# Relevistic energy problem : part 2

1. Sep 26, 2007

### Benzoate

1. The problem statement, all variables and given/known data

relavtivistic Heavy Ion collider at brookhaven is colliding fully ionized gold(Au) nuclei accelerated to an energy of 200 GeV per nucleon. each Au nucleus contains 197 nucleons . a) What is the speed of each Au nucleus just before collision b)What is the momentum of each at that instant? c) What energy and momentum would be measured for one of the Au nuclei by an observer in the rest system of the other Au nucleus

2. Relevant equations
Possible equations: For Lorentz transformation equations, E=gamma*mc^2 ;p(x)=gamma*m*u(x) , u(x) being the velocity of the Au nucleon in the x dirrection.

3. The attempt at a solution

I don't have trouble with part a, I'm only having trouble with parts b and c. I calculated the velocity to be , u= .999999999717c. and thats the correct velocity according to the back of my textbook. If you want to see how I calculated the velocity, go see my other posts titled relavistic energy problem In part b), since the value of u is extremly closed to the speed of light , I would used p=gamma*m*u. Therefore, p(x)=4152.27*(3.28e-25 kg)*(.99999999717*c)=4.09e-13 kg *m/s
E=gamma*mc^2= 4152.27*(3.28e-25 kg)*(.99999999717*c)^2 = 1.226e-4 joules
For part c) , I will have to used the Lorentz transformations for energy and momentum . I assumed the directions of y and z are zero, I think I'm supposed to find p'(x) and p'(x)= gamma*(p(x-vE/c^2) = 4152.27*(4.09e-13 kg*m/s-v(1.226e-4 joules))/c^2) and E'(x)=gamma*(E-v*p(x))= 4152.27*(1.226-4 Joules-v(4.09e-13 kg*m/s). I'm having trouble finding v. On second thought, wouldn't v=0 since the observer is in the rest system ?

Last edited: Sep 26, 2007
2. Sep 26, 2007

### dynamicsolo

I replied to your other post. These appear to be correct, but I think the velocity is somewhat less (as these things go) than you found.

I think you will want to use the relativistic velocity addition formula to find the velocity of one nucleus as seen in the rest frame of the other. Keep in mind that, since we are setting up a collision, the velocities of the two nuclei will be in *opposite* directions.