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System of particles, linear momentum (easy problem)

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2100 kg truck traveling north at 41 km/h turns east and accelerates to 51 km/h. (a) What is the change in the truck's kinetic energy? What are the (b) magnitude and (c) direction of the change in its momentum?

    2. Relevant equations
    currently using delta P = mvf - mvi, but this is not working. i then tried p = mvcom, but this doesn't work either. please help.

    3. The attempt at a solution
    first converted the speeds to 11.39 and 14.17 m/s.

    a) delta K = 74510 J. this is correct.

    b) Pi = (2100)(11.39) = 23919 kg m/s
    Pf = (2100)(14.17) = 29757 kg m/s

    delta P = Pf - Pi = 5838 kg m/s. this is incorrect.

    c) i have no idea how to do this either.
     
  2. jcsd
  3. Apr 20, 2009 #2

    LowlyPion

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    Your problem with b) is that your momentum is changing in both x and y. Remember v and hence p are vectors.
     
  4. Apr 20, 2009 #3
    wow, thanks so much! my new answer is 38178 kg m/s, which is correct.

    any help with c) (from anyone) would be appreciated, but only if you have time and whatnot. thanks!
     
  5. Apr 20, 2009 #4

    LowlyPion

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    c) is again a vector operation.

    P_final - P_initial

    Or more simply

    P_final + (- P_initial)

    In b) you found |ΔP|, now they want the angle right?
     
  6. Apr 21, 2009 #5
    arctan(vy/vx) = 38.8 degrees um... i would have said northeast. but i guess that's not right (according to the back of the book). i mean, the magnitude of the angle is right. the direction isn't.
     
  7. Apr 21, 2009 #6

    LowlyPion

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    It was originally traveling north. Then it was going east. Hence it's "change" in direction north-south must be south and of course since it moved to the east then it is South and East at 38 degrees.
     
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