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Very Quick momentum equation question

  1. Oct 20, 2008 #1
    A ball of mass 0.440 kg moving east with a speed of 3.70 m/s collides head-on with a 0.220-kg mass at rest. If the collision is perfectly elastic, what is the speed and direction of each ball after the collision?

    momentum conservation:
    m1v1 + m2v2 = m1v1¢ + m2v2¢;
    (0.440 kg)(3.70 m/s) + (0.220 kg)(0) = (0.440 kg)v1¢ + (0.220 kg)v2¢.
    1.628 kg m/s = (0.440 kg)v1¢ + (0.220 kg)v2¢

    elastic, kinetic energy is constant:
    ½ m1 v21 + ½m2 v22 = ½ m1v1’2 + ½ m2v2’2,
    ½ (0.440 kg)(3.70 m/s)2 + 0 = ½ (0.440 kg)v1’2 + ½ (0.220 kg)v2’2
    6.02 kg m2/s2 = (0.440 kg)v1’2 + (0.220 kg)v2’2

    Whats the quickest way to solve with 2 variables? This one gets a little messy because of the velocity squares.

    My teacher was talking about using a derterminant. Could I use one for this question? If not, can you please given an example of a problem in which I can use one?
  2. jcsd
  3. Oct 21, 2008 #2
    The easiest way is to use the "conservation" of the relative velocity instead of the cons of energy. If follows from conservation of energy in the general form, combined with cons of momentum.
    v1-v2 = -(v1'-v2') (all vectors). Relative velocity is reversed but it has the same magnitude.

    If you did not see this before, you'll have to solve the quadratic equation.
    Determinants are used in general for solving linear equations.
    Last edited: Oct 21, 2008
  4. Oct 21, 2008 #3
    Solve this first in the center-of-mass frame of reference, and then work back to the original frame of reference.
  5. Oct 21, 2008 #4

    wow. thank you. that was actually a very simple problem. I made it more complicated than it was.
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