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Loss of kinetic energy due to heat

  1. Nov 18, 2016 #1
    1. The problem statement, all variables and given/known data
    I have a bloc sitting on a horizontal table, and we shoot a ball through it. The speed right before entering the block is v and the speed when it exits the block is v/2. I need to prove that the fraction of the initial energy that is lost due to heat is 3/4 - γ/4, where γ is mass of ball/mass of block (γ<1)

    2. Relevant equations
    Conservation of momentum

    3. The attempt at a solution
    I know that usually kinetic energy is given by 1/2 mv^2, so then the kinetic energy of the ball once it exits the block will be 1/8 mv^2. I am not sure how to begin with this question.
     
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  3. Nov 18, 2016 #2

    BvU

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    Hi Coco,

    There is always the conservation of momentum to consider. Should have been listed as a relevant equation under item 2 of the template !
     
  4. Nov 18, 2016 #3
    Ok, I added it. I didn't think about this concept playing a role in the question
     
  5. Nov 18, 2016 #4

    BvU

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    Good. Now check your math. the 1/4 is not correct.
     
  6. Nov 18, 2016 #5
    Ok, I fixed that too. What I don't understand is how to find the heatloss from the energy or the conservation of momentum. What is the relationship between them ?
     
  7. Nov 18, 2016 #6

    BvU

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    :smile: if you keep fixing like that, my comments will make me look dumb ... never mind.

    Who mentioned heat loss ? I don't see it in the problem statement !
    Yeah, you added it in words. Now work it out in terms of ##\ v_0\ ## and ##\ \gamma\ ##,
     
  8. Nov 18, 2016 #7
    I don't quite understand what γ is
     
  9. Nov 18, 2016 #8

    BvU

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    ##\gamma = m_{\rm ball} / M_{\rm block}##
     
  10. Nov 18, 2016 #9
    Would I have to separate this into three parts, before the block, inside the block and after the block? Do I start by working with momentum or kinetic energy?
     
  11. Nov 18, 2016 #10

    BvU

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    Questions, questions, questions.... :smile:

    They tell you the ball speed before the block (##v_0##) and after (##{1\over 2}v_0##).
    And they ask for something after the block. So no need to worry about the violent path through the block :rolleyes:
    And you have the corresponding kinetic energies of the ball already

    I have to correct myself:
    It's there bright and clear, sorry o:) .

    Note that you are already very close to where you want to be: the ball goes from ##{1\over 2} mv_0^2## to ##{1\over 8} mv_0^2## so 3/4 is the maximum loss due to conversion to heat. Apparently that's overestimated and needs a correction.
     
  12. Nov 18, 2016 #11
    I don't get what the mass of the block has to do with the problem, how do i incorporate it in the forumlas? I don't think it will be used for the kinetic energy ?
     
  13. Nov 18, 2016 #12

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    Can you express the conservation of momentum in ## \ m_{\rm ball} , \ M_{\rm block},\ v_{\rm ball} , \ V_{\rm block}\ ## before the block is hit and after the ball has come out ?
     
  14. Nov 18, 2016 #13

    haruspex

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    For the KE of what, exactly? [hint]
     
  15. Nov 20, 2016 #14
    Before:
    mballv
    after
    mballv/2 + MblockVblock
     
  16. Nov 20, 2016 #15
    You have the kinetic energy of the ball before it hits the block, and the conservation of momentum before
    after it hits you have the kinetic energy of the ball with the reduced speed and for the conservation of momentum I assume now the block will come into play, but I don't know to tie all of this together.
     
  17. Nov 20, 2016 #16

    haruspex

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    Yes, and of what else?
     
  18. Nov 20, 2016 #17
    Of the block ?
     
  19. Nov 20, 2016 #18

    haruspex

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    Yes
     
  20. Nov 22, 2016 #19
    Will the block be moving at the same speed as the ball? I don't know how to set up the equations
     
  21. Nov 22, 2016 #20

    BvU

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    If the ball comes out the other end, then: no.
    Not true: you do know. :smile: You did it in your post #14. How many unknowns are there ? Remember ##\gamma = m_{\rm ball} / M_{\rm block}##
     
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