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 $m_2/4$
Best Answer
I'm shocked that there isn't a satisfactory answer on this site yet!
This can be answered in any number of dimensions with some relatively simple vector math. As follows intuitively (rigorously in the center-of-mass frame), for equal mass particles, the relative velocities of the particles are reversed along the normal direction. All that needs to be done is translate that into vector equations.
The algorithm
Calculate the vector normal of collision: $$\hat{\bf n}=\frac{{\bf r}_1-{\bf r}_2}{\|{\bf r}_1-{\bf r}_2\|}$$ Calculate the relative velocity of the particles in the direction of the normal, using a dot product: $$v_{\mathrm{rel}}=(\dot{\bf r}_1-\dot{\bf r}_2)\cdot \hat{\bf n}$$
To reverse the velocities of the particles along the normal of collision, take:
$$\dot{\bf r}_1\mapsto \dot{\bf r}_1- v_{\mathrm{rel}}\hat{\bf n}$$ $$\dot{\bf r}_2\mapsto \dot{\bf r}_2+ v_{\mathrm{rel}}\hat{\bf n}$$
That's it. Easy peasy. It works in 1D, 2D, 3D, 4D, 5D; any number of dimensions. And it's rather simple!
Seeing that the algorithm works
To see that this reverses the velocities along the normal direction, calculate $\dot{\bf r}_1\cdot \hat{\bf n}$ before and after. You'll find: \begin{align*} v_{\mathrm{rel}}= (\dot{\bf r}_1-\dot{\bf r}_2)\cdot \hat{\bf n} &\mapsto \dot{\bf r}_1\cdot \hat{\bf n}- v_{\mathrm{rel}}\hat{\bf n}\cdot \hat{\bf n}-\dot{\bf r}_2\cdot \hat{\bf n}- v_{\mathrm{rel}}\hat{\bf n}\cdot \hat{\bf n}\\ &=v_{\mathrm{rel}}-2v_{\mathrm{rel}}\\ &=-v_{\mathrm{rel}} \end{align*}