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Conservation of momentum

  1. Jul 16, 2014 #1
    1. The problem statement, all variables and given/known data

    A 55-kg girl is standing near and to the left of a 75-kg boy on the frictionless surface of a frozen pond. The girl tosses a 1.8-kg ice ball to the girl with a horizontal speed of 7.5 m/s and he catches it. What are the velocities of the boy and the girl immediately after the boy catches the ball?


    2. Relevant equations

    MgVog+MbVb=0
    Vfg=MballVb-Mb / Mg

    3. The attempt at a solution

    1.8kg x 7.5 m/s - 75kg / 55kg
    = -1.1

    Vfb=1/2 MbVb/ ( -1.1Mg-Mb)
    = -10x10^-2

    I don't think I did this correctly, can someone walk me through this?
     
    Last edited: Jul 16, 2014
  2. jcsd
  3. Jul 16, 2014 #2

    Nathanael

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    Try solving it for only the girl. If a 55 kg girl throws a 1.8kg ball at 7.5 m/s what will be her final speed? (When she starts at rest)
     
  4. Jul 16, 2014 #3

    SammyS

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    Hello Mltn12. Welcome to PF !
     
  5. Jul 16, 2014 #4
    V=Vo+at?
    A=t/v, but I don't have t.
     
  6. Jul 16, 2014 #5
    V=Vo+at?
    A=t/v, but I don't have t.
     
  7. Jul 16, 2014 #6

    Nathanael

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    What about this equation? That is conservation of momentum, right? Shouldn't it also apply to the girl and the ball?
     
  8. Jul 16, 2014 #7

    Vfg=MballVb-Mb / Mg

    1.8kg x 7.5 m/s - 75kg / 55kg
    = -1.1
     
  9. Jul 16, 2014 #8

    Nathanael

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    Where did you get [itex]V_{girl}=M_{ball}V_{ball}-\frac{M_{ball}}{M_{girl}}[/itex]?

    Use the other equation [itex]M_{girl}V_{girl}+M_{ball}V_{ball}=0[/itex]
     
  10. Jul 16, 2014 #9
    (1.8 kg)( 7.5 m/s) / (55kg)
    .25 m/s
     
  11. Jul 16, 2014 #10

    Nathanael

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    Yes correct.

    Can you do something similar for the other person?
     
  12. Jul 16, 2014 #11

    MbVb+MballVball=0
    (1.8kg)(7.5m/s) / 75
    =.18
     
  13. Jul 16, 2014 #12

    Nathanael

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    Yes, that is right.

    The reason is that the person slows down the momentum of the ball (by 1.8 kg times 7.5 m/s) and so the persons momentum must increase by the same amount
     
  14. Jul 16, 2014 #13
    So the final velocity for the girl would be .25m/s while the boys would .18 m/s?
     
  15. Jul 16, 2014 #14

    Nathanael

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    Yep!
     
  16. Jul 16, 2014 #15
    Wow really? It was that simple? I thought there was more to that problem! Thank you so much!!
     
  17. Jul 16, 2014 #16

    Nathanael

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    No problem!

    Conservation of momentum is a very useful tool
     
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