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ACuriousMind

23:00

$\theta$ is the angle that $U'_1$ has with the $x$-axis (while $\theta_1$ is the angle that $V'_1$ has with the $x$-axis)

Muhammed Ç. TUFAN

Wow, you are really clever :P

ACuriousMind

Thanks :P

Muhammed Ç. TUFAN

I haven't mentioned anything about the $\theta$ and you knew it, that amazed me :P

Let me upload the image of the collision that from the CoM frame

Yep, the $\theta$ is equal for $U'_1$ and $U'_2$

Do you understand the image of the collision from CoM frame?

It's in another language because

Sir Cumference

there's gotta be a rotation matrix for images

ACuriousMind

@MuhammedÇ.TUFAN I get it, it's fine

Muhammed Ç. TUFAN

23:08

OK, I'm keeping

Now, we have

$V'_1\cos\theta_1=U'_1\cos\theta+V$

$V'_1\sin\theta_1=U'_1\sin\theta+0$

+0 because $\vec{V}_{CM}$ has no y component

We did have $\tan\theta_1=\frac{V'_1\sin\theta_1}{V'_1\cos\theta_1}$

If we put things into equation we get $\tan\theta_1=\frac{U'_1\sin\theta}{U'_1\cos\theta+V}$

And It's equal to $\tan\theta_1=\frac{U'_1(\sin\theta)}{U'_1(\cos\theta+\frac{V}{U'_1}}$

Is that OK?

ACuriousMind

I don't get where the last equation is supposed to come from

Muhammed Ç. TUFAN

Is $\tan\theta_1$?

ACuriousMind

If you meant to cancel the $U'_1$ on the r.h.s., that should be $\tan(\theta_1) = \frac{\sin(\theta)}{\cos(\theta) + \frac{V}{U'_1}}$

Muhammed Ç. TUFAN

Yeah, absolutely

The final form of $\tan\theta_1$ is that you have said.

$U'_1$ cancelled.

Can I continue?

ACuriousMind

sure

Muhammed Ç. TUFAN

23:21

Now the speed of CoM (not velocity) is $V_{CM}=\frac{M_1}{M_1+M_2}V_1=\frac{M_1}{M_1+M_2}(U_1+V)$

$V_{CM}-\frac{M_1}{M_1+M_2}V_{CM}=\frac{M_1}{M_1+M_2}U_1$

$V_{CM}(1-\frac{M_1}{M_1+M_2})=\frac{M_1}{M_1+M_2}U_1$

ACuriousMind

Wait

We had $\vec V_1 = \vec U_1 + \vec V$

Muhammed Ç. TUFAN

Yep

ACuriousMind

But that does not imply that the speeds fulfill $V_1 = U_1 + V$

Muhammed Ç. TUFAN

From the relative speed of $\vec{U}_1=\vec{V}_1-\vec{V}_{CM}$

ACuriousMind

$V_1 = U_1 + V$ is only true if $\vec U_1$ and $\vec V$ are parallel (which they, in general, aren't)

Muhammed Ç. TUFAN

23:26

Hmm

Can we think it as they are the magnitutes of the velocities?

Because in my book it make it as the way I wrote :/

ACuriousMind

Sure - that's what speed is, but that one vector is the sum of two others doesn't mean that the length of that vector is the sum of the length of the others!

Muhammed Ç. TUFAN

That's interesting

ACuriousMind

But if I look back at what $\vec U_1$ and $\vec V_{CM}$ are, I think they're in fact parallel, so it works

Muhammed Ç. TUFAN

Oh, because we just switch these 2 vectors, do you mean that?

ACuriousMind

Well, $\vec U_1$ is the velocity in the CoM before collision in the CoM frame, and $\vec V$ is the velocity of the CoM in the lab frame. Both are just along the x-axis, so they're parallel

Muhammed Ç. TUFAN

23:33

Oh, the $\vec{V}_1$ is on the x-axis as well

Wow, cool stuff

ACuriousMind

But I'm afraid it's getting late here, so if you have a question to ask instead of leading me through mechanics exercises I've done long ago, better ask it now ;)

Muhammed Ç. TUFAN

Yeah therr are just a little more thing left

As I said at the start, the aim of this example is finding the $\theta_1^{\max}$ in 3 situations: $M_1=M_2$, $M_1<M_2$ and $M_1>M_2$

My question is about the technique that used in the last strps of the fiding maxiumum value

Please stay with me if you can

I approximately give you maximum 20 minutes to reach what i want to ask

I will simplify the $\tan\theta_1$ one more time and i will take the derivative of it and everything is done

May I keep, please?

I mean continue :P

ACuriousMind

Just go ahead, if I don't answer it right now, I promise I'll come back to it when I have time (and someone else might take over in the meantime if they feel like it)

Muhammed Ç. TUFAN

Okay, thank you so much! We have $\frac{M_2}{M_1+M_2}V_{CM}=\frac{M_1}{M_1+M_2}U_1$

Is equal to $M_2V_{CM}=M_1U_1$

We didd have $\tan\theta_1=\frac{sin\theta}{\cos\theta+\frac{V}{U'_1}}$

We get $V_{CM}=\frac{M_1}{M_2}U_1$

Put that into tangent equation and we get

$\tan\theta_1=\frac{\sin\theta}{\cos\theta+\frac{M_1}{M_2}}$

Say that $\frac{M_1}{M_2}=C$ (constant).

Now take the derivative of $\tan\theta_1(\theta)$

$\frac{d}{d\theta}=0\Rightarrow \cos\theta=-\frac{M_1}{M_2}$

All is done!

We have found a maximum $\tan\theta_1$ by taking the derivative of it and equal it to $0$

Now my question comes

If $M_1=M_2, \cos\theta=-\frac{M_1}{M_2}=-1$

And $\tan\theta_1=\frac{\sin\theta}{\cos\theta+1}$

My teacher did something like this:

$\tan\theta_1 \to \infty$

$\cos\theta+1=0$

$\cos\theta=-1$

$\theta=\pi$

I don't understand how he found $\theta=\pi$

ACuriousMind

23:52

Yeah, that's a hack. Note that the way we got $\tan(\theta_1) = \dots$ was by cancelling $U'_1$, i.e. assuming it is not zero. But for the case $M_1 = M_2$, we in fact have $U'_1 = 0$, so we've not allowed to use that.

Muhammed Ç. TUFAN

So, you say that the firdt object stops in CoM frame?

ACuriousMind

Hm, no - in the CoM frame, the two objects just interchange velocities. $\theta = \pi$ is correct there - from the CoM frame, the first object reverse velocity

Muhammed Ç. TUFAN

The thing that I can't understand is the taking the limit of $\tan\theta_1\to \infty$ and finding the $\theta=\pi$

ACuriousMind

We don't have $U'_1 = 0$, we have $V'_1 = 0$, but anyway, our derivation still fails because of that (because you multiplied with $\frac{V'_1}{V'_1}$ the step before)

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