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Kinetic Energy/Momentum from change in direction and speed

  1. Oct 22, 2006 #1
    From a calculus based physics course:

    A 2100 kg truck that travels north at 41 km/hr is turning east and accelerating to 60 km/hr
    -What are the magnitudes and direction of change in the trucks momentum?
    - What is the change in the trucks kinetic energy?

    I need some help on steps - because I'm lost
  2. jcsd
  3. Oct 22, 2006 #2
    momentum p = mv. Use this to work out the magnitudes of the vectors, then resolve the vector system.
    K.E. = 1/2mv^2
  4. Oct 22, 2006 #3
    would the change in KE be:
    KE = .5mv^2
    KE1 = .5(2100)(41km/hr * 0.277777778 m/s)^2
    KE2 = .5(2100)(60km/hr * .277777778 m/s)^2
    KE1-KE2 = 136192.132-291666.671
  5. Oct 22, 2006 #4
    and the momentum would be
    p =mv
    p1 = 2100*41km/hr
    p2 = 2100*60km/hr
    p1-p2 =-39900 kg*km/hr
    Are those the right units? Or does it need to be in m/s?
    Would the direction change be east? thats where I'm lost, or would it be the negative direction?
    Last edited: Oct 22, 2006
  6. Oct 22, 2006 #5
    so am I right, or really wrong?
  7. Oct 22, 2006 #6
    KE is right since it's just a scalar. Momentum is a vector so you need to resolve it.
  8. Oct 22, 2006 #7
    but how exactly would you do that unless you multiply the mass to the velocity?

    I tried the average of the 2 velocities, but thats not right
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