Center of Mass Coordinates
In this section we introduce
a third coordinate system to add to the two we have already looked at:
the lab (or inertial system, r1
and r2) and
the Rr (R is the center of mass
coordinate and r is the relative
coordinate).
For this third system, we define the center of mass
coordinates, r1i
and r2i, as:
r1i = r1 – R and
r2i = r2 – R .
It follows from this that
v1i = v1 – V and
v2i = v2 – V
.
In this particular section,
we will limit our lab or inertial system to the special case where the second particle,
m2, is initially at rest in the lab: v2I
= 0.
We will use the subscripts I for initial and F for final in all of the
coordinate systems.
We will also assume that the external
forces are zero, and so the center of mass velocity, V, is constant.
With this, we can then get a nice expression for V
(see the next paragraph).
Since R = (m1r1+m2r2)/(m1+m2),
V = (m1v1+m2v2)/(m1+m2) = m1v1I /(m1+m2) since V
is constant and v2I = 0.
Also, recall that μ = m1m2/(m1+m2). This leads to:
V =
(μ/m2)v1I .
Note that v1I and V
are in the same direction.
Also, from the definition of v1i = v1 – V , we see that v1iI
is also in that same direction.
Also in this section we will consider only elastic collisions.
This means that vI = vF, that is, the magnitude of the relative velocity finally is the same as it was
initially.
This comes from the fact that for initial and final values, KE = ½MV2
+ ½μv2 = constant in an elastic collision, and V is constant in our case.)
From the previous section, we know that for an inverse square law force, the direction change for the relative
velocity is Θ, where
tan(Θ/2) = K/μvI2s .
Instead of writing the center
of mass coordinates in terms of R (for example, r1i
= r1–R), we can write them in terms of r.
To do
this, we write:
r1i = r1 – R = r1 –
(m1r1+m2r2)/(m1+m2)
= {(m1+m2)r1 - (m1r1+m2r2)} / (m1+m2) =
m2(r1-r2) /(m1+m2) = (μ/m1)r
.
Similarly, r2i = -(μ/m2)r
.
This also means that v1i = (μ/m1)v and v2i = -(μ/m2)v .
Since the magnitude of v for initial and
final doesn’t change for elastic scattering, the magnitudes of v1i
and v2i for initial
and final values also don’t change.
Since the direction of v changes by the
angle Θ, the directions of v1i
and v2i also both
change by the angle, Θ.
Below is a diagram showing the initial and final velocities in all three
systems.
As mentioned on the previous
page, the magnitudes of v1iI
and v1iF
are the same, and their direction changes by Θ.
The center of mass velocity, V, is a
constant.
We also see from the diagram that (in the lab frame) the magnitude of v1F is less than that of v1I .
This should not be surprising, since v2I
is zero, and after the collision m2 should have some velocity and so
should take some kinetic energy away from the incoming particle, m1. We also see from the diagram that the angle
between v1F and v1I, which we call θ1, should be less than the
angle, Θ.
We now try to determine expressions for the magnitude of v1F and the angle, θ1.
From the diagram, we see
that:
v1Fx
= v1iFx
+ Vx = v1iF
cos(Θ) + V , and
v1Fy
= v1iFy
+ Vy = v1iF
sin(Θ) + 0 .
Therefore, we get:
tan(θ1)
= v1Fy / v1Fx = v1iF
sin(Θ) / {v1iF cos(Θ) + V}.
Now, recall from the previous
page that v1i = (μ/m1)v and V = (μ/m2)v1I , so that
tan(θ1)
= (μ/m1)vF
sin(Θ) / {(μ/m1)vF cos(Θ) + (μ/m2)v1I
(m1/m1)(vF/vF)}
= sin(Θ) / {cos(Θ) + (m1/m2)(v1I/vF)}.
We can further simplify
this.
Recall that v = v1 – v2
(from the definition of relative velocity), so vI
= v1I – v2I, and that v2I = 0 (m2 is initially stationary in the
lab frame), and so vI = v1I .
Further, for the elastic scattering we are considering, the magnitudes of vI and
vF are the same (v1I
= vI = vF).
Therefore, we get:
tan(θ1)
= sin(Θ) / {cos(Θ) + (m1/m2)}. ***
Special Cases: tan(θ1) =
sin(Θ) / {cos(Θ) + (m1/m2)}.
If m1>>m2, then θ1 ≈ 0 (as expected, since m1 will
simply push the lighter mass away).
If m1<<m2, then θ1 ≈ Θ (as expected, since m2 will act
essentially as a fixed point).
If m1 = m2,
then we can use some trig identities:
sin(θ) =
sin(θ/2 + θ/2) = 2 sin(θ/2)cos(θ/2)
cos(θ) = cos(θ/2 + θ/2) = cos2(θ/2)
– sin2(θ/2) = cos2(θ/2) – {1-cos2(θ/2)}, or
cos(θ) + 1
= 2cos2(θ/2)
to get:
tan(θ1)
= sin(Θ) / {cos(Θ) + (m1/m2)} =
sin(Θ) / {cos(Θ) + 1}
= {2
sin(Θ/2)cos(Θ/2)} / {2cos2(Θ/2)} =
sin(Θ/2) / cos(Θ/2)
=
tan(Θ/2) ; or
θ1
= Θ/2 (which splits the difference
between the two extremes).
Speed of incoming particle
To see how the speed of the
incoming particle changes in the lab frame, we start with the square of the
final speed in the lab frame (and use expressions we derived on the previous
page when looking at the angles):
v1F2
= v1Fx2 +
v1Fy2 = {v1iF
cos(Θ) + V}2 + {v1iF
sin(Θ)}2
= v1iF2
cos2(Θ) + 2 v1iF V cos(Θ)
+ V2 + v1iF2 sin2(Θ)
= v1iF2
+ V2 + 2 v1iF V cos(Θ) .
We now recall that v1i = (μ/m1)v and V =
(μ/m2)v1I to get
v1F2
= (μ/m1)2vF2
+ (μ/m2)2
v1I 2 + 2(μ/m1)vF
(μ/m2)v1I cos(Θ) .
We also recall (from the
previous page) that v1I = vI
= vF so that we get:
v1F2
= (μ/m1)2vF2
+ (μ/m2)2 vF
2 + 2(μ/m1)vF (μ/m2)vF
cos(Θ)
= (μ2/m1m2) {m2/m1
+ m1/m2 + 2 cos(Θ)} v1I2 ***
Special Cases: v1F2 =
(μ2/m1m2) {m2/m1
+ m1/m2 + 2 cos(Θ)} v1I2
If m1>>m2, then
μ
= m1m2/(m1+m2) ≈ m1m2/m1
= m2 , so (μ2/m1m2)
≈ (m2/m1)
and this leads to
v1F2 = (m2/m1)
{m2/m1 + m1/m2 + 2 cos(Θ)} v1I2
= {(m2/m1)2 +
1 + 2(m2/m1)cos(Θ) } v1I2
,
but (m2/m1) ≈ 0, so v1F2 ≈
v1I2 , and
hence v1F ≈
v1I (and θ1
≈ 0).
This should be expected, since the heavy m1 will
push the very light particle out of the way without much loss of speed (or
change in direction).
If m1<<m2, then
μ
= m1m2/(m1+m2) ≈ m1m2/m2
= m1 , so (μ2/m1m2)
≈ (m1/m2)
and this leads to
v1F2 = (m1/m2)
{m2/m1 + m1/m2 + 2 cos(Θ)} v1I2
= {1 + (m1/m2)2
+ 2(m1/m2)cos(Θ) } v1I2
,
but (m1/m2) ≈ 0, so v1F2 ≈
v1I2 , and
hence v1F ≈
v1I (and θ1
≈ Θ).
This should also be expected, since the light m1
will bounce off of the heavy m2 which will remain essentially
stationary.
If m1 = m2 (recall that for this case θ1
= Θ/2), then (with m1 =
m2 = m)
μ = m1m2/(m1+m2)
= m/2, and so (μ2/m1m2)
= ¼, and this leads to
v1F2 =
(1/4) {1+ 1 + 2 cos(Θ)} v1I2 =
½{1 + cos(Θ)} v1I2 .
If
Θ=0, then v1F = v1I
(and θ1=0). This
corresponds to a miss.
If
Θ=180o , then v1F = 0. This corresponds to a direct hit in which m2
continues on with the initial speed of m1, and m1 remains
where m2 initially was.
Application:
If a satellite flies by a
planet, can the fly-by be considered a “collision”? If so, can the satellite pick up speed in
that “collision”?
Since the satellite comes in
with an initial velocity and goes out with a final velocity at an angle
different from the incoming angle, and if we assume the satellite does not lose
energy by friction with the planet’s atmosphere, then we can assume that there
is an elastic collision of the satellite with the planet. And since the satellite has a mass much
smaller than the mass of the planet, we can use the results above for m1/m2 ≈
0, so in the lab frame of the planet (m2) not moving, v1F
= v1I and q1
= Θ (where the sub 1 refers to the
satellite). In this case, the satellite
comes in and goes out with the same speed (even though it will speed up as it
approaches the planet, it will slow down an equal amount as it recedes from the
planet).
However, if the planet is
moving (as they all do around the sun), then we need to change coordinates from
the one with m2 being stationary to m2 moving with speed
v2* in the q2* direction. This requires that we add a speed of v2*
cos(q2*) to the x components of both v1I and v1F
and add a speed of v2* sin(q2*) to the
y components of both v1I and v1F . Recall that initially, v1I is
moving only in the x direction and finally v1F is moving in the q1 = Θ direction.
For the initial speed in the
new (*) frame:
v1Ix* = v1I + v2*
cos(q2*) and v1Iy* = v2*
sin(q2*)
so that
v1I*2
= {v1I + v2*
cos(q2*)}2
+ {v2* sin(q2*)}2 = v1I2
+ v2*2 + 2v1I v2* cos(q2*) .
For the final speed:
v1Fx* = v1Fx + v2*
cos(q2*) = v1I cos(Q) + v2* cos(q2*) and
v1Fy* = v1Fy + v2*
sin(q2*) = v1I sin(Q) + v2*
sin(q2*)
so that
v1F*2 = {v1I cos(Q) + v2* cos(q2*)}2 + {v1I
sin(Q) + v2*
sin(q2*)}2
= v1I2 + v2*2
+ 2v1I v2* cos(Q) cos(q2*) + 2v1I v2* sin(Q) sin(q2*) .
We can use the trig
identity: cos(Q-q2*) = cos(Q) cos(q2*) + sin(Q) sin(q2*) to get:
v1F*2
= v1I2 + v2*2
+ 2v1I v2* cos(Q-q2*) .
To see if we pick up or lose
speed, we can subtract the v1F*2 from the v1I*2 :
v1F*2 - v1I*2 = {v1I2
+ v2*2 + 2v1I v2* cos(Q-q2*) } – {v1I2
+ v2*2 + 2v1I v2* cos(q2*) } = 2v1I
v2* {cos(Q-q2*) - cos(q2*) }.
From this we can see that it
is possible to pick up speed and to lose speed in a “collision” with a
planet. What is the best angle to pick
up the maximum speed? How do we
determine a maximum? We do it be taking
the derivative with respect to the variable we want to maximize and setting it
equal to zero:
d[2v1I v2*
{cos(Q-q2*) - cos(q2*) }]/dq2* = 0 gives
us the condition:
+ sin(Q-q2*) + sin(q2*) = 0
which gives us the condition:
sin(q2*) = - sin(Q-q2*) = sin(Q-q2* +/- 180o)
which gives us the condition:
q2* = (Q-q2* +/- 180o
) , or q2* = (Q/2) +/- 90o for either
a maximum or minimum difference in final versus initial speed.
To see what angle gives the
maximum speed versus the minimum speed, we take the second derivative and
evaluate it at the equilibrium values and see if it is negative (for a maximum)
or positive (for a minimum):
d2[2v1I v2*
{cos(Q-q2*) - cos(q2*) }]/dq2*2 = d[2v1I v2*
{sin(Q-q2*) + sin(q2*)} ]/dq2* = 2v1I v2* [-cos(Q-q2*) + cos(q2*)] .
Evaluating this at the
equilibrium values of: q2* = (Q/2) +/- 90o gives (after taking away the positive
constant of 2v1I v2* ):
-cos(Q-q2*) + cos(q2*) =
-cos(Q-{(Q/2) +/- 90o}) + cos{(Q/2) +/- 90o } .
Note that cos(a +/- 90o) = -/+ sin(a) so that we get:
-cos(Q-{(Q/2) +/- 90o}) + cos{(Q/2) +/- 90o } = -cos{(Q/2) -/+ 90o}
+ cos{(Q/2) +/- 90o }
= -/+ sin(Q/2) -/+ sin(Q/2) .
For the + case of q2* = (Q/2) + 90o we get a
negative value, so we get a maximum increase in speed; and for the case of q2* = (Q/2) - 90o , which is
opposite in direction of the max case, we get a minimum which means a maximum
decrease in speed.
If you look at a diagram of
this, the satellite has to come in approaching an oncoming planet to pick up
speed, and it has to appproach a receding planet to reduce speed. This is just like a baseball approaching a
bat swung at the ball will leave the bat going faster than when it came in.
