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Collision Problems
2-body
R & r coordinates
We have already defined the center of mass coordinate, R, as:
For a two-body problem, this becomes:
R = Σmiri / Σmi .
R = (m1r1 + m2r2) / (m1 + m2) .
If we take the second derivative of each side, we get:
R’’ = (m1r1’’ + m2r2’’) / (m1 + m2), or
MR’’ = (m1r1’’ + m2r2’’) ,
where M = m1 + m2 .
We will now define the relative coordinate, r, as:
r = r1 – r2 .
To see how these coordinates are useful, we start with Newton’s Second Law applied to
each particle:
F1internal + F1external = m1r1’’
F2internal + F2external = m2r2’’
Where F1internal is the internal force on particle 1 due to particle 2, and F1external is the sum
of all external forces on 1 (not including the force from particle 2).
If we now add the two equations of motion above, we get:
F1internal + F1external + F2internal + F2external = m1r1’’ + m2r2’’ .
We now note that Newton’s Third Law of Motion says that F1internal = - F2internal , and
we recall that (m1r1’’ + m2r2’’) = MR’’. This simplifies the above equation to become:
F1external + F2external = MR’’ .
***
This says that the external forces cause the center of mass to accelerate, but that the
internal forces do not.
To get an equation for the relative coordinate, r, we need to multiply the equation of
motion for m1 by m2, and multiply the equation of motion for m2 by m1. We can then
subtract the two equations to get:
m2F1internal + m2F1external – m1F2internal - m1F2external = m1m2r1’’ - m1m2r2’’ .
We again use Newton’s Third Law, F1internal = - F2internal , and if we can assume that
m2F1external ≈ m1F2external , then we get:
(m1 + m2) F1internal = m1m2 (r1’’ – r2’’) = m1m2r’’ .
If we further define a quantity we will call the reduced mass as: μ = m1m2/(m1+m2) ,
we get:
F1internal = μr’’ .
***
This indicates that the motion of particle 1 as seen from particle 2 (that is what the vector,
r, indicates) is the same as if particle 2 were stationary and particle 1 had the reduced
mass, μ, instead of its regular mass, m1. This is just a single particle equation of motion,
and we can apply all the techniques we learned in parts 1 and 2 to this type of problem.
Please remember that to obtain the above equation we had to assume that
m2F1external ≈ m1F2external .
If we are able to solve the problem using the Rr system, we should be able to go back to
the inertial system, r1 and r2, to get a solution in this system. To do this, we need to
develop the inverse transformation equations. We start with our definitions of R and r:
R = (m1r1 + m2r2) / (m1 + m2)
and
r = r1 – r2 .
Knowing R and r, we have two equations for our two unknowns: r1 and r2 .
Start with
r2 = r1 – r,
and use this in the R equation to get:
R = {m1r1 + m2(r1 – r)}/(m1+m2) = r1 – m2r/(m1+m2),
or
r1 = R + m2r/(m1+m2),
or
r1 = R + (μ/m1)r,
In a similar way, we get
r2 = R – m1r/(m1+m2), or
r2 = R – (μ/m2)r .
You should now be able to do Homework #26 (see below) which asks how KE, L, and p
look when expressed in the Rr coordinates. From this homework problem the kinetic
energy is
KE = ½m1v12 + ½m2v22 = ½μv2 + ½MV2 .
If there are no external forces, V is constant and so the only energy available in an
inelastic collision between m1 and m2 is the ½μv2 amount.
Homework Problem #26:
a) Starting from KE = ½ m1 v1² + ½ m2 v2² show that KE = ½ M V² + ½  v²
where M is the total mass and  is the reduced mass; V is the
center-of-mass velocity and v is the relative velocity.
b)
Starting from L = m1 (r1 x v1) + m2 (r2 x v2) show that L = MR x V + r x v.
c)
Starting from p = m1 v1 + m2 v2 show that p = M V.
Homework Problem #27: In part 2, the period of orbit, , for an inverse square law
force was expressed by the relation: 2 = 42a3m/K where a is the semimajor
axis distance and K for gravity is GMm. (This is a generalization of one of
Keplar’s laws.) This applies as long as m is much smaller than M. Work out a
correction for the case where M = m.
Scattering
From Part 2, in the section on Scattering from an inverse square law force, we obtained
the expression:
tan(Θ/2) = [mK2/2EL2]½ = [mK2/{2(½mvo2)( m2vo2s2)}] = K/mvo2s .
Since the equation of motion for the relative position, r, is the same as for the above case
with the one change that the m of the particle becomes the reduced mass, μ =
m1m2/(m1+m2), we can see that the relative velocity, r’ = v, will be deflected in the
scattering by an angle, Θ, where Θ can be determined from the relation:
tan(Θ/2) = K/μvI2s .