9781118230725.pdf

for the motion of a single particle. However, the three quantities that
appear in Eq. 9-14 must be evaluated with some care:

1. is the net force of all external forcesthat act on the system. Forces on one
   part of the system from another part of the system (internal forces) are not in-
   cluded in Eq. 9-14.


2. Mis the total massof the system. We assume that no mass enters or leaves the
   system as it moves, so that Mremains constant. The system is said to be closed.


3. is the acceleration of the center of massof the system. Equation 9-14 gives
   no information about the acceleration of any other point of the system.

Equation 9-14 is equivalent to three equations involving the components of

and along the three coordinate axes. These equations are

Fnet,xMacom,x Fnet,yMacom,y Fnet,zMacom,z. (9-15)

Billiard Balls.Now we can go back and examine the behavior of the billiard
balls. Once the cue ball has begun to roll, no net external force acts on the (two-
ball) system. Thus, because 0, Eq. 9-14 tells us that 0 also. Because
acceleration is the rate of change of velocity, we conclude that the velocity of the
center of mass of the system of two balls does not change. When the two balls col-
lide, the forces that come into play are internalforces, on one ball from the other.
Such forces do not contribute to the net force , which remains zero. Thus, the
center of mass of the system, which was moving forward before the collision,
must continue to move forward after the collision, with the same speed and in the
same direction.
Solid Body.Equation 9-14 applies not only to a system of particles but also
to a solid body, such as the bat of Fig. 9-1b. In that case,Min Eq. 9-14 is the mass
of the bat and is the gravitational force on the bat. Equation 9-14 then tells us
that In other words, the center of mass of the bat moves as if the bat
were a single particle of mass M, with force acting on it.
Exploding Bodies.Figure 9-5 shows another interesting case. Suppose that at
a fireworks display, a rocket is launched on a parabolic path. At a certain point, it
explodes into fragments. If the explosion had not occurred, the rocket would have
continued along the trajectory shown in the figure. The forces of the explosion are
internalto the system (at first the system is just the rocket, and later it is its frag-
ments); that is, they are forces on parts of the system from other parts. If we ignore
air drag, the net externalforce acting on the system is the gravitational force on
the system, regardless of whether the rocket explodes. Thus, from Eq. 9-14, the ac-
celeration of the center of mass of the fragments (while they are in flight) re-
mains equal to This means that the center of mass of the fragments follows the
same parabolic trajectory that the rocket would have followed had it not exploded.
Ballet Leap.When a ballet dancer leaps across the stage in a grand jeté, she
raises her arms and stretches her legs out horizontally as soon as her feet leave the

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9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES 221

Figure 9-5A fireworks rocket explodes in
flight. In the absence of air drag, the center
of mass of the fragments would continue to
follow the original parabolic path, until
fragments began to hit the ground.

The internal forces of the

explosion cannot change

the path of the com.