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Problem 22 Medium Difficulty

At what speed will the momentum of a proton (mass 1 u) equal that of an alpha particle (mass 4 u) moving at $0.5 c ?$

Answer

0.918c

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Video Transcript

So when we are looking at an alpha particle and the proton, which has a massive one year, and the offer particles amounts of for you and the off court was moving at half see and we had to find the speed of the proton if we know that they're Momentum's air equal. So it's just write down. What were given were given. The mass of the proton is equal to one you, so it's equal to you. The mass of the alpha particle is gonna be equal to for you. The the loss of the proton is what we're finding in the mass or the velocity of the Alfa particle is half See. So what do we know given this information? So we know that MP, the momentum of the proton is gonna be equal to U P Times MP, which is adjust our Newtonian momentum. And then we have toe Adam Lawrence factor and this is going to be equal to the momentum of the awful particle where we have our mass of the alpha particle of times that lost it off a particle just again times Allred's factor so intuitively. Let's think about this room. So we know that the momentum of our alpha particle is going to be traveling is gonna be the mass sorry of our offer. Particle is gonna be four times larger than our mo mentum. So when we have to accommodate for it, what we're going to see is that the velocity and over the Lawrence factor of our proton is gonna have to be equal to is gonna be four times larger than the velocity over the Lawrence factor of our alpha particles. So now that's gonna be what we get when we add the Masters and everything in. So we're gonna have that u p over the squared of one minus u p squared over C squared is gonna be four times larger, then the velocity of our alpha particle, which is 0.5 c. So we can just add that in now. So 0.5 c over the squared of one minus 0.5 squared, So one minus 0.25 and squared of one minus 0.25 is gonna be equal to 0.866 So when we get that out of the way, it's gonna be zero point 5/0 0.0.866 So just add for a single completion Could just add it in now and then when we get from here is that you pee over sward of one minus u p squared overseas squared is gonna be equal to 0.5 divided by zero point 866 See, So it's gonna be called a 0.577 seat. So the next thing that we can do is we can square both sides. So with that in mind we get Let's just bring it over to this side. We're gonna get you P squared over one minus u P squared over C squared is gonna be acquitted. What is 0.577 squared? Well, that's gonna be equal to 1/3. So when we square, both sides were actually going to get sorry. We're gonna get 16 c squared over three, and then, with that in mind, we can go ahead and multiply both sides by this one minus u P squared over C squared and then eventually we're gonna have multiplied about one money's U P Squared overseas square. And then let's delete a few things to make it easier. The next thing that we're gonna do is transition to distribute this. When we distribute, we're gonna get U P squared is gonna be equal to 16 c squared over three minus 16 U P squared overseas squared times three. So then, with that in mind, we can try to get our common factors on one side so that it will be a U. P. Square plus 16 u p squared over three square C squared is gonna be quite a 16 c square over three. From here, we can kind of make this one fraction, so it's gonna be 19 u p squared over three c squared. It's gonna be equal to 16 c squared over three. So what we see is gonna be it's gonna be 19/3 U P. Squared is equal to 16/3 c squared. So then we can set up finally isolate our velocity of the proton. It's gonna be equal to the square root of 16 C scored times 3/3 times 19 so 16 c squared over 19. So the square root of that, which is gonna be equal to approximately 0.918 sir, velocity for the proton is going to be 0.918 c

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Top Physics 102 Electricity and Magnetism Educators
Farnaz M.

Simon Fraser University

Zachary M.

Hope College

Aspen F.

University of Sheffield

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Physics 102 Electricity and Magnetism Bootcamp

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