11.2.2 Isolated Systems

The second implication of Equation 11.23, and the more important one from the perspective of this chapter, applies to an isolated system. In Chapter 10, we defined an isolated system as one that is not influenced or altered by external forces from the environment. For momentum, that means a system on which the net external force is zero: $\vec{F}_{net} = \vec{0}$ . That is, an isolated system is one on which there are no external forces or for which the external forces are balanced and add to zero.

For an isolated system, Equation 11.23 is simply

\frac{d \vec{P}}{d t} = \vec{0} (isolated system)

(11.24)

In other words, the total momentum of an isolated system does not change. The total momentum $\vec{P}$ remains constant, regardless of whatever interactions are going on inside the system. The importance of this result is sufficient to elevate it to a law of nature, alongside Newton’s laws.

The total momentum after an interaction is equal to the total momentum before the interaction. Because Equation 11.25 is a vector equation, the equality is true for each of the components of the momentum vector. That is,

\begin{matrix} (p_{f x})_{1} + (p_{f x})_{2} + (p_{f x})_{3} + \dots = (p_{i x})_{1} + (p_{i x})_{2} + (p_{i x})_{3} + \dots \\ (p_{f y})_{1} + (p_{f y})_{2} + (p_{f y})_{3} + \dots = (p_{i y})_{1} + (p_{i y})_{2} + (p_{i y})_{3} + \dots \end{matrix}

(11.26)

The x-equation is an extension of Equation 11.18 to N interacting objects.

NOTE It is worth emphasizing the critical role of Newton’s third law. The law of conservation of momentum is a direct consequence of the fact that interactions within an isolated system are action/reaction pairs.

EXAMPLE 11.2 A glider collision

A 250 g air-track glider is pushed across a level track toward a 500 g glider that is at rest. FIGURE 11.12 shows a position-versus-time graph of the 250 g glider as recorded by a motion detector. Best-fit lines have been found. What is the speed of the 500 g glider after the collision?

The figure illustrates the best-fit lines for a position-versus-time of an air-track glider pushed across a level track to calculate the speed of the glider after it collides with another glider.

FIGURE 11.12 Full Alternative Text

MODEL Let the system be the two gliders. The gliders interact with each other, but the external forces (normal force and gravity) balance to make $\vec{F}_{net} = \vec{0}$ . Thus the gliders form an isolated system and their total momentum is conserved.

VISUALIZE FIGURE 11.13 is a before-and-after pictorial representation. The graph of Figure 11.12 tells us that the $250 g$ glider initially moves to the right, collides at $t = 1.0 s,$ then rebounds to the left (decreasing x). Note that the best-fit lines are written as a generic $y = \dots$ , which is what you would see in data-analysis software.

The figure consists of two parts labeled Before and After, that illustrate the before-and-after collision between two gliders.

FIGURE 11.13 Full Alternative Text

SOLVE Conservation of momentum for this one-dimensional problem requires that the final momentum equal the initial momentum: $P_{f x} = P_{i x}$ . In terms of the individual components, conservation of momentum is

(p_{f x})_{1} + (p_{f x})_{2} = (p_{i x})_{1} + (p_{i x})_{2}

Each momentum is $m v_{x}$ , so conservation of momentum in terms of velocities is

m_{1} (v_{f x})_{1} + m_{2} (v_{f x})_{2} = m_{1} (v_{i x})_{1} + m_{2} (v_{i x})_{2} = m_{1} (v_{i x})_{1}

where, in the last step, we used $(v_{i x})_{2} = 0$ for the $500 g$ glider. Solving for the heavier glider’s final velocity gives

{(v_{f x})}_{2} = \frac{m_{1}}{m_{2}} [{(v_{i x})}_{1} - {(v_{f x})}_{1}]

From Chapter 2 kinematics, the velocities of the $250 g$ glider before and after the collision are the slopes of the position-versus-time graph. Referring to Figure 11.12, we see that $(v_{i x})_{1} = 0.75 m / s$ and $(v_{f x})_{1} = - 0.21 m/s$ . The latter is negative because the rebound motion is to the left. Thus

{(v_{f x})}_{2} = \frac{250 g}{500 g} [0.75 m/s - (- 0.21 m/s)] = 0.48 m/s

The $500 g$ glider moves away from the collision at $0.48 m / s$ .

REVIEW The 500 g glider has twice the mass of the glider that was pushed, so a somewhat smaller speed seems reasonable. Paying attention to the signs—which are positive and which negative—was very important for reaching a correct answer. We didn’t convert the masses to kilograms because only the mass ratio of 0.50 was needed.

11.2.2 Isolated Systems

EXAMPLE 11.2 A glider collision

FIGURE 11.12 Position graph of the 250 g glider.

FIGURE 11.13 Before-and-after representation of a collision.