Elastic collision of point particle and rod 
A 1 meter long rod on the ice with mass $m_2=1$ kg is perpendicularly hit on one end by a point particle with mass $m_1=0.1$ kg. The collision is elastic and the point particle is bounced back in the same direction. After the collision the rod's frequency is $\nu =2$ Hz. What was the initial velocity of the point particle?

My attempt:
Since the collision is elastic, the kinetic energy of the system is the same before and after the collision:
$$0.5m_1v_1^2=0.5J_2 \omega_2^2+0.5m_2v_2^2+0.5m_1v_3^2$$
Where $v_3$ is the velocity of the point particle after the collision.
Now, in the case of a rod:
$$J=\frac{1}{12}L^2m$$
And, we also know:
$$\omega_2=2 \pi \nu$$
And there are also no external forces, therefor the momentum of the system is the same before and after the collision:
$$m_1\vec{v_1}=m_1 \vec{v_3}+m_2\vec{v_2}$$
Here $v_1$ is the quantity we're looking for, $v_3$ is the point particle's velocity after the collision and $v_2$ is the velocity of the rod's center of mass. It follows:
$$\vec{v_2}=\frac{m_1 \vec{v_1}-m_1 \vec{v_3}}{m_2}$$
From this it follows:
$$0.5m_1v_1^2=\frac{1}{24}L^2m_2 4 \pi^2 \nu^2+0.5m_2 \left|\frac{m_1 \vec{v_1}-m_1 \vec{v_3}}{m_2}\right|^2+m_1v_3^2$$
This is 1 equation with 2 unknowns, and this is where I get stuck.
Any help is appreciated.
 A: 
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 $m_2/4$
A: 
Using conservation of energy:
$$\left[\frac 12m_1v_i^2\right]_{particle}=\left[\frac 12m_1v_f^2\right]_{particle}+\left[\frac 12I\omega^2+\frac 12m_2v_2^2\right]_{rod}\\\text{where }I=\frac{ml^2}{12}=\frac1{12}$$
$$\frac {v_i^2}{20}=\frac {v_f^2}{20}+2\frac {\pi^2}{3}+\frac {v_2^2}2$$
$$v_i^2-v_f^2=\frac{40\pi^2}3+10v_2^2$$
(Alternative):Using coefficient of restitution=$1$
$$v_{i}=\frac{lw}{2} + v_2 +v_f$$

Using conservation of Momentum:
$$m_1v_i=m_2v_2+m_1(-v_f)$$
$$ \frac{v_i}{10}=\frac{-v_f}{10}+v_2$$
$$ v_i+v_f=10v_2$$

Using conservation of angular Momentum:
$$m_1v_i\left(\frac l2\right)=m_1\left(-v_f\right)\left(\frac l2\right)+I\omega$$
$$\frac{v_i}{20}=\frac{-v_f}{20}+\frac{\pi}3$$
$$v_i+v_f=\frac{20\pi}3$$

So, $$10v_2=\frac{20\pi}3\implies v_2=\frac{2\pi}3$$
$$(v_i-v_f).\frac{20\pi}3=\frac{40\pi^2}3+10\left(\frac{2\pi}3\right)^2=\frac{160\pi^2}9\implies v_i-v_f = \frac{8\pi}3$$
Also, $$v_i+v_f=\frac{20\pi}3$$
So:
$$v_i=\frac{14\pi}3,v_f=2\pi$$ 

A: The collision of a rod with a point mass is the similar to the collision of two masses but with the effective mass of the rod being
$$ m' = m_{rod} \frac{I_{rod}}{I_{rod}+m_{rod} r^2} $$ where $r$ is the distance between the point of impact and the center of mass, and $I_{rod}$ is the mass moment of inertia about the center of mass. If the rod is slender with length $\ell$ then $$ I_{rod} = \frac{m}{12} \ell^2 \\ r = \frac{\ell}{2} \\ m' = m_{rod} \frac{1}{4} $$
So the momentum exchanged is $$ J =  \frac{(1+\epsilon)\, v}{\frac{1}{m_{point}} + \frac{1}{m'}} $$ where $v$ is the impact speed and $\epsilon$ the coefficient of restitution.
The final velocity of the point mass is $$v_{point} = v - \frac{J}{m_{point}}$$
The final velocity of the rod center of mass is $$v_{rod} = \frac{J}{m_{rod}} \\ \omega_{rod} = \frac{r J}{I_{rod}} $$
NOTE: that the rod will rotate about a point with distance $c$ from the center of mass (in the opposite side from the impact point) located at $c =\frac{I_{rod}}{m_{rod} r} = \frac{\ell}{6}$. This is known as the instant center of rotation, and the point of impact is the center of percussion of the rotation center C.
A: One thing you can do is use conservation of angular momentum instead of linear - the point mass has angular velocity relative to the center of rotation of the rod.  (If you are confused about this, imagine watching a car pass you on a road.  The car moves in a straight line, but you rotate your head to follow it, giving it a [constantly changing] angular velocity.)  Find the angular velocity of the particle the instant before the collision, apply conservation of angular momentum, and solve the problem from there.
A: The stroke of genius
The laws of Nature are wonderful: extremely simple - make a tremendously complex world, a few symbols can be extremely powerful and beautiful
all roads lead to Rome, you may go from London to Rome following many different routes, only a fool would get there via Moscow. But an eagle can fly over the Alps in a straight-line.
I'll try to explain what many readers do not understand:

....if you know shortcuts and are confident about applying them, by all
means do. In my case, I could not understand exactly what he was doing
so I chose to write a "pedestrian" answer for the benefit of other
visitors

Visitors have not benefitted.
The standard solution
was fully and clearly explained last year, if there had not been a banal, trivial typo we would not be here now.
If the initial velocity is known, the unknown is $\omega$

and we need 3 equations:
$$ \left[\frac 12m_1v_i^2\right]_{m}=\left[\frac 12m_1v_f^2\right]_{m}+\left[\frac 12I\omega^2+\frac 12m_2v_2^2\right]_{rod}\tag1$$
$$ m_1 v_1 = m_1 v_3 + m_2 v_2 \tag2$$
$$ m_1 v_1 *r(=L/2) = m_1 v_3*r (=L/2) + I\omega \tag3$$
because there are 3 unknowns: $v_3$,  $v_2$ and $\omega$
The solution of this problem
If $\omega$ is hnown ( = $4\pi$)

we do not need 3 equations because there are only 2 unknowns, 1. : $v$  and 2.:  $v_3$, and the equations 2, 3 become obviously identical:
$$ m_1 v_1 = m_1 v_3 + m_2 v_2 \rightarrow .1x= .1y + \pi2/3\tag2$$
$$ r*m_1 v_1  = r*m_1 v_3  + L \rightarrow .1x = .1y +(L/r)\pi2/3\tag3$$
since the linear momentum of the rod is equal to its angular momentum divided by the radius (and mass which is = 1Kg)
$$m_2v_2 = I\omega /r$$
This was reminded in a few comments, but no correction was made to the post
Using 3 equations instead of two is an unhealthy practice, and doing a monumental useless work is not only pedestrian but ... ehm... let's say: non-rational, and encouraging inexperienced or credulous visitors to do so is misleading
The simplest solution
The correct standard solution has been offered as a first choice, but was apparently rejected by the vast majority of the 'visitors'.
The ingenious solution has probably been overlooked, not-understood or considered not-worth-examining-at-all.
But it so easy to understand: any schoolboy knows that the velocity of a body at rest after an impact is $v = 2p/ M$

Considering the effective point-mass at the tip of the rod ( $m' =m_2/4$ ) we can determine the initial velocity of the point-particle just reversing the simple formula of the head-on collision
