Question

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 0.690c, and the speed of each particle relative to the other is 0.940c. What is the speed of the second particle, as measured in the laboratory?

Answer 1

The speed of one particle, as measured in the laboratory, u = 0.640 c
The speed of each particle relative to the other V = 0.930 c

The speed of the second particle, as measured in the laboratory v = ?

We know V = ( u + v ) / [ 1+(uv/c ² ) ]

          0.93 c = (0.64 c + v ) / [ 1+ (0.64cv / c ² )]

                     = (0.64 c + v ) / [ 1+(0.64v / c ) ]

                     = (0.64 c + v) / [ (c+0.64 v) / c ]

                     = c (0.64c + v) / (c + 0.64 v )

            0.93 = (0.64c + v) / (c + 0.64 v )                 

0.93 (c+0.64 v ) = 0.64 c + v

0.93 c + 0.5952v = 0.64 c + v

0.4048 v = 0.29 c

            v = 0.7164 c