As mentioned in the previous lesson on Newton's three laws of motion, Newton's 2nd law only deals with the forces acting on a single 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 or the impulse received by the object.
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 is represented by the variable J and has units of N sec.
J = (net F)t = Δp where p = mv.
This equation can be presented graphically as


Since the impulse that an object receives equals the change in its momentum, the units for impulse and momentum must be equivalent.
impulse

Δmomentum

N sec

kg m/sec

(kg m/sec^{2}) sec


kg m/sec

Remember by Newton's Third Law that during a collision, the impulse gained by one object is exactly equal to the impulse lost by the second object  that is,
F_{AB}t = F_{BA}t
.
In calculus terms the area under a graph is expressed as an integral. This allows us to restate the impulse as Let's work an example using this relationship. 
Refer to the following information for the next two questions.
Suppose a force, F(t) = 6t^{2}  3t +1, acts on an 7kg mass for three seconds.

Force as the rate of change of momentum
The impulse equation
J = (net F)t = Δp where p = mv
can be rearranged to state that the applied net force applied to an object equals the rate of change of the its momentum.
That is, the net force acting on an object can be calculated as the slope of a momentum vs time graph. In terms of the calculus, this result equates to taking the derivative.
Notice that force must expressed as a function in terms of time, not displacement. Calculus will allow us to determine expressions for instantaneous, nonconstant forces and thus is applicable to a wider range of situations. Let's work an example using this relationship. 
Refer to the following information for the next three questions.
Using the graph provided below, determine the instantaneous force acting on the 7kg mass at each of the specified times: t = 2 seconds, t = 0 seconds, t = 3.5 seconds.

Notice in the force vs time graph shown below of F(t) = 12t^{2} + 10t  20 that not only are all of the force values negative but so is the area bounded by the graph and the taxis.
This qualitatively agrees with the previous momentum vs time graph since both graphs correctly depict that the object is continuously losing momentum.