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Continuous Inelastic Collisions

  1. Apr 8, 2012 #1
    Hi,
    I am trying find equations for continuous "stretchy" collisions, in other words, I have two perfectly round objects of known mass, radius, and velocity, and want to collide them and be able to have them squish together and then bounce apart. I am aware of the method of solving for the intersections of momentum and KE equations, but I need collisions which take actual time, that method is instantaneous. So what I want is a force function of displacement which I can integrate and evaluate at a time t. So, something like this:
    [itex]\int_{0}^{t}{f\left( x \right)\; dx}[/itex]
    For f(x), I have been using Hooke's law of linear deformation--it works beautifully for elastic collisions. So [itex]f\left( x \right)=\frac{1}{\frac{1}{k1}+\frac{1}{k2}}[/itex], where k1 and k2 are the spring constants of the objects. Now, I want to expand this function for the general case: anywhere from inelastic to elastic depending on the objects. So what would f(x) be in that case? I'm sure that there are messy solutions involving pre-computing the time-length of the collision and then solving for percent loss based on that and percent elasticity, but what is the neatest, most generalizable integrand for this purpose? Hopefully one that just involves that instant's variables and no attempts to predict the future?
    Thanks!
     
  2. jcsd
  3. Apr 10, 2012 #2
    It seems to me that one could argue that the concept of a coefficient of restitution doesn't work here because if the percent loss is constant for a material, then, because the duration of the collision is variable depending on masses and velocities, the instantaneous energy loss would also have to be variable, which makes no sense.

    So perhaps there is a constant instantaneous energy loss? Because there are infinite instants in the finite duration of the collision, perhaps this instantaneous energy loss would have to be infinitesimal? How would it fit into calculations?
     
  4. Apr 10, 2012 #3

    K^2

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    If you are using harmonic potential, use harmonic damping term for energy loss.

    [tex]f(x)=-kx-c\dot{x}[/tex]
     
  5. Apr 11, 2012 #4
    Perfect! Thanks so much!
     
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