Speed of second proton relative to first in colliding beam apparatus

[ilovejava]

Homework Statement

In a colliding beam apparatus, two beams of protons are aimed at each other. The first proton moves with a speed of 0.82c to the right; the second moves with a speed of 0.86c to the left. Both speeds are measured relative to the laboratory frame of reference. What is the speed of the second proton relative to that of the first?

Homework Equations

u^'=(u-v)/(1-(uv/c²))

The Attempt at a Solution

u = speed of second object -0.86c
v = speed of first object +0.82c

Plugging into the equation u and v I get -1.68C but this doesn't make sense. So, what I did was just take the absolute value of the speed of the second object (0.86c) and now plugged this new value and got u^' = 0.136c . Did I approach this problem correctly?

Thanks.

 

[phyzguy]

You just did the algebra wrong. Try again to calculate:
$$\frac{-0.86 - 0.82}{1-(-0.86)*(0.82)}$$