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ACuriousMind
ACuriousMind
11:00 PM
$\theta$ 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
Muhammed Ç. TUFAN
Wow, you are really clever :P
 
ACuriousMind
ACuriousMind
Thanks :P
 
Muhammed Ç. TUFAN
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
Sir Cumference
there's gotta be a rotation matrix for images
 
ACuriousMind
ACuriousMind
@MuhammedÇ.TUFAN I get it, it's fine
 
Muhammed Ç. TUFAN
Muhammed Ç. TUFAN
11:08 PM
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
ACuriousMind
I don't get where the last equation is supposed to come from
 
Muhammed Ç. TUFAN
Muhammed Ç. TUFAN
Is $\tan\theta_1$?
 
ACuriousMind
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
Muhammed Ç. TUFAN
Yeah, absolutely
The final form of $\tan\theta_1$ is that you have said.
$U'_1$ cancelled.
Can I continue?
 
ACuriousMind
ACuriousMind
sure
 
Muhammed Ç. TUFAN
Muhammed Ç. TUFAN
11:21 PM
OK
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
ACuriousMind
Wait
We had $\vec V_1 = \vec U_1 + \vec V$
 
Muhammed Ç. TUFAN
Muhammed Ç. TUFAN
Yep
 
ACuriousMind
ACuriousMind
But that does not imply that the speeds fulfill $V_1 = U_1 + V$
 
Muhammed Ç. TUFAN
Muhammed Ç. TUFAN
From the relative speed of $\vec{U}_1=\vec{V}_1-\vec{V}_{CM}$
 
ACuriousMind
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
Muhammed Ç. TUFAN
11:26 PM
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
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
Muhammed Ç. TUFAN
That's interesting
 
ACuriousMind
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
Muhammed Ç. TUFAN
Oh, because we just switch these 2 vectors, do you mean that?
 
ACuriousMind
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
Muhammed Ç. TUFAN
11:33 PM
Oh, the $\vec{V}_1$ is on the x-axis as well
Wow, cool stuff
 
ACuriousMind
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
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
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
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
ACuriousMind
11:52 PM
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
Muhammed Ç. TUFAN
So, you say that the firdt object stops in CoM frame?
 
ACuriousMind
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
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
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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