Mechanics 2: Linear and Rotational Dynamics - 3D Math Primer for Graphics and Game Development - page 591

in momentum of the two objects. A force that is treated in this way is
known as an impulsive force, or more simply, an impulse. Since an impulse
is an immediate change in momentum, it has the same units as momentum:
ML/T. Note that an impulse is a very different beast from a regular force,
which has units ML/T 2 . It is very common programmer mistake to use
impulses and forces incorrectly, so be sure to watch your units.
When two objects collide, many things can happen, even if we assume
they remain intact. A likely scenario is for them to bounce off of each
other, changing the signs of both v 1 and v 2 . Or they may stick together.
The former is known as an elastic collision and the latter an inelastic col-
lision. (Actually, only a “perfect bounce” is considered truly elastic. The
terms “perfectly inelastic” and “perfectly elastic” are used to refer to the
two extremes, while an intermediate collision is described as simply “inelas-
tic.” Section 12.4.2 defines these terms a bit more precisely by using the
coe cient of restitution, but to fully understand the distinction requires
an understanding of kinetic energy. As we mentioned at the beginning of
Chapter 11, energy is certainly an important concept in physics, but it ac-
tually doesn't play a central role in the Newton-Euler dynamics used by
most real-time simulations, and it isn't discussed much in this topic.) The
velocity (and momentum) of each object is likely to change, but the law of
conservation of momentum says that the total momentum of the system of
both objects must remain constant. That is, p = p

′

.
In general, we cannot predict the individual velocities v 1 and v 2 using
just the law of conservation of momentum, since the conservation of mo-
mentum law gives us one equation (p = p ′ ) and there are two unknowns.
Before we consider what other bit of information we need, let's look at some
simpler cases of collisions and conservation of momentum. Assume for the
moment that the collision is perfectly inelastic, that is, the objects stick
together upon impact. This gives us the other equation we needed to solve
the system of equations: v ′

1 = v ′ 2 .

12.4.1 Perfectly Inelastic Collisions

A classic example of an inelastic collision is a gun firing a bullet into a block.
Assume that, as illustrated in Figure 12.10, a block of wood weighing 2.00 kg
is at rest, hanging from a wire whose mass is neglected. We fire a gun at
the block, and the bullet, which has a mass of 10 g, strikes the block with
a speed of 350 m/s. The bullet remains stuck in the block. What is the
horizontal speed of the block immediately after the impact?
First, we compute the initial momentum of the system, which is all
contained in the bullet:

p = m 1 v 1 + m 2 v 2 = (2.00 kg)(0) + (10.0 g)(350 m/s) = 3.50 kg m/s.