Environmental Engineering Reference
In-Depth Information
“frames” here because so far we did not specify the location of the origin, so there
exist an infinite number of zero momentum frames all moving parallel to each other
and differing only in the location of the origin. We are at liberty to specify the
location of the origin and shall usually choose it to be the position of the centre of
mass
R
and we refer to this particular zero-momentum frame as the centre-of-mass
frame (although very often the two are used synonymously). While the distinction
is not important for the discussions in this chapter, it will be so in Chapter 4 when
we come to consider rotating bodies and will sometimes want to insist that the
origin is coincident with the position of the centre of mass.
We can compute the momenta of the colliding particles in the centre-of-mass
frame as follows:
p
1
c
=
m
1
v
1
c
m
1
v
1
−
m
1
v
1
+
m
2
v
2
=
m
1
+
m
2
m
1
m
2
m
1
+
=
m
2
(
v
1
−
v
2
).
(3.36)
The result is a product of two terms, the reduced mass
m
1
m
2
m
1
+
µ
=
m
2
,
(3.37)
and the relative velocity,
v
=
v
1
−
v
2
.
(3.38)
Both
µ
and
v
feature heavily in two-body calculations. We could perform a similar
calculation for
p
2
c
but the result is clear from the fact that the total momentum
must be zero:
p
1
c
=
µ
v
,
p
2
c
=−
µ
v
.
(3.39)
The total kinetic energy in the lab frame can be expressed in terms of velocities in
the centre-of-mass frame:
1
2
m
1
v
1
+
1
2
m
2
v
2
=
1
2
m
1
(
v
1
c
+
1
2
m
2
(
v
2
c
+
K
=
V
)
·
(
v
1
c
+
V
)
+
V
)
·
(
v
2
c
+
V
).
The simplification of this expression requires the evaluation of, e.g.
v
1
c
+
V
2
,
(
v
1
c
+
V
)
·
(
v
1
c
+
V
)
=
2
v
1
c
·
V
+
which leads to
1
2
m
1
v
1
c
+
1
2
m
2
v
2
c
+
1
2
(m
1
+
m
2
)V
2
.
K
=
(m
1
v
1
c
+
m
2
v
2
c
)
·
V
+
(3.40)
