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Change in Momentum

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A biker, who is 72.4 kg, rides down a hill at 4.76 m/s [tex][3.83^\circ[/tex] down from south], when he hits a bump which makes him go 2.00 m/s [tex][1.80^\circ[/tex] up from south]. What is the change in momentum?


    2. Relevant equations
    [tex]p=mv[/tex]


    3. The attempt at a solution
    Here's what I did:

    p=(72.4)(2.00+4.76)
    p=489 kg m/s

    I know this is probably wrong, but I didn't know how to accomodate the degrees. Can someone help me?
     
  2. jcsd
  3. Sep 20, 2008 #2

    Doc Al

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    Staff: Mentor

    Momentum and velocity are vectors, so direction counts. I suggest that you find the horizontal and vertical components of the velocity before and after he hits the bump, then use those to find the change in velocity.
     
  4. Sep 20, 2008 #3
    So is this what I have to do?

    [​IMG]

    And then the answer would be 490 kg m/s [[tex]2.17^\circ[/tex] down from south]?
     
    Last edited: Sep 20, 2008
  5. Sep 20, 2008 #4

    Doc Al

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    Staff: Mentor

    Your first two diagrams, where you found the components of the velocities, are perfect. But to find the change in velocity you need to subtract the two vectors, which means subtracting the components. Change = Final velocity - initial velocity. So do that part over and you should be OK. (Pay attention to the signs of the components; let downward = negative.)
     
  6. Sep 20, 2008 #5
    So you're saying that I must subtract the final sum of the components? But how is it possible to subtract x and y components when they are considered separately? I've never done this before in class.
     
  7. Sep 20, 2008 #6

    Doc Al

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    Staff: Mentor

    To find the components of the change in velocity, you subtract the components of the final and initial velocities. You subtract the x and y components separately, not their sum.
     
  8. Sep 20, 2008 #7
    Oh ok, I understand what you're trying to say. Just give me a minute, I'll fix my work, and post it.
     
  9. Sep 20, 2008 #8
    Here's the new work:

    [​IMG]

    So is the answer 201 kg m/s [[tex]7.88^\circ[/tex] up from north]?
     
  10. Sep 20, 2008 #9

    Doc Al

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    Staff: Mentor

    Perfect!
     
  11. Sep 20, 2008 #10
    Thanks Doc Al, you've been a great help.
     
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