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Gun-Projectile Elastic Collision

  1. Jul 8, 2017 #1
    IMG_4684.JPG IMG_4685.JPG 1. The problem statement, all variables and given/known data
    A projectile of 45kg has a muzzle speed of 656.6 m/s when fired horizontally from a gun held in rigid support (no recoil). What will be the muzzle speed (relative to the ground) of the same projectile when fired from a gun that is free to recoil? The mass of the gun is 6600 kg. (Hint: The kinetic energy of the gun-projectile system is the same in both cases.

    2. Relevant equations
    (Pictures below) IMG_4684.JPG
    IMG_4685.JPG
    3. The attempt at a solution
    image.jpg
     
  2. jcsd
  3. Jul 8, 2017 #2
    I'm not really sure if I'm on the right track or not....
     
  4. Jul 8, 2017 #3

    scottdave

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    One way to approach this is that the total momentum created by firing the weapon will be the same, whether there is recoil or not. The first situation is simpler and you can find out the value of this momentum
     
  5. Jul 8, 2017 #4
    So would that be as simple as adding the momentums together and calling it the momentum of the system?
     
  6. Jul 8, 2017 #5

    scottdave

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    Yes, that will give you a second equation. Then you have 2 equations and two unknowns.
     
  7. Jul 8, 2017 #6

    berkeman

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  8. Jul 8, 2017 #7
    Image1499564269.615716.jpg

    So that's where I'm at now....(ignore that first line at the top)
     
  9. Jul 8, 2017 #8
    It looks to me like you are trying to conserve the momenta between the two scenarios... this is not what conservation of momentum means.

    In the recoil-less scenario, momentum is not conserved because outside forces are exerted (to hold the gun in place). We don't need to find any variables in this case though, because all the information is given!

    In the case with recoil, all forces are taken to be absent, so then the momentum of the gun+bullet will be the same before and after firing.
     
  10. Jul 8, 2017 #9
    I feel like that's similar to the very last line, p = p'. I didn't think of it in that extent, though.

    I'm getting really confused with this system of equations. Maybe I've just had too much calculus and I've forgotten my algebra, but I keep getting some type of proof, like v(projectile) = v(projectile).
     
  11. Jul 9, 2017 #10
    In that last line you are equating the net momentum of two different systems... why should they be equal??

    Conservation of momentum is about comparing a single system at two different times (provided no outside forces act on the system... because that is the definition of force; the rate of change of momentum).


    All the algebra in the world won't help if the physics isn't right.
     
  12. Jul 9, 2017 #11
    Now I'm really confused. Haha I may need someone to explicitly tell me what to do in this case cause I'm just getting super thrown off at this point :( I have no idea how I've been passing this class
     
  13. Jul 9, 2017 #12

    ehild

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    The mechanical energy supplied by the explosion of the gun-powder in the gun is the same in both scenarios. The momentum is conserved during the shot, but the system is the earth+gun+projectile in the first case, and the gun+projectile in the second case. You can use conservation of momentum in the second case (total momentum is the same before and after the shot) , and you get the energy of the system from the muzzle speed in the first case.
     
  14. Jul 9, 2017 #13
    Image1499624231.412472.jpg

    So far so good?
     
  15. Jul 9, 2017 #14
    So wait, if the total momentum is the same before and after, wouldn't it just be zero? Because I thought the momentum before the shot would be zero because nothing is moving.
     
  16. Jul 9, 2017 #15

    ehild

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    Yes. Now you write the relation between v1 and v2 using conservation of momentum. And yes, the total momentum is zero before the shot.
     
  17. Jul 9, 2017 #16

    berkeman

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    Correct, the sum before and after is zero. But remember that momentum is a vector quantity, having both magnitude and direction... :smile:

    EDIT -- beaten out by ehild :smile:
     
  18. Jul 9, 2017 #17
    So then for the momentum after the collision, I would have an equation looking like:
    m(sub1)v(sub1) = -m(sub2)v(sub2)

    ?
     
  19. Jul 9, 2017 #18

    ehild

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    Yes. m1v1= -m2v2.
    (Use x2 and x2 for subscript and superscript)
     
  20. Jul 9, 2017 #19
    So long story short, I think I finally figured out the answer (with much pain and agony). From that last equation you put in, I solved for v1, plugged that equation in the kinetic energy equation and solved for the new velocity of the projectile, which came out to be 655.7 m/s. Seems like a valid answer. I definitely won't try to type out the work because the app doesn't support the math functions, but should I post a picture of my ugly work? Haha just thinking for future people who might stumble across this thread or for myself to see later.
     
  21. Jul 9, 2017 #20

    ehild

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    Just post a picture of your work, ugly or not, it is useful. :smile:
     
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