- #1
beecher
- 15
- 0
Homework Statement
A proton and an antiproton are moving toward each other in a head-on collision. If each has a speed of 0.8c with respect to the collision point, how fast are they moving with respect to each other?
Homework Equations
Ux = [U'x + V] / [1 + (v/c^2) Ux']
The Attempt at a Solution
I believe I have solved it correctly, but I am unsure of one assumption which I made, and want to make sure that it is alright. Instead of looking at it as the collision point being stationary, and the proton moving towards it at 0.8c, I assume that the proton is stationary, and the collision point moves towards it at 0.8c. Then the anti-proton moves towards the collision point at 0.8c as measured in the frame of the collision point.
This gives me Ux = the speed of the antiproton in the proton frame
Ux' = the speed of the anti-proton in the collision point frame (0.8c)
V = Speed of collision point in the proton frame (0.8c)
Thus, Ux = [0.8c + 0.8c] / [1 + (0.8c/c^2) 0.8c] = 1.6c / 1.64 = 0.976c
Does this look correct?
Thanks