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Symbolic Kenetic Energy

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    One car has one and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 8.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars?

    2. Relevant equations
    ke = (1/2)mv^2

    3. The attempt at a solution
    m1 = 1.5m2
    k1 = k2/2
    (1/2)(1.5m)(v1+8)^2 = (1/2)(m)(v2+9)^2
  2. jcsd
  3. Feb 6, 2010 #2


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    This question makes my brain hurt.

    But I would add to your attempt:

    ke1 = 1/2*m1*v1^2 = 1/4*m2*v2^2 = k2/2
    getting rid of the fractions yields:
    2m1*v1^2 = m2*v2^2
    adding in m1 = 1.5m2
    3m2*v1^2 = m2*v2^2
    the m2's cancel leaving
    3v1^2 = v2^2
    v2 = ?

    replacing v2 in your 3rd equation should yield one easily solved quadratic equation.
    (1/2)(1.5m)(v1+8)^2 = (1/2)(m)(v2+8)^2
  4. Feb 6, 2010 #3


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    I think that is it, except (v2+9) should be (v2+8). Multiply out all the terms and solve via the quadratic equation, see:

  5. Feb 6, 2010 #4
    Before their speeds increase:
    KE of car 1 = 1/2(Ke of car 2)
    .5*1.5m*v1^2 = .5(.5*m*v2^2)
    .75mv1^2 = .25mv2^2
    v2^2 = 3v1^2
    v2 = 1.73v1

    After the increase:
    KE1 = KE2
    .5*1.5m*(v1 + 8)^2 = .5*m*(v2 + 8)^2
    .75m(v1 + 8)^2 = .5m(v2 + 8)^2
    .75m(v1 + 8)^2 = .5m(1.73v1 + 8)^2

    Hope I did that right, I'm kind of tired. There might be an easier way too, but you should be able to go from there.
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