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Conservation of Momentum and Energy

  1. Oct 31, 2004 #1
    This one is giving me fits for some reason:

    A proton of mass m undergoes a head-on collision with a stationary carbon nucleus of mass 12m. The speed of the proton is 300 m/s. Find the velocity of the proton after the collision.

    Ok, I know that both momentum and energy are conserved. So:

    mTvT = m1*v1 + m2v2


    1/2mTvT^2 = 1/2 m1*v1^2 + 1/2 m2*v2^2

    I tried to solve for v2 in both equations and then set them equal to each other:

    v2 = (mTvT - m1v1) / m2

    v2 = sqrt((1/2mTvT^2 - 1/2 m1v1^2)/m2)

    and then solved for:
    (mTvT - m1v1) / m2= sqrt((1/2mTvT^2 - 1/2 m1v1^2)/(.5*m2))

    but this gives me -678.4 m/s which isn't right, apparently I'm doing something wrong here.

  2. jcsd
  3. Oct 31, 2004 #2
    any ideas?
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