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  =   r1R     and      r2i  =   r2R  .

It follows from this that

v1i  =   v1V     and      v2i  =   v2V .

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  =   v1V  , 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 = r1R), we can write them in terms of r.
To do this, we write:

r1i   =   r1R   =   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 = v1v2 (from the definition of relative velocity), so vI = v1Iv2I,  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   =   2/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  2/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.