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Bullets Per Minute

  1. Mar 16, 2008 #1
    This was a two part question, the first part I was able to calculate.

    Question Part 1:
    A machine gun fires a stream of bullets into a block that is free to move on a horizontal frictionless tabletop. Each bullet has mass 66 grams; their speed is 930 m/sec, and the block a mass of 7.36 kg. After 15 bullets, the speed of the block is?

    Calculation Part 1:
    [tex]\frac{(0.066 kg * 15 bullets)}{(0.066 kg * 15 bullets) + 7.36 kg}=110.263 m/sec[/tex]

    Question Part 2:
    If a man in the previous statement can exert an average force of 180 N against the gun, determine the maximum number of bullets he can fire per minute.

    Attempt to Calculate:
    I am unsure of how I would go about this. I attempted to think of it as a reverse of the previous question, but what didn't work out was getting bullets per minute without figuring out a basic equation then using sample bullets/minute numbers to see where the breaking point of 0 m/sec is. Also, I kept in mind the need to change the velocity of m/sec to m/min if needed.

    The equation I attempted to use was:

    Any help with where to start?
  2. jcsd
  3. Mar 16, 2008 #2
    it is basically just a simple problem of conservation of momentum.
  4. Mar 16, 2008 #3
    You have the "force equation": F*t=m*v
    the total momentum during a minute is m*v*k, where k is the number of bullets per minute.
  5. Mar 16, 2008 #4

    So I get...

    [tex]F \times t=m \times v[/tex]

    [tex]F \times t=m \times v \times k[/tex]

    [tex]\frac{F \times t}{m \times v}=k; k=bullets/min[/tex]

    [tex]\frac{180 N \times 60 sec}{0.066 kg \times 930 m/sec}=175.953 bullets/sec[/tex]

    Thank you both.
  6. Mar 17, 2008 #5
    Never knew Bullets went to the target in fractions.!!!!!
  7. Mar 17, 2008 #6
    ever heard of splinters/shotguns/shells
  8. Mar 17, 2008 #7

    Andrew Mason

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    Science Advisor
    Homework Helper

    Part 1 is correct, although your expression is not correct. You need the muzzle velocity in there:

    [tex]m_{bullet}v_{muzzle} N_{bullets} = \Delta P = (M_{block} + m_{bullet}N_{bullets}) v_{block/bullets}[/tex]

    The analysis for Part 2 is not clear.Since the force applied by the gun to the shooter is the time rate of change of momentum of the gun/bullet system F = dp/dt:

    [tex]F = \frac{dm}{dt}v_{muzzle}[/tex]

    [tex]\frac{dm}{dt} = F/v = 180/930 = .194 \text{kg/sec}[/tex]

    Therefore, the number of bullets per second is .194/.066 = 2.94 or 176 bullets/minute, but this is only if you want to provide an average force of 180 N. The actual peak force that must be applied will be greater than this, so actually firing this number of bullets will cause the force on the shooter to exceed 180N.

    Last edited: Mar 17, 2008
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