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Momentum of a bullet

  1. Sep 4, 2016 #1
    1. The problem statement, all variables and given/known data
    a bullet 15 g travels 300 m/s then hit the plastic sheet that is thick 2 cm and pass through it. After the crash, the bullet travels 90 m/s ( continue to lose the velocity ) How much the average force acting on the bullet?

    2. Relevant equations
    Momentum = Mass x Velocity

    3. The attempt at a solution
    ddg.png
    From the diagram, While the bullet is travelling, it has momentum (15x10-3)(300) = 4.5 kg(m/s) and after the crash, it has momentum (15x10-3)(90) = 1.35 kg(m/s)
    4.5 - 1.35 = 3.15 kg(m/s)
    the change in momentum is equal to Impulse
    Impulse = 3.15 kg(m/s)
    F(t) = 3.15 kg(m/s)

    I found the time with S=ut+1/2at2
    2x10-2 = 300(t) + 1/2[(90-300)/t](t)2
    t is about 1x10-4 s

    F(10-4) = 3.15 kg(m/s)
    F = 3.15x104 N
     
  2. jcsd
  3. Sep 4, 2016 #2
    Are you asking a question ?

    The only error I can see is that you have force as kg(m/s) when it should be kg(m/s2)
     
  4. Sep 4, 2016 #3
    You're making the assumption that the acceleration is constant during the process, which is unlikely. Try not to assume anything about what happens inside the material, except that all the reaction forces acting on the bullet can be averaged into a single force that slows it down.

    I'll suggest the best and cleanest way to solve this problem is through energy. Is the energy of the bullet conserved?
     
  5. Sep 4, 2016 #4

    PeroK

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    These problems that ask for the "average" force are ambiguous: does that mean the average over time or the average over distance? If you want to calculate the average over time, then you do have to assume constant acceleration, as otherwise the time is variable.

    Otherwise, you can calculate the average over distance: using energy, as you suggest.
     
  6. Sep 4, 2016 #5

    haruspex

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    That's being kind. Seems to me that, unless average over distance is specified, average over time is implied. Average acceleration is well defined as ##\frac{\Delta v}{\Delta t}##. Average force should equal average acceleration times mass, so ##\frac{\Delta p}{\Delta t}##.
    Questions which imply to the student that it is ok in general to calculate average force as ##\frac{\Delta E}{\Delta s}## should be erased from the textbooks.
     
  7. Sep 4, 2016 #6

    SammyS

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    Furthermore, the sketch implies that the plastic sheet is displaced somewhat as the bullet passes through it. Thus the deceleration occurs over a distance that's greater than the 2cm thickness of the plastic sheet. By the way, that's very thick for what I would refer to as a plastic sheet.
     
    Last edited: Sep 4, 2016
  8. Sep 4, 2016 #7
    My teacher said I can actually use energy equation to solve this but since I am learning about momentum :/
     
  9. Sep 4, 2016 #8

    haruspex

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