Environmental Engineering Reference
In-Depth Information
Since we are working in a zero-momentum frame
(m
1
v
1
c
+
m
2
v
2
c
)
=
0
and so
1
2
(m
1
+
m
2
)V
2
,
K
=
K
c
+
(3.41)
where the kinetic energy in the centre-of-mass frame is
1
2
m
1
v
1
c
+
1
2
m
2
v
2
c
.
K
c
=
(3.42)
So we see that the kinetic energy measured in the lab is a sum of two components:
the energy of the particles as measured in the centre-of-mass frame, and a kinetic
energy term
m
2
)V
2
that we can ascribe to the motion of the total mass
of the system at the centre-of-mass velocity. Since the collision involves a system
of two particles for which there are no external forces, Eq. (2.28) implies that
V
is constant. The energy contained in the motion of the centre of mass is therefore
unchanged during the collision and we can think of this contribution to the kinetic
energy as being 'locked in' to the motion of the system as a whole; it cannot be
used to alter the internal state of the system. Since
V
2
1
2
(m
1
+
0, Eq. (3.41) also shows
that the zero-momentum frame is the inertial frame in which the system has the
lowest possible total kinetic energy.
≥
3.3.2
Elastic and inelastic collisions
During the interaction stage (see Figure 3.8),
K
c
will generally change as a
result of the forces acting on the particles. If we assert that all of these forces are
conservative, then we can represent them by a potential
U
. The mechanical energy
in the centre-of-mass frame is then
E
c
=
K
c
+
U
(3.43)
and this is a conserved quantity. We shall define the potential energy to be zero
when the particles are infinitely separated in which case
U
is zero both before
and after the collision and hence
K
c
has the same value both before and after the
collision (even though it can change during the collision). We distinguish quantities
measured after the collision from their counterparts beforehand by the use of a
prime (
). In this notation we can write that
K
c
.
K
c
=
(3.44)
1
2
(m
1
+
m
2
)V
2
Since
is a constant, this result is also true in the lab frame:
K
.
K
=
(3.45)
Furthermore, we can describe the collision by specifying the momentum vectors of
the two particles in the centre-of-mass frame (both before and after the collision).
They are always back-to-back since the total momentum is zero in this frame. The
collision will generally alter the direction of the momentum vectors, so that
p
1
c
and
