1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    and

    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

    and
    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.

    ~Lyuokdea
     
  2. jcsd
  3. Oct 31, 2004 #2
    any ideas?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?