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Collision of two vehicles

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A drunk driver strikes a parked car. During the collision the cars become entangled and skip to a stop together. The drunk driver's car has a total mass of 880 kg, and the parked car has a total mass of 804 kg. If the cars slide 39.5 m before coming to rest, how fast was the drunk driver going? The coefficient of sliding friction between the tires and the road is 0.37.

    2. Relevant equations

    Δ(K + U) = Wnc


    3. The attempt at a solution

    I am having some difficulties setting up the problem correctly. Now, I know that the drunk driver's care will have KE but the parked car will not. I know that the Wnc = Fd which means that it will be the displacement of the vehicles (39.5 m) times the Force (0.37) of the sliding friction on the road. What I am a bit confused on is how does potential energy (U) play into this equation. Any assistance or suggestions will be greatly appreciated.
     
  2. jcsd
  3. Oct 10, 2009 #2
    Because the cars become entangled, some energy is wasted as heat, thus you cannot use that equation.
    A collision like this is called inelastic collision. Use the momentum equation for this.
    http://en.wikipedia.org/wiki/Momentum has a short section on inelastic collisions
     
  4. Oct 10, 2009 #3

    ideasrule

    User Avatar
    Homework Helper

    By the way, the force of friction isn't 0.37; that's the coefficient of friction.
     
  5. Oct 10, 2009 #4
    Well, I was able to finally figure it out.

    It seems as though I must first find the velocity of both the cars after the collision in order to solve for the initial velocity of the drunk driver's vehicle. Thus,

    -uk(m)(g)(d) = mgh - 0.5mv^2

    uk = sliding friction
    m = mass
    g = gravity
    d = displacement

    We are given the sliding friction, gravity, and displacement and mass cancels each other and PE will equal 0. Thus...

    -(0.37)(9.8)(39.5) = 0 - 0.5v^2

    v = 16.9 m/s

    This is the velocity of both of the vehicles after the collision. So, I used this velocity to find the initial velocity of the drunk driver by the following process:

    MaVa + MbVb = MaVa' + MbVb'

    MaVa = mass and velocity of the drunk driver's car
    MbVb = mass and velocity of the parked car

    We are given Ma, Mb, and we just solved for Va and Vb. We also know that since the parked car is indeed parked that MbVb' is equal to 0. Thus:

    (880)(16.9) + (804)(16.9) = (880)Va' + (0)

    I solved for Va' and got 32.3 m/s which is the answer.

    Thank you for all of your help. I do appreciate it.
     
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