11.2 Conservation of Momentum

The momentum principle was derived from Newton’s second law and is really just an alternative way of looking at single-particle dynamics. To discover the real power of momentum for problem solving, we need also to invoke Newton’s third law, which will lead us to one of the most important principles in physics: conservation of momentum.

FIGURE 11.9 shows two objects with initial velocities $(v_{i x})_{1}$ and $(v_{i x})_{2}$ . The objects collide, then bounce apart with final velocities $(v_{f x})_{1}$ and $(v_{f x})_{2}$ . The forces during the collision, as the objects are interacting, are the action/reaction pair $\vec{F}_{1 on 2}$ and $\vec{F}_{2 on 1}$ . For now, we’ll assume that the motion is one dimensional along the x-axis.

The figure has three parts labeled Before, During, and After that illustrate a collision between two spherical objects.

FIGURE 11.9 Full Alternative Text

NOTE The notation, with all the subscripts, may seem excessive. But there are two objects, and each has an initial and a final velocity, so we need to distinguish among four different velocities.

Newton’s second law for each object during the collision is

\begin{array}{l} \frac{d {(p_{x})}_{1}}{d t} = {(F_{x})}_{2 on 1} \\ \frac{d {(p_{x})}_{2}}{d t} = {(F_{x})}_{1 on 2} = - {(F_{x})}_{2 on 1} \end{array}

(11.15)

We made explicit use of Newton’s third law in the second equation.

Although Equations 11.15 are for two different objects, suppose—just to see what happens—we were to add these two equations. If we do, we find that

\frac{d {(p_{x})}_{1}}{d t} + \frac{d {(p_{x})}_{2}}{d t} = \frac{d}{d t} [{(p_{x})}_{1} + {(p_{x})}_{2}] = {(F_{x})}_{2 on 1} + (- {(F_{x})}_{2 on 1}) = 0

(11.16)

If the time derivative of the quantity ${(p_{x})}_{1} + {(p_{x})}_{2}$ is zero, it must be the case that

{(p_{x})}_{1} + {(p_{x})}_{2} = constant

(11.17)

Equation 11.17 is a conservation law! If ${(p_{x})}_{1} + {(p_{x})}_{2}$ is a constant, then the sum of the momenta after the collision equals the sum of the momenta before the collision. That is,

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

(11.18)

Furthermore, this equality is independent of the interaction force. We don’t need to know anything about $\vec{F}_{1 on 2}$ and $\vec{F}_{2 on 1}$ to make use of Equation 11.18.

As an example, FIGURE 11.10 is a before-and-after pictorial representation of two equal-mass train cars colliding and coupling. Equation 11.18 relates the momenta of the cars after the collision to their momenta before the collision:

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

The figure illustrates the before-and-after pictorial representations of two equal-mass train cars colliding and coupling.

FIGURE 11.10 Full Alternative Text

Initially, car 1 is moving with velocity $(v_{i x})_{1} = v_{i}$ while car 2 is at rest. Afterward, they roll together with the common final velocity $v_{f}$ . Furthermore, $m_{1} = m_{2} = m$ . With this information, the momentum equation is

m v_{f} + m v_{f} = m v_{i} + 0

The mass cancels, and we find that the train cars’ final speed is $v_{f} = \frac{1}{2} v_{i}$ . That is, we can predict that the final speed is exactly half the initial speed of car 1 without knowing anything about the complex interaction between the two cars as they collide.

11.2 Conservation of Momentum

FIGURE 11.9 A collision between two objects.

FIGURE 11.10 Two colliding train cars.