Physics

Given: Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.580c, and the speed of each particle relative to the other is 0.920c.

Problem:What is the speed of the second particle, as measured in the laboratory?

I'm not even sure where to start...

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1. In the frame of reference of the slower particle (as seen in the lab) the speed of the other particle is +0.920c. To get the lab speed of the faster particle, you have to do a transformation back to lab coordinates, with a relative velocity of 0.580c

Th relativistic formula for adding relative velocities in this way is

V' = (V1 + V2)/[1 + V1*V2/c^2]
= 1.500 c/[1.5336] = 0.978c

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posted by drwls
2. I tried this and it said it was incorrect. Is there another way?

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posted by Chelsea

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