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Conservation of momentum of a rifle

  1. Sep 14, 2007 #1
    1. The problem statement, all variables and given/known data
    THe expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has a mass of 0.00720kg and a speed of 601 m/s relative to the muzzle when fired from a rifle that has a mass of 2.50kg. The loosely held rifle recoils at a speed of 1.85m/s relative to the earth. Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

    2. Relevant equations
    If the external forces is zero The Total momentum of the system is constant:
    P = p_{A} + p_{B} .....


    3. The attempt at a solution
    i just used the formula:
    R = rifle; B = bullet
    Px = mRVR + mBVB
    Px = (2.50kg)(-1.85m/s) + (0.00720kg)(601m/s)
    Px = -0.2978kg m/s
     
  2. jcsd
  3. Sep 14, 2007 #2

    learningphysics

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    What does that tell you about the momentum of the gases?
     
  4. Sep 14, 2007 #3
    The gas is the total momentum of the system? and there are no external forces... it is conserved
     
  5. Sep 14, 2007 #4

    learningphysics

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    yes momentum is conserved... momentum before = 0... momentum after must also equal 0. ie: momentum of rifle + momentum of bullet + momentum of gases = 0
     
  6. Sep 15, 2007 #5
    okay... ty so much

    Kindly check if the signs are correct
    G = gas
    So what ill find here is the momentum of the gases:
    0 is the momentum of the rifle before; same as after
    0 = mRVR + mBVB + mGVG
    P(gas) = -mRVR - mBVB
    P(gas) = -(2.50kg)(-1.85m/s) - (0.00720kg)(601m/s)
    P(gas) = 0.2978 kg * m/s
     
  7. Sep 15, 2007 #6

    learningphysics

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    Looks good to me!
     
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