# Inelastic collision and conservation of linear and angular momentum

Is it possible for two spheres (a & b) to have an inelastic collision with BOTH the total linear and angular momentum preserved? I'm doing some physics simulation of some spheres attracting each other like gravity and and a initial net angular momentum making them spin around a center. I want to have inelastic collisions while keeping the same total linear and angular momentum.

The simulation is in 3 dimensions. Therefore, all velocities are 3d vectors. I can calculate the velocity the particles after an completely inelastic collision by solving the following equations for $v$:

$$m_a \vec v_{a_0} + m_b \vec v_{b_0} = \vec v (m_a + m_b)$$ This gives velocity (v) zero degrees of freedom, yet I have not taken into account the formula for preservation of angular momentum (around origin):

$$m_a (\vec p_a \times \vec v_{a_0})+m_b (\vec p_b \times \vec v_{b_0}) = m_a (\vec p_a \times \vec v_{a_1})+m_b (\vec p_b \times \vec v_{b_1})$$ And for an inelastic collision $\vec v_{a_1} = \vec v_{b_1} = \vec v$ $$m_a (\vec p_a \times \vec v_{a_0})+m_b (\vec p_b \times \vec v_{b_0}) = \vec v \times (m_a \vec p_a+m_b \vec p_b)$$ Where $\vec p_a$ and $\vec p_b$ is the position vector of the spheres. The problem here is that the spheres also have a radius, so in the instant of a collision $p_a$ and $p_b$ is not equal. The only way I could see both angular and linear momentum being preserved is if the radius changes. Is this at all possible?

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The second term of the second equation is wrong IMHO. You can either express it as that of the velocities of the two masses (they are NOT the same as $\vec v$, the velocity of the center of mass, since the two masses start spin around each other) $m_a(\vec p_{a_1}\times\vec v_{a_1})+m_b(\vec p_{b_1}\times\vec v_{b_1})$ or as the sum of the angular momentum $M(\vec p \times \vec v)$ of the center of mass with the angular momentum of the two masses in the center of mass reference frame. –  Ferdinando Randisi Oct 29 '12 at 11:00
It would be a real problem if it wasn't possible, since it's not possible for either conservation law to be violated... it would proove that inelastic collisions are impossible in the first place! –  leftaroundabout Oct 29 '12 at 11:41
So $\vec v$ is not equal to $\vec v_{a_1}$ and $\vec v_{b_1}$ ?? If so, I am thoroughly confused. After a completely inelastic collision the velocity of the to spheres will be the same right? –  bofjas Oct 29 '12 at 12:08

"Is it possible for two spheres (a & b) to have an inelastic collision with BOTH the total linear and angular momentum preserved?"

More than that. It is not possible to have any collision in which they are not preserved.

However, you are not completely off-base here. Let's think about just what we mean by saying that "energy is not conserved" in a inelastic collision in the first place.

We don't actually mean that energy disappears, we mean that it disappears from the bulk kinetic terms (i.e. $\frac{1}{2} m_a v^2_{a1}$ and it's friends), and ends up in some other forms that we don't consider in our kinematics (mostly sound and heat for classroom demos).

Likewise some of the angular momentum could disappear from the bulk terms like $\vec{v}_{a1} \times m_a \vec{p}_{a1}$ into a channel that you are not writing down: the internal angular momentum of the product mass(es).

To be completely correct you should be attaching a moment of inertia and a angular velocity figure to each of your masses.

Next question for a simulation. When to almost-but-not-quite-point-masses with moments of inertia $I_{a,b}$ collide and stick what should the moment of inertia $I$ of the combined mass be and why?

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But if each of the masses are a perfect sphere and there is no friction, then all contact forces on the masses would go through the center of the sphere (the center of mass) leaving the spin unchanged? –  bofjas Oct 29 '12 at 13:19
No. In the center of mass frame for the new combined mass there is a angular moment at the moment of contact that derives from the relative velocity and the offset (from a square on collision). By construction those forces can't affect that quantity, so it is preserved as the internal angular momentum of the new mass. –  dmckee Oct 29 '12 at 13:24
I know what I did wrong. I was thinking that I didn't incorporate friction, however, the formula $m_a \vec v_{a_0} + m_b \vec v_{b_0} = \vec v (m_a + m_b)$ can only be true IF there is friction OR there is a head on collision. This means that I should remove friction from the first equation OR include interal spin in the second. Correct? –  bofjas Oct 29 '12 at 14:01
Well, "friction" is kind of a slippery concepts here. You need a mechanism to allow them to stick. This might be a combination of friction, deformation and simply stickyness. –  dmckee Oct 29 '12 at 15:20
I actually don't want them to "stick". I guess what I wanted was an inelastic collision as possible without friction. But without friction, unless there is a head on collision, the spheres will just slide of each other. In other words, the first equation is a wrong assumption from my part. –  bofjas Oct 29 '12 at 15:30

In any collision ,the quantities angular momentum, linear momentum and total energy always remains conserved. This is so for an inelastic collision also, only the kinetic energy is not conserved in case of an inelastic collision.

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