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348 CHAPTER 11 Collisions

Online Online 11.3 INELASTIC COLLISIONS
13 14
Concept Concept IN ONE DIMENSION
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If the collision is inelastic, kinetic energy is not conserved, and then the only conser-
vation law that is applicable is the conservation of momentum.This, by itself, is insuf-
ficient to calculate the velocities of both particles after the collision. Thus, for most
inelastic collisions, one of the final velocities must be measured in order for momen-
tum conservation to provide the other. Alternatively, we must have some independent
knowledge of the amount of kinetic energy lost. However, if the collision is totally
totally inelastic collision inelastic, so a maximum amount of kinetic energy is lost, then the common velocity of both
particles after the collision can be calculated.
In a totally inelastic collision, the particles do not bounce off each other at all; instead,
the particles stick together, like two automobiles that form a single mass of interlocking
wreckage after a collision, or two railroad boxcars that couple together. Under these
conditions, the velocities of both particles must coincide with the velocity of the center
of mass. But the velocity of the center of mass after the collision is the same as the
velocity of the center of mass before the collision, because there are no external forces
and the acceleration of the center of mass is zero [see Eq. (10.40)]. We again consider
a stationary target, so that before the collision the velocity of the target particle is zero
(v 0) and the general equation [Eq. (10.37)] for the velocity of the center of mass
2
yields
m v
1 1
v (11.19)
CM
m m
1 2
This must then be the final velocity of both particles after a totally inelastic collision:
m v
1 1
final velocities in totally inelastic v' v' v CM (11.20)
2
1
collision with stationary target m m 2
1
We have already come across an instance of this formula in Example 3 of Chapter 10.

Suppose that the two boxcars of Example 4 couple during the
EXAMPLE 5
collision and remain locked together (see Fig. 11.7). What is
the velocity of the combination after the collision? How much kinetic energy is
dissipated during the collision?

SOLUTION: Since the boxcars remain locked together, this is a totally inelastic
collision. With m 20 tons, m 65 tons, and v 5.0 m s, Eq. (11.19) gives us
1 2 1
the velocity of the center of mass:
m v 20 tons 5.0 m s
1 1
v 1.2 m s
CM
m m 20 tons 65 tons
1 2
and this must be the velocity of the coupled cars after the collision.
The kinetic energy before the collision is that of the moving boxcar,

1 2 1 2 5
2
2 m v 20 000 kg (5.0 m s) 2.5 10 J
1 1
and the kinetic energy after the collision is that of the two coupled boxcars,