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Momentum/law of conservation of energy - collision of two cars

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data


    A car with a mass of 1875 kg is travelling along a country road when the driver sees a
    deer dart out onto the road. The driver slams on the brakes and manages to stop before hitting the deer. The driver of a second car (mass of 2135 kg) is driving too close and does not see the deer. When the driver realizes that the car ahead is stopping, he hits the brakes but is unable to stop. The cars lock together and skid another 4.58 m. Allof the motion is along a straight line. If the coefficient of friction between the dry concrete and rubber tires is 0.750, what was the speed of the second car when it hit the stopped car?


    2. Relevant equations

    Notation: "A" is the stopped car and "B" is the second car. v' is the resulting velocity after the two cars have collided and locked.
    mA = 1875 kg
    mB = 2135 kg
    uF = 0.750
    d= 4.58m

    The velocity of car B before the collision and the final velocity are not given.
    vB = ?
    v' = ?

    So momentum is conserved, and the resulting equation would be:
    mB * vB = (mA + mB) * v' <--- Eqn 1

    Also, the energy is conserved. The kinetic energy of the second car (B) right before the collision should equal the energy of friction (thermal energy?) used to stop the two locked cars (I think this is right...). So:
    Ek(before) = Ef (after)
    0.5 * mB * vB ^2 = uF * Fn * d <--- Eqn 2


    3. The attempt at a solution

    So basically my method was to use Eqn 2 to isolate "vB" , then sub this rearranged equation into Eqn 1:

    Eqn 2:

    0.5 * mB * vB ^2 = uF * Fn * d
    0.5 * mB * vB ^2 = uF * (mA + mB) * g * d
    vB = SQRT ([2* uF * (mA + mB) * g * d] / mB)
    vB = SQRT ([2* 0.750 * (1875 + 2135) * 9.8 * 4.58] / 2135)
    vB = 11.25 m/s ---> 40.48 km/h


    Eqn 1:

    mB * vB = (mA + mB) * v'
    v' = (mB * vB)/ (mA + mB)
    v' = (2135 * 40.48) / (2135 + 1875)
    v' = 21.5 km/h

    The answer is 55.5km /h ...can someone check over my method for me please?
     
  2. jcsd
  3. Jul 17, 2010 #2
    is energy conserved - the collision is perfectly inelastic isn't it. I would say that your method is wrong.
     
  4. Jul 17, 2010 #3
    Oh right, I can't know for sure if its elastic.

    So... :/ now I have no idea how to solve this !
     
  5. Jul 17, 2010 #4
    you do know for sure that its inelastic though.... if the two stick together energy is not conserved in the system. the only thing to remember is that you cant use energy methods unless the question explicitly states that it was a perfectly elastic collision.

    to solve:
    momentum is always conserved
    the speed of the 1st car is 0 upon collision - mass is 1875Kg
    speed of the second car is unknown - mass is 2135kg
    the combined mass is (1875 +2135)kg and the speed immediately after is unknown but calcuable(excuse the word i just made up)

    to find this out
    F=ma m=(1875 +2135)kg
    F=uN
    F=(1875 +2135)*9.81*0.75
    Find a
    now the final velocity is 0, you know the mass and the deceleration
    use kinematics to solve for the velocity after the collision

    now you have everything needed use pi=pf and solve!!!


    Hope this was helpful????
     
  6. Jul 17, 2010 #5
    Oh that makes sense, thank you!

    One last question though.

    To find acceleration, it's:

    F friction = m * a
    (1875 +2135)*9.81*0.75 = m * a

    You're saying this "m" on the right is also the sum of the masses of both cars right?
    Since they both lock and go forward for a stretch together?
     
  7. Jul 17, 2010 #6
    Yes thats right.
     
  8. Jul 17, 2010 #7
    Thank you!
     
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