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Conservation of Momentum and energy problem help

  1. Jan 25, 2006 #1
    SOmeone explain to me this because im lost. i got 2 masses one velecity. the equation is like is mass times volocity + mass times velocity=0

    122kg*22m/s+14kg*v=0 I made these numbers up what i do is multiply and add then move my anwser to the other side becoming negative but that doesnt seem to be the right anwser.
     
  2. jcsd
  3. Jan 25, 2006 #2
    ki even tried multipling then dividing by the remaining number still gives me the wrong anwser
     
  4. Jan 25, 2006 #3

    Hootenanny

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    Generally;
    [tex]M_{1}V_{1} + M_{2}V_{2} = 0 [/tex]
    Then Rearranging
    [tex]V_{2} = - \frac{M_{1}V_{1}}{M_{2}}[/tex]
     
  5. Jan 25, 2006 #4
    Ok i worked the first problem out now i have a nother problem




    A 125kg pile driver falls from a hieght of 10m to hit a piling.

    From this i see mass is 125kg and distance is 10m and it says find the speed it hits the piling? and the momentum how can i do this with just mass and distance or is distance really velocity?
     
  6. Jan 25, 2006 #5

    Hootenanny

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    Use conservation of energy. At 10m up the piledriver will have a potential energy of mgh. At the instant it hits the pile all the potential energy will be converted to kinetic [itex] \frac{1}{2}mv^2 [/itex]. You should be able to work out the velocity by rearranging and hence the momentum.
     
  7. Jan 25, 2006 #6
    i did 1/2*m moved it to the other side negative and ttok the square root and 7.91m/s
     
  8. Jan 25, 2006 #7
    and the right anwser is 14m/s
     
  9. Jan 25, 2006 #8

    Hootenanny

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    I got a different answer. What value did you get for the potential energy?
     
  10. Jan 25, 2006 #9

    Hootenanny

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    Yes, I got 14m/s.
     
  11. Jan 25, 2006 #10
    12,250Newtons
     
  12. Jan 25, 2006 #11
    how?................
     
  13. Jan 25, 2006 #12

    Hootenanny

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    Check you units. Newtons for energy? The value is correct. Now show me how you re-arrange the kinetic energy equation.
     
  14. Jan 25, 2006 #13
    do u have msn IM??
     
  15. Jan 25, 2006 #14

    Hootenanny

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    No..............
     
  16. Jan 25, 2006 #15
    1/2 times 125 = 62/1/2 and i took the square root gives me 7.9
     
  17. Jan 25, 2006 #16

    Hootenanny

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    [tex]E_{k} = \frac{1}{2} mv^2 \Rightarrow v = \sqrt{\frac{2E_{k}}{m}} [/tex]
     
  18. Jan 25, 2006 #17
    Whats Ek........
     
  19. Jan 25, 2006 #18
    kinetic energy
     
  20. Jan 25, 2006 #19
    potential got it
     
  21. Jan 25, 2006 #20

    Hootenanny

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    The potential energy you calculated it would all be converted into kinetic energy. The same equation could be written
    [tex]v = \sqrt{\frac{2mgh}{m}} \Rightarrow v = \sqrt{2gh}[/tex]
     
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