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Linear Conservation of Momentum of electron

  1. Nov 4, 2007 #1
    An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision? (hydrogen / electron)

    Momentum is conserved, so

    m1v1i +m2v2i = m1v1f + m2v2f

    I can rearrange this to get

    v2f = [ 2m1 / (m1 + m2) ] v1i

    So this pretty much gives me the ratio I need (final hydrogen in comparison to initial electron). Since m2 = 1837(m1), I get

    v2f = [ 2m1 / m1 + 1837m1 ] v1i

    Canceling out mass, I get

    v2f = [ 2 / 1 + 1837 ] v1i
    v2f = [ 2 / 1838 ] v1i

    So the hydrogen has more than 900 times the velocity as the electron does? How does this go into KE? This is where I fall apart. Any help is appreciated.
  2. jcsd
  3. Nov 4, 2007 #2
    you state that momentum is conserved. What else is conserved in an elastic collision?
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