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Minimum possible kinetic energy in an interaction

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Object A of mass 10 kg moving at 10 m/s [E] interacts with (but does not touch) Object B of mass 8 kg moving at 12 m/s [N]. What is the minimum possible kinetic energy of the system after the collision?


    2. Relevant equations
    M1V1o + M2V2o = M1V1f + M2V2f
    KE = (.5)(m)(v2)
    Kinetic energy is minimized in a completely inelastic collision.


    3. The attempt at a solution

    10(10) + 8(12) = (10 + 8)v
    v = 10.89 m/s
    KE = (.5)(10)(10.89)2 + (.5)(8)(10.89)2
    KE = 1067 J
    Correct answer: 534 J (which is one-half of the answer I got.)

    Help please?
    Thanks.
     
  2. jcsd
  3. Nov 17, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Realize that momentum is a vector and must be added as such. (One momentum points east and the other points north. Find their correct vector sum.)
     
  4. Nov 17, 2012 #3
    Oh...Thanks! Previous problems like this had the second object initially at rest, so I used that way without thinking.
     
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