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Merging bullets final velocity

  1. Jan 13, 2005 #1
    two bullets (bullet A and bullet B) are fired at a 45 degree angle to the horizontal. They hit each other and merge. If air resistance is not negligable, the x velocty would be zero since the initial velocites and mass of the two bullets are the same. This cancels each other out. What happens to the y velocity? Do I just add the y velocity of A with y velocity of B to get the final y velocity?
     
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  3. Jan 13, 2005 #2

    dextercioby

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    I didn't understand anything...Please post your problem in the original form,maybe attach a drawing...

    Daniel.
     
  4. Jan 13, 2005 #3
    object A:
    mass=m
    x velocity=v
    y velocity=v

    object b:
    mass=m
    x velocity=-v
    y velocity=v

    The two object collide. object A from the left, object b from the right. Their mass now is 2m. Their final x velocity is 0. What is the y velocity?
     
  5. Jan 13, 2005 #4

    dextercioby

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    Since i'm still not clear with the problen,then all i can do i advise you to apply the law of momentum conservation in a proper way.U'll get your result very easily.

    Daniel.
     
  6. Jan 13, 2005 #5
    yes,that's what I did. When the two bullets collided inelastically in the x direction, one having a momentum of mv and the other -mv the final velocity would be 0. However, the bullets were not traveling in just the x direction, but also the y direction. If I applied the law to the y-direction then... mv+mv=2mv2

    The final velocity in the y directions is 2v.

    However, I do not know if this is correct.
     
  7. Jan 13, 2005 #6

    dextercioby

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    It can't be correct.Write the conservation of momentum on the "y" axis under the form
    [tex] m\frac{v}{\sqrt{2}}+m\frac{v}{\sqrt{2}}=(2m) v' [/tex]

    Find v'."v" is the modulus of the initial velocity for every bullet.

    Daniel.
     
  8. Jan 14, 2005 #7

    HallsofIvy

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    If they both have the same mass and speeds but come from opposite directions, at 45 degrees to the horizontal, then their final motion together will be directly up.
    dextercioby's calculation is the right one for the speed.
     
  9. Jan 15, 2005 #8
    why is it over radical 2? and not just mv?
     
  10. Jan 15, 2005 #9

    HallsofIvy

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    The bullets are traveling at 45 degrees to the horizontal. If the magnitude of the momentum vector is mv, then the x and y components are each [itex]\frac{mv}{\sqrt{2}}[/itex].

    If each bullet has speed v, at 45 degrees to the horizontal, toward each other, and mass m, then one has momentum vector [itex]\frac{mv}{\sqrt{2}}i+ \frac{mv}{\sqrt{2}}j[/itex] and the other has momentum vector [itex]-\frac{mv}{\sqrt{2}}i+ \frac{mv}{\sqrt{2}}j[/itex]. Their total momentum, which is the momentum of the "joined" bullets is [itex]0i+ \frac{2mv}{/sqrt{2}}j= 0i+ \sqrt{2}mvj[/itex[. As you say, the two x- components of momentum cancel. The y component of velocity is [itex]\sqrt{2}v[/itex].
     
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