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Conservation of Momentum and Energy

  • Thread starter Lyuokdea
  • Start date
147
0
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
 
147
0
any ideas?
 

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