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Explosion of firecracker on pond

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data A firecracker (mtot = 0.70 kg), sitting on a frozen pond, explodes into three pieces, each of which moves horizontally. Piece 1 (m1 = 0.20 kg; v1 = 100 m/s) moves at a right angle to piece 2 (m2 = 0.20 kg; v2 = 125 m/s). What is the speed of the third piece?



    2. Relevant equations mv=p=j=ft



    3. The attempt at a solution m(tot)v1+v2+v3=ft i think t=v(final)/a and a would be equal to gravity but iam just not sure. thanks for help!
     
  2. jcsd
  3. Oct 15, 2012 #2

    tiny-tim

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    hi omc1! :smile:
    what is this supposed to be? :confused:

    what principle do you think applies here?

    (and what topics have you been lectured about this week?)​
     
  4. Oct 15, 2012 #3
    we have only talked bout impulsion and momentum so far i think thurseday we begin collision....
     
  5. Oct 15, 2012 #4
    the formula applies to mas total times the sum of the velocities is the force times time....
     
  6. Oct 15, 2012 #5

    tiny-tim

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    an explosion is a type of collision (in reverse!)

    what happens to momentum in a collision? :smile:

    (and do you need to answer this before thursday?!)
     
  7. Oct 15, 2012 #6
    momentum is conserved in a collision, yes my hw is due wednesday night.
     
  8. Oct 15, 2012 #7

    tiny-tim

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    ok, the initial momentum is zero :wink:

    (because the question says the firecracker is "sitting")

    sooo … ​
     
  9. Oct 15, 2012 #8
    mmmm ok so m(tot)v1+v2+v3=0 iam not sure
     
  10. Oct 15, 2012 #9

    tiny-tim

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    no, that's multiplying the total mass by the sum of the velocities

    you need to add the individual momentums (ie individual mass times individual velocity) :wink:
     
  11. Oct 15, 2012 #10
    soo m1v1+m2v2+m3v3=0 ??
     
  12. Oct 16, 2012 #11

    tiny-tim

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