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?
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.
Thanks for the timely help
To find the speed of the second particle in the laboratory, we can use the relativistic velocity addition formula. According to special relativity, the velocity of an object relative to an observer is not simply the sum of the velocities but rather a more complex calculation.
The relativistic velocity addition formula is given by:
v' = (v1 + v2) / (1 + (v1 * v2) / c^2)
Where:
v' is the velocity of the second particle as measured in the laboratory,
v1 is the velocity of the first particle as measured in the laboratory,
v2 is the velocity of the second particle relative to the first particle,
and c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s).
In this case, we are given:
v1 (velocity of the first particle in the laboratory) = 0.700c
v2 (velocity of each particle relative to the other) = 0.920c
Substituting these values into the formula, we have:
v' = (0.700c + 0.920c) / (1 + (0.700c * 0.920c) / (3.00 x 10^8 m/s)^2)
To simplify the calculation, we can convert the speed of light to units of c (the speed of light). Dividing the speed of light by c, we have:
c^2 = (3.00 x 10^8 m/s) / (3.00 x 10^8 m/s) = 1
Now, we can substitute the values into the formula:
v' = (0.700c + 0.920c) / (1 + (0.700c * 0.920c) / 1)
= (1.620c) / (1 + (0.644c^2))
= 1.620c / (1 + 0.644)
= 1.620c / 1.644
Calculating this expression, we find:
v' ≈ 0.984c
Therefore, the speed of the second particle, as measured in the laboratory, is approximately 0.984 times the speed of light.