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Energy Conservation and Coefficient of Restitution

  1. Sep 23, 2014 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    A 1300-kg car is backing out of a parking space at 5.0 m/s . The unobservant driver of a 1700-kg pickup truck is coasting through the parking lot at a speed of 3.2 m/s and runs straight into the rear bumper of the car.

    (a). What is the change in internal energy of the two-vehicle system if the velocity of the pickup is 1.5 m/s backward after they collide?
    (b). What is the coefficient of restitution for this collision?

    2. Relevant equations
    [itex] p_{total, \, initial} = p_{total, \, final} = m_{c} v_{ci} + m_{p} v_{pi} = m_{c} v_{cf} + m_{p} v_{pf} [/itex]
    Conservation of momentum, you can use it to solve for the velocity of the car after the collision.
    [itex] e = -\frac{v_{12, \, f}}{v_{12, \, i}} [/itex]
    where v12 is the car's velocities relative to each other. e is the coefficient of restitution.
    [itex] K = \frac{1}{2} mv^{2} [/itex]
    [itex] \Delta K = -\Delta E [/itex]


    3. The attempt at a solution
    To start the problem, I drew a diagram and defined my coordinate system such that the initial velocity of the pickup is in the positive direction. Since the author of this textbook ignores two dimensional motion for 10 chapters, the collision is one-dimensional therefore the truck drives directly into the rear of the car. So, the initial velocity of the car is -5.0 m/s, because if is coming toward the pickup. The pickup is going backward after the collision, therefore having negative velocity. So I set up my momentum equation,
    [itex] (1300 kg)(-5.0 m/s) + (1700 kg)(3.2 m/s) = (1300 kg)v_{c, \, f} + (1700 kg)(-1.5 m/s) [/itex]
    I got the initial total momentum to be -1060 and used algebra (adding 2550 kg m/s to both sides and dividing by 1,300) to arrive at my answer of ~2.78 m/s for the velocity of the car. Using my calculator I confirmed it agreed with the conservation of momentum (same magnitude, just different directions). I started the problem with the easiest question, which was (b). and got 0.52, incorrect. I set up my problem:

    [itex] e = \frac{2.78+1.5 (m/s)}{8.2 (m/s)} [/itex]

    The denominator is flipped because I distributed the negative sign, but got the same answer if I didn't so it doesn't matter much. And I used the real, not rounded, value for the velocity of the car so that is not the source of my error.
     
  2. jcsd
  3. Sep 24, 2014 #2

    haruspex

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    If the posterior velocities are -2.78 and -1.5, what is the posterior relative velocity?
     
  4. Sep 25, 2014 #3

    B3NR4Y

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    ~-1.25 m/s, I would say
     
  5. Sep 25, 2014 #4

    haruspex

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    Well, 1.28, not caring about the sign for now; not 2.78+1.5, as in your original post.
     
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