
#1
Jan2207, 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. 



#2
Jan2207, 10:23 PM

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#3
Jan2307, 06:21 PM

P: 4

ah. could you show me how you would set about solving it? stepbystep, if you can, because i need to understand how to do this myself.




#4
Jan2307, 07:11 PM

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constant resisting force 



#5
Jan2307, 07:29 PM

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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. 



#6
Jan2307, 10:55 PM

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vf =0, and vi = 6.4. 



#7
Jan2407, 06:02 PM

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400*t=(500/9.8)*6.4
t=.815s, which isn't the correct answer. what else did i mess up? lol. 



#8
Jan2407, 06:55 PM

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