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Average Bullet Velocity Inside the barrel

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    The M61 Vulcan cannon is a Gatling Gun-type weapon deployed on an aircraft. It fires a 100g, 20mm diameter projectile at a rate of up to 6,000 shots per minute. Barrel length is 170cm, and muzzle velocity (speed of the projectile leaving the barrel) is 1030 m/s.

    -Assuming constant acceleration, what is the mean (average) velocity of the projectile in the barrel?

    (It has been 8 years since a physics class...not looking for an answer but an arrow in the correct direction. Thank you)

    2. Relevant equations
    V=D/T

    [tex]\overline{V}[/tex] = [tex]\Delta[/tex] Distance/Time

    [tex]V^{2}[/tex]=[tex]V^{2}_{0}[/tex] + 2as

    3. The attempt at a solution

    Where do I start...I began with the second equation. V = 170cm / T Well if it shoots 6000rpm then it would be broken down to 100rps. Given that I would say that fires a bullet every .01 seconds. Plug the variables into the equation.

    V = 170 cm / .01 sec or 17000cm or 170m/s

    If you see the muzzle velocity of 1030m/s then you would know right off the bat that 170m/s is wrong.

    So I used the second equation...and came up with some crazy answer that I know is definitely wrong. See below...

    10302 = [tex]1030^{2}_{0}[/tex] + 2a170

    1060900 = 0 (Initial Bullet Velocity is zero) + 2a170 (Solve for A)

    a = V2/2*170cm

    a = 1060900/340cm or 3120.29cm or 312,029 meters

    Plug back into the equation

    1060900 = 0 + 2*312,029*1.7

    And they magically equal. Some how I think I might be using the wrong equation but I need someone to help me.
     
  2. jcsd
  3. Feb 13, 2010 #2
    To achieve this answer you need to know how much time the bullet spends in the barrel.

    [tex] V^{2} = V^{2}_{0} + 2as [/tex] [1]
    [tex] V = V_{0} + at [/tex] [2]

    Divide 1 by 2 to eliminate a, as we dont know it. V0 is 0

    [tex] V = \frac{2s}{t} [/tex]
    [tex] t = \frac{2s}{V} [/tex]

    which is t = 2 * 1.7 m/ 1030 m/s = 0.0033 seconds.


    See where you can go from there...:wink:
     
    Last edited: Feb 13, 2010
  4. Feb 13, 2010 #3
    So it would be fair to say that the bullet travels the length of the barrel in .0033 seconds.

    Solve for Avg Velocity...

    [tex]\bar{V}[/tex] = [tex]\Delta[/tex]d / [tex]\Delta[/tex]t

    [tex]\bar{V}[/tex] = 170cm/.0033 secs = 51515.15cm/s or 515.15m/s

    Would you have to change the time to correlate with cm and not meters?

    [tex]\bar{V}[/tex] = 170cm/.000033 secs = 5151515.15cm/s or 51515m/s :confused:

    Wouldn't the bullet be more close to it's terminal velocity? Or because this is the "average" velocity, you wouldn't have such a large number...???

    I'm trying to learn via a few MIT/Stanford University lec. videos...
     
  5. Feb 13, 2010 #4
    the average velocity should be the middle point, of your velocities, because acceleration is constant, therefore it accelerates from 0 to 1030 in the length of the barrel.
    (1030+0)/2 = 515m/s is your average velocity.
    i give the answer because you have already stumbled upon it, but hopefully my view of things allowed your mind to take an easier path to the answer :)
    hit back with any more questions and i can go into further detail
     
  6. Feb 13, 2010 #5
    Not quite sure what you mean here. 51500 cm/s is exactly the same as 515 m/s

    Get this idea out of your head quickly. This is confusing and illogical.

    This is correct but I prefer my reasoning as the mathematical derivation from the equation provides a mathematical basis, in addition to physical intuition (which can sometimes be flawed).
     
  7. Feb 13, 2010 #6
    Well good to know I was on the right track. I understand mean as average and only average. If I wanted the middle I would take the length of the barrel and divide it by half to find the velocity at that point.

    When I was in my algebra class I was taught to simplify. I guess it isn't wrong but is not fully relative to the muzzle velocity which is in m/s.

    Oh and totally forgotten the seconds issue. The answer was in seconds not m/s or cm/s...

    THANK YOU!! I'm starting to understand now!
     
  8. Feb 13, 2010 #7
    Then to find the acceleration of the bullet inside the barrel I would use the following formula.
    a = Δv/Δt= (vfinal - vinitial) / (tfinal - tinitial)

    Initial thought was to use 1030m/s which is the muzzle velocity given above. And .0033 is the time it takes the bullet to travel the length of the barrel.

    a = (1030-0) / (.0033-0) = 312121 m/s

    For some reason I believe that this one is correct below. [tex]\downarrow[/tex]

    For Vfinal I calculated 170cm/.0033 = 51515cm

    a = (515.15-0)/(.0033-0) = 156106 m/s

    Correct?
     
  9. Feb 13, 2010 #8
    This one is indeed correct. The unit of acceleration is [itex] ms^{-2} [/itex]
    You already nearly computed this correctly from [itex] v_f^2 = v_0^2 + 2as [/itex]
    in your first post (except for the units)

    The difference in the numbers is round-off error. If you round off the numbers to 3
    significant figures, you'll get 3.12 * [itex]10^4 [/itex] [itex]ms^{-2} [/itex] from both calculations.

    Here you use the average speed instead of the final speed.
     
  10. Feb 13, 2010 #9
    So is this
    the correct way to write acceleration?
     
  11. Feb 13, 2010 #10
    Yes. Oops, it's [itex] 3.12 * 10^5 [/itex] [itex]ms^{-2} [/itex]
     
  12. Feb 13, 2010 #11
    Excellent. I've been watching a few more lecture videos so they seem to be helping too. Thank you! Mark this one as solved!
     
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