00:01
For this problem, we are told that two particles with rest masses m1 and m2 collide and stick together, creating a new particle.
00:08
We're asked to calculate the rest mass of this new particle if the relativistic energies of initial particles are given by epsilon 1 and epsilon 2, and their initial relativistic momenta are in part a, parallel to each other.
00:26
So, one moment here.
00:28
So one thing that we need to keep in mind here is that we are dealing with four momenta here.
00:39
So we would have that the initial, let's see here, for part a, especially if we are starting parallel to each other, let's say p1, going to be equal to epsilon 1.
00:58
So that's, okay, yeah, relativistic energy.
01:01
So that's epsilon 1 over c, epsilon 1 over c for the first 4 momentum.
01:12
And we would have that the relativistic correction.
01:19
Actually, sorry, let me correct myself here.
01:21
That is the relativistic correction.
01:23
So we could call that p1 gamma.
01:29
And p1 would be equal to just mc 0 -0.
01:34
Zero and that should be m1c really and we had to have the same deal for p2 including its relativistic correction so what we what we have to have is that the momentum the fore momentum is conserved so that means that the length must be conserved as well or the length of the four momentum vector must be conserved oh also i'll note that for p2 gamma this would have to be equal to see here that would be epsilon 2 over c zero epsilon 2 over c since they have to be going perpendicular to each other or uh pardon me no initially they're parallel to each other so they'd be they'd have the same exact same direction so what we have to have then is that p we can call it p gamma plus p this here we have that the initial momentum of the system is going to be p1 gamma plus p1 plus p2 gamma plus p2 that equals our initial momentum total so if we take the dot product of that expression with itself.
03:30
We'll get an expression for what the total length of the vector must be.
03:37
So actually i'll just one moment here.
03:42
We also have that p dotted with itself is going to be equal to the final mass squared times c squared.
03:56
That's what we must have on the right hand side.
03:59
And then we'd have, let's see here, p, p1 gamma dot p1 gamma plus p1 gamma dotted with p1 plus p2 gamma actually will be a little bit easier to write this a slightly different way we would have we can express the total initial momentum so we'll just call it p bar 1 prime will be equal to epsilon 1 over c plus m1c...