Relativity and Lorentz Transformations

[mhz]

Homework Statement

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9500c as measured in the laboratory.

What is the magnitude of the velocity of one particle relative to the other?

Homework Equations

$v_x^'\frac{v_x-u}{1-\frac{uv_x}{c^2}}$

The Attempt at a Solution

I've considered the laboratory to be moving in addition to one particle having twice the given speed and the other zero, but I don't know what I'm actually doing. I really would look a nice, concise explanation of what is going on. Thank you.

Additionally, there is a similar problem that I am completely lost with:

Two protons are moving away from each other. In the frame of each proton, the other proton has a speed of 0.615c.

In the rest frame of the Earth the protons are moving in the opposite directions with equal values of speed. What does an observer in the rest frame of the Earth measure for the speed of each proton?

If you are interested in previous responses to a similar question, see here: https://www.physicsforums.com/showthread.php?t=481467

 

[ehild]

A particle travels with velocity v_x along the x-axis in the laboratory frame of reference. In an other frame of reference that travels with velocity u with respect to the laboratory, the velocity of the particle is v_x'. If you are an observer, sitting in that new fame of reference, you would see the particle traveling with v_x'.

$$v_x'=\frac{v_x-u}{1-\frac{u v_x}{c^2}}$$.

Substitute the velocity of one particle for u and the velocity of the other one for v.

ehild