University Physics I - Classical Mechanics, 2019

4.2. “CONVERTIBLE” AND “TRANSLATIONAL” KINETIC ENERGY 77
Figure 4.5 shows that the greatest loss of kinetic energy happens for the totally inelastic collision,
which, as we will see in a moment, is, in fact, a general result. That being the case, the figure
also shows that it may not be always be possible to bring the total kinetic energy down to zero,
even temporarily. The reason for this is that, if momentum is conserved, the velocity of the center
of mass cannot change, so if the center of mass was moving before the collision, it must still be
moving afterwards; and, as mentioned in this chapter’s introduction, as long as there is motion in
a system, its total kinetic energy cannot be zero.
All of this suggests that it should be possible to break up a system’s total kinetic energy into two
parts: one part associated with the motion of the center of mass, which cannot change in any
momentum-conserving collision, and one part associated with the relative motion of the parts that
make up the system. This second part would vanish irreversibly in a totally inelastic collision,
whereas it would recover its original value in an elastic collision.
The way to see this mathematically, for a system of two objects with masses m 1 and m 2 ,isto
introduce the center of mass velocity v cm [Eq. (3.10)]
v cm = m 1v 1 + m 2 v 2
m 1 + m 2
and the relative velocity v 12 = v 2 − v 1 (Eq. (4.3) above), and observe that the velocities v 1 and v 2
can be written, respectively, as
m 2
v 1 = v cm − v 12
m 1 + m 2
m 1
v 2 = v cm + v 12 (4.10)
m 1 + m 2
Substituting the equations (4.10) into the expression K sys = 1 2 m 1v1 2 + 1 2 m 2v2 2 , one finds that the
cross-terms vanish, and all that is left is
K sys = 1 2 (m 1 + m 2 )vcm 2 + 1 m 1 m 2 2 + m 2m 2 1
2 (m 1 + m 2 ) 2 v12
2
Afactorof(m 1 + m 2 ) may be canceled in the last term, and the final expression takes the form
K sys = K cm + K conv (4.11)
where the center of mass kinetic energy (or translational energy) is just what one would have if the
whole system was a single particle of mass M = m 1 + m 2 moving at the center of mass speed:
K cm = 1 2 Mv2 cm (4.12)
and the “convertible energy” K conv is the part associated with the relative motion, which can be