Hi, I\'m working on Young and Freedman\'s \"University Physics\" Ch 37Problem 19

Question

Hi, I'm working on Young and Freedman's "University Physics" Ch 37Problem 19E. The problem is this: in a laboratory, two particlesare sent off in opposite directions. The speed of the firstparticle, as measured in the laboratory, is 0.65 times the speed oflight (c). The speed of each particle relative to the other is0.95c. We have to set up two frames of reference, S and S', and weneed to use one of these two formulas:

v' = ( v - u ) / (1 - uv/c^2)
v = (v' + u) / (1 + uv'/c^2)

The first is velocity of an object in the frame S', and the secondis the velocity of that object in S (I think).

Now I think I'm supposed to use the laboratory as frame S, and thefirst particle as S'. That means u = 0.65 (the speed of thereference frame S'). Here's where I'm confused. The problemsolution says that v is 0.95, and that we need to find v'. Thismakes no sense to me. I would think that since the relativevelocity of the particles is 0.95, that would mean v' (not v) is0.95. Actually it seems like it should be -0.95, since theparticles are moving in opposite directions. Then the problem wouldbe to find v. But when I solve for v I get the wrong answer. Theway that cramster does it gets the right answer, but I just don'tunderstand why that way of setting up the problem is correct, whilemy way is incorrect.

Any help would be greatly appreciated!

Explanation / Answer

r = relative speed = 0.95c u = speed of one particle relative to the lab frame is0.65c relative speed = (w-u)/(1-uw/c2) = r where w is the speed of the second particle in the labframe. if v is taken positive in the direction of u. So, 0.95c = (w-0.65c)/(1-0.65w/c) So, w - 0.65c = 0.95c - 0.65*0.95w So, w[1+0.65*0.95] = 1.6c So, w = c(1.6/1.6175) = 0.9892c Here, we've considered the lab-frame as the restframe(S), measuring w and u relative to it. We've considered the frame attached to the secondparticle as the second frame (S' frame) measuring r relative to it. If we consider a universal positive direction, along thedirection of motion of the first particle relative to the labframe(S) the problem is much easier to solve. If you consider w' positive in the direction opposite tou, relative speed would become : r' = (w+u)/(1+uw/c2) And thus in the S' frame, r would be negative (-0.95c)since the first particle is travelling opposite to w. So, -0.95c = (w+0.65c)/(1+0.65w/c) So, w + 0.65c = -0.95c - 0.95*0.65w So, w' = -0.9892c Since we assumed w' opposite to u, the answer remains unchanged, giving w = -w' =0.9892c

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