How do linear momentum and angular momentum divide on collision?
The key here is that the object experiences a vertical impulse from the ground. The description of how the impulse divides up into linear momentum and angular momentum is given by applying forces and torques over a period of time. If $S=F \Delta t$ is the impulse applied in the vertical, we can look at how linear and angular momentum change:
- $m\dot{y}'=m\dot{y}+S$ (linear momentum)
- $I\omega'\hat{z}=I\omega \hat{z}+\vec{r}\times (S\hat{y})$ (angular momentum - $\vec{r}$ is the vector from the object's center of mass to the object's point of contact)
If you plug these two equations into an equation that demands energy conservation, you can solve for the impulse $S$, along with $\dot{y}'$ and $\omega'$, and this gives a complete description of the impact. I work it out completely below.
General solution to a rigid body colliding with the ground
So, to deal with a rotating rigid body, I'll work in 3D space (though we really only need two dimensions) and with a rectangle, as shown in your picture. I'll make use of angular momentum. The keys here are the vector cross product and the concept of an impulse!
On to definitions: $(x,y)$ are the coordinates of the center of mass of the object. I'll assume the object is moving in a 2D plane, so $z=0$. The object has an $x$ velocity $\dot{x}$, a $y$ velocity $\dot{y}$, an angular velocity $\omega$, a mass $m$, and a moment of inertia about the $\hat{z}$ axis, $I$.
The block contacts the ground somewhere and receives a vertical "punch" - an impulse - from the ground. To describe this punch, let $\vec{r}$ be the vector from the center of mass of the object, to the corner where it contacts the ground. The object has energy $E=\frac{1}{2}m\dot{x}^2+\frac{1}{2}m\dot{y}^2+\frac{1}{2}I \omega^2$ and angular momentum $\vec{L}=I\omega\hat{z}$. By writing out how these quantities change (linear momentum, angular momentum, energy), we solve the problem.
After the collision the object has velocities $\dot{x}'$ and $\dot{y}'$, and angular velocity $\omega'$. The problem is to determine these quantities. It will help to imagine the collision as not being instantaneous, but instead taking place over a small period of time $\Delta t$.
For the duration of the impact, there is an upwards directed normal force $F\hat{y}$. Over the whole period of time $\Delta t$ this provides an impulse $F\Delta t=S$. The change in vertical velocity from this impulse is just $S/m$, so that $\dot{y}'=\dot{y}+S/m$. The force $F$ provides a torque $\vec{r}\times (F\hat{y})$, where $\vec{r}$ is the vector from the center of mass of the object to its point of contact with the floor. This gives a total change in angular momentum: $I\omega'\hat{z}=I\omega \hat{z}+\vec{r}\times (F\hat{y})$
So we have three equations:
- $\frac{1}{2}m\dot{y}^2+\frac{1}{2}I \omega^2=\frac{1}{2}m\dot{y}'^2+\frac{1}{2}I \omega'^2$ (energy conservation. $\dot{x}$ cannot change because the force is directed vertically)
- $I\omega'\hat{z}=I\omega \hat{z}+\vec{r}\times (S\hat{y})$ (change in angular momentum due to a vertical impulse)
- $m\dot{y}'=m\dot{y}+S$ (change in momentum due to a vertical impulse)
This gives three equations in three unknowns: $\omega'$, $\dot{y}'$, and $s$. If you plug equations two and three into equation 1, you get a quadratic equation in $s$ and you can solve it. A quadratic equation has two solutions.
The first solution is $\omega'=\omega$, $\dot{y}'=\dot{y}$, and $s=0$. Nothing happening is consistent with energy conservation!
The second solution is, where $\vec{r}=(r_x, r_y,0)$:
$$\dot{y}'=\frac{m r_x^2 \dot{y}-I (2 \omega r_x+\dot{y})}{I+m r_x^2}$$
$$\omega'=\frac{I \omega-m r_x (\omega r_x+2 \dot{y})}{I+m r_x^2}$$
$$S=-\frac{2 I m (\omega r_x+\dot{y})}{I+m r_x^2}$$
Note that the vertical velocity of the collision point is $\omega r_x$, and that quantity appears a few times in the formula!
Best Answer
In Newtonian mechanics, the momentum of a particle of mass $m$ is given by
$$\mathbf p = m\frac{d {\mathbf r}}{dt} = m \mathbf v$$
where $\mathbf r$ is the position vector and $t$ is a universal parameter. However, in relativistic mechanics, $t$ is a coordinate, not a parameter, and is thus a component of a four-vector, the four-position $\mathbf R = (ct, \mathbf r)$.
The four-velocity is then defined as
$$\mathbf U = \frac{d \mathbf R}{d \tau} = \frac{d \mathbf R}{d t}\frac{dt}{d\tau} = \frac{d \mathbf R}{d t}\gamma = \gamma (c, \mathbf v) $$
where $\tau$ is the proper time parameter. In analogy with Newtonian mechanics, the four-momentum is then
$$\mathbf P = m \mathbf U = \gamma m(c, \mathbf v)$$
and then we see that the relativistic momentum is simply the spatial part of the four-momentum.