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Linear collision

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A 870 kg sports car collides into the rear end of a 2500 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.9 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.
    Part A
    What was that speed?

    2. Relevant equations
    law of conservation momentum

    3. The attempt at a solution
    so , we have
    mivii + 0 = m1v1f + m2v2f
    and since final velocities are same(inelastic collision)
    m1v1i = vf( m1 +m2 )
    or m1( v1i - vf ) = m2vf - (i)

    kinetic enrgy is not conserved
    0.5m1v1i2 = 0.5vf2( m1 + m2 )
    taking the mivf2 to the left and since 0.5 cancel
    m1(v1i2 - vf2) = m2vf2
    m1 (v1i + vf)(v1i - vf ) = m2vf2 -(ii)

    dividing (i) and (II)
    we get, v1 + vf = vf
    so from here do i conclude that the final velocities are different otherwise answer makes no sense and instead use subscripts for different final velocities . but final velocity should be same ????

    and why is that distance and coefficient of kinetic friction given need help here..:confused:
  2. jcsd
  3. Nov 30, 2008 #2
    (i) is better rewritten as vi= vf(m1+m2)/m1 since the question asks for the initial speed of the sports car

    You stated that KE is not conserved, which is true, which means that the equivalence of KE does not hold and (ii) cannot be used.

    Looking back at (i), you would realise that the only unknown is vf which is the velocity immediately after the collision. The combined vehicles came to a halt, right? We have the distance, and the final velocity of the wreakage = 0. And friction provides deceleration. So, we are looking for initial velocity. Hmm... an equation comes to mind...
  4. Nov 30, 2008 #3
    so like u said , using kinematics equation vf2 = vi2 + 2ad
    where , vf = o (final velocity of the wreckage)
    vi = velocity immediately after the collision (unknown)
    a = [tex]\mu[/tex]mg/m ( m = total mass of two cars)
    d = given

    so here a would have a negative sign in front of it right , cause decelerating? and from this equation formed , the value for vi goes as vf in equation - (i) and so on n so forth u find the initial velocity of the car.

    right ?
    thanks for ur help n time..
  5. Dec 1, 2008 #4
    Yes, you are approaching the question correctly. a must have a negative sign, otherwise, the wreakage is accelerating!
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