Last section we studied head on
collisions, in which both objects move on a line. Most natural collisions,
however, are not head on, instead causing objects to move at an angle to their
original trajectory. Consider a game of pool, in which balls are frequently hit
at an angle to get them in the pockets. These kinds of collisions, though more
complicated, can be solved using the same methods as those used in one
dimension. An elastic collision still conserves kinetic energy and, of course,
any collision conserves linear momentum. We shall examine the elastic and
completely inelastic case, and show how each of these cases can be solved.
Elastic Collisions in Two Dimensions
Since the theory behind solving two dimensional collisions problems is the same
as the one dimensional case, we will simply take a general example of a two
dimensional collision, and show how to solve it. Consider two particles, m1
and m2, moving toward each other with velocity v1o and v2o,
respectively. They hit in an elastic collision at an angle, and both particles
travel off at an angle to their original displacement, as shown below:
Figure %: Two particles collide at point A, then move of at angles to their
original motion
To solve this problem we again use our conservation laws to come up with
equations that we hope to be able to solve. In terms of kinetic energy, since
energy is a scalar quantity, we need not take direction into account, and may
simply state:
v1o2 + v2o2 = v1f2 + v2f2
Whereas in the one dimensional problem we could only generate one equation for
the conservation of linear momentum, in two dimensional problems we can generate
two equations: one for the x-component and one for the y-component.
Let's start with the x-component. Our initial momentum in the x direction is
given by: m1v1o - m2v2o. Note the minus sign, as the two particles are
moving in opposite directions. After the collision, each particle maintains a
component of their velocity in the x direction, which can be calculated using
trigonometry. Thus our equation for the conservation of linear momentum in the
x-direction is:
pox
=
pfx
m1v1o - m2v2o
=
m1v1fcosθ1 + m2v2fcosθ2
Regarding the y-component, since both particles move initially in the x
direction, there is no initial linear momentum in the y direction. The final
linear momentum again can be found through trigonometry, and used to form
another equation:
poy
=
pfy
0
=
m1v1fsinθ1 + m2v2fsinθ2
We now have three equations: conservation of kinetic energy, and conservation of
momentum in both the x and y directions. With this information, is this problem
solvable? Recall that if we are given only the initial masses and velocities we
are working with four unknowns: v1f, v2f, θ1 and θ2.
We cannot solve for four unknowns with three equations, and must specify an
additional variable. Perhaps we are trying to make a pool shot, and can tell
the angle of the ball being hit by where the hole is, but would like to know
where the cue ball will end up. This equation would be solvable, since with the
angle the ball will take to hit the pocket we have specified another variable.
Completely Inelastic Collisions
Surprisingly enough, the completely inelastic case is easier to solve in two
dimensions than the completely elastic one. To see why, we shall examine a
general example of a completely inelastic collision. As we've done previously,
we will count equations and variables and show that it is solvable.
The most general case of a completely inelastic collision is two particles m1
and m2 moving at an angle of θ1 to each other with velocities v1
and v2, respectively. They undergo a completely inelastic collision, and
form a single mass M with velocity vf, as shown below.
Figure %: Two particles collide at point A, Forming a single particle
What equations can we come up with to solve this type of problem? Clearly
because the collision is inelastic we cannot invoke the conservation of energy.
Instead we are limited to our two equations for conservation of linear momentum.
Observe that we have conveniently oriented our axes in the figure above such
that the path of m1 is entirely in the x direction. With this in mind, we
can generate our equations for the conservation of momentum in both the x and y
directions:
x component:
m1v1 + m2v2cosθ1 =
Mvfcosθ2
y component:
m2v2sinθ1 =
Mvfsinθ2
Though we only have two equations, we also only have two unknowns, vf
andθ2. Thus we can solve for any completely inelastic collision in two
dimensions.
Conclusion
Our entire study of collision can be seen as simply an application of the
conservation of linear momentum. So much time is spent on this topic, however,
because it is such a common one, both in physics and in practical life.
Collisions occur in particle physics, pool halls, car accidents, sports, and
just about anything else you can think of. A thorough study of the topic will
be well rewarded in practical use.
v1o2 + 
