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Momentum conservation law in shooting gun problem

  1. Jun 11, 2007 #1
    1. The problem statement, all variables and given/known data

    There is a gun which mass is M = 4 kg. The length of the gun's barrel is l = 50 cm. I need to find the average recoil force while shooting the bullet which mass is m = 5g. The speed of bullet when it leaves the gun's barrel is v = 930 m/s.

    3. The attempt at a solution

    Acceleration of the bullet is:

    a = v^2/(2*l);​

    To get out of the barrel it takes time which is:

    t = v/a = 2*l/v;​

    The average momentum of the bullet when it is in barrel is equal:

    p(ave bullet) = mv/2;​

    But the gun gets the same average momentum (momentum conservation law):

    p(ave gun) = mv/2;​

    So this means that the average gun's recoil force is:

    F = p(ave gun)/t = mv/2 : 2l/v = mv^2/(4*l);​

    I am really a little bit confused. I haven't used gun's mass M in the solution. Isn't here any mistakes in the solution?

    Thanks in advance.
  2. jcsd
  3. Jun 11, 2007 #2


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    Homework Helper

    I don't think you need to go much further than this

    a = v^2/(2*l);

    equation of yours.

    If you multiply this by the mass of the bullet you have the average force that the "gun" exerts on the bullet. According to Newton's third law the "bullet" will exert the same average force in the opposite direction on the gun.
  4. Jun 11, 2007 #3
    Thanks for help. But if I just use the acceleration equation and multiply it by the mass of the bullet, I get the answer: mv^2/(2*l). And in my solution there is an answer mv^2/(4*l), so twice less. I think that this happens because the speed of the bullet is not steady. So the recoil force is not steady also - at the beginning of the process it is 0, and when the bullet leaves the barrel, recoil force reaches its maximal value. So the average recoil force is maximum recoil force (mv^2/(2*l)) divided by 2, and I get the answer mv^2/(4*l).

    Are these ideas correct? :)
  5. Jun 11, 2007 #4


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    Homework Helper

    Well, I thought your acceleration equation assumes a constant (average) force acting on the bullet while it progresses down the barrel (it is a constant acceleration equation).

    What happens in reality is that the force will at first be a maximum - when the explosion occurs and the intial amount of gas pushes the bullet out of the cartridge. As the gas expands the force will decrease down the barrel, also due to leakage of gas past the bullet.
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