Mass times velocity, or, equivalently, a quantity whose time rate of change is equal to the net force applied to a system. |
Forces are actions which cause a change in the velocity of an object, but a given application of force will have very different results when applied to objects of very different mass. Consider the force imparted by a baseball player swinging a bat. When delivered to a baseball, the change in velocity is dramatic. A 95 mph fasball might be completely reversed and exit the bat moving 110 mph in the other direction. When delivered to a car, however, the change in velocity is miniscule. A car moving 95 mph will not be slowed noticeably by the action of a bat. Thus, although the change in velocity of a system is proportional to the force applied, it is not equal to the force applied. To define a quantity whose rate of change is equal to the force applied, we must include both the mass and velocity of the system subject to the force.
The momentum (p) of a point particle with mass m and velocity v is defined as:
\begin{large}\[ \vec{p} \equiv m\vec{v}\]\end{large} |
For a system composed of N objects which are approximated as point particles with their position specified by the objects' centers of mass, the system momentum is defined as the vector sum of the momentum of the constituents:
\begin{large}\[ \vec{p}^{\rm \: sys} = \sum_{j=1}^{N} m_{j}\vec{v}_{j} \]\end{large} |
This definition is completely equivalent to
\begin{large}\[ \vec{p}^{\rm \: sys} = M^{\rm sys} \vec{v}^{\rm \: CM} \]\end{large} |
where Msys is the total mass of the system and vCM is the velocity of the system's center of mass.
One way of stating Newton's Second Law is that the rate of change of a system's momentum is equal to the vector sum of the forces applied to the object:
\begin{large}\[ \frac{d\vec{p}^{\rm \: sys}}{dt} = \sum_{k=1}^{N_{F}} \vec{F}_{k} \] \end{large} |
By Newton's 3rd Law, internal forces cancel from the vector sum above, leaving only the contribution of external forces:
\begin{large}\[ \frac{d\vec{p}^{\rm \:sys}}{dt} = \sum_{k=1}^{N_{F}} \vec{F}^{\rm ext}_{k} \] \end{large} |
The integrated change in momentum can be found explicitly by using the net external impulse (Jext):
\begin{large}\[ \vec{p}^{\rm \:sys}_{f} - \vec{p}^{\rm \:sys}_{i} = \int_{t_{i}}^{t_{f}} \sum_{k=1}^{N_{F}} \vec{F}_{k}^{\rm ext} \:dt \equiv \sum_{k=1}^{N_{F}} \vec{J}_{k}^{\rm ext} \]\end{large} |
In the absence of a net external force, the momentum of a system is constant:
\begin{large}\[ \vec{p}_{f}^{\rm \:sys} = \vec{p}_{i}^{\rm \:sys}\]\end{large} |
This equation is normally broken up to explicitly show the system constituents and the vector components:
\begin{large}\[ \sum_{j=1}^{N} p^{j}_{x,f} = \sum_{j=1}^{N} p^{j}_{x,i} \] \[ \sum_{j=1}^{N} p^{j}_{y,f} = \sum_{j=1}^{N} p^{j}_{y,i} \] \[ \sum_{j=1}^{N} p^{j}_{z,f} = \sum_{j=1}^{N} p^{j}_{z,i} \]\end{large} |
Because the change in momentum is proportional to the impulse, which involves a time integral, for instantaneous events:
\begin{large}\[ \lim_{t_{f}\rightarrow t_{i}} \int_{t_{i}}^{t_{f}} \vec{F}^{\rm ext} \:dt = 0 \]\end{large} |
For approximately instantaneous events such as collisions, it is often reasonable to approximate the external impulse as zero by considering a system composed of all the objects involved in the collision. The key to the utility of this assumption is that often during collisions the change in momentum of any individual system constituent being analyzed is dominated by the internal collision forces (the external forces make a negligible contribution to that constituent's change in momentum during the collision).
Before the collision occurs and after the collision is complete, the collision forces will usually drop to zero. Neglecting external impulse can only be justified during the collision. It is also completely incorrect to say that the momentum of each object is conserved. Only the system momentum is (approximately) conserved. |