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(III) The potential energy of the two atoms in a diatomic (two-atom) molecule can be written$$U(r)=-\frac{a}{r^{6}}+\frac{b}{r^{12}}$$where $r$ is the distance between the two atoms and $a$ and $b$ are positive constants. $(a)$ At what values of $r$ is $U(r)$ a minimum? A maximum? $(b)$ At what values of $r$ is $U(r)=0 ?(c)$ Plot $U(r)$ as a function of $r$ from $r=0$ to $r$ at a value large enough for all the features in $(a)$ and $(b)$ to show. (d) Describe the motion of one atom with respect to the second atom when $E<0,$ and when $E>0 .(e)$ Let $F$be the force one atom exerts on the other. For what values of $r$ is $F>0, F<0, F=0 ?$ (f) Determine $F$ as a function of $r .$

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$$\begin{array}{l}\text {a) } U_{min} \text { at } r=(\frac{2 b}{a})^{1 / 6}, U_{max} \text { at } r=0 \\\text {b) } U(r) = 0 \text { at } (\frac{b}{a})^{1 / 6} \\\text {c) GRAPH} \\\text {d) Bounded when } E <0 \text { , unbounded when } E>0 \\\text {e) } F > 0 \text { for } r < (\frac{2 b}{a})^{1 / 6} \\F = 0 \text { for } r = (\frac{2 b}{a})^{1 / 6} \\F < 0 \text { for } r > (\frac{2 b}{a})^{1 / 6} \\\text {f) } F(r) = \frac{12b}{r^13} - \frac{6a}{r^7}\end{array}$$

Physics 101 Mechanics

Chapter 8

Conservation of Energy

Work

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

University of Washington

Hope College

University of Winnipeg

Lectures

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

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in this problem were given the potential energy function as a distance between two atoms in a di atomic molecule, and we have a few different questions about the potential energy minimum. Maxima zeros the force on the molecule on the atoms in the molecule and the motion of the atoms. Eso some concepts to use in this problem, we need to keep in mind. When do minima or Maxima uh, extreme a occur? And that will be when the first derivative of a function is equal to zero. When we're thinking about the force, the relationship between force and potential energy we can use, the relationship force is equal to negative, the derivative of potential energy with respect to position. When you think about total energy, want to think about genetic plus potential equals U Plus K and want to keep in mind dysfunction and how it relates to the physical context. So what values makes sense for what's actually happening in reality? And so our function is you of our so potential energy as a function of our the distance between these two atoms in this molecule, and it's equal to minus a over our to the sixth must be over our to the 12th and eso well, first will do some basic looking at extra MMA zeros and minimum and maximum of the function. Then we can look at the graph Thio, verify, and to see that that makes sense. And let's see, Part A is asking us what values of our do we have a minimum or Maxima for So for this we just want to again. So to find minimum and maximum values, we just need to look for the points where the derivative equals to zero. So where does do you d are equal to zero and so first derivative of our function, you of our is gonna going to give us a six a over our to the seventh minus 12 B over our to the 13th. So we want that to equal zero. And if we solve that, we will find that occurs at r equals the sixth route of to be over a eso we could So this could be a positive or a negative on DWI could have extreme at either point. But we cannot have in our lesson zero that would mean are our atoms have less than negative distance between them, which doesn't make physical sense. Um, so we'll be looking So we have an extra MMA at this point. So at this point, let's usually at this point, it's good to look at a graph of our function. So if we insert that and besides that a little bit mhm. Mhm. Uh huh. Okay, So here's our graph, and so we can see the We have a minimum here, and this is for, um for B s. So this is a graph of you of our for b and A both equal to one. So being are are just physical constants describing the particular molecule. So we see we've got that at the sixth root of two, we're gonna have a minimum. And are you of our is just going to be increasing, Aziz, We unbounded as we get closer to zero. Right, So we're gonna have a minimum right here at the at the point are equal thesixties of to be over a and the value of you would be a maximum at r equals zero. Although we will. We won't reach that in this graph. So are you Minimum? It's gonna occur at the sixth route of to be over A and our maximum would occur at our equal to zero Mhm. And then part B asks us, um, what where are the zeros for? There's an r for ah, where does U of r equals zero? And, uh, just looking at the graph we can see that that occurs at are equal to one for this A and B eso Let's solve that. And so that would just, uh, that's pretty simple. We'll just say, um, a over art of the sixth equals B over art of the 12th, right. And from there we get our solution for this R is the six through of Be over a right and so and we can verify that with our graph when be an A r equal to one that's six throat will be one. Right, So that will be are part B okay, and part C. It asks us to, uh, plot the graph. And so, yeah, let's just take a look A to this graph so we'll see that it has an ah has a horizontal ascend tote at you equal to zero. So as our separation, uh, increases our potential energy become approaches zero. And so it has that local minimum. Well, on also global minimum at are equal to that six through. Okay, Now, Part D is asking us to describe the motion of one atom in terms of energy. So we need to we want to keep in mind here that energy equals two potential plus kinetic. So our we're looking at potential so weakened draw a total energy graph and then draw some conclusions about what kinetic energy will be. So if our potential energy or our total energy that is less than zero, we can look at it like this. So let's say this blue line here I'll say e one eyes less than zero. So our kinetic energy can never go. We can't have negative kinetic energy. So are we can't go above this line. So if the this is our total energy is equal to the one, we have to move back and forth somewhere in this valley between these two points. So this Adam is going to oscillate back and forth somewhere in between here, right? So it will be abounded. A Silla Torrey motion. And then, if we have our e greater than zero, let's say this was our potential energy e to now our kinetic energy eyes going to be We're always gonna have some as we go to the right on this graph here, we can go. We can't go to the left of this point because then we would have negative kinetic energy. But anywhere to the right of here will have positive kinetic energy. So our Adam is going to have continually have some positive kinetic energy. So it's going to continue to have a positive velocity and it's going to continue to move right s So we can say that for for this if r e is less than zero and that would be a Z example are pointy one we were, So we'll have our bounded, oscillate or emotion. And if our E is greater than zero, as in our point E to will have unbounded motion. Okay, Part E asks us if for what are for what values of our are we going to get in f greater than zero f equal to zero and f less than zero. Okay, so we want to keep in mind that the relationship between force and potential energy so say force is negative, the derivative of potential energy. So in this case, F equals two minus. Do you d r? And so, looking at this graph, um, we can just look at the slope so we see anywhere to the left of this minimum. Our slope of you is negative. So that means f will be greater than zero. Right, So half greater than zero means that are slope of you is less than zero for F equal to zero. We want that slope of you to be equal to zero. That's going to occur at that minimum point. And for an af lesson zero, we will want the slope of you to be greater than zero. Right? And so that's going to occur. Ah, anywhere to the right of our minimum point. Okay. And we found that minimum point in part A. It was Thesixties route of to be over a eso We're gonna have a positive force for our are ah for our less than that sixth fruit of to be over a We'll have a zero force for our exactly equal to that six through of to be over a and we'll have a negative force for our greater than that sixth route point. Okay, so that would be our part E. And then part F is just asking us to find a function for f of our right. And so this one, we simply say the an expression for F is minus the derivative of you. And so we've got Are you waas minus a over art of the sixth plus B over our to the 12th. So just the first derivative of that function, and we'll get a 12 b over our to the 13th, minus six a over our to the seventh, and that will be our answer for part F.

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