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Conservation of Momentum - Finding Initial Velocity

  1. Nov 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1500-kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200-kg SUV traveling from east to west. The two cars become embellished due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.39 m west and 6.43 m south of the impact point. How fast was each car traveling just before the collision?

    2. Relevant equations
    m1v1+m2v2=MV -->Conservation of momentum
    Kinematics

    3. The attempt at a solution
    So I started by using the equation
    μmg=ma
    and the equation
    v^2=v^2+2ad
    to find the velocity of the two cars once they have crashed. It came to 11.108m/s

    From here I used the conservation of momentum:
    1500(Va)+2200(Vb)=(1500+2200)(11.108)
    1500Va+2200Vb=41099.6
    so then Va=27.4-1.47Vb

    Because I have two variables I need a second equation so I was trying to use Kinetic Energy.
    ½ma(Va)^2+½mb(Vb)^2=½MV^2 +Work Done by Friction

    To find force friction I took the normal force of the two cars stuck together
    3700⋅9.8=36260
    Friction=μNormal
    Friction=(0.75)(36260)=27195N
    Force⋅distance --> 27195(8.39)=228166.1J

    from here we have
    750Va^2+1100Vb^2=210172.2+228166.1
    Input Va from the previous expression
    750(274.-1.47Vb)^2+1100Vb^2=429238.25
    cleaning that up a bit
    2720.75Vb^2-60417Vb+133831.75=0

    I plugged this into a quadratic equation, but I isn't giving me the correct answer... Any ideas where I went wrong?
     
  2. jcsd
  3. Nov 16, 2015 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    Careful! Remember that momentum is a vector quantity--direction matters.
     
  4. Nov 17, 2015 #3
    oooh thank you. I will try that!
     
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