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Conservation of momentum with initial velocity of 0

  1. Jan 16, 2010 #1
    I'm having a bit of trouble understanding the concepts of momentum conservation. Lets say I'm standing on a frictionless surface and I throw a 3kg brick horizontally with a velocity of 4m/s. In this scenario I can see that I'm going to be repelled backwards at a velocity of 12/my mass in kg but lets say instead of throwing a brick I punch a wall. The normal force acting on my fist will repel me backwards and I will have a new momentum but how is momentum conserved here if the velocity of the wall remains 0? Does all the kinetic energy go into vibrating the molecules of the wall + the tiny amount of heat and sound produced or something?
     
  2. jcsd
  3. Jan 16, 2010 #2

    A.T.

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    It doesn't remain 0. You know that if Chuck Norris makes push-ups, he doesn't push himself up but instead the Earth down? That is not a joke.

    Not all but some kinetic energy. But total momentum as a vector has to be preserved on it's own, regardless of energy.
     
  4. Jan 17, 2010 #3
    Since you are attached to the Earth by your feet, this creates a torque that cancels the one you deliver to the wall.

    However, even if you were not attached by your feet you can consider the following thought experiment. You are floating in outer space, just you and a ball with a rod sticking radially out of it. You push the rod, the normal force repels you backwards. What does the ball do? It picks up a rotation and translation. Now, if you make the ball bigger and heavier, the translation and rotation are less noticable. It is more or less fair to consider the Earth/wall system as an extreme limit of this.

    BANG!
     
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