Since momentum is a vector, all objects in a closed system can individually have momentum while the system as a whole have a net momentum of zero. Consider, for example,  the following system.

The 3 kg mass has a positive momentum of 6 kg m/sec and
the 2 kg mass has a negative momentum of 6 kg m/sec.
Each one is moving, but the total momentum of the system equals 0.

To further our study of the properties of momentum, we will revisit Newton's 2nd and 3rd laws and restate each in terms of momentum. 

For the remainder of this lesson we will focus on Newton's 2nd Law, to learn about Newton's 3rd Law and its relationship to momentum and energy, read this additional lesson.

As before, Newton's 2nd law deals only with forces acting on one object - that is, the net force. When one or more forces act on an object, the product of the sum of those forces times the duration of time over which they act equals the change in the object's momentum. If the forces are balanced, that is, the object is in a state of either dynamic or static equilibrium, then the net F equals zero and there is no change in the object's momentum - it maintains a constant velocity. However, if there is an unbalanced force, then there will be a nonzero net force and subsequently a nonzero impulse and the object will experience an acceleration allowing us to witness a change in the object's velocity and momentum.

Impulse can be represented by the variable J and has units of N sec. Since the  impulse that an object receives equals the change in its momentum, the units for impulse and momentum must be equivalent.

N sec		kg m/sec
[ kg m/sec2 ] sec
kg m/sec

Note that in the equation for impulse, if the amount of momentum being changed remains constant, that the net F can be maximized by decreasing the time of impact.  That is, a short jab is much more effective than a long, drawn-out swing when striking someone for maximum force.  Conversely, when a large amount of momentum has to be lost, doing so over a longer period of time, decreases the net force applied to the object.  For example, when you jump off of a high tower, bending your knees when you land, decreases the impact force by increasing the time of impact.  Moreover, if the amount of force remains constant, the change in momentum can be maximized by increasing the time of impact.  For example, in sports, coaches are always encouraging athletes to "follow through."  This allows the force to act on the ball for a greater amount of time, therefore increasing the time of impact and the net impulse delivered to the ball.  Subsequently, the ball receives a greater change in its momentum and has a greater final velocity.

Accompanying this restatement for Newton's 2nd law are two important graphical relationships.

Since the area under a graph can be re-expressed in terms of calculus as an integral, impulse can be restated as

Notice that force must expressed as a function in terms of time, not displacement. The calculus will allow us to determine the impulse for non-constant forces and thus is applicable to a wider range of situations.

Let's work an example using this relationship. Suppose a force, F(t) = 6t2 - 3t +1, acts on an 7 kg mass for three seconds. If the mass starts from a state of rest, what will be its final velocity?  To solve this problem, we will begin by determining the impulse that the force applies to the 7 kg mass.  Graphically, this impulse is represented by the area bounded by the graph F(t) = 6t2 - 3t +1 and the t-axis between the values of t = 0 and t = 3.

Impulse =	=
=
=	43.5 nt sec

To calculate the final velocity of the 7 kg mass, remember that the impulse, J, delivered to the 7 kg mass equals the change in its momentum. Your solution should look similar to

J = mv_f - mv_o
43.5 = 7(mv_f ) - 7(0)
v_f = 43.5/7 = 6.2 m/sec

Even when the forces are not given, we can still calculate impulse by examining the change in an object's momentum. As an example, how would the impulses compare for an object that strikes a wall and sticks to it, versus, an identical object bouncing off of a wall with no loss of kinetic energy?

Note that when the ball bounces, the impulse delivered to the ball is twice as great as when the ball sticks.

Since the slope of a graph can be re-expressed in terms of calculus as a derivative, the instantaneous value of the net applied force can be restated as

Notice that force must expressed as a function in terms of time, not displacement. The calculus will allow us to determine expressions for non-constant forces and thus is applicable to a wider range of situations.

Let's work an example using this relationship.

To determine the instantaneous force acting on the 7 kg mass at each of the specified times (t=-2 seconds, t = 0 seconds, t = 3.5 seconds) you would take the derivative of p(t) = -4t³+5t²-20t and evaluate it at each moment.

F(t) = = -12t² + 10t - 20

F(-2) = -12(-2)² + 10(-2) - 20 = -88 N
F(0) = -12(0)² + 10(0) - 20 = -20 N
F(3.5) = -12(3.5)² + 10(3.5) - 20 = -132 N

Notice in the graph shown below of force vs time that not only are all of the force values negative but so is the area bounded by the graph and the t-axis. This qualitatively agrees with the previous momentum vs time graph since both graphs correctly depict that the object is continuously losing momentum.