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Momentum in a Many-Particle System

  1. Oct 31, 2007 #1
    A Maxim machine gun fires 450 bullets per minute. Each bullet has a mass of 14 g and a velocity of 630 m/s.
    (a) What is the average force that the impact of these bullets exerts on a target? Assume the bullets penetrate the target and remain embedded in it.
    (b) What is the average rate at which the bullet deliver their energy to the target?

    2. Relevant equations
    We probably need the fact that [tex]\frac{dp}{dt} = F_{ext} [/tex]. (i'm not sure if the system is isolated).

    3. The attempt at a solution
    Well, momentum = mass * velocity
    So, I can find the momentum of one bullet easily:
    [tex] p = mv [/tex]
    [tex] p = (0.014 kg) * (630 \frac{m}{s}) [/tex]
    [tex] p = 8.82 kg * \frac{m}{s} [/tex] <--- momentum of a single bullet.

    Now, I can set up the equation:
    [tex] p_{total} = 450 (0.014) (630) t [/tex] <--- because there's 450 bullets.. and I put the t in there because you don't know how many minutes.

    Now, you want force, and since [tex] \frac{dp}{dt} = F_{ext} [/tex], you can differentiate the above expression:
    [tex] \frac{dp}{dt} = 450 (0.014) (630) [/tex]
    [tex] F_{ext} = 3969 N [/tex]

    Well, the kinetic energy of the bullets is [tex] K = 450 * \frac{1}{2} * m * v^2 * t [/tex]
    With numbers, [tex] K = 450 * \frac{1}{2} * (0.014) * (630)^2 * t [/tex]

    Now to get the rate at which the bullets deliver their energy (rate of change of energy), we differentiate this expression:
    [tex] \frac{dK}{dt} = 450 * \frac{1}{2} * 0.014 * (630)^2 [/tex]
    [tex] \frac{dK}{dt} = 1250235 J / min [/tex]

    Is this correct?

    Thanks in advance.
    Last edited: Oct 31, 2007
  2. jcsd
  3. Oct 31, 2007 #2

    I havent checked the calculations, but reasoning part sounds good to me!

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