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Conservation of Momentum Basics

  1. Jan 24, 2007 #1
    I have a fundamental question regarding conservation of momentum (not homework).

    So if we have a system of 2 balls, identical in size and weight, one of which is initially stationary until hit by the other ball. We have the following:

    m1v1 + 0 = 0 + m2v2

    or v2 = v1

    and therefore there is conservation of momentum. Now what happens when say a biker hits the pavement with his helmet... how does momentum conserve? Is this a system of earth vs bikers head? How can I quantify the conservation of momentum? having a hard time understanding...
  2. jcsd
  3. Jan 24, 2007 #2


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    Note that this is only true in a head on collision i.e. one dimensional case, in general both balls will have a non zero velocity after the collision.
    The system would be the biker (not just his head) and the earth. What do you mean your having hard time quantifying conservation of momentum?
  4. Jan 24, 2007 #3
    Hi Hootenanny, thanks for the reply :)

    So basically how do I demonstrate conservation of momentum for a system in which there is collision between a biker and pavement (earth)? Is it safe to consider the earth as stationary given that it has rotational and translational movements? What happens to the kinetic energy of the biker once he hits the pavement... where does that energy go?

  5. Jan 24, 2007 #4
    not really the same case is here, because mass of earth and of helmet is not same therefore velocity would differ.
  6. Jan 24, 2007 #5


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    No problem :smile:
    You can consider that earth stationary before the collision and just consider the velocity of the biker (which for the sake of simplicity we can model as a particle). After the collision however, you must consider the velocity of both the earth and the biker. You can ignore the rotational momentum of both the biker and the earth as this is not going to significantly affect the result. To be honest this calculation is going to produce a very small change in velocity for the earth, so I don't know how accurate this is going to be.
    The kinetic energy is dissapated as heat, sound, deformation of helmet etc.
  7. Jan 24, 2007 #6
    Makes sense :)

    Thanks for the input Hootenanny.

    All the best!
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