Lorentz Transformation Question

[phisci]

Homework Statement

2 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 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second particle, as measured in the laboratory.

Homework Equations

Lorentz Velocity Transformation Equation.

V_x= V_x^'+u/(1+(uV_x^'/C²))

The Attempt at a Solution

I let S be the frame of reference of the laboratory and S^, be the frame of reference of the first particle.

From the question, u=+0.650c and the velocity of the 2nd particle in the S' frame is
Vx'= -0.950c since it is approaching the first particle.

Hence when i plug in the values into the formula, I should get

Vx=[(-0.950+0.650)]c/(1+(-0.950x0.650)) = -0.784c

the speed is hence 0.784c.

However the solution given puts a negative sign to the 0.650 in the denominator. i.e.
Vx=[(-0.950+0.650)]c/(1+(0.950x-0.650))

Why did they negate the 0.650 and not the 0.950? Where have i gone wrong in my reasoning? Thanks!

 

[vela]

Probably just a typo. Your solution is fine.

 

[Rasalhague]

You can also use (0.950-0.650)c/1-0.650*0.950) = 0.784c. Since the question doesn't label one direction as positive, and only asks for speed (the magnitude of velocity), it doesn't really matter. I don't know why they chose to deploy the minus signs as they did; maybe it's a trivial difference in convention: which formula they started off with, and how they assigned the variables in this particular problem to those of the general formula etc., or maybe just a typo as vela suggests. I wouldn't lose too much sleep over it ;-)

 

[phisci]

okay noted. thanks a lot! :)