11.1.1 Momentum

The product of an object’s mass and velocity is called the momentum of the object:

m o m e n t u m = \vec{p} \equiv m \vec{v}

(11.3)

The units of momentum are $kg m / s$ .

Momentum $\vec{p}$ is a vector—specifically, a vector parallel to the velocity vector $\vec{v}$ . FIGURE 11.2 shows that $\vec{p}$ , like any vector, can be decomposed into x- and y-components:

\begin{array}{l} p_{x} = m v_{x} \\ p_{y} = m v_{y} \end{array}

The figure illustrates the decomposition of a momentum vector arrow labeled vector p into the x and y components.

FIGURE 11.2 Full Alternative Text

NOTE One of the most common errors in momentum problems is a failure to use the appropriate signs. The momentum component $p_{x}$ has the same sign as $v_{x}$ . Momentum is negative for an object moving to the left (on the x-axis) or down (on the y-axis).

An object can have a large momentum by having either a small mass but a large velocity or a small velocity but a large mass. For example, a 5.5 kg (12 lb) bowling ball rolling at a modest $2 m/s$ has momentum of magnitude $p = (5.5 kg) (2 m / s) = 11 kg m / s$ . This is almost exactly the same momentum as a 9 g bullet fired from a high-speed rifle at $1200 m/s$ .

Newton actually formulated his second law in terms of momentum rather than acceleration:

\vec{F} = m \vec{a} = m \frac{d \vec{v}}{d t} = \frac{d (m \vec{v})}{d t} = \frac{d \vec{p}}{d t}

(11.4)

This statement of the second law, saying that force is the rate of change of momentum, is more general than our earlier version $\vec{F} = m \vec{a}$ . It allows for the possibility that the mass of the object might change, such as a rocket that is losing mass as it burns fuel.

Returning to Equation 11.2, you can see that $m \vec{v}_{i}$ and $m \vec{v}_{f}$ are $\vec{p}_{i}$ and $\vec{p}_{f}$ , the object’s momentum before and after the collision. Further, $\vec{p}_{f} - \vec{p}_{i}$ is $Δ \vec{p}$ , the change in the object’s momentum. Thus Equation 11.2 is a statement about the object’s change in momentum:

Δ \vec{p} = {\vec{p}}_{f} - {\vec{p}}_{i} = \int_{t_{i}}^{t_{f}} \vec{F} (t) d t

(11.5)

Now we need to examine the right-hand side of Equation 11.5.