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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 bindrogen

atom is $1.67 \times 10^{-27} \mathrm{kg}$ .

$2.40 \times 10^{4} \mathrm{m} / \mathrm{s}$

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{'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"}

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