Energy and momentum conservation of two balls

Question

I was wondering what principles (axiom or laws?) of physics is broken in classical mechanical sense.

Say in a 1-dimensional world, I have ball 1 (denoted as $B_1$) and ball 2 (denoted as $B_2$) in an ideal physics world. Initially, $B_1$ has rest velocity ($v_1 = 0$) and $B_2$ has velocity $-v$ moving towards $B_1$ from the right. Let $v_1', v_2'$ denote the velocity of $B_1, B_2$ respectively, after the collision.

Momentum and energy are the invariants:

$$\text{Momentum: } p_1 + p_2 = -mv \implies p_1' + p_2' = mv_1' + mv_2' = -mv$$ $$\text{Energy: } E_1 + E_2 = \frac{1}{2}mv^2 \implies E_1' + E_2' = \frac{1}{2}mv^2$$

Working out maths:

\begin{align*} \frac{1}{2}mv^2 = E_1' + E_2' &= \frac{1}{2}m(v_1'^2 + v_2'^2) \\ &= \frac{1}{2}m([-(v+v_2)]^2 + v_2'^2) \\ &= \frac{1}{2}m(v^2+2vv'_2+v_2'^2 + v_2'^2)\\ &= \frac{1}{2}m(v^2+2v_2'^2+2vv_2') \end{align*}

Hence, we are left with: $$0 = v_2'(v_2' + v)$$ meaning our new velocity $v_2'$ is either $0$ or $-v$.

If I say $v_2' = 0$, then everything is put in a perfect harmony: all momentum from $B_2$ is transferred to $B_1$ while conservering the energy. Perfect. However, if I claim $v_2' = -v$, then it means that $B_2$ somehow passed through $B_1$ and keeps going.

Question 1. For the second scenario in which $B_2$ passes through $B_1$, both momentum and energy is conserved. What principles are violated? I could sound dumb, but it means that the collision has been ignored. On a flip side however, suppose we lose the information about the system at the moment of collision. Hence, we do not know whether collision would happen a priori, and only know that $B_2$ was moving at velocity $-v$ before the missing event and $B_2$ is still moving at velocity $-v$ after the missing event. Based on the given information, I need to determine whether there is an inconsistency in the system or not. In this case, both momentum and energy are conserved and how would I logically deduce that the seemingly absurd event "$B_2$ went through $B_1$" cannot happen?

Question 2. In a non-perfect physics world, at the moment of the collision, $B_2$ either is knocked off slightly to the positive direction (right) or nudges slightly to the negative direction (left). Why does this happen in contrast to $B_2$ completely stopping in the ideal world?

Answer 1

1. How $B_1$ and $B_2$ interact is not determined purely by the equations for conservation of energy or conservation of momentum. Perhaps they are two non-interacting particles that really do just pass through each other. This is why those two equations alone do not determine a single solution. You have to specify some additional knowledge of the collision interaction. Of course, the only two interactions that preserve both momentum and energy are: perfectly elastic and no interaction at all.

2. In a non-ideal system the masses are rarely truly identical, and some kinetic energy is lost to internal energy in the collision, such as to heat. With those effects, the equations you have used are only approximate, and so the ideal system solution of all of the momentum of $B_2$ being transferred to B1 is also only approximate. That leaves $B_2$ with a non-zero momentum.

Answer 2

The answer to question (1) seems straightforward. The equations you write are not just valid at the moment of collision - they are valid at all times. If B1 and B2 are in contact, then the $v_2 = 0$ solution is the appropriate choice. If not, then the $v_2 = -v$ solution is more appropriate.

Often in simulation, we have to make a choice about which bodies are allowed to collide or be in contact and which ones aren't. I don't know if it's a law like the laws of motion but common sense dictates that two rigid bodies of finite volume and density cannot occupy the same space at the same time. Unless you decide that they can in your problem definition. Sort of like deciding on what the ball material is.

As an aside, there are several different tiers of "laws" that are taught in undergraduate mechanics and the differences are not always appreciated. Conservation of momentum and angular momentum are fundamental and extremely accurate for non-relativistic everyday applications. Equations for material like springs and dampers $F = kx$ etc. or $\sigma = \epsilon E$ for mechanical stress and strain are known as "Constitutive Laws" and are also pretty accurate for describing the macroscopic behavior of materials. Then there are the lowest level of "laws" like $F = \mu N$ which describe friction and other material interaction phenomena. These laws are convenient but could be described as over-simplifications. They are at best accurate to maybe +/- 10% and can be wildly wrong in some cases. But they are usually presented alongside the first two classes as if they are all equally fundamental.

As to the second question I am not certain. I would imagine that this non-ideal result comes from the effect of friction and the fact that there is some tiny but finite compression and contact area at the moment of collision. If the moving ball has angular momentum (i.e. rolling), there could be some small rotation aspect to the collision. There are also residual compression waves in each ball after the collision that may have some small effect.