🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Oh no! Our educators are currently working hard solving this question. In the meantime, our AI Tutor recommends this similar expert step-by-step video covering the same topics. Numerade Educator ### Problem 108 Easy Difficulty # When two hydrogen atoms of mass$m$combine to form a diatomic hydrogen molecule$\left(H_{2}\right),$the potential energy of the system after they combine is$-\Delta,$where$\Delta$is a positive quantity called the binding energy of the molecule. (a) Show that in a collision that involves only two hydrogen atoms, it is impossible to form an$\mathrm{H}_{2}$molecule because momentum and energy cannot simultaneously be conserved. (Hint: If you can show this to be true in one frame of reference, then it is true in all frames of reference. Can you see why?) (b) An$\mathrm{H}_{2}$molecule can be formed in a collision that involves three hydrogen atoms. Suppose that before such a collision, each of the three atoms has speed$1.00 \times 10^{3} \mathrm{m} / \mathrm{s}$, and they are approaching at$120^{\circ}$angles so that at any instant, the atoms lie at the comers of an equilateral triangle. Find the speeds of the$\mathrm{H}_{2}$molecule and of the single hydrogen atom that remains after the collision. The binding energy of$\mathrm{H}_{2}$is$\Delta=7.23 \times 10^{-19} \mathrm{J},$and the mass of the bindrogenatom is$1.67 \times 10^{-27} \mathrm{kg}$. ### Answer ##$2.40 \times 10^{4} \mathrm{m} / \mathrm{s}\$

#### Topics

Moment, Impulse, and Collisions

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##### Christina K.

Rutgers, The State University of New Jersey

##### Jared E.

University of Winnipeg

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### Video Transcript

{'transcript': "okay. In the first part part, they were using momentum conservation. So initially you have Maybe a on B is a rest. That's a call to em. Avery Prime Final spirit of a plus M B v b prime final speed of be, um a self M B B prime is equal to m a times via a minus. V a prime. Now we know that for elastic collisions via and minus V, B is equal to negative V a prime minus P b v b prime. This implies that v a prime ah must be And knowing that the V b here zero hey, primed, uh, must be you'd be prime minus V s. Right. So therefore, we have m b you be prime is equal to m a times V A minus B b prime plus via so we have view be prime time's M A plus M b physical to to m a V a. All right, So view prime as a result is uh uh the A times to Emma, uh, over M A plus m b. Uh, we multiply both sides by divide both sides by M A. We get V eight times, uh, to over a, uh, over whoops one that should be one plus m a over m b. And so let's go all the way up here. Uh, since m a is less than and be the ratio, I mean, over m b is, uh, always less than one. Uh, so the denominator is less than two. Therefore, viewed be prime will be less than two v A over to. In other words, be prime will be less than ah, our baby prime will be greater than I should say, uh, via because it's greater than two v a over to. Since the over to part, the denominator is less than two. Okay, um so for part B, we, uh, is a similar self when c is so when Hey, is, uh, incident on CD we have the same thing is before we have VC Prime is two times, uh, m a V a over a plus M c. Um, in the case that the C c m c is incidental, be we have again the same idea of the sea we have BB prime is equal to two is equal to two m c ah v c prime, um, to, um CVC prime. That's correct. Over m A over and B I should sy plus m c uh and so this is just too m c over and B plus I'm c times we see prima's and you're so that's times to m a V a over m a plus m c. All right. Therefore be be prime simplifies to four times m eight times m c times v a over m a class M c times m b plus m. C. Okay, Andi in. So that's part bute in part. See, we have in part c we want to find the maximum of obstinate across that I but in part See, we, uh, the maximum value of M. C is where is is where d V pride are is where the derivative of the V b prime With respect, I'm see another one's devi Prime DMC is equal to zero And so the derivative eyes actually Ah, whole complicated expression. It's four v I'm just gonna write it down for via at many times, M A plus m c uh Times m b plus m c minus m c u Times M A plus and a plus and B plus two Mmm soup, Uh, over over May plus M c quantity squared times and B plus M c quality squared. We said that he pulled a zero, and after some painstaking algebra will find Got, uh, m A and B minus m c squared will be zero. Therefore, m c will be square root of m a times I'd be So that's part C. And finally, for part, do. Here is the plot. We have, uh, the Prada Vehbi prime of plotted against M. C so that we can see from the plot. The maximum, uh, V B prime is at 4.5 meters per second. Uh, this occurs at M c. Recall six kilograms on, And so, uh, and so we can reconvene to be contested out. Uh, numerically, um, So I'm seeing from the values given here, um, is route M eight times and be so that's 18 kilograms times two kilograms, 36 giving us six kilograms. So that checks out. And then b v b Prime Max will just be we'll just we'll plug in M c m a v a in there. And that will also give us, um, 4.5 meters per second just to check the new miracle results against a graphical. And that's it"}

New Mexico State University

#### Topics

Moment, Impulse, and Collisions

##### Christina K.

Rutgers, The State University of New Jersey

##### Jared E.

University of Winnipeg

Lectures

Join Bootcamp