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.
Best Answer
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 $m_2/4$