Lecture 2
Wednesday - January 11, 2006
Written or last updated: January 11, 2006
P442 – Analytical Mechanics - II
Elastic Collisions
c
Alex
R. Dzierba
Collisions - A Linear Algebra Approach
In the last lecture we found an expression relating final velocities (u1 , u2 ) to initial velocities (v1 , v2 ) for a
one-dimensional elastic collision. The relations can be expressed in matrix form:
u1
u2
=
1
m1 + m2
m1 − m2
2m1
2m2
m2 − m1
v1
v2
(1)
or more compactly as u = Mv where:
1
M=
m1 + m2
m1 − m2
2m1
2m2
m2 − m1
(2)
You can easily show that the matrix M is it own inverse, so that MM = 1. The consequence of this is that
Mu = MMv = v or −v = M(−u) which is a statement of time-reversal invariance.
This matrix has eigenvalues ±1 and it should be possible to make a similarity transformation that diagonolizes
M. The interpretation of all this is discussed in the attached paper by Yuk Ling Yung.
Collisions in the Center of Mass and Lab Frames
Consider the collision of two particles of mass m1 and m2 colliding in their center of mass system (CMS)
where the momenta of the two particles sum to zero. We will also view the same collision in the LAB frame
where before the collision m2 is at rest. Figure 1 shows the collision in the CMS. Note that we use a star
to denote CMS quantities, such as the scattering angle θ∗ . Figure 2 shows the same collision in the LAB
frame. And FIgure 3 shows how the velocity vectors in the two frames are related. We will considering only
elastic collisions so kinetic energy is conserved.
Before the collision in the LAB frame the two particles are located by vectors ~r1 and ~r2 and the center of
mass is located by:
~ = m1~r1 + m2~r2
R
m1 + m2
and the velocity of the center of mass is given by:
1
(3)
center of mass
m1 u*1
m1 v*1
θ∗
m2 v*2
m2 u*2
BEFORE
AFTER
Figure 1: In the CMS frame the net momentum before and after collisions is zero. Since the velocities before
the collision are collinear and they are also collinear after the collision only one scattering angle is required.
For an elastic collision v1∗ = u∗1 and v2∗ = u∗2 .
lab frame
m1 u1
m1 v1
θ1
θ2
m2 v2 = 0
AFTER
BEFORE
m2 u2
Figure 2: The same collision as above viewed in the LAB frame.
center of mass
lab frame
u*1
u1
u1sinθ1 = u*1sinθ∗
u*1sinθ∗
θ∗
θ1
u*1cosθ∗
u1cosθ1 = u*1cosθ∗ + V
Figure 3: Relation between the CMS and LAB frames.
~ =
V
m1~v1
m1 + m2
(4)
which follows since m2 is at rest in the LAB.
In the CMS the momenta of the particles before the collision sum to zero and this is also the case after the
collision. If the collision is elastic the before and after velocities are the same. So the collision only changes
the angle in the CMS.
You can see from Figure 3 that you can go from the CMS to the LAB by simply adding the velocity vector
~ , to the CMS velocities. We can see that:
of the CMS, V
tan θ1 =
u∗1
u∗1 sin θ∗
cos θ∗ + V
2
(5)
Dividing through by u∗1 and making use of the fact that u∗1 = v1∗ :
tan θ1 =
sin θ∗
cos θ∗ + V /v1∗
(6)
Now you should realize that before the scattering the velocities of m1 in the CMS and LAB frames are
related like so: v1 = v1∗ + V . Using this in equation 4 you can show that: V /v1∗ = m1 /m2 and so:
tan θ1 =
sin θ∗
cos θ∗ + m1 /m2
(7)
Now recall that in the CMS any angle θ∗ is allowed. What about in the LAB for θ1 ? You can see from
equation 7 that the denominator is never zero if m1 > m2 so θ1 < π/2 in the LAB. See the plots on the next
page.
3
2.5
2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
3
Figure 4: Relation between θ1 (vertical axis) versus θ∗ (horizontal axis) for m1 /m2 equal to 0.5 (highest
curve), 1.0 (next lowest) and 2.0 (lowest).
3