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Homework Help: Conservation of momentum problem

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Two boats move parallel each other inertially in opposite directions. They both have a speed of 6m/s. When they are near we move a load from one boat to the other. After this the second boat continues in the same direction with a speed of 4m/s. Find the mass of the second boat if the first boat weighs 5000N and the load weighs 600N. Water resistance should not be taken into consideration

    2. Relevant equations



    3. The attempt at a solution
    I tried a few ways but none of them takes me to the same solution as the book. It should be 300kg but I keep getting other answers.
     
  2. jcsd
  3. Jan 30, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi nejikun! Welcome to PF! :wink:

    Show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
     
  4. Jan 30, 2010 #3
    i thought that it should be :
    [m(1)+m(l)]*V(1) + m(2)*V(1)= m(1)*V + [m(2)+m(l)]*V(2)
    the only problem is that i don't know the speed of the first boat after moving the load.
    the other ways are logically wrong as i now understand. I have no clue for any other way to solve it.
     
  5. Jan 30, 2010 #4

    tiny-tim

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    Hi nejikun! :smile:

    You don't need the final speed of the first boat (though you can find it, if you want :wink:)

    Hint: how much momentum does the load lose? :smile:
     
  6. Jan 30, 2010 #5
    Well you confused me even more with the hint :S. I read again all the theory about momentum but i am still missing something. I dont understand what you mean by the load losing momentum.
     
  7. Jan 30, 2010 #6

    tiny-tim

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    The load was going at 6 m/s one way, now it's going at 4 m/s the other way …

    how much was the change in its momentum? :smile:
     
  8. Jan 30, 2010 #7
    oh so all the problem was because i didn't take into consideration the direction.
     
  9. Feb 2, 2010 #8
    sorry to disturb you again but i still can't solve this. i can't figure out anything
     
  10. Feb 2, 2010 #9

    tiny-tim

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    ok, let's do it in stages …

    i] The load was going at 6 m/s one way, now it's going at 4 m/s the other way …

    how much was the change in its momentum? :smile:
     
  11. Feb 2, 2010 #10
    i think it is 120 kg*m/s
     
  12. Feb 2, 2010 #11

    tiny-tim

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    (that's with g = 10 m/s2 ?)

    It would help if you wrote your reasons as well …

    change in momentum = change in velocity times mass …

    so what is the change in velocity, and what is the mass?

    No, it's not 120, that would be if the load slowed down from 6 km/s to 4km/sec, both in the same direction.

    Try again! :smile:
     
  13. Feb 2, 2010 #12
    yes g=10m/s^2. So the change is velocity is 10m/s and the mass is 60 kg so the change in momentum is 600 kg*m/s
     
  14. Feb 2, 2010 #13

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    Yup! :biggrin:

    ok, now …

    ii] apply the conservation of momentum equation for the load and the second boat combined …

    what do you get? :smile:
     
  15. Feb 2, 2010 #14
    i tried this : m(2)*V(1)+600=[m(2)+m(L)]*V(2) but it is not correct.
     
  16. Feb 2, 2010 #15

    tiny-tim

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    uhh? :confused:

    what numbers?
     
  17. Feb 2, 2010 #16
    it's m(2)*6+ 600 = [m(2)+ 60]*4 and when i solve for m(2) it comes out negative
     
  18. Feb 2, 2010 #17

    tiny-tim

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    ah, you've counted some of the load's momentum twice

    do you see that the 60*4 is already included in the 600? :smile:

    Look at it this way … the momentum gained by the load must equal the momentum lost by the second boat. :wink:
     
  19. Feb 2, 2010 #18
    I finally sovled it :D thank you veeery much tiny-tim for all your time ;)
     
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