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Momentum delivered to a piston

  1. Jan 23, 2014 #1
    Okay so please forgive the simple question - I am slowly educating myself but find that sometimes the intuitive view evades me.

    In Feynmans Chapter 38 of Vol 1 - I am reading through kinetic theory of gases about the momentum of a particle hitting a static piston. The section after 39.2. He states that in the case of a particle with mass and velocity hitting the piston with a an elastic collison the particle is reflected. OKay. And goes on to say the the total momentum delivered to the piston is the momentum 'in' and the momentum 'out'. So I understand that the total momentum in the system is 2mv but I'm having difficulty understanding why the momentum 'delivered' to the piston is not just mv. (Also I am unsure if the momentum 'out' refers to the particle reflected and its momentum or the momentum of the piston after the collison - I assumed the reflected particle because I think the piston is locked in place).
    I know I'm just missing something stupid as all sources do the same work up - but I still intuitively can't see why the Force acting on the piston is not just the mv. Please help me reach enlightenment!
     
  2. jcsd
  3. Jan 23, 2014 #2

    A.T.

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    Because the change of the particles momentum is -2mv. The piston receives the same change of momentum as the particle, just in opposite direction.
     
  4. Jan 23, 2014 #3
    Whilst I 'understand' that I don't have it clear in my mind. The particle has an initial momentum going in before it hits and a final momentum coming out after it hits - and since it is the same mass and the same speed the momentum going 'in' is the same as the momentum going 'out'. So momentum doesnt change its value.

    Or the only way I can think of it is

    if the particle has mass 2 and velocity 2 then the momentum in is 4
    it is reflected and the vector changes direction to -2
    now the mass 2 and velocity of -2 gives a momentum of -4
    and now the difference between 4 and -4 is 8 or 2 times momentum???
     
  5. Jan 23, 2014 #4

    Nugatory

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    Momentum is a vector not a number, so it has both a magnitude and a direction just as does velocity. So you're right except for the part that I marked in bold, which is just plain wrong.

    It's a worthwhile exercise to rerun the numbers as they would look to you if you were traveling at the same speed in the same direction as the particle before the reflection. Then the momentum before hitting the wall would be zero and the momentum after hitting the wall would be -8. The difference between 0 and -8 is 8, but it's not "twice" anything else.
     
  6. Jan 23, 2014 #5

    A.T.

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    So the particles momentum was changed by -8. The sum of momentum changes in a collision is 0 (momentum conservation). Which means that the particle must have transferred a momentum of +8 to the piston.
     
  7. Jan 23, 2014 #6
    Thanks guys - its starting to sink in now.
     
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