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Impulse review

Overview of key terms and equations related to impulse, including how impulse can be calculated from a force vs. time graph.

Key terms

Term (symbol)Meaning
Impulse (FΔt)Product of the average force exerted on an object and the time interval during which the force is exerted. Impulse is equal to the change in momentum (Δp) and is sometimes represented with the symbol J. Vector quantity with SI units of Ns or kgms.

Equations

EquationSymbolsMeaning in words
Δp=FnetΔtΔp is the change in momentum, Fnet is the net force, and Δt is the time period of the net forceImpulse is proportional to the constant net force acting on an object and the time period that the net force acts.

How force changes momentum

If we take the impulse equation and solve for force, another relationship of the equation presents itself:
FΔt=ΔpF=ΔpΔt
When a net force is exerted on an object, it changes that object's momentum over the time of the force exertion. In other words, force is the rate at which momentum changes. For example:
  • If an object experiences a large momentum change (Δp) over a short time duration (Δt), then there must have been a large net force (F) applied to it.
  • Conversely, if an object experiences a small momentum change (Δp) over a long time duration (Δt), then there must have been a small net force (F) applied to it.

How to find impulse from a force vs. time graph

Impulse is the area under the curve of the force vs. time graph. Areas above the time axis are positive Δp and areas below the axis are negative Δp. If the force is not constant, we can divide the graph into sections and add up the impulse in each section.
For example, to find the total impulse on the object in the force vs. time graph in Figure 1 over t1+t2, the areas of A1 and A2 can be added together.
Figure 1. Impulse is the area under a force vs. time graph. This graph can be analyzed as two separate areas to find the total impulse over time t1+t2.
A1 is a rectangle of height F0 and width t1. A2 is a triangle of height F0 and base t2. The total impulse on the object over t1+t2 is
Δp=A1+A2=F0t1+12F0t2
For worked examples of finding impulse or change in momentum from a force vs. time graph, watch our video about calculating impulse from force vs. time graphs.

Common mistakes and misconceptions

  1. People forget what the sign of impulse means. Impulse is a vector, so a negative impulse means the net force is in the negative direction. Likewise, a positive impulse means the net force is in the positive direction.
  2. People mistake impulse with work. Both impulse and work depend on the external net force, but they are different quantities. The properties of impulse and work are compared in the table below.
ImpulseWork
Product of net force and...timedisplacement
Changes an object’smomentumenergy
Quantity typevectorscalar

Learn more

For deeper explanations of impulse, see our video about momentum and impulse.
To check your understanding and work toward mastering these concepts, check out our exercise on calculating change in momentum and speed from force vs. time graphs.

Want to join the conversation?

  • blobby green style avatar for user Lisa Wolf
    In the practice problems using a force v time graph to find change in momentum, when a negative impulse was applied, a negative final velocity results, and the 'correct' answer states velocity decreases. But velocity increasing in the negative direction is not the same as a decrease velocity. In fact velocity can be increasing in the negative direction, and a negative impulse can result in an increase in velocity. If an object starts from rest and accelerates in the negative direction, would you say its velocity decreases??
    (4 votes)
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  • blobby green style avatar for user shaul753
    During a safety test, a car hits a wall and stops in 0.55,
    The net force on the car is 6500 N during the collision.

    What is the magnitude of the change in momentum of the car?


    Shouldn't the answer be divided by two?
    If the net force during the collision is 6500 N, and the car stops after 0.55 s, doesn't that mean the force goes from 6,500 N to 0 in 0.55 s, and the area under the force time graph (the impulse) is a triangle?
    (0 votes)
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    • hopper cool style avatar for user obiwan kenobi
      No, the force does not change. The force is constant during those 0.55 seconds. The force is equal to the change in momentum over the change in time. You know the force, you know the time, so you can solve for the change in momentum:
      Net force = (∆p)/(∆t)

      6500 N = (∆p)/(0.55 s)

      ∆p = 3575 kg m/s. Hope this helps!
      (4 votes)
  • stelly green style avatar for user Addy
    How do u get final velocity from acceleration,time,and impulse?
    (1 vote)
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  • blobby green style avatar for user James
    In the practice problem:

    A tennis player hits a ball and gives it a change in momentum of 3.3kgm/s over 5.0ms

    What is the magnitude of the net force on the tennis ball?

    Round answer to two significant digits.

    I did Net force = (∆p)/(∆t)

    to get 0.66 but the answer was wrong. Apparently there is a 10^-3s I should have multiplied the ∆t by.

    Where does that come from?
    (1 vote)
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  • blobby green style avatar for user mohamad
    the article says"When the force is not constant, we use the average force to find the impulse."but in previous article was written that"For a constant force over time, the net external force is the same as the average external force over the time period."!
    (0 votes)
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