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Speed and direction of Velocity Vector - Inelastic

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data
    A 55kg man is running in the positive y direction at 4 m/s. He is tackled by a 115kg man running 8.5 m/s at a 140 degree angle. What is the speed and direction after the inelastic collision?


    2. Relevant equations

    p=mv
    p = sqrt of x^2 + y^2

    3. The attempt at a solution

    The final velocity is 7.04 m/s, but the angle keeps coming out at 130 degrees, which I know is wrong.
     
  2. jcsd
  3. Mar 13, 2013 #2

    TSny

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    Hello.
    What do the x and y stand for here?
    Can you please show steps for how you got these answers? Then maybe we can track down where you're having difficulty. It will also help if you state what physics principle(s) you are using.

    When you say that the angle is 140 degrees, that's not really very clear. Is the 140 degrees measured counterclockwise from the +x axis?
     
  4. Mar 13, 2013 #3
    the x is the velocity in the x direction, and same for y

    I multiplied each mass times the velocity to get momentum. Then I set that P = MV1 + MV2 and I solved for the missing V. The angle is 140 degrees from 0. Counterclockwise.
     
  5. Mar 13, 2013 #4

    TSny

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    You still haven't shown much detail. :frown: How did you use the angle of 140 degrees? Did you only set up one equation to solve for V?
     
  6. Mar 13, 2013 #5
    I haven't used the angle yet, I'm not sure how to use it to find the direction after the collision. I set up a momentum equation, and, since it's an elastic collision, I set up a KE equation with it. I had trouble when I ran through the equations: I solved for x in the MV equation, but I got a y value in the answer. Now I'm supposed to plug it into the KE equation for it's x, then solve to find two possible answers.
     
  7. Mar 13, 2013 #6

    TSny

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    The collision is given to be inelastic, not elastic. So, you won't be able to use an energy equation to help you find the final velocity.

    Momentum is a vector quantity. It has both x and y components: px and py. When momentum is conserved, the total x component of momentum will be conserved and the total y component of momentum will be conserved. So, you can set up two equations: one equation for conservation of the x component of momentum and a separate equation for the y component of momentum.

    So, to get started, see if you can find the following:

    1. the x-component of momentum of the 55 kg person before the collision

    2. the x-component of momentum of the 115 kg person before the collision

    3. the y-component of momentum of the 55 kg person before the collision

    4. the y-component of momentum of the 115 kg person before the collision

    You're going to need to use that 140 degrees to get answers for 2 and 4.
     
  8. Mar 13, 2013 #7
    1. 0
    2. -6.5
    3. 4
    4. 5.46

    from what I gathered.
     
  9. Mar 13, 2013 #8
    Actually, I found the components of the velocity by mistake.
     
  10. Mar 13, 2013 #9
    1. 0
    2. -748.8
    3. 220
    4. 628.32
     
  11. Mar 13, 2013 #10

    TSny

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    OK. Good. Suppose you let Vx be the x-component of the final velocity and Vy the y-component of the final velocity.

    What would the equations for conservation of the x and y components of momentum look like?
     
  12. Mar 13, 2013 #11
    I don't fully understand the question.
     
  13. Mar 13, 2013 #12
    I think I solved it on my own. Is the new angle 131.4 degrees from the +x ccw?
     
  14. Mar 13, 2013 #13

    TSny

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    Great!
    Yes, I think so.
     
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