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I Momentum of a System and External Forces

  1. Oct 13, 2018 #1
    So Pearson is telling me that, basically, the ratio of internal to external forces and the briefness of the time interval is what determines whether the external forces on a system whose momentum we're studying will affect whether we can obtain a decent approximation of the momenta of the objects using the conservation of momentum principle.

    Specifically, in a car collision situation where the drivers are pushing their breaks when the cars hit: "The collision between the cars involves brief forces that are much stronger than the forces of friction exerted on the cars by the road. Thus if we apply conservation of momentum to a very thin “slice” of time surrounding the collision, the total momentum of the two cars will not change very much and will be approximately conserved."

    This makes intuitive sense to me, that the tiny force of rolling friction for say a pool ball collision isn't going to affect momentum conservation much. But I'm trying to find a F(internal)/F(external) expression of some kind to mathematically back up this idea, and I can't find anything.

    Does J = F(net average) * delta t = delta p have any bearing on this?

    I mean, in the sense of the fact that the breaking won't have a huge impact on the cars' velocities, this makes sense, but I'm looking for a mathematical expression to back it up. Is there one?
     
  2. jcsd
  3. Oct 14, 2018 #2

    vanhees71

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    The very fact that cars do not simply bounce off each other without any deformations shows you that car crashes are usually not described by elastic scattering processes. This would be great since then crashes wouldn't damage the cars at all ;-))).

    [Edit: This argument is wrong. Momentum is indeed conserved in the inelastic collision, but of course not energy. @hilbert2 is right!]
     
    Last edited: Oct 14, 2018
  4. Oct 14, 2018 #3

    hilbert2

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    But isn't the elasticity of a collision related to the conservation of energy, not that of momentum?
     
  5. Oct 14, 2018 #4

    vanhees71

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    That's indeed true.
     
  6. Oct 14, 2018 #5

    hilbert2

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    Thanks for correcting.
     
  7. Oct 14, 2018 #6
    The ratio between braking distance and crush zone should be a good approximation.
     
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