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Collisions in 2 dimensions

  1. Oct 26, 2007 #1
    how do you if the collision is in 2 dimensions?
     
  2. jcsd
  3. Oct 26, 2007 #2
  4. Oct 26, 2007 #3
    so if there is an angle present the collison is 2d. So collision of ball or pucks are 1d.
    i have a difficulty in this question A 3000 kg space capsule travelling in outer space with a velocity of 200 m/s. In an effort to alter its course, it fires a 25.0 kg projectile perpendicular to its original direction of motion at a speed of 2000 m/s. What is the new speed of the space capsule and by what angle has its direction changed? I just dont understand the question. could you explain it to me?
     
  5. Oct 26, 2007 #4
    ok so this has got to do something with impulse, right? I can find P1 of the capsule and p1 of the projectile...and i think i can find F1 by F=mg? and find out P2 by using P2=P1+Ft
    But i dont have t..
     
  6. Oct 26, 2007 #5

    Doc Al

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    what's conserved?

    Hint: In an "explosion" (which this can be treated as) what quantity is conserved?
     
  7. Oct 26, 2007 #6
    momentum is conserved
     
  8. Oct 26, 2007 #7

    Doc Al

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    Yes. Apply momentum conservation. That's all you need. You have the initial total momentum; and you have the final momentum of one piece. Remember that momentum is a vector, so direction is important.
     
  9. Oct 26, 2007 #8
    So inititial momentum of the capsule+ im of the projectile= fm of capsule+fm of projectile
    im of projectile is 0
    then we find an equation for final v of the casule
    so now we apply this to the x direction right?
     
  10. Oct 26, 2007 #9

    Doc Al

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    OK
    No. Initially the projectile was part of the capsule, so treat them as one piece at that point.
    I suggest that you call the initial direction of the capsule to be in the +x direction. And call the direction of the projectile to be the -y direction. (Draw yourself a picture.)
     
  11. Oct 26, 2007 #10
    ok so that means, the initial momentu of the capsule includes the projectile as well,
    So it will be 3000v= 25*v of projectile+3000*Vf of capsule
    Now we isolate Vf.
    Vf= (3000v-25*V of projectile)/3000
    k so i tried to draw the diagram, but i still dont understand
     
  12. Oct 26, 2007 #11
    k so i got, Vfx=(3000*Vx-25*Vy of projectile)/3000

    here Vx will be 200, Vy will be 2000, right? So now how do i got about finding the angle?
     
  13. Oct 26, 2007 #12

    Doc Al

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    Remember to treat momentum as a vector. First find the x and y components of the initial momentum of capsule + projectile:

    Px = (M + m) v = (3000 + 25)*200
    Py = 0

    (Although its not clear from the problem statement, I'm treating the capsule without the projectile as having mass 3000 kg and the projectile as an additional 25 kg.)

    We know the final momentum of the projectile:
    P_Proj_x = 0
    P_Proj_y = mv = 25*(-2000) = -50,000

    Apply momentum conservation to figure out the x and y components of the capsule's final momentum. That will tell you its speed and direction.
     
  14. Oct 26, 2007 #13
    Ok so,
    mv+Mv= mV1+MV2
    So now we apply this to the x direction,
    (3000+25)200= 0+3000V2
    And we can find V2 (x)
    For the y direction,
    0=-50000+3000V2
    we can find V2(y)
    So now using the phythagorean theorem, we can find V2 and the angle..sounds k..?
     
  15. Oct 26, 2007 #14

    Doc Al

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    Sounds good.
     
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