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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)
 
Wow, you are really clever :P
 
Thanks :P
 
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
 
there's gotta be a rotation matrix for images
 
@MuhammedÇ.TUFAN I get it, it's fine
 
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?
 
I don't get where the last equation is supposed to come from
 
Is $\tan\theta_1$?
 
If you meant to cancel the $U'_1$ on the r.h.s., that should be $\tan(\theta_1) = \frac{\sin(\theta)}{\cos(\theta) + \frac{V}{U'_1}}$
 
Yeah, absolutely
The final form of $\tan\theta_1$ is that you have said.
$U'_1$ cancelled.
Can I continue?
 
sure
 
11:21 PM
OK
Now the speed of CoM (not velocity) is $V_{CM}=\frac{M_1}{M_1+M_2}V_1=\frac{M_1}{M_1+M_2}(U_1+V)$
$V_{CM}-\frac{M_1}{M_1+M_2}V_{CM}=\frac{M_1}{M_1+M_2}U_1$
$V_{CM}(1-\frac{M_1}{M_1+M_2})=\frac{M_1}{M_1+M_2}U_1$
 
Wait
We had $\vec V_1 = \vec U_1 + \vec V$
 
But that does not imply that the speeds fulfill $V_1 = U_1 + V$
 
From the relative speed of $\vec{U}_1=\vec{V}_1-\vec{V}_{CM}$
 
$V_1 = U_1 + V$ is only true if $\vec U_1$ and $\vec V$ are parallel (which they, in general, aren't)
 
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 :/
 
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!
 
That's interesting
 
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
 
Oh, because we just switch these 2 vectors, do you mean that?
 
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
 
11:33 PM
Oh, the $\vec{V}_1$ is on the x-axis as well
Wow, cool stuff
 
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 ;)
 
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
 
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)
 
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 $\theta=\pi$
 
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.
 
So, you say that the firdt object stops in CoM frame?
 
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
 
The thing that I can't understand is the taking the limit of $\tan\theta_1\to \infty$ and finding the $\theta=\pi$
 
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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