 'em out.

bobman1235

Man, I just got anxious reading that.

I do NOT miss college.

wingnut4772

I think the answer is 5.

petrym

Make that 5 newton meters per second and you've got something.

Either that or 5 furlongs per fortnight - I can't remember the conversions...

Actually I have no idea, sorry,

Serendipity

Hmmm - that's what I am thinking. The problem is missing too many variables.

I read the problem again, and I think I missed a sentence:

"A soldier on a firing range fires an 8-shot burst from an assault weapon at a full automatic rate of 1000 rounds per minute. Each bullet has a mass of 7.45g and a speed of 293 m/s relative to the ground as it leaves the barrel of the weapon. Calculate the AVERAGE recoil force exerted on the weapon during that burst."

The fact that the weapon is an automatic makes some difference - although I know nothing about weapons.

Is that sentence more useful in helping to solve the problem?

Again, thanks for your help!

beardog03

green


the answer is green...!!

unc2701

OK- It's a physics question, so they don't give a **** what kind of weapon it is. In fact, under the equal and opposite reaction, what they're really asking is how much force is applied to move those bullets, since that will be equal to the recoil (for the purposes of the question- normally a good bit of the force is used to cycle the action and the friction between the soldiers feet and ground enter the problem...)

So, figure out the time needed to fire 8 rounds from a 1000 rpm gun, then everything else should fall straight into your formulas. The only ambiguity is whether the time includes the action cycling on the last round- I'm going with "Yes".

bvette94

After reading this i have never felt do dumb in my life.

MrNightly

bvette94 wrote: »

After reading this i have never felt do dumb in my life.

"Do dumb. Do dumb. Do dumb, do dumb, do dumb, do dumb do dumbbbb... do do do dumb." - What the pink panther said when asked a calculus problem.

ben62670

Without knowing the weight of the weapon I would assume that the answer would be the same for every shot fired. I would also write up a little thesis presenting all the missing variables to this equation just to show off:D I would also end it with a quote " I'm a mathematician not a magician"

Ben

PolkThug

The average recoil on the weapon itself would be the same as one bullet. The average recoil in your hands would be slightly more, because, once you jolt back a little, the next bullet is pushing you back before you get your body back to "zero position".

jdhdiggs

Good lord, you yahoos are going to eff up a perfectly straight forward question. The answer is obviously green.

Getting serious, this is pretty easy, espeically since the prof counter acted most of the yahoo's comments in the thread with the "exerted on the weapon" phrase.

Caluculate the volume of the impulse curves (Pretty straight forward F=MA and work calculations) then you calculate the time it took to fire the eight shots and ta-da, your done. Just find out the work done on one shot, multiply by eight, divide by time and Bob's your uncle.

beardog03

wwaaaaaaaaaaaaaaaa

and Bob`s your uncle !

Priceless

ben62670

jdhdiggs wrote: »

Getting serious, this is pretty easy, espeically since the prof counter acted most of the yahoo's comments in the thread with the "exerted on the weapon" phrase.

.

If said weapon is moving rearward do to recoil then the second shot would put less force on the weapon do to rearward travel. Goes back to not enough info given for question. There is no true answer for this question. Mass of weapon, mass of shooter, strength of shooter, even what angle is the shooter shooting from all can have an effect on force applied to weapon.

I think this is a trick question

appadv are you messing with us? Or is this a school type of question?

Serendipity

This is for my Calculus class. The problem is due on Monday.

beardog03

I think you should listen to JD and see how you do..

Whatever you come up with, please post it...and the answer on monday, if ya would..






I still say the answer is GREEN...!!

mrbigbluelight

appadv wrote: »

This is for my Calculus class. The problem is due on Monday.

No, the ANSWER is due on Monday.

However, if the instructor incorrectly stated, "The problem is due on Monday", than give her back the problem.

If he/she says, "I was looking for the answer", than merely retort with

"Look, I'm not a mind reader ! If you wanted an answer to your GD problem, than you should have told us you were looking for an answer to your GD problem !!!

Than stomp away from his/her desk in a sulking manner.

Serendipity

The answer is due on Monday.

JD - I understand your reasoning but I am not sure about the distance. Since Work = Force * distance, but you don't know the distance, and I'm not sure if the bullet travels in the path of a projectile, then...

Do I use the equation for projectile motion x = x0 + v0*t + (1/2)at^2

I'm totally confused at this point.

mrbigbluelight

appadv wrote: »

I'm totally confused at this point.

Than it appears that my work here has not been in vain.

hearingimpared

All I know is all that recoil would hurt!!!

wallstreet

Check this out, http://en.wikipedia.org/wiki/F%3Dma#Newton.27s_second_law.E2.80.94historical_development and look down to Newtons second law. It appears to have all the variables that you need.

Serendipity

Just read the article on Newton's 2nd Law. Thanks for the link.

I understand that F = m*a and that F = d(mv)/dt, but I'm unsure as to how to find the distance that the bullets traveled in the work equation. Since W = F*d and F = m*a, then Work should equal W = m*a*d.

The problem is that I'm unsure as to how to find the distance. If the bullet travels in the path of a projectile, how do I determine the distance?

Thanks for all your help so far!

RSTERN

What about gravity? You know the weight of the bullet, you know the speed of the bullet. The pull of gravity should determine how far the bullet goes.
Rob