01:00 - 23:0023:00 - 00:00

1:40 AM
oh hey winterbash started
2

4 hours later…
5:27 AM
Hats! :-)
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4 hours later…
9:35 AM
morning

9:58 AM
2

Today in our chemistry class we derived the pressure-volume work done on an ideal gas. Our assumption was that $$P_{ext}=P_{int}+dP$$ so that all the time the system remains (approximately) in equilibrium with the surrounding and the process occurs very slowly (its a reversible process). Now $$W_... @JohnRennie I understand why getting a measurement from experiment that is different from what is expected from a scientific law does not necessarily disprove the law. I'm curious how the laws are tested experimentally to higher degrees of precision in general. And since laws are commonly expressed as mathematical equation with various variables, how do we repeat the experiments to cover as much possible values of the variables as possible? Is it impossible to test the law under all possible values for the variables (there would be infinitely many tests)? It is impossible, yes. There are infinitely many different physical models we can come up with to fit experimental data To narrow it down we use a variety of heuristics Which are, if not truthier, at least more practical to use @yh05 in principle getting a measurement from experiment that is different from what is expected from a scientific law does disprove the law. In practice we always worry that there might be some experimental error in our measurement. 10:14 AM Well that's part of the whole bit of scientific law that nobody ever teaches in class The ~correspondence rules~ Also the whole business of predictability and epistemology and all that nonsense Unless we're lucky and we get a measurement that is blatantly different usually something gets disproved gradually by multiple measurements that disagree. In practice we don't have enough informations about the initial conditions of our experiment, nor the means to compute it We don't know the position and momentum of every particle in the universe So we only get a degree of confidence We just have a few informations on what we hope is relevent But in real experiments, 1) we only have limited control and knowledge of the experimental setup (ie we can't prevent the temperature from varying somewhat) and 2) we don't necessarily know everything relevent to the measurements A good example is that whole experimental error of the LHC For a while at the LHC there was a weird measurement discrepency happening every day around the same time Turned out that it was caused by a train passing nearby Re your second question, there are usually key features of a theory and we test those first. If our experiments agree with the key features then experimenters may or may not test the more subtle predictions or they may lose interest. There's also the whole LIGO business being disturbed by trucks a few miles away~ Hm I can't find that train for the LHC business again what was it Ah there it is This is probably the best thing to show if someone ever wonders why it's a good idea to wait a while before interpreting experimental results 10:34 AM Where can I read about how scientific laws are tested? Depends how you want to read about it Scientific laws or laws of science are statements, based on repeated experiments or observations, that describe or predict a range of natural phenomena. The term law has diverse usage in many cases (approximate, accurate, broad, or narrow) across all fields of natural science (physics, chemistry, biology, geology, astronomy, etc.). Laws are developed from data and can be further developed through mathematics; in all cases they are directly or indirectly based on empirical evidence. It is generally understood that they implicitly reflect, though they do not explicitly assert, causal relationships... I read it. I'm looking for a detailed description of how scientific law is tested. which law? 10:50 AM I don't really know any standard texts on the topic but look into epistemology and whatever else It's a mix of rather abstract philosophy and practical concerns sort of deal rly I can't find any relevant one with google search. =( The ideal case is that you have an experimental setup. You input all the initial conditions that you are using for that experiment, your theory gives you what you should get in return. You perform the experiment, and the results from the experiment should match up with what you predicted. this is rarely the case due to a variety of reasons For instance : 1) Measurement errors : the way you measure both the initial conditions and the final ones is unlikely to be perfect, due to 1a) Systematic errors : Errors that always happen during the measurement (for a start, no measuring apparatus has infinite precision, the influence of the measuring apparatus on the value itself, etc) 1b) Random errors : Due to our lack of perfect knowledge, we can only get fairly broad measurements (ie : temperature) and this is likely to change in a random manner during the experiment @yh05 have you seen this? A somewhat common experimental error that I have heard about is The scientist himself I have heard of a few experiments in which the source of error was the scientist approaching the setup to read the results because the scientist's body emitted heat, reflected light/sound/neutrons, etc those are hard things to predict 11:18 AM It is not possible to test a law under all possible values for the variables present in the law. How do we tackle this problem? Can I have a detailed explanation, please? 11:46 AM @yh05 there isn't a single detailed explanation. It depends on what the law is. If you can come up with a specific example then we could discuss how it has been tested, but there isn't going to be an answer that applies to all laws. 1 hour later… 12:57 PM omg let's make an AI where it will decode what a person on a keyboard is typing by just recording the audio where it will have learned with first listening and looking at what the person is typing (supervised, labeled data) #hacking #bless that's been done already 1:40 PM @Slereah Then I would point out that a negative second derivative is not sufficient to lower the badness. You can decrease the rate of rising of the badness a president did come up with a new proof of the Pythagorean theorem 1:57 PM also "U.S. President Richard Nixon, when campaigning for a second term in office announced that the rate of increase of inflation was decreasing, which has been noted as "the first time a sitting president used the third derivative to advance his case for reelection."" @Slereah I assume he didn't explicitly talk about "derivatives" and instead talked about them as rates of change. @Slereah Right, but I would much rather the first derivative actually become negative that was an explicit demo of a jerk using the jerk to advance his case for reelection... 63 I was doing some physics problems for homework and, while procrastinating, I came up with a theoretical scenario that I couldn't figure out the result of. The following is from a side view and in a frictionless, ideal environment: A ball moves toward a wall at a constant velocity. At the... come on, one more vote @AaronStevens Saw it on HNQ the other day and made sure to cast mine. 2:04 PM @JMac It is a great physics question. It's just the way it has been asked makes it off topic. @AaronStevens Besides not showing any analysis, the diagrams uses comic sans. It deserves a custom close reason for that. @skillpatrol Garfield's proof is ok, but it's not a major breakthrough. It's essentially an older proof cut in half. ;) But I will admit I thought it was cute when I first encountered it, several decades ago. Paul Painlevé (French: [pɔl pɛ̃ləve]; 5 December 1863 – 29 October 1933) was a French mathematician and statesman. He served twice as Prime Minister of the Third Republic: 12 September – 13 November 1917 and 17 April – 22 November 1925. His entry into politics came in 1906 after a professorship at the Sorbonne that began in 1892. His first term as prime minister lasted only nine weeks but dealt with weighty issues, such as the Russian Revolution, the American entry into the war, the failure of the Nivelle Offensive, quelling the French Army Mutinies and relations with the British. In the 1920s... did you catch my Nixon comment? 11 mins ago, by Slereah "U.S. President Richard Nixon, when campaigning for a second term in office announced that the rate of increase of inflation was decreasing, which has been noted as "the first time a sitting president used the third derivative to advance his case for reelection."" 7 mins ago, by skillpatrol that was an explicit demo of a jerk using the jerk to advance his case for reelection... @JMac On an unrelated note, two users were removed, and someone removed the accepted status on one of my answers, so I have lost 45 reputation points fairly quickly. I think I will make a meta post complaining that someone is out to get me @skillpatrol Yes. I was trying to come up with a good pun involving the word checkers 2:09 PM @skillpatrol Too bad Nixon was not talking about the third derivative of position @AaronStevens That is definitely a government conspiracy. They are out to get you where it hurts - imaginary internet points. @JMac I shouldn't expect fairness on a site that punishes critical thinking @AaronStevens I think we need a banner on top of the site that reminds users that they will be punished for critical thinking. It would really clear up a lot of confusion. 2:12 PM @AaronStevens Closed. @PM2Ring Finally... haha @JMac I have been raising custom flags on posts that use too much critical thinking. For some reason they keep getting declined though RIP @AaronStevens It's because the moderators and experienced users work together to make sure new users like yourself have a bad time. @JMac No, they are threatened by my critical thinking and insight et al 2:34 PM @skillpatrol Facebook and youtube are the two key weapons in fighting the heliocentric enthusiasts who poison our children's minds with things like "outer space" and "science". indeed 3:18 PM @JMac imho, youtube has more positive applications for the heliocentrics than facebook @skillpatrol Youtube has a lot of anti-flat Earth stuff. I'm sure FB has some; but the format doesn't work the same. yeah, and twitter is just plain "fake news" @PM2Ring, hi sir. I gone through the same experience again but this time a different one. @yuvrajsingh Haven't there been issues of pinging users before? 3:35 PM let's not be overly "critical" here :P let's \mathbb{R}^n has an irreducible representation of O(n) Hi may I ask a question about physics here? But how many reducible representations does it have Is it also one 3:39 PM @Koray yes, of course The representation with n trivial groups Thank you :) I hope this question makes sense. According to Einstein's equivalence principle, which one has an acceleration: an object free falling to a non-atmosphere planet, or an object staying still on the surface of the planet? The first one travels along a geodesic while the second does not @Koray you have to define what you mean by acceleration, and you have to do it in an observer independent way i.e. a way that all observers can agree on. The second one is the only one that has an invariant definition of acceleration (it's the acceleration produced by the electrostatic interaction with the earth) 3:41 PM I'm talking about the same acceleration mentioned in the twin-paradox @Koray typically we use the proper acceleration as this is a well defined quantity that all observers can agree on. @Koray the twin paradox isn't really about acceleration. I'm asking this to understand, if the travelling twin turn using gravitational pull, would there be a difference. @JohnRennie It kinda is If you're interested in understanding the twin paradox I wrote an answer on the main site specifically to explain exactly what causes the twin paradox (it isn't really a paradox of course). Sir I've seen your answer and I is great. But I could not able to understand it. I'll try harder. Thank you very much.. It is great I mean :) 3:44 PM @Koray it doesn't matter whether the travelling twin turns round by using a rocket or by slingshoting round a massive object. Sir thank you very much again, I really appreciate it. Have a good day :) Hello :) @JohnRennie @user8718165 hi :-) @AaronStevens didn't, t get you in what context u are referring. @yuvrajsingh You had trouble with dacoits again? 3:59 PM @PM2Ring not completely. Now I am OK, I thought to share with you. 4:19 PM 0 Winter Bash is on! I am very excited about hats though I don't own one yet :-( I was just hanging around clicking on random cards on the Winter Bash page when I came across this after I clicked on the "Where in the World" card. Shouldn't it be hats instead of has? I mean is that really a ty... 4:44 PM In software engineering and computer science, abstraction is: the process of removing physical, spatial, or temporal details or attributes in the study of objects or systems in order to focus attention on details of higher importance, it is also very similar in nature to the process of generalization; the creation of abstract concept-objects which are created by mirroring common features or attributes from various non-abstract objects or systems of study — the result of the process of abstraction.Abstraction, in general, is a fundamental concept to computer science and software development. The... For a universe to be simulatable, it is actually demanding something very strict: It is saying that every thing can be reduced into instructions that can be combined in a well defined manner @Secret Pls, I hate that, This was a chapter in our OOP C++ class. I almost got bored This is only usefull if you are more towards abstraction kind of programming. perhaps, he is :-) well, considering that I just spent the last whole week reading and going through the JDK 8 section on classes, I think that is where I am heading @Secret :) In fact, my whole way of comprehending this world relies on something very similar to that 4:48 PM @Secret You must have a deep understanding of Assembly to know it's real purpose. The reason why I can talk about highly subjective things like the impression of some given art had on me or other people, or a mystical experience, or can classify up to 50 different kinds of nothingness is because they are reproducible enough that I can treat them as data types even though there is no way to speak of an ineffable, they can be identified when they are experienced again and that is more than enough to use them as data because many ineffable, are actually quite specific in how they behave As for assembly language, well I once touch on that back in my high school, but I don't think I really knew the details of it including how each keyword is converted into a binary string @Secret very ezy 5:02 PM @Secret Sure, if anything requires uncomputable numbers you aren't going to simulate it with a system that can only produce computable numbers. The simulation hypothesis is fun to think about, but IMHO it's a bit silly. Why would anyone waste vast amounts of computing power simulating a world as large & as detailed as this one appears to be... unless they're cheating, and it doesn't really contain as much information as it appears to. @PhysicsMeta I deleted this from here and posted it on the Main Meta site. well, what I am hypothesising is that if our universe is really a simulation, it is build on a very strange programming language that already contain any level of infinities, and so unless for inconsistency reasons that are imposed by some kind of incompleteness theorem, what we think are uncomputable numbers are actually computed relatively straightforward in that alien programming language. Thus the philosophically interesting thing bout the simulation hypothesis, IMO, is it makes you think about what really makes programming and computation, programming and computation So unlike most people, the reason I am interested in simulation hypothesis is not because of existential stuff about creators, but because it makes me think really hard on the question: What really is a programmable thing 5:29 PM thinking really hard is hard work 6:16 PM @PM2Ring The issue is that I think it would be very hard for us to try to understand the motivations of anything capable of simulating our universe. Although it may seem incomprehensible to us to put that amount of effort into a simulation, we also don't really have any idea how whatever creates the simulation would even perceive it. Maybe we're a side effect of a 5-D toaster. 6:27 PM You are So Mean @Aaron — Saikat Das 54 mins ago Lol simulation hypothesis is probably bogus. there's troubles enough with It peeps these days working on a single computer, but tying an ever-increasing network of billions of minds together without a single hiccup? though i'm sure SH advocates would chalk it up to "well it's a really advanced computer" i believe there also issues with deterministic evolution of systems @AaronStevens Every one of that users posts have basically frustrated me since I've noticed them a few days ago. Especially their formatting. The overuse of bold and italics is frustrating enough, but the overuse of emojis (overuse being any number > 0) is what really gets me. That and they straight up reposted an accepted answer to the same question and just added bold. @JohnRennie how'd you get a glasses-with-number-on-top hat? > post a message in chat within ±12 hours of the UTC New Year’s begin that gets starred That's the most frivolous use for a time machine I can think of 2 @EmilioPisanty so everyone's gonna star something in chat? Who can jump higher than the highest mountain? 6:43 PM Perhaps we can make a room for people to go get their stars? Answer: anyone, mountains can't jump @EmilioPisanty More seriously, see meta.stackexchange.com/q/339893/263383 @KyleKanos nice starbait =) @EmilioPisanty actually, i was going to type that anyway. reading my son's 104 Story Treehouse book & it's littered with terrible jokes speaking of creating new chatrooms 0 The create a new chat-room page is probably one that doesn't get a lot of traffic, and this is starting to show: The text reads: All conversations here are public, and they are logged and licensed under the same cc-wiki license as The Stack Exchange Network. and yes, that is a link t... that's an interesting quirk 6:54 PM @JMac well, at least they are not mean like I am 7:10 PM I invented a space that is isometric to the Minkowski plane. What are some other examples of spaces that are isometric to the Minkowski plane? 7:43 PM who downvoted? no idea, votes are anonymous @KyleKanos do you know other examples of spaces that are isometric to the minkowski plane? not even sure i have rep enough to downvote on math.se @Ultradark nope. 0 Minor issue .. I asked this question: http://physics.stackexchange.com/questions/223610 Currently the question has the callout: "Highly active question. You have enough reputation to answer this question. The reputation requirement helps protect this question from spam and non-answer activi... 7:48 PM @Ultradark You can get an infinity of examples of spaces isometric to Minkowski by taking any diffeomorphism from \mathbb{R}^2 to some target and defining a metric on the target by demanding that the diffeomorphism be an isometry. @ACuriousMind what do you think of the question? -1 Consider the following matrix which transforms points in \Bbb R \cap (0,1)^2, according to a parameter {\mathfrak S}.$$ \mathfrak T_{\mathfrak S} = \begin{bmatrix} (e^{e^{\mathfrak S}})^{-1} & 0 \\\ 0 & (e^{e^{\mathfrak{ -S}}})^{-1} \end{bmatrix}$$I want to prove that this space is a P... this one It doesn't make any sense to me. What doesn't make sense, so I can improve it? the matrix acts on points in the unit square, and the flow is an image of hyperbolic rotation @Ultradark Everything, really. What is this matrix, and how does it "transform" the points? Are you just applying it as a linear map? You suddenly talk about "this space" but you didn't define any space (if you mean the image of the linear map, say so and declare what metric you're using on it), nor have you explained why you think it should be isometric to Minkowski space. I guess I'll delete it, it's not a very good question the matrix transforms the points by letting the parameter change 8:08 PM I don't know what you mean by that. Mappings are easiest to understand if you write them in the standard notation (0,1)^2 \to ?, (x,y)\mapsto ?. it's a rotation matrix math.stackexchange.com/questions/3469153/… Made some edits, hopefully it's better now It's not a rotation matrix (except for when the parameter is zero, then it's the trivial rotation). how is it not a rotation matrix? Rotation matrices by definition are of the form \begin{pmatrix} \cos(\phi) & \sin(\phi) \\ -\sin(\phi) & \cos(\phi) \end{pmatrix}, this one is not. that's for circular rotation 8:15 PM hyperbolic rotations are the same form for \cosh, \sinh instead of \cos,\sin, which this one also is not. you can express a hyperbolic rotation in the form of$$\begin{pmatrix} e^s & 0 \\ 0 & e^{-s} \end{pmatrix}$$Ah, right, sorry. So what you're saying is that you're looking at the map (0,1)^2\to (0,e^s)\times (0,e^{-s}), (x,y) = (e^sx, e^{-s}y). @ACuriousMind well I was looking at the family of level set curves that the points flow across I'm not sure if these curves are geodesics @ACuriousMind so this matrix$$\begin{pmatrix} e^s & 0 \\ 0 & e^{-s} \end{pmatrix}$$acts on points to make them flow along rectangular hyperbolas. And this matrix$$\mathfrak T_{\mathfrak S} = \begin{bmatrix} (e^{e^{\mathfrak S}})^{-1} & 0 \\\ 0 & (e^{e^{\mathfrak{ -S}}})^{-1} \end{bmatrix}
acts on points to make them flow along a function known as $f(x)=\exp(\frac{1}{\log(x)})$ for $x,f\in(0,1)$
So I'm pretty sure this latter matrix is mapping the former matrix to another space, and hence the latter matrix is an image of hyperbolic rotation, which intrinsically must be the same as hyperbolic rotation
and following this line of thought and with some help from the people in the math chat, we came to the conclusion that the space is Pseudo-Euclidean and isometric to the Minkowski plane

9:31 PM
Should questions that belong on the Main stack exchange meta be closed on Physics Meta?

@AaronStevens No - after all, meta is about "the software that powers Physics Stack Exchange". I know that SE monitors at least on all child metas (even if they don't act on it)

@ACuriousMind Oh ok. It is just that when I have asked questions on Physics Meta before that is more about the general SE site, I am told it belongs there and not on Physics Meta
Or maybe I am conflating that with the cases where the question has already been asked on the main meta site

Well, it would be better on main meta, but it's not wrong to post it to ours.

Sounds good, thanks

We generally try to be rather lenient about what goes on meta, since it is the first (and sometimes only) place for support users go to.

9:49 PM
Hello
May I ask you a question?

Oh, thanks :)

Do you know about the center of mass frame?

Yes
That was an easy question ;)

9:55 PM
@ACuriousMind Teach me more!

Now that's harder!

Sorry, I'm back
Now consider we have 2 objects that have mass M1 and M2
The first object has V1 vector velocity
And the second object has V2=0
You can imagine this collision
But after the collision the objects seperate
On the top, it shows the before collision and on the bottom it shows after the collision.
Is everything OK until here?
The collision is elastic by the way.

Everything's fine. Note that you can use MathJax in chat (meta.stackexchange.com/a/220976/263383) to typeset formulae

Yeah, but I'm sorrily I can't see MathJax rendered and I'm using mobile phone :(
sorry*

Ah, yeah, on mobile you have to render the LaTeX in your head ;)

10:07 PM
Yeah, I'm good at it but it's hard to write formulas :/
You can imagine the collision by the center of mass frame
The center of mass would have a constant velocity
Now, the question is find the $\theta_1$ by examining the system with center of mass frame.

@MuhammedÇ.TUFAN In the center of mass frame, the center of mass would have zero velocity by definition!

My teacher made the wrote all velocities (velocities relative to CM velocity, $\vec{U}_1, \vec{U}_2, \vec{U}_1^{'}, \vec{U}_2^{'}$ in center of mass frame,
Oh, no.
It has a constant velocity.
$\vec{R}_{CM}=\frac{M_1\vec{V}_1+M_2\vec{V}_2}{M_1+M_2}$
The position vector of CM.

What is your definition of "center of mass frame"?

$\vec{V}_2=0$ so, the $\vec{R}_{CM}=\frac{M_1}{M_1+M_2}\vec{V}_1$
It's like we sit down at the center of mass of the system and watch everything.
And the total momentum is equal to $\vec{0}$ before and after thr collisions.
Property of the center of mass frame

Yeah, and that means that, in the center of mass frame, $R'_{CM} = 0$ and $V'_{CM} = 0$. (Primed now means "as seen in the center of mass frame")

10:18 PM
Mmm

And since total momentum vanishes, you have $m_1 V'_1 = - m_2 V'_2$.

The CM has no velocity relative to itself but it has a velocity relative to laboratory frame

Yes, that's true

Yeah yeah you are right at the momentum.

I'm just saying that, in the CoM frame, the "constant" velocity of the CoM is a very special constant - zero.

10:20 PM
Oh, then I mean the CM has a velocity relative to the laboratory frame.

Okay

Alright
As I derived at the top side, the position vector of CM is equal to $\vec{R}_{CM}=\frac{M_1}{M_1+M_2}\vec{V}_1$
Is that OK?

The dimensions don't work out - $R$ has dimension of length, but your r.h.s. has dimension of mass times length per time.

Mmm
I'm not sure why it has that dimension but the derivation is true as I know, my teacher and the book that I use is use like that.

I'm afraid then your teacher and book are wrong. Equalities need to have the same dimensions on both sides.

10:25 PM
What is the dimension of position vector, I don't know :/

It's a position, it has just the dimension of length.

By the way, the R.H.S has dimensions of $\frac{[M]}{[M]}{[L]{[T]}^{-1}=[L]{[T]}^{-1}$ lenght per time as I think

Oh, yes, sorry, you're right

Mmm, I'm not really sure why It's not work out

for $V_2 = 0$, what you've got there is not the expression for the position of the CoM, but for the velocity (in the lab frame)

10:28 PM
Yeah but I thinj It's ok
When we think about the general definition of position vector of the CoM it has this dimensions

Well, you've correctly said that the r.h.s. has dimensions of length per time, i.e. it's a velocity. So the l.h.s. needs to be a velocity, too

Yeah I got you but
Ohhhhhhh
I'm a big dumb haha
I have wrote it wrong!

It happens :)

$\vec{R}_{CM}=\frac{M_1\vec{r}_1+M_2\vec{r}_2}{M_1+M_2}$
That is now OK :P
Sorry, forgive me for that mistake :P

Yeah, that seems more like it

10:31 PM
Now, I take the derivative of the $\vec{R}_{CM}$
So, it will give me the velocity of the CoM
$\vec{V}_{CM}=\dot{\vec{R}_{CM}}=\frac{M_1\vec{V}_1+M_2\vec{V}_2}{M_1+M_2}$

yup

If $\vec{V)_2=\vec{0}$ the velocity of CoM is
$\vec{V}_{CM}=\frac{M_1}{M_1+M_2}\vec{V}_1$
Is that OK?

yup

Alright
Now, I will write the velocities relative to CoM frame
$\vec{V}_{1,CM}=\vec{U}_1=\vec{V}_1-\vec{V}_{CM}=\vec{V}_1-\frac{M_1}{M_1+M_2}\vec{V}_1=\vec{V}_1(1-\frac{M_1}{M_1+M_2})=\vec{V}_1\frac{M_2}{M_1+M_2}$
I can write all the other 3 velocities but It will take a long time, you know I use mobile phone :(
Is that OK for you?

Yeah, it's fine. Just ask what you're going for and we'll figure it out

10:39 PM
Okay.
Thanks.
Now, you see that $\tan\theta_{1}=\frac{\sin\theta_1}{\cos\theta_1}$
Multiply it with $\frac{{\vec{V}_1}^{'}}{{\vec{V}_1}^{'}}$
so $\tan\theta_1=\frac{{\vec{V}_1}^{'}\sin\theta_1}{{\vec{V}_1}^{'}\cos\theta_1}$
Is everything OK?
${\vec{V}_1}^{'}$ is the velocity of the first object after the collision btw.

alrighty

Okay
Now, you see that ${\vec{V}_1}^{'}\cos\theta_1$ is the after collision velocity of the first object on the $x$ dimension.
and the ${\vec{V}_1}^{'}\sin\theta_1$ is on the $y$ dimension
I will make a little transformation here.
${\vec{U}_1}^{'}={\vec{V}_1}^{'}-\vec{V}_{CM}$
${\vec{V}_1}^{'}={\vec{U}_1}^{'}+\vec{V}_{CM}$
Is that OK?

yup

Okay, now i will seperate ${\vec{V}_1}^{'}$ into It's components
${{{V}_1}^{'}}_{x}={{U}_1}^{'}\cos\theta+V_{CM}$
${{{V}_1}^{'}}_{y}={{U}_1}^{'}\sin\theta+0$
Is that OK?

10:54 PM
Yes - except I want to go back one step - how do you know that $V'_1\cos(\theta_1)$ is the after collision $x$-velocity?

It is showed on the image of the collision. Let me upload it one more time.

Common sense from watching billard balls collide heads-on (where $M_1 = M_2$ and $\theta = 0$) tells us that when the angle is zero, the ball 1 will stop and ball 2 will have all the velocity.

That picture does not model the case $\theta = 0$ well.

Yeah, I get what you mean. But in this collisions there is no information about the masses, but I don't know why the collision doesn't keep moving on the same dimension :P

10:57 PM
Oh, wait, sorry, I got confused - in my degenerate case, $V'_1 = 0$, so everything's fine. Please continue.

I think this is something like a general definition or proof of 2 mass colliding
OK, thanks.
At the last relative velocities do you know what is $\theta$ angle?
(not $\theta_1$)

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