Impulse is the magnitude we use in dynamics to relate the force applied on a body to the time the force has been applied. It let us understand, for example, the mechanism of takeoff of the space shuttles, but also why football players put the ball behind their heads for the throw-in. In this chapter we are going to study it through the following points:

Gather momentum!, let´s get started!


Have you ever ask yourself, why football players put the ball behind their heads to throw-in? That gesture does not increase considerably the value of the force to throw the ball but, however, it will let the players to exert the same force during more time. Players do what we know as gather momentum.

So, it seems clear that if we want to give a specific velocity to a body we have two options: to apply a bigger force during a shorter interval of time or, a smaller force during a longer interval of time. The longer the force applied is, the higher the speed we can get is.

goalkeeper giving impulse to a ball

Goalkeeper giving impulse to a ball

The goalkeeper moves the arm backwards as much as possible to start a movement forward it let him to apply the force on the ball for longer. This will make the ball to get further.

impulse is the magnitude that let us quantify those ideas. Let´s define it formally.


Impulse   is a vector magnitude that relates the force to the time its action takes.



  • J: It´s the impulse of the force. Some times it is also abbreviated Imp. Its unit in the S.I. is the newton per second ( N·s ).
  • F: It is the force we are considering, supposedly constant. Its unit of measure in the S.I. is the newton (N).
  • t: It is the interval time according to wich the force acts. Its unit of measure in the S.I.  is the second (s).

Observe that, from the previous definition we can deduce that the impulse vector of a force possess the same direction than the force to which it´s associated. 

Impulse-momentum theorem

Taking into account that the variation of linear momentum can be related to the resultant force acting on a body according to:

F=ptF·t = p

The product is the own definition we have given for the impulse, so that is related to the variation of the linear momentum of the body. It is the impulse-momentum theorem.

Impulse-momentum theorem  establishes that the impulse of the resultant force acting on a body is equal to the variation of its linear momentum:

J=F·t =p


  • J: It is the total impulse the body is subdued, the impulse of the resultant force. Its unit in the S.I. is the newton per second ( N·s ).
  • F: It is the resultant force or total force to what the body is subdued, supposedly constant. Its unit of measure is the newton (N).
  • t: It is the time interval during which the force is acting. Its unit of measure is the second ( s ).
  • p: Represents the variation of the linear momentum produced in the considered time interval. It can be calculated as the difference between its final value and its initial value. Remember its unit of measure in the S.I. is the kg·m/s.

Observe that the previous expression brings to light the statement we did about that by giving a determined velocity to a body (to increase its linear momentum) we can act in two ways: by acting over the force or acting over the time on which it acts. So, in the space shuttles the ship get the desired speed due to the continuous effect of the force the motor drive provides.

They are closely related though, you should not confuse the linear momentum with the impulse. Impulse can be related to the variation of the first one, but they are magnitudes conceptually different.

It is normal to get confused. Bear in mind that their ecuations and dimensions are the same...


... and the units of measure in the S.I. are equivalent...


Variable force

In the definition we have made of the linear impulse we assumed the force remains constant during the time interval ∆t acting. This is not like that in general, the force is variable though. We can, then, find the impulse acting  on the time interval infinitly small (diferential). That impulse would be a diferential impulse, and the force it would act on that time interval it would be constant, indeed. So:


So the impulse transferred during a finite time interval is obtained adding the finite diferential impulses by integrating:


Regarding the impulse theorem, we can get to the same statement already introduced for constant forces, considering this time as variable forces. In order to check it we must consider the differential version of the Newton´s second Law, that is F=dpdt. So:



Impulse Charts

Probably you know now that the defined integral between two values of a function coincides numerically in value with the area set under that function. So, if we represent in the horizontal axis the time, and, the force in the vertical axis, either constant or variable, the area of the curve between  ti and tf coincides with the value of the impulse: 

Cálculus of impulse as area under the force graph

Impulse calculus

In the chart it is represented how the force acting on a body varies in time. The area locked under the curve between the instants ti and tf, in red, coincides numerically with the value of the impulse of that force in the interval tf - ti,  thus it coincides with the value of the variation of the linear momentum the body on which that interval will be applied is going to experience.

From this idea, watch that it is always possible to find an average constant force whose value of impulse in that time interval coincides with the one of the variable force.

Cálculus of impulse as area under the average force graph

Average constant force.

Due the area in blue and the area in green are equal, the area locked under the average force curve that represents the real and variable force coincides with the area under the average force, represented in black. As it is a rectangle area, the calsulation of that is reduced to a simple multiplication, opposite the use of integrals the first one would require.

Thus, most of the times on this level, when we talk about impulse force we are referring to that same assumed average constant force.

Now... ¡Test yourself!

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