Since the collision is elastic, the kinetic energy of the system is
the same before and after the collision: $$0.5m_1v_1^2=0.5J_2
\omega_2^2+0.5m_2v_2^2+0.5m_1v_3^2$$
This kind of problem has usually
3 equations: conservation of: 1. Ke, 2. p, 3. L, and
3 unknowns: $y= v_3, z =v_2, x = \omega$, when the initial velocity $v_1$ is known.
But in this case the unknown parameter is $v_1 = x$ and you know that $\omega (2\pi\nu) = 4\pi$, angular momentum $L (I\omega) =\pi/3$ and $Ke (L\omega) = 2\pi^2/3$. This simplifies the problem, because that means that also the linear velocity of the rod is known $v_2 (L/r[m_2])=\frac23 \pi$
Based on this, your KE equation becomes:
$$x^2=y^2 + \frac{1}{m_1} \left[I\omega^2+\left(\frac{2\pi}{3}\right)^2\right]\rightarrow
x^2=y^2+10\frac{(12+4)\pi^2}{9} \tag1$$
the second equation can regard p (or L): $$m_1x = m_1y + \frac{2\pi}{3} \rightarrow y= x-\frac{2\pi}{3m_1} \tag2$$
There are 2 unknowns and 2 equations:
$$\left\{\begin{align}x^2&=y^2+\frac{160\pi^2}{9} \\
y&= x-\frac{20\pi}{3}\end{align}\right.$$
and you may solve that simple system for $x$.
$$[\x^2]= \left[[\x^2] -x\frac{40\pi}{3} + \frac{400\pi^2}{9} \right]+ \frac{160\pi^2}{9}\rightarrow x = \frac{[3]}{[40 \pi]} * \frac{14\pi* [40\pi]}{3*[3]}$$
Knowing the rules of collisions, the solution can be found even more quickly, since the linear velocity of the rod: $v_2=2/3\pi$ summed to its rotational velocity: $v_\omega(\omega r)=2\pi$ is the velocity of the rod $v_{m'}= 8/3\pi$ considered as a point-mass $m'$ at the tip of the rod, and you know its value is $m'=m_2/4$ *
The initial velocity $x$ can be found in a very simple way with the trivial 1-D formula (using the velocity of CoM) : $x=v_{m'}*1.75$:
$$v_i =v_{m'} \frac{m_1+m'}{2(m_1 )}=\pi\frac{8}{3}\left[\frac{.35}{.2}\right]$$
$x = 14.66076... =\pi14/3$
Note:
* linear momentum is of course the same: $m_2*v_2=m'*v_{m'} \rightarrow m' =( v_2/v_{m'}= 2/3*3/8) = 0.25$, but It is not even necessary to calculate it, since its value at CM, CoP, tip varies linearly (1, 3/4, 1/4), and therefore at the tip it is always $m_2/4$
If you consider the disks from the center-of-mass frame, there is zero initial (and therefore final) total linear momentum. Thus, in this frame you will end up with the two disks stuck together, rotating about the origin (the center of mass).
To determine how rapid this rotation is, you can conserve angular momentum about the origin. The initial angular momentum includes that of each disk about its own center of mass, plus an additional part due to the translational motion of each disk relative to the origin. The final angular momentum is the same, and is equal to the rotation rate of the combined disks times the rotational moment of inertia of the combination.
You can determine the moment of inertia as the sum of the individual moments of inertia of each disk plus $m_i r_i^2$, where $r$, for each disk, is the distance from its own center of mass the the center of mass of the combination.
Best Answer
When the balls collide , draw a line connecting their centers. Break the relative speed of the balls into two components, along this line and perpendicular to it. The balls will keep the perpendicular speed component without passing it to each other. Then all you have left is to solve the direct collision at the speed of the direct speed component, which is a known easy problem.