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Conserved momentum (car accident)

  1. Aug 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A blue car with a mass of 2000 kg and a yellow truck with mass 9000 kg enter an intersection. The truck comes from the south at vtruck = 7 m/s, and the car comes from the west. When the two collide they stick together and slide at an angle of q = 43° north of east.

    What is the speed of the pair after the collision?

    2. Relevant equations
    how can i do anything with question without speed of the blue car???
    i know p=63000, and angle plays into it, but cant figure this out
    please help

    3. The attempt at a solution
  2. jcsd
  3. Aug 7, 2007 #2


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    Use conservation of momentum... momentum is conserved along any axis... so momentum is conserved in the north/south direction... and momentum is also conserved in the east/west direction... Write the equations out for conservation of momentum in the north/south direction, and also in the east/west direction... you'll have to use some trigonometry... You'll have two equations with two unknowns...
  4. Aug 7, 2007 #3
    OK, so north south is y and east west is x
    p in y before collision is 63000
    i dont understand p is conserved in both directions???
    the combined mass cant chaneg (=11000kg) and it moves with only one speed and p is the same after collision as it was before.
    So i dont understand this
  5. Aug 7, 2007 #4


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    Yes, the combined mass is 11000kg. And there's an unknown final velocity call it vf... and it's moving at an angle of 43 degrees. So the momentum after the collision has a magnitude of 11000*vf... Draw a picture of this vector... Can you divide up this vector into a horizontal(east/west) and vertical component(north south)? Use cos and sin...
  6. Aug 7, 2007 #5
    yes i understand about getting the components of vectors, but what components are they, speed or momentum. and how can that tell me the speed?
  7. Aug 7, 2007 #6


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    Since you CAN get components of vectors, I would suggest you start doing so. They are velocity AND momentum. And as learningphysics points out momentum is conserved along both axes. What do you get from the north-south axis? Your doubts you will get anything are stopping you from continuing.
  8. Aug 8, 2007 #7


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    :smile: Yes. The first step is writing out the equations mujadeo. If you're having trouble writing out the 2 conservation of momentum equations, then let us know where you're getting stuck.
  9. Aug 8, 2007 #8
    K= 1/2mv^2
    i am sorry i am not getting what the components are?
    should i get x and y components of the total momentum (63000) befor the collision and use that to someone get v??
  10. Aug 8, 2007 #9
    what i mean is, take total momentum (which is same cause its conserved) and then multiply by cos43, which will give v in x dire
    then mult by sin43 giving v in y
    then use trig to find the hyp side (velocity at 43deg)

    edit -- i forget the divide by the mass step
    (i am using p/m = v )
  11. Aug 8, 2007 #10


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    Yes, that's the right idea... I wouldn't worry about the velocity vector... I'd just deal with the momentum vectors...

    So the east-west componenet of momentum would be mf*vf*cos43 and the north-south component would be mf*vf*sin43... where mf is the total mass of the clump after the collision, and vf is the final velocity.

    So the total momentum in the east west direction before the collision = total momentum in the east west direction after the collision... and the same idea with the north-south direction...
  12. Aug 8, 2007 #11
    OK I got this one nailed
    thank you for the help!
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