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Momentum in explosions

  1. Dec 3, 2003 #1
    Need help on this problem - exactly as stated

    Homer the human powder keg, initially at rest, suddenly explodes into three pieces each with equal mass. One piece moves east at 30 m/s and a second piece moves at 30 m/s southeast.

    Caculate the speed of the third piece and calculate the direction in which it travels

    Thanks for the help
     
  2. jcsd
  3. Dec 3, 2003 #2
    I'm going to go with 55.4m/s at 22.5degrees North of East.

    Hopefully someone can verify that answer.
     
  4. Dec 4, 2003 #3
    i agree with the direction but the speed wud be equal would it not
     
  5. Dec 4, 2003 #4
    Why? Momentum is conserved, not 'speed'. Before the explosion momentum was zero. It must be afterwards too.
     
  6. Dec 4, 2003 #5

    selfAdjoint

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    Since the three pieces had equal mass, you can factor the mass out of momentum and just use speed, if you want. The point is that the vector sum of the three vectors has to be 0, and so you have to do some trig.
     
  7. Dec 4, 2003 #6
    yeah, add the x components up = 8.8 m/s East
    add the y components up = 21.2 m/s South

    so 8.8 m/s West and 21.2 m/s North will give you the opposite vector, thereby conserving momentum

    answer = 22.95 m/s @ 22.5 deg West of North
     
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