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A glass marble is dropped down an elevator shaft

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data

    A glass marble is dropped down an elevator shaft and hits a thick glass plate on top of an elevator that is descending at a speed of 2.10 m/s. The marble hits the glass plate 7.5 m below the point from which it was dropped. If the collision is elastic, how high will the marble rise, relative to the point from which it was dropped?

    2. Relevant equations
    v 2 - u 2 = 2gh
    v 2 = u 2 + 2gh

    H = v 2 / 2g
    H-h= required height
    3. The attempt at a solution
    v^2=(2.10 m/s)^2+ 2(9.81)(7.5m)
    v= 12.31
    12.31^2 / 2(9.81)
    7.72-7.5= .22
    not right answer though!!
     
  2. jcsd
  3. Apr 16, 2015 #2

    BvU

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    Hello Salk, welcome to PF :smile: !

    There are a few things you must explain to me a bit further before I can help sensibly.

    What is it the relevant equations are saying ?
    Not ##v_2 = u_2 + 2 gh## because, at least if ##v_2## and ##u_2## are speeds, with the dimension of m/s, that doesn't fit dimensionally.
    Perhaps you mean ##v^2 = u^2 + 2 gh## ? But what are ##v## and ##u## ?

    From the next one I almost gather ##u^2 = 0## but I find it hard to believe that's what you mean.

    So a bit more verbose, please, and we can get going.
    Nice exercise, this is.
     
  4. Apr 16, 2015 #3

    BvU

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    Oh, and do you get a green bar above the tekst you are editing ? With an x2 for subscripts and an x2 for exponents ? Very nifty !
     
  5. Apr 16, 2015 #4

    haruspex

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    That would be right if the marble were descending at 2.1 m/s initially, and fell a further 7.5m before hitting the glass. But the marble starts from rest - it's the 'ground' that's moving.
     
  6. Apr 16, 2015 #5
    BvU sorry for the confusion on equation I meant to say v[2] = u[2] + 2(a)x with x=7.5 u= speed of elevator and v= velocity of ball before impact

    haruspex okay I thought about finding the speed of the marble before impact by using conservation of energy. I am having trouble with getting the velocity of the marble with respect to the "ground" so i can calculate height using v[2] equation

    BvU i tried using the green bar for exponents but they arent working so those are v squared
     
  7. Apr 16, 2015 #6

    haruspex

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    That's good - what answer do you get?
    From energy you can calculate the speed relative to the lift shaft. If the elevator is descending at 2.1 m/s, what's the marble's speed relative to that?
     
  8. Apr 16, 2015 #7
    I got 12.13 m/s relative to the shaft. Then, using relative motion, the marble should be 14.23 m/s relative to ground.
     
  9. Apr 16, 2015 #8

    haruspex

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    Are the marble and the glass roof travelling in the same direction or opposite directions before bounce?
     
  10. Apr 16, 2015 #9
    they are travelling in the same direction
     
  11. Apr 16, 2015 #10

    haruspex

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    If the marble is falling at 12.13m/s and the glass plate at 2.1 m/s, what's the relative velocity?
     
  12. Apr 16, 2015 #11
    14.23 m/s
     
  13. Apr 16, 2015 #12

    haruspex

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    No.
    Consider this, if they were both descending at 2.1m/s, what would the relative velocity be? Hint: it wouldn't be 4.2m/s.
     
  14. Apr 16, 2015 #13
    it would be 0. I would use the equation Vac= Vab+Vbc with a= marble, b=elevator and c= ground. If they are both falling to the ground, then Vac=Vbc=2.1 m/s, leaving the relative velocity to 0.
     
  15. Apr 17, 2015 #14

    haruspex

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    Right, so do the same with the marble falling at 12.13 m/s instead of 2.1m/s.
     
  16. Apr 17, 2015 #15
    The relative velocity would be 10.03 m/s
     
  17. Apr 17, 2015 #16

    SammyS

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    Yes.

    That's much better.
     
  18. Apr 17, 2015 #17

    haruspex

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    Right, so what would the relative velocity be immediately after the bounce?
     
  19. Apr 17, 2015 #18
    10.03 m/s because the collision is elastic and kinetic energy is conserved
    based on that would height be 5.127m ?
     
  20. Apr 17, 2015 #19

    SammyS

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    Height is relative to what ?
     
  21. Apr 17, 2015 #20

    vela

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    If you're talking about the height asked for in the problem statement, the answer is no.
     
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