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Exchange of momentum

  1. Dec 27, 2013 #1
    Imagine a block of wood resting on two supports at each end. You point a gun upward under the block and shoot the block upwards.

    One scenario is you shoot the block upwards in the exact center. Another scenario is you shoot upwards near one of the support ends. The block is thick enough that in both cases the bullet comes to rest inside the block.

    Question: in which case does the centre of mass of the block rise the highest?

    I could quantify the first case (both for velocity and distance rise of cog) using conservation of momentum and conservation of energy as follows:
    mu = (m+M)V
    1/2(mu^2) = (m+M)gh

    Can somebody help me in the second case?
     
  2. jcsd
  3. Dec 27, 2013 #2

    tiny-tim

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    hi pukb! :smile:

    hint: in the second case, why does 1/2 mu2 = (m+M)gh not work? :wink:
     
  4. Dec 27, 2013 #3
    yes, it does apply - what i meant was the heights to which cog is lifted are not same when force is applied at center and at corner because the latter case will cause some rotation as well. The problem is to quantify that
     
  5. Dec 27, 2013 #4

    tiny-tim

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    hi pukb! :smile:

    (btw, it's better to say centre of mass, com, than centre of gravity, cog :wink:)
    yes! … so the kinetic energy will be the "translational" kinetic energy 1/2 mv2 plus the rotational kinetic energy 1/2 Iω2 :wink:
     
  6. Dec 27, 2013 #5
    There's a video of this very problem on YouTube. The results are not exactly what you'd expect. The video does compare two wooden blocks, one shot thru the CG and the other shot off-center. Both went the same height into the air. Of course, this raises the question of what happened to the conservation of energy. The block that's spinning should not go as high into the air due to energy being transferred as angular momentum. But it does. What gives?

    The video proposed an answer that both blocks must go the same height in order to conserve linear momentum. The difference in energy is due to the bullet not traveling as far into the wood (doing work in the process) and that difference in work is getting translated into angular energy.

    Yes, it's not a comforting explanation. But it does seem to be correct.
     
  7. Dec 28, 2013 #6

    tiny-tim

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    Hi Ryoko! :smile:

    Yes, I didn't think this through fully :redface:: the energy equation which I said didn't work …
    should be 1/2 mu2 + 1/2 Iω2 = (m+M)gh + 1/2 Iω2,

    which of course is the same equation, with the same u (since u is fixed by conservation of momentum, irrespective of ω and of the slight difference in I).

    So h (the height) will be the same in both cases.

    Thanks, Ryoko. :smile:
     
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