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Speeds of a Person

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    A person is standing on a sheet of ice so slippery that friction may be ignored. This individual fires a gun parallel to the ground. When a standard cartridge is used , a 17-g bullet is shot forward with a speed of 280 m/s, and the person recoils with a speed of vc. When a blank cartridge is used , a mass of 0.13g is shot forward with a speed of 53 m/s , and the recoil speed is vb. Find the ratio vb/vc.

    mass of bullet = 0.017 kg
    mass of blank = 1.3e-4 kg

    2. Relevant equations
    KE = 1/2mv^2 ?


    3. The attempt at a solution
    I tried using that equation and solved for KE of the bullet (666.4 J) and the KE of the blank (0.1825 J) but I don't know if that's useful or not. I'm not given the mass of the person, and I don't just want to assume 60 kg. I know I'm supposed to attempt to do the problem before asking for help, but I don't even know where to start.
     
  2. jcsd
  3. Sep 21, 2008 #2
    Here, remember that momentum is always conserved. Because of this, you can say that the total momentum change is 0.

    Also keep in mind that the mass of the man never changes in either situation. Think about this.
     
  4. Sep 21, 2008 #3
    The kinetic energy (KE) would not be helpful here, becase chemical energy is being converted into kinetic energy when the gun is fired. Momentum, however, is helpful.

    Think of the system (man plus pistol plus bullet) splitting in two. The two parts (man plus pistol) and (bullet) will have equal but opposite momenta. Their speeds will therefore be in the inverse ratio of their masses. In practice, the bullet would be so small, that the man and his paraphernalia may be assumed to have the same mass throughout the experiment.
     
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