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Pushing off on ice

  1. Mar 9, 2009 #1
    Hi
    I would like to put up an old post that was put here in 2005 by vexxa. I noticed there was disagreement among the answers given last time and I would really like to know the correct answer because it is bugging me too!
    Thanks very much.

    Question:
    Okay, this is something that's been bugging me for a long time, and I should have asked my physics teacher this past year way back when we were discussing mechanics, but I never could remember to do it.

    Let's say I have an identical twin, and we're both on a big frozen pond, standing still relative to the pond and the earth. I push him with a force F, and through Newton's 2nd law I'm pushed back with a force of F. He goes one way with velocity V, I go with velocity -V (relative to the earth). But what if he has special boots, or is nailed to the ground, or is stuck in the ice, or through whatever means is fixed to the earth? Would I still have velocity -V relative to the pond, or would it be -2V or something else entirely?

    Back when I first started turning this over in my head, I thought that I exert F on him (and he F on me from Newton's 2nd), but since he's attached to the ground, he exerts F on the ground, which exerts F on him, which gets transferred to me, which means 2F total is exerted on me, and I'd move backward at -2V.

    But I was thinking today about conservation of momentum... in the first scenario, since our masses are equal, our speeds would be equal, but in the opposite direction. But in the second scenario, the only difference is that I'm not just pushing him, I'm pushing him and the earth, which has a bit more mass and wouldn't have any significant change of motion - and I'd still have just the regular old F put back on me, and still move -V relative to the earth.

    Now my problem is that the first solution doesn't seem right with all the transfers of force and whatnot, but the second seems kind of counterintuitive since I'm exerting a force on the frame of reference itself, and it seems that my speed should be different because of the different setup. But since the frame (the earth) won't be moving much due to the force, it can still be used as a frame of reference... Sorry if this all sounds kinda confusing, it's hard to put into words, especially considering I've only had one year of high school physics, but I'm hoping someone here can shed some light on the subject...

    EDIT: It's the 3rd law, not the second... yeah, it's been a while for me.
     
  2. jcsd
  3. Mar 9, 2009 #2
    The complication is that "same force" isn't a sufficient constraint on the question (change in momentum also depends on how long the force is applied for).

    If you push for the same time, then it stays V (what you said about forces being transferred doesn't make sense). If the same amount of work is expended in each instance (or if you push for the same mutual-distance, say two arm-lengths) then the answer is [itex]\sqrt{2}[/itex]V.
     
  4. Mar 9, 2009 #3

    rcgldr

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    Assume you apply the same force over 1 arms length in both cases. To simplify this, imagine a compressed spring is pushing between you and the other person. From the initial point in 3d space, in the first case, an opposing force is applied between both you and the other person, 1/2 arm length each, and the result is V for other person, -V for you. In the second case, from that iniitial point in 3d space, the other person is fixed to the earth, so almost all the motion is in the -V direction since the other person and earth only move a tiny distance, so the force is applied against you for almost 1 arm length, work done is doubled, so - [itex]\sqrt{2}[/itex]V for you and a very tiny fraction of +V/c (c is very large) for the person and earth.

    Linear and angular momentum of the system, you, other person, and earth is conserved sinced it's a closed system with no external forces involved (just internal forces). Total energy = spring potential energy + kinetic energy of you, other person, and earth is also conserved.
     
    Last edited: Mar 9, 2009
  5. Mar 9, 2009 #4

    rcgldr

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    I left out the sqrt, should be - [itex]\sqrt{2}[/itex]V (work on "you" is doubled not quadrupled). It's fixed now.
     
  6. Mar 9, 2009 #5
    Thanks cesiumfrog and Jeff, thats starting to make some sense now.
    Could you please tell me something about what would be happening during a pushing contest of 2 people on land ?
    I assume as 2 people push against each other, both people will experience the same force (but in opposite directions, from Newtons 3rd Law), and the "winner" will be the person who can withstand the developed forces (without moving backwards) the best.
    So is it true to say that the winner is not really the person who could push the strongest but the one who could stay rigid and develop the most friction with the ground?
    Thanks
     
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