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Machine gun conservation of momentum problem

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A machine gun held by a soldier fires bullets at the rate of three per second. Each bullet has a mass of 30 g and a speed of 1200 m/s. Find the average force exerted on the soldier.


    2. Relevant equations
    FΔt = mgf-mgi


    3. The attempt at a solution
    I attempted to do this by plugging it into the above equation 1200(.03)/ 3 = 12 to find the average force. The actual answer is 108. I know that the 3 should actually be multiplied to find the force but I am not sure why. Wouldn't that be finding the average momentum per second?
     
  2. jcsd
  3. Jun 6, 2014 #2

    SteamKing

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    Isn't the impulse equation:

    FΔt = mΔv ?

    What Δt are you using?

    Remember, the cyclic rate of the gun is 3 rounds per second.
     
  4. Jun 6, 2014 #3
    Yes it is the title was a typo. I am using 3 as the change in time.
     
  5. Jun 6, 2014 #4

    Nathanael

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    Why did you divide by 3?
     
  6. Jun 6, 2014 #5
    because I thought that was the Δt.
     
  7. Jun 6, 2014 #6

    Nathanael

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    You could make Δt 3 if you want to, but then would the "1200*0.3" be correct? (To me that implies Δt=1/3 since 0.3 is the mass of one bullet and it takes 1/3 of a second for 1 bullet to be shot.)

    Edit:
    What is the mass of all the bullets shot in Δt=3 seconds?
     
    Last edited: Jun 6, 2014
  8. Jun 6, 2014 #7
    OK I see the time would be 1/3 because it takes 1/3 of a second to fire a bullet therefore you have to divide by 1/3. Thanks!
     
  9. Jun 6, 2014 #8

    Nathanael

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    Yes, correct.

    But you could use any Δt you want. If Δt=3 then instead of 1200*0.03 you'll have 1200*0.27 (because 0.27 is the mass of all of the bullets shot in 3 seconds).

    Then divide by 3 and you'll get the same answer.
     
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