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Mechanical Energy Lost In A One-dimensional Collision

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data


    A 1200-kg car traveling initially with a speed of 25 m/s in an easterly direction crashes into the rear end of a 9000-kg truck moving in the same direction at 20 m/s. The velocity of the car right after the collision is 18 m/s to the east. (a) What is the velocity of the truck right after the collision? (b) How much mechanical energy is lost in the collision? How do you account for this loss in energy?


    2. Relevant equations


    [tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
    [tex]K = \frac{1}{2}mv^2[/tex]


    3. The attempt at a solution


    I was able to solve part (a) as follows ...

    [tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
    [tex]\Rightarrow v_{2f} = \frac{m_1v_{1i} + m_2v_{2i} - m_1v_{1f}}{m_2}[/tex]
    [tex]\Rightarrow v_{2f} = \frac{(1200 kg)(25 m/s) + (9000 kg)(20 m/s) - (1200 kg)(18 m/s)}{9000 kg} = 20.9 m/s east[/tex]


    For part (b), however, I didn't get very far ...

    I know the energy loss is due to heat (thermal energy), but I could only calculate the loss in kinetic energy, not mechanical energy:

    [tex]K_f - K_i = \frac{1}{2}m_1{v_{1f}}^2 + \frac{1}{2}m_2{v_{2f}}^2 - \frac{1}{2}m_1{v_{1i}}^2 - \frac{1}{2}m_2{v_{2i}}^2[/tex]
    [tex]\Rightarrow K_f - K_i = \frac{1}{2}(1200 kg)(18 m/s)^2 + \frac{1}{2}(9000 kg)(20.9 m/s)^2 - \frac{1}{2}(1200 kg)(25 m/s)^2 - \frac{1}{2}(9000 kg)(20 m/s)^2 = -14,955 Joules = -15.0 kJ [/tex]

    The problem is that the answer in the back of the book is 8.74 kJ.
     
  2. jcsd
  3. Apr 25, 2008 #2

    dynamicsolo

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    Homework Helper

    Check your calculation again: working with your numbers, I don't quite get the book's result, but it's close (-9310 J). (Be careful about how much you round-off by early in the calculation. I may not be getting the book's answer because I used 20.93 m/sec for the truck's later speed, rather than 20.9.)

    As for where the energy went, the term "internal energy" is a catch-all for lots of processes. Some of the energy could well go into heating, but some could also go into "physical deformation" of the bodies of the car and truck (crash damage), into frictional dissipation with the road surface because the rolling of tires briefly became sliding, and (a rather tiny bit) into sound (acoustical) energy.
     
    Last edited: Apr 25, 2008
  4. Apr 25, 2008 #3
    I can't see anything wrong with your calculation. Mechanical energy is kinetic energy that isn't thermal energy.
     
  5. Oct 7, 2008 #4
    i get -8680 no rounding
     
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