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Bullets Bouncing off Superman

  1. Feb 20, 2006 #1
    I cant figure this one out. Please help.

    It is well known that bullets fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with bullets of mass m = 1.8 g at a rate of R = 200 bullets/min. The speed of each bullet is v = 460 m/s. Suppose the bullets rebound with no change in speed. What is the average force exerted by the stream of bullets on Superman's chest?


    THANK YOU!
     
  2. jcsd
  3. Feb 20, 2006 #2

    Galileo

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    Use Newton's 2nd and 3rd laws. Force equals the rate of change of momentum.
     
  4. Feb 20, 2006 #3
    Well I have tried using Impulse (change in p = F * time) and I am not getting it right. The change in momentum is 1.656, right? And then you should be able to just plug in...using 60 sec for time, right?
     
  5. Feb 20, 2006 #4

    Galileo

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    Correct, 1.7 kgm/s is the change in momentum of one bullet. Now just calculate how many bullets hit per second for the total change in momentum per second.
     
  6. Feb 20, 2006 #5
    I got it! Thanks!
     
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