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 $m_2/4$}

The problem simply is not integrable and thus we cannot generally trace the evolution of all phase-space variables analytically. The easiest way to describe it is via a Hamiltonian in cylindrical coordinates
$$H = \frac{(p_\phi - c A_{\phi})^2}{2m \rho^2} + \frac{1}{2m} (p_\rho^2 + p_z ^2)$$
where $A_{\phi}=B(t) \rho /2$ (you can easily see that there is no need for electric potential $\Phi$). The obvious symmetries are rotational leading to the conservation of $p_\phi= m\dot{\phi}\rho^2 + c A_\phi$, and translational, leading to the consevation of $p_z=\dot{z}$. Unfortunately, there are still two degrees of freedom, $\rho$ and $t$ which means that general initial conditions may lead even to chaotic scattering.

If the system truly exhibits chaotic scattering, it is a proof of the fact that you cannot find a general analytical formula. However, sometimes it happens that a system has a "hidden" additional integral. There is no easy way to discern between two cases. I think the easiest thing you can do is resort to some kind of approximation such as assuming $\Delta t$ is small, or on the other hand, that $\Delta t$ is large and you can thus integrate the energy loss as adiabatically evolving through the orbits in the time-independent system.

## Best Answer

As the comments said, this is a basic application of the Lorentz force. Written in your notation, we have the following.

$$\vec{F} = q[\vec{E} + (\vec{v} \times \vec{B})]$$

$$\vec{v} = \left( 0, v_2, 0 \right)$$ $$\vec{E} = \left( 0, E, 0 \right)$$ $$\vec{B} = \left( 0, 0, B \right)$$

The most likely thing you had trouble with would be the cross product I think. You can use a formula for this or your right hand. For the latter, in my own head, the magnetic field is "up" and the y-axis is toward my chest when I hold out my right hand. My middle finger points toward me, my index finger points up, and my thumb points to the right. So the answer is in the positive x-axis!

$$\vec{v} \times \vec{B} = \left( v_2 B , 0, 0 \right)$$

I will stop here because any more would certifiably be me doing it "for you". The idea of something being trivial is relative to the person, but everything that remains to answer your question should be trivial unless you are missing some fundamental part of physics instruction.