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More momentum

  1. Mar 30, 2006 #1
    A car (1500 kg) travels east at 25 m/2. Van (3000 kg) travels south at 16.7 m/s. what is their final velocity after they collide? completely inelastic.

    I can't figure it out at all. none of my formulas seem to fit.
     
  2. jcsd
  3. Mar 30, 2006 #2
    deleted this because I didn't read the post properly :redface:
     
    Last edited: Mar 30, 2006
  4. Mar 30, 2006 #3

    nrqed

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    Completely inelastic means that the objects move with the same velocity after the collision (they get "stuck" together of you will). You have to impose [itex] \sum p_{x,before} = \sum p_{x,after} [/itex] and likewise for the y direction. The two unknowns will be [itex] v_{x,after}[/itex] and[itex] v_{y,after} [/itex].
     
  5. Mar 30, 2006 #4
    damn!!...really sorry about that one...yes they are stuck together (read inelastic as elastic :redface: )
     
  6. Mar 30, 2006 #5
    But don't I need to know theta in order to figure out how much of x and how much of y there are all together?
     
  7. Mar 31, 2006 #6

    Hootenanny

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    You know the inital y-momentum (due to the car travelling south). You also know the x-momentum (due to the car travelling east). As they combine on impact, there is only one resultant path, the x and y components of momentum must be equal to the inital x and y components.

    I'll try and set it up for you...
    (this is as nrqed said, but written in a more explicit way)

    [tex]P_{xi} = m_{1}v_{xi}[/tex]

    [tex]P_{yi} = m_{2}v_{xy}[/tex]

    [tex]P_{xf} = (m_{1} + m_{2})v_{xf} = P_{xi}[/tex]

    [tex]P_{yf} = (m_{1} + m_{2})v_{yf} = P_{yi}[/tex]

    If you solve these for [itex]v_{yf}[/itex] and [itex]v_{xf}[/itex], you have two velocities which are perpendicular. These can be resolved into a single velcoity using trig. The angle can also be found using trig.

    Hope this helps
    -Hoot:smile:
     
    Last edited: Mar 31, 2006
  8. Mar 31, 2006 #7
    Lemme try and redeem myself a little :redface: ...You have two vehicles, and you know their weights and speeds...if you look at one vehicle individually you will notice that by considering east and south as the components of velocity this vehicle really only has one component (either east or south)...from this you can say what both vehicles momentums are in either direction (note that if you are focusing on south, the car can be considered at rest, ie: it has no component of it's velocity in this direction)...after collision this momentum is conserved (for each component) but you now have 2 vehicles that move as one (ie: one object), this one object weighs as much as both objects before collision and will now move both east and south. The final speeds in these directions will allow you to find the resultant velocity and hence find [tex] \theta [/tex]

    edit...sorry hootenanny didn't see your post
     
  9. Mar 31, 2006 #8

    Hootenanny

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    Don't worry about it, I'd have left it for you if I'd have known you were still online. Atleast we're all saying the same thing :biggrin:
     
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