00:01
In this question, we have a proton beam which is directed towards another to produce a head -on collision.
00:06
We're told in such an accelerator, protons move with a speed relative to the lab reference frame of 0 .9972c.
00:16
We want to calculate the speed of approach of one proton with respect to another when they're about to collide head -on.
00:22
We want to find the kinetic energy of each of the protons in mega -electron volts in the laboratory reference frame and then in the rest frame of the other proton.
00:30
So what we know is the speed of the proton with respect to another is given by u prime plus v over 1 plus u prime v over c squared.
00:40
We know the relativistic kinetic energy is given by gamma minus 1 mc squared, where gamma is the lorentz factor as shown here.
00:48
So what we need to do is calculate the speed of one proton with respect to another.
00:55
So doing this, we can see that ux and we're given that the speed of the speed.
01:03
Of each proton with respect to the laboratory frame is 0 .997 .2c.
01:08
So we can say that this is equal to 0 .9972c plus 0 .9972c, as both of the protons have the same speed, over 1 plus 0 .9972c, and we can square this because it's multiplied by itself here over c squared, which gives us an answer of 0 .9999 -9 -9 -6c, which is the speed of the protons with respect to the laboratory frame.
02:09
The next thing we want to find out is the kinetic energy off each proton beam in the laboratory frame.
02:15
We know the kinetic energy is given by this expression here.
02:19
So we can substitute in our numerical values...