Two protons (each with rest mass M = 1.67 ×10^-27 kg) are initially moving with equal speeds in

step 1 This is chapter 37 problem number 38. We have two protons on an undergoing a collision. So let's

Two protons (each with rest mass M = 1.67 ×10^-27 kg) are...

Two protons (each with rest mass $M$ = 1.67 $\times$ 10$^{-27}$ kg) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that also produces an $\eta_0$ particle (see Chapter 44). The rest mass of the $\eta_0$ is m = 9.75 $\times$ 10$^{-28}$ kg. (a) If the two protons and the $\eta_0$ are all at rest after the collision, find the initial speed of the protons, expressed as a fraction of the speed of light. (b) What is the kinetic energy of each proton? Express your answer in MeV. (c) What is the rest energy of the $\eta_0$, expressed in MeV? (d) Discuss the relationship between the answers to parts (b) and (c). 37.39 .

Transcript

00:01 This is chapter 37 problem number 38.

00:03 We have two protons on an undergoing a collision.

00:07 So let's say the velocities are v for both, and the masses are capital m.

00:15 We know the rest mass of protons are given to us.

00:20 So after the collision, actually, so let's say this is v4.

00:25 And after the collision, we have both protons at rest, both protons and a new particle is now generated by this collision, and they're all at rest after the collision.

00:54 And we are, has to calculate the initial speed of protons.

01:00 You're asked to calculate v in part a.

01:04 So actually we can solve this problem easily using the conservation of energy, right? so initially, like total energy that we have in the system initially has to be equal to the total energy of the system after the collision.

01:19 So let's say initial and final to refer to before and after collision.

01:25 Initially, let's look at what we have.

01:27 We have two protons.

01:28 They're moving with a certain speed, which means that we're going to have kinetic energy of the first proton, plus the kinetic energy of the second proton, plus the rest mass, you know, the energy, let's call this e not one, to refer to the rest energy of the first proton and the rest energy of the second proton.

01:56 That's all we have before the collision.

01:58 So this is the total energy of the system initially.

02:02 After the collision, we have three particles now.

02:05 Both of them are protons and the other one is a new particle.

02:09 And they're all at rest.

02:10 So no kinetic energy, turn them on the right hand side.

02:13 All we need to have is e .0.

02:16 One again, the rest energy of the first proton, the rest energy of the second proton, plus the rest energy of this new particle.

02:25 Okay? so as you can see from this equation, we know the rest mass for a proton would be m times c squared, but we don't have to calculate it because we have the same terms on both sides.

02:39 They can cancel out.

02:40 Now we have twice the kinetic energy of the protons equals to the rest energy of a mu particle.

02:51 So let's write what kinetic energies would be.

02:54 So the kinetic energy of the first proton equals kinetic energy of the second proton, since it's given to us that they're moving with the same speed initially.

03:03 So we call this k -e.

03:07 And relatively speaking, this equals to gamma factor minus one times mc squared, right? so this is what we have on the left side, twice of this term here.

03:19 So we can put it there, twice gamma minus 1 mc squared.

03:25 This capital m here is the mass of proton, obviously.

03:31 And on the right -hand side, we have the rest mass for, let's, instead of mu, let's just put new there to refer to the rest mass of this new particle times c square, right? so we're going to solve this equation for gamma vector because it has, the information regarding the speed in it.

03:58 Now, c squares are going to cancel each other out.

04:02 Let me continue this.

04:03 Well, 2 gamma vector minus 1 is going to be equal to m of the new particle, new particle divided by the mass of a proton.

04:16 So if we divide both sides by 2, these are going to cancel out.

04:21 Gemma minus 1 is going to be equal to m new over, to m.

04:27 So let's add one to both sides to isolate gamma on the next page.

04:32 Gamma then is going to be equal to m new over to m of the proton plus one.

04:41 Now we know that the gamma looks like this...

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