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Elastic Collisions

  1. Oct 30, 2005 #1
    Two particles of masses M and 3M are moving towards each other along the +x and -x directions with the same initial speed 3.29 m/s. They undergo a 'head-on' elastic collision and each rebounds along the same line as it approached. What is the final speed of the particle of mass M in m/s?

    so i used mv + mv = mv + mv so
    3.29M -3.29(3M) = M(v) + 3M(v):
    -6.58 = v + 3v
    v = -1.645

    now im confused, i know this isn't the right answer. I don't know what to do next. thanks.
     
  2. jcsd
  3. Oct 31, 2005 #2

    daniel_i_l

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    Gold Member

    The final speeds of M and 3M aren't equal.
     
  4. Oct 31, 2005 #3
    ok so we have m1v1_ + m2v2_i = m1v1_f + m2v2_f

    and

    .5m1v1_i^2 + .5m2v2_i^2 = .5m1v1_f + .5m2v2_f

    now lets manipulate it some and we get

    v1_i + v1_f = v2_f + v2_i

    and

    v1_i - v2_i = -(v1_f - v2_f)

    since we know the mass and initial velocity

    v1_f = [(m1 - m2) / (m1 + m2)) * v1_i + [(2 * m2) / ( m1 + m2)) * v2_i

    and

    v2_f = [(2 * m1) / (m1 + m2)] * v1_i + [( m2 - m1) / (m1 + m2)] * v2_i

    so lets see what you get
     
  5. Oct 31, 2005 #4
    i got v1_f is -6.58 m/s
    and v2_f is 0 m/s

    is the sign on v1_f correct?
     
  6. Oct 31, 2005 #5
    does this sound correct to you??

    i think that v2 should have a final vel since it had an initial vel, but i havent calc'd it out

    and as far as the sign goes, it depends which way it was going initially
     
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