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.
QUESTION: What is the speed of the second particle, as measured in the laboratory?
ANSWER: v= ???? c
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To find the speed of the second particle, as measured in the laboratory, we can use the concept of relative velocity.
Let's denote the speed of the first particle as v1, the speed of the second particle as v2, and the speed of each particle relative to the other as v_rel. We are given that v1 (speed of the first particle) as measured in the laboratory is 0.580c and v_rel (speed of each particle relative to the other) is 0.920c.
To find the speed of the second particle in the laboratory, we can use the relativistic velocity addition formula:
v2_lab = (v2_rel + v1_lab) / (1 + (v2_rel * v1_lab) / c^2)
where c is the speed of light.
Plugging in the given values:
v2_lab = (0.920c + 0.580c) / (1 + (0.920c * 0.580c) / c^2)
Next, we can simplify the equation:
v2_lab = (1.5c) / (1 + (0.5336c^2) / c^2)
v2_lab = (1.5c) / (1 + 0.5336)
v2_lab = (1.5c) / 1.5336
v2_lab ≈ 0.977c
Therefore, the speed of the second particle, as measured in the laboratory, is approximately 0.977c.