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of uncoupled equations involving R and r . This is similar to the approach used
to study two-body collisions in Section 3.3. The non-acceleration of the centre of
mass is a consequence of momentum conservation for an isolated system, i.e.
m 1 ¨
x 1 +
m 2 ¨
x 2 =
0
(9.30)
and we look for a point R which does not accelerate, i.e. the centre of mass satisfies
the equation
m 2 ) R
(m 1 +
=
0 .
(9.31)
Equations (9.30) and (9.31) have the solution
m 1 x 1 +
m 2 x 2
R
=
.
(9.32)
m 1 +
m 2
Subtracting the two equation in Eq. (9.29) yields
1
m 1 +
F ( r ).
1
m 2
¨
r
=
(9.33)
Notice that equations (9.32) and (9.33) are entirely equivalent to the pair of
equations (9.29) but have the virtue that they are decoupled from each other. Our
attention is now focussed upon Eq. (9.33) and once we have solved it we will have
solved the general motion of our two particles since
m 2
m 1 +
x 1 =
R
m 2 r ,
m 1
m 1 +
x 2 =
R
+
m 2 r .
(9.34)
We shall conclude this section by noting that Eq. (9.33) can be written in the form
F ( r )
=
µ
r ,
¨
(9.35)
where
m 1 m 2
m 1 +
µ
=
(9.36)
m 2
is none other than the 'reduced mass' of the system that we encountered in Section
3.3.1. This way of writing the equation of motion makes explicit the fact that the
motion of this two body system is mathematically equivalent to the motion of a
single particle of mass µ under the action of the force F . That is, the problem
in hand reduces to a problem whose mathematical analysis is exactly the same
as the analysis of the motion of a single particle. Of course we should always
remember that there are really two particles and that we are solving for their
relative position.
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