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Constant resisting force

by moondust0109
Tags: constant, force, resisting
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moondust0109
#1
Jan22-07, 09:45 PM
P: 4
1. The problem statement, all variables and given/known data

a 500 lb gun fires a 2lb projectile with a muzzle velocity of 1600 ft/s. if the recoil is against a constant resisting force of 400 lb, find the time taken to bring the gun to rest and the distance it recoils.

a) .20s, .80ft b) .25 s, .80 ft c) .25 s, .85 ft d) .15 s, .75 ft
2. Relevant equations

p=mv

3. The attempt at a solution

initial recoil is -6.4 ft/s (500*-6.4=2*1600)

i'm not sure where to start on this...the constant resisting force means what? i've never come across that before.
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PhanthomJay
#2
Jan22-07, 10:23 PM
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Quote Quote by moondust0109 View Post
1. The problem statement, all variables and given/known data

a 500 lb gun fires a 2lb projectile with a muzzle velocity of 1600 ft/s. if the recoil is against a constant resisting force of 400 lb, find the time taken to bring the gun to rest and the distance it recoils.

a) .20s, .80ft b) .25 s, .80 ft c) .25 s, .85 ft d) .15 s, .75 ft
2. Relevant equations

p=mv

3. The attempt at a solution

initial recoil is -6.4 ft/s (500*-6.4=2*1600)

i'm not sure where to start on this...the constant resisting force means what? i've never come across that before.
When the gun is fired, it recoils back against, say, the shoulder of the person holding the firearm. The shoulder provides an assumed constant force of 400 pounds to bring the gun to a stop in a certain distance over a certain period of time. Think along the lines of impulse =momentum change.
moondust0109
#3
Jan23-07, 06:21 PM
P: 4
ah. could you show me how you would set about solving it? step-by-step, if you can, because i need to understand how to do this myself.

PhanthomJay
#4
Jan23-07, 07:11 PM
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Constant resisting force

Quote Quote by moondust0109 View Post
ah. could you show me how you would set about solving it? step-by-step, if you can, because i need to understand how to do this myself.
We can help you through it, if you show some of your work. I've already suggested Impulse = momentum change (where Impulse =F*t); and you've already correctly calculated the initial recoil velocity of the rifle....you should be able to solve for the time with that formula...give it a try ...watch your units...
moondust0109
#5
Jan23-07, 07:29 PM
P: 4
it still does not fully register with me. i'm not sure how to use the impulse formula in this case.
(this isn't right, but here's what i did)
400lb*t=(500lb*-6.4ft/s)-0
but that gives me a time of -8 seconds

i used the vf^2=vi^2+2ad to find that (6.4/t)(d)=20.48
i know that the right answer is b) .25 s, .80 ft already, but i just need to understand how to plug everything in to the impulse equation because it's just not sinking in.
PhanthomJay
#6
Jan23-07, 10:55 PM
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Quote Quote by moondust0109 View Post
it still does not fully register with me. i'm not sure how to use the impulse formula in this case.
(this isn't right, but here's what i did)
400lb*t=(500lb*-6.4ft/s)-0
but that gives me a time of -8 seconds

i used the vf^2=vi^2+2ad to find that (6.4/t)(d)=20.48
i know that the right answer is b) .25 s, .80 ft already, but i just need to understand how to plug everything in to the impulse equation because it's just not sinking in.
Oh I think it's sinking in, but remember, a common error in the States, you can't use weight for mass. You're using 500 pounds as the gun's mass. This is not correct. That's the gun's weight. What is the formula that relates weight and mass? then, you will have got it right. Except for that tricky minus sign, it's m(vf -vi) for momentum change, where
vf =0, and vi = -6.4.
moondust0109
#7
Jan24-07, 06:02 PM
P: 4
400*t=(500/9.8)*6.4
t=.815s, which isn't the correct answer.
what else did i mess up? lol.
PhanthomJay
#8
Jan24-07, 06:55 PM
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Quote Quote by moondust0109 View Post
400*t=(500/9.8)*6.4
t=.815s, which isn't the correct answer.
what else did i mess up? lol.
units, units, units! You used g=9.8m/s/s, which is correct in the SI international system of measure (metric units); but when you're using the Imperial system of measure, used primarily in the USA only, that is, when dealing with pounds of force instead of newtons, feet versus meters, and slugs versus kilograms, then you must use the value of 'g' in units of ft/s/s. Surely you know what that value is?


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