4.7: Elastic Collisions in 2-3 Dimensions

As we can see, elastic collisions in one dimension are “good” because we can completely solve them using only kinematics – we don’t care about the details of the interaction between the colliding entities; we can find the final state from the initial state for all possible elastic forces and the only di erences that will depend on the forces will be things like how long it takes for the collision to occur.

In 2+ dimensions we at the very least have to work much harder to solve the problem. We will no longer be able to use nothing but vector momentum conservation and energy conservation to solve the problem independent of most of the details of the interaction. In two dimensions we have to solve for four outgoing components of velocity (or momentum), but we only have conservation equations for two components of momentum and kinetic energy. Three equations, four unknowns means that the problem is indeterminate unless we are told at least one more thing about the final state, such as one of the components of the velocity or momentum of one of the outgoing masses. In three dimensions it is even worse – we must solve for six outgoing components of velocity/momentum but have only four conservation equations (three momentum, one energy) and need at least two additional pieces of information. Kinematics alone is simply insu cient to solve the scattering problem – need to know the details of the potential/force of interaction and solve the equations of motion for the scattering in order to predict the final/outgoing state from a knowledge of the initial/incoming state.

The dependence of the outoing scattering on the interaction is good and bad. The good thing is that we can learn things about the interaction from the results of a collision experiment (in one dimension, note well, our answers didn’t depend on the interaction force so we learn nothing at all about that force aside from the fact that it is elastic from scattering data). The bad is that for the most part the algebra and calculus involved in solving multidimensional collisions is well beyond the scope of this course. Physics majors, and perhaps a few other select individuals in other majors or professions, will have to sweat blood later to work all this out for a tiny handful of interaction potentials where the problem is analytically solvable, but not yet!

Still, there are a few things that are within the scope of the course, at least for majors. These involve learning a bit about how to set up a good coordinate frame for the scattering, and how to treat “hard sphere” elastic collisions which turn out to be two dimensional, and hence solvable from kinematics plus a single assumption about recoil direction in at least some simple cases. Let’s look at scattering in two dimensions in the case where the target particle is at rest and the outgoing particles lie (necessarily) in a plane.

We expect both energy and momentum to be conserved in any elastic collision. This gives us the following set of equations:

for kinetic energy conservation.

We have four equations, and four unknowns, so we might hope to be able to solve it quite generally. However, we don’t really have that many equations – if we assume that the scattering plane is the x − y plane, then necessarily p0z = p1z = p2z = 0 and this equation tells us nothing useful. We need more information in order to be able to solve the problem.

Let’s see what we can tell in this case. Examine figure 55. Note that we have introduced two angles: θ and φ for the incident and target particle’s outgoing angle with respect to the incident

θ p

m1 1x

p p

1 1y

Figure 55: The geometry for an elastic collision in a two-dimensional plane.

direction. Using them and setting p0y = p0z = 0 (and assuming that the target is at rest initially and has no momentum at all initially) we get:

In other words, the momentum in the x-direction is conserved, and the momentum in the y- direction (after the collision) cancels. The latter is a powerful relation – if we know the y-momentum of one of the outgoing particles, we know the other. If we know the magnitudes/energies of both, we know an important relation between their angles.

This, however, puts us no closer to being able to solve the general problem (although it does help with a special case that is on your homework). To make real progress, it is necessarily to once again change to the center of mass reference frame by subtracting ~vcm from the velocity of both particles. We can easily do this:

Since p′i1 = p′i2 = p′i (the magnitudes are equal) we can simplify this a bit further:

collision. All that can change, then, is the direction of the incident momentum in the center of mass frame. In addition, since the total momentum in the center of mass frame is by definition zero before and after the collision, if we know the direction of either particle after the collision in the center of mass frame, the other is the opposite:

p~′f 1 = −p~′f 2 (445)

We have then “solved” the collision as much as it can be solved. We cannot uniquely predict the direction of the final momentum of either particle in the center of mass (or any other) frame without knowing more about the interaction and e.g. the incident impact parameter. We can predict the