A proton is traveling to the right at 2.0 x 10⁷ m/s...

Question

A proton is traveling to the right at 2.0 x 10⁷ m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?

Accepted Answer

Hey, everyone in this problem, an alpha particle moves at a speed of 4.5 times 10 to the exponent six m per second due north. It collides elastically with a neon atom that's initially at rest. OK. That neon atom is five times the mass of the alpha particle. And we are asked to determine the velocity of the particles after the collision. OK. We're told to take the initial direction of the alpha particle as positive. Now, we're given four answer choices A through D and each of them contain a different combination of velocities for the neon and alpha particles. OK? And we're gonna come back to those as we work through this problem. Now, we have a collision, we know that conservation of momentum will apply. OK. So let's think about this momentum and conservation of momentum tells us that the initial momentum P is gonna be equal to the final momentum P F. Now each of these momentum is made up of two components. OK. We have the alpha particle and we have the neon atom. And so what we're gonna have is peanut a the initial momentum of the alpha particle plus P and the initial momentum of the neon atom is equal to P fa the final momentum of the alpha particle plus P F and the final momentum of the neon atom now recall that momentum is equal to mass multiplied by velocity. So for each of these terms, we have the mass multiplied by the corresponding velocity ma V not A was M N we not and is equal to ma V fa plus M N V F N. Now filling in what we know we know that the neon atom is initially at rest. So its initial velocity is zero, the momentum initially for that neon atom goes to zero. OK. So on the left hand side, we're left with just ma multiplied by V. Not a substituting in what we know. Now we have the mass of the alpha vertical ma we don't know what that mass is equal to multiplied by its initial velocity, which we're told is 4.5 times 10 to the exponent six meters per second. This is gonna be equal to the mass ma multiplied by the velocity V fa plus the mass of the neon atom. Now, we don't know the mass of the neon atom, but we do know that it's related to the mass of the alpha particle. OK. It's gonna be five times and so M N is gonna be equal to five ma. The mass is five times the mass of the alpha particle. And this is multiplied by V F N. Now, what you'll see is that we have this mass of the alpha particle in each of these terms. OK. So we can divide by ma on both sides of the equation and that's gonna divide out. And so what this tells us is that the mass of that particle actually doesn't matter. The only thing that mattered was the relationship between the mass of that alpha particle and the mass of the neon atom. OK. All right. So now on the left hand side, we have 4.5 times 10 to the exponent six meters per second. On the right hand side, we have V FA plus five V F N. And we are trying to find V FA and V F N the velocity of the alpha particle and of the neon atom at the end after this collision. OK. So we want to find both of those, let's rewrite one in terms of the. Now V FA does not have a coefficient. OK. The coefficient is just one. So let's isolate for V fa. It's gonna be a little bit simpler. We have the V FA is equal to 4.5 times 10 to the exponent six meters per second minus five V F N. And we're gonna call this equation one. OK. So we've done our conservation of momentum. Now we want to switch gears. OK? We still have two unknowns. We don't know how to solve for them. So what else do we know? And we know that we have an elastic collision. We're told that in the problem, what that means is that we also have a conservation of kinetic energy. So if we look at our conservation of kinetic energy, and in this case, we have just kinetic energy, we don't have to worry about the any potential energy because these particles are not at a particular height where we have gravity acting. We don't have any springs or anything like that. OK. So we're looking just at the kinetic energies when we talk about the total mechanical energy. So we have that knot, it's gonna be equal to K F. The initial kinetic energy is equal to the final kinetic energy. Just like with the momentum, we have two components, the alpha particle and the neon atom. So for each of these, we have to break it down. We have K not A plus K not N is equal to K fa what's K F N recall that the kinetic energy is given by one half M V squared. OK. So we have one half mass A V, not a squared lost one half M N V, not N squared is equal to one half ma V fa squared plus one half M N V F N squared. Now again, the neon atom is initially at rest, its initial velocity is zero. And so the initial kinetic energy of neon is gonna go to zero. So on the left hand side, we're left with just one half ma V, not A squared. Now, we have this one half in each of these terms. So we can actually divide by that as well. And then we don't have to worry about that coalition. Now substituting in what else we know we have MA V not A squared is equal to M N V fa squared. Who was whoops. And that should be ma not M N A plus M N. Now again, M N is five ma. So we have five ma multiplied by V F N squared. Just like with the conservation of momentum, we can now divide by ma both sides of the equation. It's gonna divide out. It does not matter what that mass is. All that matters is the relationship between the two different masses. OK. Substituting in some values now the initial velocity of our alpha particle 4.5 times 10 to the exponent six m per second squared. That's gonna equal V fa squared. OK? Well, look at equation one, OK. We've written V fa in terms of V F N. So let's substitute that in. OK. So we have that V fa is gonna be 4.5 times 10 to the exponent six m per second minus five V F N all squared. And then we still have plus five V F N squared. All right. So let's try to expand and simplify on the left hand. Side, we get 2.25 times 10 to the exponent 13 meter squared per second squared. On the right hand side, and we have a binomial squared expanding that we have 2.25 times 10 to the exponent 13 m squared per second squared. OK. That's multiplying that first term by itself. Then we have our middle term which is gonna be minus 4.5 times 10 to the exponent seven meters per second multiplied by V F N. And then that final term 25 V F N squared. And we can't forget to add this final five V F N squared. Yeah, we've done that down below just because we were running out of room. So let's go ahead and simplify. We're gonna try to move everything to one side so that we have zero on the other side, moving our constant of 2.25 times 10 to the exponent 13 m squared per second squared to the right hand side by subtracting, we get zero is equal to and that constant is actually gonna cancel with the constant we have on the right hand side, they're the exact same. So when we take one and subtract the other, we get zero, what we're left with is 30 V F N squared minus 4.5 times 10 to the exponent seven m per second V F F OK. We have this common term of V F N So we can factor that out. We get V F N multiplied by 30 V F N minus 4.5 times 10 to the exponent seven m per second. Now, we have two factors multiplied together that equals zero. If either one of them is zero, then the entire right hand side will be zero, then the equation will be satisfied. So from this first factor we get that V F N is equal to zero m per second. And that's actually that initial case we had, OK. We had that velocity of the neon atom was zero m per second. OK. After this collision, we expect that it's not going to be zero. So let's use this other term. And from this other term, we get that 30 V F N minus 4.5 times 10 to the exponent seven m per second is equal to zero. We can move 4.5 times 10 to the exponent seven m per second to the right by adding it, then we divide by 30. And we get that V F N is equal to 1.5 times 10 to the exponent six m per second. OK. So now we have the final velocity of the neon app. Now this is positive and that makes sense. OK. We have the neon atom not moving the alpha particle is going to be moving towards it and collide with it. OK. That's in the positive direction. So it makes sense that that neon atom is then going to be moving in the direction that that alpha particle was initially moving in. All right. So we have that final velocity of the neon atom. We need to do the final velocity of the alpha particle as well. And we can use equation one. OK. So let's get back to equation one. OK. Remember that we had V FA is equal to 4.5 times 10 to the exponent six m per second minus five V F N. Now that we know V F N, we can substitute that in 4.5 times 10 to the exponent six m per second minus five times 1.5 multiplied by 10 to the exponent six meters per second. And if we work all of this out, we get that final velocity of the alpha particle is a negative three multiplied by 10 to the exponent six m per second. And again, that makes sense. OK. The neon atom is much greater than the alpha particle in terms of mass. So when it collides with that neon atom that's not moving, this alpha particle is gonna bounce off and then move backwards in the opposite direction. It was initially moving. OK. That makes sense. Now, looking at our answer choices, we found that the final velocity of the neon atom was 1.5 times 10 to the exponent six m per second. What that means is we can eliminate B OK. Option. AC and D all have that 1.5 times 10 to the exponent six m per second. And then we moved to the final velocity of the alpha particle. We found that that was negative three times 10 to the exponent six m per second, which corresponds with answer choice. A OK. Answer C had the same magnitude but the wrong direction, an answer D had both the wrong direction and the wrong magnitude. So the correct answer here is a thanks everyone for watching. I hope this video helped see you in the next one.

A proton is traveling to the right at 2.0 x 10⁷ m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?

Key Concepts

Conservation of Momentum

Elastic Collision

Mass Ratio and Velocity Relationship

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A proton is traveling to the right at 2.0 x 107 m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?

Key Concepts

Conservation of Momentum

Elastic Collision

Mass Ratio and Velocity Relationship

Watch next

A proton is traveling to the right at 2.0 x 10⁷ m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?