An example I was thinking of is two balls of equal mass M, where the first ball has an initial velocity of v_1 m/s, and the second ball has 0 velocity. The final velocities are v_2 and v_3 respectively. By conservation of momentum, $$Mv_1 = Mv_2 + Mv_3\,,$$ so $$v_1 = v_2 + v_3$$ and by conservation of energy, $$\frac{1}{2} M v_1^2 = \frac{1}{2} M v_2^2 + \frac{1}{2} M v_3^2\,,$$ so $$ v_1^2 = v_2^2 + v_3^2\,. $$ Based on these two equations, it seems like we have 2 possible solutions: $$ v_1 = v_2, v_3 = 0 $$ and $$ v_1 = v_3, v_2 = 0 $$ Logically we know that the first solution would be valid before a collision/if a collision never occurs, and the second solution is valid after a collision, but does that logic rely on some underlying physical law that defines the behavior of the system? Like, how do we "know" that the first ball doesn't just pass through the second and keeps going at its original velocity, leaving the second ball stationary, as the first solution might suggest? I'm not sure if this question makes sense or if I'm just poking at common sense, but I just want to better understand the deterministic behavior of an elastic collision more generally.
4 Answers
In 1D, there is only head-on collision that can happen, and then the solution is unique, provided you disallow passing through each other. Otherwise, both passing through solution and reflected solution can work together.
From 2D onwards, you can have glancing collisions, and so the solution is not fixed. It is fixed up to that, though.
To fully express the system's equations of motion, you will also need the initial positions, any non-collision forces (e.g. friction to make colliding balls roll instead of slide), and the shape and initial orientation and angular velocities of the colliding bodies. Initial position and velocity tell you where and if there will be a collision. Shape and rotation tell you how much translational kinetic energy will end up as rotational kinetic energy or vice versa.
No, it doesn't, unless your motion is one-dimensional. By conservation of momentum, we have 3 equations in 3D (since momentum is a vector), and conservation of energy gives us one other equation. That is 4 equations for 6 quantities (both velocities that are vectors). So you need other information. For rigid bodies, consevation of angular momentum often does the trick. In 2D, you have 3 equations for 4 quantities, so that's not sufficient either.
Here is a concrete example: You have to balls that come from opposite directions, have equal mass and speed and then collide. By the conservation of momentum you know that the velocity of one ball is exactly minus the velocity of the other ball ($\vec v_1 = - \vec v_2$), so they move with the same speed but in opposite directions. The conservation of energy tells you that the speed they have now is the same they had before colliding (say that was $V_0$, then conservation of energy gives you $\frac{1}{2} m (V_0^2 + V_0^2) = \frac{1}{2} (v_1^2 + v_2^2) = \frac{1}{2} (v_1^2 + v_1^2) \Rightarrow v_1 = V_0$). But only by those conservation laws you don't know the direction of the balls, you only know that they move in opposite directions and you know their speed.
the collision equations for 1D case are:
conservation of the linear momentum
$$m_1\,v_1+m_2\,v_2=m_1\,u_1+m_2\,u_2=\text{constant}\tag 1$$
conservation of the energy
$$\frac 12\left(m_1\,v_1^2+m_2\,v_2^2\right)=\frac 12\left(m_1\,u_1^2+m_2\,u_2^2\right)=\text{constant}\tag 2$$
where $~u_i~$ are the velocities bevor the collision and $~v_i~$ are the velocities after the collision.
the solutions $~v_1=u_1~,v_2=u_2~$ are " trivial solution $~(\text{const.}=\text{const.})~$ thus, this is not the solutions that we are looking for.
the solutions
$$v_1=\frac{1}{m_1+m2}\left((m_1-m_2)\,u_1+2\,m_2\,u_2\right)$$
$$v_2=\frac{1}{m_1+m2}\left((m_2-m_1)\,u_2+2\,m_1\,u_1\right)$$
for $~m_1=M~,m_2=M~$ you obtain $~v_1=u2=0~,v_2=u_1~$
