Asked by sandhya
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 0.700 c , and the speed of each particle relative to the otheris 0.920 c . What is the speed of the second particle, as measured in the laboratory?
Answers
Answered by
drwls
At those high speeds, you must use a relativistic formula for adding relative velocities.
At nonrelativistic speeds, the speed of one particle (#2) relative to the other (#1), when moving apart would be the sum of the two speeds in lab coordinates. That is not the case here. Instead,
0.92 c = [0.700c + V2]/[1 + V2*0.7c/c^2]
= [0.7 c + V2]/[1 + 0.7V2/c]
0.92c + 0.644V2 = 0.7c + V2
0.356 V2 = 0.22c
V2 = 0.618c is the speed of #2 in lab coordinates, in the opposite direction from that of #1.
At nonrelativistic speeds, the speed of one particle (#2) relative to the other (#1), when moving apart would be the sum of the two speeds in lab coordinates. That is not the case here. Instead,
0.92 c = [0.700c + V2]/[1 + V2*0.7c/c^2]
= [0.7 c + V2]/[1 + 0.7V2/c]
0.92c + 0.644V2 = 0.7c + V2
0.356 V2 = 0.22c
V2 = 0.618c is the speed of #2 in lab coordinates, in the opposite direction from that of #1.
Answered by
sandhya
Thanks for the timely help
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