Elastic collision of three balls

[skrat]

Homework Statement

The first ball collides with $v_1$ collides with the second two as shown in the picture. All balls are identical and the second balls have no speed at the beginning.
Find the transform matrix and find the covariation matrix after the collision if it is $\begin{bmatrix} \sigma ^2 & 0\\ 0 & 0 \end{bmatrix}$ before the collision.

Homework Equations

The Attempt at a Solution

The momentum is conserved: $$mv_1=2m{v_2}'cos\varphi$$ where luckily the balls are identical, therefore $cos\varphi =\frac{1}{\sqrt 2}$. And also kinetic energy is conserved $$\frac 1 2 v_1^2={v_2}'^2$$ This brings me to transform matrix $$
\begin{bmatrix}
{v_1}'\\
{v_2}'
\end{bmatrix}=A\begin{bmatrix}
{v_1}\\
{v_2}
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0 & \frac{1}{\sqrt 2}
\end{bmatrix}\begin{bmatrix}
{v_1}\\
{v_2}
\end{bmatrix}$$ and finally also to covariance matrix $${M}'=AM_0A^T=
\begin{bmatrix}
0 &0 \\
0&0
\end{bmatrix}$$ The last step was done using mathematica. And the most confusing part here is that all the components are completely correlated - I am having some troubles to interpret this absolute correlation. Could somebody help me with that? It is also possible that my result is completely wrong in that case, please let me know.

 

[mfb]

where luckily the balls are identical, therefore $cos\varphi =\frac{1}{\sqrt 2}$.

Why?
And why do you think ball 1 stops?

And the most confusing part here is that all the components are completely correlated - I am having some troubles to interpret this absolute correlation. Could somebody help me with that? It is also possible that my result is completely wrong in that case, please let me know.

How could they be independent (and independent of what)?

 

[skrat]

How would I then find $\varphi$?

If ball 1 doesn't stop, than from the conservation of momentum $$v_1=2{v_2}'cos\varphi + {v_1}'$$and from conservation of kinetic energy$$v_1^2=2{v_2}'^2+{v_1}'^2$$

Now I have two equations for three unknown parameters ${v_1}'$,${v_2}'$ and $\varphi$.

 

[skrat]

Should I work in the CMS system?

$u_1=v_1+v_{CMS}$ and $u_2=v_2-v_{CMS}=-v_{CMS}$. Conservation of momentum than brings me to $mu_1-mu_2=0$ meaning $v_{CMS}=-\frac{v_1}{2}$ and conservation of energy to $u_1^2=2{u_2}'^2+{u_1}'^2=3{u_1}'^2$.

Now ${u_1}'=\sqrt 3({v_1}'+v_{CMS})=u_1=v_1+_{CMS}$ mening $${v_1}'=v_1\frac{1+\sqrt 3}{2\sqrt 3}$$ and similar for $${v_2}'=v_1\frac{\sqrt 3 -1}{2\sqrt 3}$$.

Would this be ok?

 

[mfb]

How would I then find $\varphi$?

Momentum transfer is along the line "contact point <-> center of balls".

If ball 1 doesn't stop, than from the conservation of momentum $$v_1=2{v_2}'cos\varphi + {v_1}'$$and from conservation of kinetic energy$$v_1^2=2{v_2}'^2+{v_1}'^2$$

Now I have two equations for three unknown parameters ${v_1}'$,${v_2}'$ and $\varphi$.

You can get the angle from the geometry of the problem.

I would not go to the cms. Also, your formulas look wrong there.

 

[skrat]

Hmmm, from the geometry?

So, since the balls are identical, I can say that $\varphi = \pi /4$ ?

 

[mfb]

Why pi/4? Where does that number come from?

Yes from geometry, see the first part of my post.

 

[skrat]

Blaaah, that's a mistake, I mean't to write $\varphi =\pi /3$

 

[mfb]

Keeping guessing numbers does not make it better. Sure eventually you'll hit the right one (not yet), but that is not a solution.

 

[skrat]

I wasn't guessing, the geometry is clear. Equilateral triangle with all three sides $2R$ if $R$ is the radius of each ball.

The angles in the equilateral triangle are $\pi /3$. However, you are right, If I wouldn't be so careless I would see that the angle is actually half of that, therefore $\pi /6$.

 

[mfb]

Right.