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This page defines impulse and describes its relationship to momentum.
<big><b>Edwyn Torres — Spring 2026</b></big>


==The Main Idea==
=Impulse and Momentum=


Impulse is a vector quantity describing both the nature and duration of a force. It is defined as the time integral of the net force vector: <math>\vec{J} = \int \vec{F}_{net}dt</math>. Impulse is represented by the letter <math>\vec{J}</math>. The most commonly used metric unit for impulse is the Newton*second.
==Introduction==


People are interested in impulse primarily because of its relationship to momentum, as described by the impulse-momentum theorem. The theorem states that if an impulse is exerted on a system, the change in that system's momentum caused by the force is equal to the impulse: <math>\Delta \vec{p} = \vec{J}</math>. Often, the system consists of a single particle whose motion we want to predict. This works out dimensionally because the units for impulse are equivalent to the units for momentum. For example, the Newton*second is equivalent to the kilogram*meter/second because a Newton is defined as a kilogram*meter/second^2.
Impulse and momentum are important ideas in physics because they help explain how forces change the motion of objects. Momentum describes how difficult it is to stop or change the motion of an object. Impulse describes how much a force changes an object's momentum over a period of time.


===A Mathematical Model===
A large force acting for a short time can create the same impulse as a smaller force acting for a longer time. This idea is important in collisions, sports, car safety, and many other real-world situations.


The impulse-momentum theorem is a consequence of the momentum principle (see [[Newton's Second Law: the Momentum Principle]]). Below is its derivation:
==Main Idea==


<math>\vec{F}_{net} = \frac{d\vec{p}}{dt}</math>
Momentum is a vector quantity that depends on mass and velocity:


can be arranged to <math>d\vec{p} = \vec{F}_{net}dt</math>.
<math>\vec{p} = m\vec{v}</math>


Integrating both sides yields <math>\int d\vec{p} = \int \vec{F}_{net}dt</math>
Impulse is also a vector quantity. It is the effect of a net force acting over a time interval:


which simplifies to <math>\Delta \vec{p} = \int \vec{F}_{net}dt = \vec{J}</math>.
<math>\vec{J} = \vec{F}_{net,avg}\Delta t</math>


In the above derivation, <math>\vec{F}_net</math> is the instantaneous force acting on the system.
The impulse-momentum theorem says that impulse equals the change in momentum:


<math>\vec{J} = \Delta \vec{p}</math>


According to calculus,
This can also be written as:


<math>\int \vec{F(t)}dt = \vec{F}_{avg} \Delta t</math>, where <math>\vec{F}_{avg}</math> is the time average of the force over the time interval <math>\Delta t</math>. When the force is constant, the average force is equal to that constant force, so the impulse exerted by constant force <math>\vec{F}</math> is <math>\vec{F} \Delta t</math>.
<math>\vec{p}_f = \vec{p}_i + \vec{F}_{net}\Delta t</math>


===A Computational Model===
This equation is useful because it lets us predict how an object's momentum changes when a force acts on it.


Often in computational simulations of particles, a momentum variable is assigned to each particle. Such simulations usually occur in "time steps," or iterations of a loop representing a time interval. In each time step, the particles' momenta are updated according to the impulse-momentum theorem. Usually in this kind of simulation, the time steps are assumed to be small enough that the force does not significantly vary over the time step, allowing the force and time interval to be multiplied to yield the impulse. By adding this small impulse each time step, the sum approximates the time integral of force over the course of the simulation, like a Riemann sum.
==Key Equations==


The following is an example of a line of vPython responsible for updating the momentum of a particle according to the impulse-momentum theorem:
* <math>\vec{p} = m\vec{v}</math>
* <math>\vec{J} = \vec{F}_{net,avg}\Delta t</math>
* <math>\vec{J} = \Delta \vec{p}</math>
* <math>\Delta \vec{p} = \vec{p}_f - \vec{p}_i</math>
* <math>\vec{p}_f = \vec{p}_i + \vec{F}_{net}\Delta t</math>


p = p + fnet*deltat
Where:


The simulation below is an example of a program that uses the momentum principle. It simulates the motion of a cart, represented by a rectangle, being blown by a gust of wind:
* <math>\vec{p}</math> = momentum
* <math>m</math> = mass
* <math>\vec{v}</math> = velocity
* <math>\vec{J}</math> = impulse
* <math>\vec{F}_{net}</math> = net force
* <math>\Delta t</math> = time interval


https://trinket.io/glowscript/ce43925647
==Units==


For more information, see [[iterative prediction]].
The unit for momentum is:


==Examples==
<math>kg \cdot m/s</math>


===Simple===
The unit for impulse is:


A 2kg particle is travelling in a northerly direction at 4m/s. After an impulse is applied to it, the particle is travelling in an easterly direction at 3m/s. The impulse consisted of a constant force applied for .02s. What was the magnitude of the force?
<math>N \cdot s</math>


Let us define the northerly direction as the +y direction and the easterly direction as the +x direction.
These units are equivalent because:


<math>\vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i = <6, 0, 0> - <0, 8, 0> = <6, -8, 0>.</math>
<math>1N = 1kg \cdot m/s^2</math>


Since the force was constant, the impulse is simply the product of that force and the time interval over which it was applied.
So:


<math>\vec{F} = \frac{\vec{J}}{\Delta t} = <6, -8, 0> / .02 = <300, -400, 0> </math>
<math>N \cdot s = kg \cdot m/s</math>


<math>|\vec{F}| = \sqrt{300^2 + (-400)^2 + 0^2} = 500</math>
This makes sense because impulse equals change in momentum.


===Middling===
==Conceptual Explanation==


This problem combines the impulse-momentum theorem with a number of kinematic equations that take advantage of the constant acceleration.
Impulse depends on two things: force and time. If the force is larger, the impulse is larger. If the force acts for a longer time, the impulse is also larger.


Standing on Earth, you throw a small rock with a mass of 0.5 kg into the air. At the instant it leaves your hand, the rock's velocity is <math>{\vec{v}=<0.1,4.0,0.3> m/s}</math> Ignore air resistance.
For example, when catching a baseball, a person usually moves their hand backward as the ball arrives. This increases the time over which the ball slows down. Since the ball still needs the same change in momentum to stop, increasing the stopping time decreases the average force on the hand.
* Initial Momentum?
<math>{m=0.5, \vec{v} = <0.1,4,0.3>}</math><br>
<math>{\vec{p} = m\vec{v} = <0.05,2,0.15> kg*m/s}</math><br><br>


* Rock's momentum after 0.25 seconds?
This is why catching a ball with stiff hands hurts more than catching it while moving your hands backward.
<math>{\vec{p}_f=\vec{p}_i+\vec{F}_{net}(\Delta{t})}</math>


<math>{\vec{p}_f = <0.05,2,0.15> + <0,(-9.8)(0.5),0>*0.25 = <0.05,2,0.15> + <0,-1.225,0>}</math><br>
==Worked Example 1: Finding Final Momentum==
:<math>{= <0.05,0.775,0.15> kgm/s}</math><br><br>


* Calculate the average velocity of the rock from just after it leaves your hand to 0.25 seconds later.
A 2 kg cart is moving with an initial velocity of <math>\langle 3,0,0 \rangle m/s</math>. A constant net force of <math>\langle 4,0,0 \rangle N</math> acts on the cart for 2 seconds. What is the final momentum of the cart?
<math>{\vec{p}=m\vec{v}, \vec{v}=\vec{p}/m}</math>


<math>{\vec{v}_f=(\vec{p}_f/m)=(1/0.5)*<0.05,0.775,0.15>}</math><br>
First find the initial momentum:
:<math>{=<0.1,1.55,0.3> m/s}</math><br>
<math>{\vec{v}_{avg}=(\vec{v}_i+\vec{v}_f)/2 = (0.5)*[<0.1,4,0.3> + <0.1,1.55,0.3>]= (0.5)<0.2,5.55,0.6>}</math><br>
:<math>{=<0.1,2.775,0.3> m/s}</math><br><br>


* If a rock's initial position just as it leaves your hand is <0,1.2,0>m, find the vector position of the ball after 0.25 seconds.
<math>\vec{p}_i = m\vec{v}</math>


<math>{\vec{r}_i=<0,1.2,0> m}</math><br>
<math>\vec{p}_i = 2\langle 3,0,0 \rangle = \langle 6,0,0 \rangle kg \cdot m/s</math>
<math>{\vec{r}_f=\vec{r}_i+\vec{v}_{avg}(\Delta{t})= <0,1.2,0> + <0.1,2.775,0.3>*0.25}</math><br>
:<math>{= <0,1.2,0> + <0.025,0.694,0.075>}</math><br>
:<math>{=<0.025,1.894,0.075> m}</math>


===Difficult===
Now find the impulse:


==Connectedness==
<math>\vec{J} = \vec{F}_{net}\Delta t</math>


==History==
<math>\vec{J} = \langle 4,0,0 \rangle(2) = \langle 8,0,0 \rangle N \cdot s</math>


== See also ==
Now use the impulse-momentum theorem:


===Further reading===
<math>\vec{p}_f = \vec{p}_i + \vec{J}</math>


===External links===
<math>\vec{p}_f = \langle 6,0,0 \rangle + \langle 8,0,0 \rangle</math>
 
<math>\vec{p}_f = \langle 14,0,0 \rangle kg \cdot m/s</math>
 
The final momentum of the cart is:
 
<math>\boxed{\langle 14,0,0 \rangle kg \cdot m/s}</math>
 
==Worked Example 2: Finding Average Force==
 
A 0.15 kg baseball is moving at 40 m/s before being hit. After contact with the bat, it moves in the opposite direction at 50 m/s. If the contact time is 0.01 seconds, what is the average force on the ball?
 
Let the original direction be positive. Then:
 
<math>v_i = 40 m/s</math>
 
<math>v_f = -50 m/s</math>
 
Find the change in momentum:
 
<math>\Delta p = m(v_f - v_i)</math>
 
<math>\Delta p = 0.15(-50 - 40)</math>
 
<math>\Delta p = 0.15(-90)</math>
 
<math>\Delta p = -13.5 kg \cdot m/s</math>
 
Now use:
 
<math>F_{avg} = \frac{\Delta p}{\Delta t}</math>
 
<math>F_{avg} = \frac{-13.5}{0.01}</math>
 
<math>F_{avg} = -1350 N</math>
 
The average force is:
 
<math>\boxed{1350N}</math>
 
The negative sign means the force acts opposite the ball's original direction.
 
==Common Mistakes==
 
* Forgetting that momentum and impulse are vectors.
* Forgetting to include direction when velocity changes.
* Confusing momentum with impulse.
* Using force instead of net force.
* Forgetting to multiply force by time.
* Thinking a bigger force always means a bigger impulse, even though time also matters.
* Forgetting that <math>N \cdot s</math> and <math>kg \cdot m/s</math> are equivalent units.
 
==Real-World Applications==
 
===Airbags===
 
Airbags reduce injury during car crashes by increasing the time it takes for a person's momentum to change. The person still needs to be brought to rest, so the change in momentum is the same. However, because the stopping time is longer, the average force on the person is smaller.
 
===Sports===
 
Impulse is important in sports such as baseball, tennis, soccer, and football. When a bat, racket, or foot stays in contact with a ball for a longer time, the impulse can increase. This changes the ball's momentum and can make it leave with a greater speed.
 
===Landing from a Jump===
 
When someone lands from a jump, bending the knees increases the stopping time. This lowers the average force on the person's legs. Landing with locked knees creates a shorter stopping time and a larger force.
 
==Computational Model==
 
In a computer simulation, momentum can be updated step by step using the momentum principle:
 
<pre>
p = p + Fnet*deltat
</pre>
 
This means that during each small time interval, the object's momentum changes because of the net force acting on it.
 
A simple GlowScript simulation for this topic could show a ball moving forward while a force acts on it for a short time. The simulation could display the ball's changing momentum after the impulse is applied.
 
==Practice Questions==
 
# A 3 kg object moves with velocity <math>\langle 2,0,0 \rangle m/s</math>. What is its momentum?
# A force of <math>\langle 10,0,0 \rangle N</math> acts on an object for 4 seconds. What impulse is delivered?
# A cart has initial momentum <math>\langle 5,0,0 \rangle kg \cdot m/s</math>. An impulse of <math>\langle -2,0,0 \rangle N \cdot s</math> acts on it. What is its final momentum?
 
==See Also==
 
* [[Linear Momentum]]
* [[Newton's Second Law: the Momentum Principle]]
* [[Conservation of Momentum]]
* [[Collisions]]
* [[Iterative Prediction]]


==References==
==References==


[[Category:Which Category did you place this in?]]
* OpenStax. ''College Physics 2e'', Momentum and Impulse.
* The Physics Classroom. ''Momentum and Impulse Connection.''
* HyperPhysics. ''Impulse and Momentum.''
* Khan Academy. ''Impulse and Momentum.''

Latest revision as of 16:42, 27 April 2026

Edwyn Torres — Spring 2026

Impulse and Momentum

Introduction

Impulse and momentum are important ideas in physics because they help explain how forces change the motion of objects. Momentum describes how difficult it is to stop or change the motion of an object. Impulse describes how much a force changes an object's momentum over a period of time.

A large force acting for a short time can create the same impulse as a smaller force acting for a longer time. This idea is important in collisions, sports, car safety, and many other real-world situations.

Main Idea

Momentum is a vector quantity that depends on mass and velocity:

[math]\displaystyle{ \vec{p} = m\vec{v} }[/math]

Impulse is also a vector quantity. It is the effect of a net force acting over a time interval:

[math]\displaystyle{ \vec{J} = \vec{F}_{net,avg}\Delta t }[/math]

The impulse-momentum theorem says that impulse equals the change in momentum:

[math]\displaystyle{ \vec{J} = \Delta \vec{p} }[/math]

This can also be written as:

[math]\displaystyle{ \vec{p}_f = \vec{p}_i + \vec{F}_{net}\Delta t }[/math]

This equation is useful because it lets us predict how an object's momentum changes when a force acts on it.

Key Equations

  • [math]\displaystyle{ \vec{p} = m\vec{v} }[/math]
  • [math]\displaystyle{ \vec{J} = \vec{F}_{net,avg}\Delta t }[/math]
  • [math]\displaystyle{ \vec{J} = \Delta \vec{p} }[/math]
  • [math]\displaystyle{ \Delta \vec{p} = \vec{p}_f - \vec{p}_i }[/math]
  • [math]\displaystyle{ \vec{p}_f = \vec{p}_i + \vec{F}_{net}\Delta t }[/math]

Where:

  • [math]\displaystyle{ \vec{p} }[/math] = momentum
  • [math]\displaystyle{ m }[/math] = mass
  • [math]\displaystyle{ \vec{v} }[/math] = velocity
  • [math]\displaystyle{ \vec{J} }[/math] = impulse
  • [math]\displaystyle{ \vec{F}_{net} }[/math] = net force
  • [math]\displaystyle{ \Delta t }[/math] = time interval

Units

The unit for momentum is:

[math]\displaystyle{ kg \cdot m/s }[/math]

The unit for impulse is:

[math]\displaystyle{ N \cdot s }[/math]

These units are equivalent because:

[math]\displaystyle{ 1N = 1kg \cdot m/s^2 }[/math]

So:

[math]\displaystyle{ N \cdot s = kg \cdot m/s }[/math]

This makes sense because impulse equals change in momentum.

Conceptual Explanation

Impulse depends on two things: force and time. If the force is larger, the impulse is larger. If the force acts for a longer time, the impulse is also larger.

For example, when catching a baseball, a person usually moves their hand backward as the ball arrives. This increases the time over which the ball slows down. Since the ball still needs the same change in momentum to stop, increasing the stopping time decreases the average force on the hand.

This is why catching a ball with stiff hands hurts more than catching it while moving your hands backward.

Worked Example 1: Finding Final Momentum

A 2 kg cart is moving with an initial velocity of [math]\displaystyle{ \langle 3,0,0 \rangle m/s }[/math]. A constant net force of [math]\displaystyle{ \langle 4,0,0 \rangle N }[/math] acts on the cart for 2 seconds. What is the final momentum of the cart?

First find the initial momentum:

[math]\displaystyle{ \vec{p}_i = m\vec{v} }[/math]

[math]\displaystyle{ \vec{p}_i = 2\langle 3,0,0 \rangle = \langle 6,0,0 \rangle kg \cdot m/s }[/math]

Now find the impulse:

[math]\displaystyle{ \vec{J} = \vec{F}_{net}\Delta t }[/math]

[math]\displaystyle{ \vec{J} = \langle 4,0,0 \rangle(2) = \langle 8,0,0 \rangle N \cdot s }[/math]

Now use the impulse-momentum theorem:

[math]\displaystyle{ \vec{p}_f = \vec{p}_i + \vec{J} }[/math]

[math]\displaystyle{ \vec{p}_f = \langle 6,0,0 \rangle + \langle 8,0,0 \rangle }[/math]

[math]\displaystyle{ \vec{p}_f = \langle 14,0,0 \rangle kg \cdot m/s }[/math]

The final momentum of the cart is:

[math]\displaystyle{ \boxed{\langle 14,0,0 \rangle kg \cdot m/s} }[/math]

Worked Example 2: Finding Average Force

A 0.15 kg baseball is moving at 40 m/s before being hit. After contact with the bat, it moves in the opposite direction at 50 m/s. If the contact time is 0.01 seconds, what is the average force on the ball?

Let the original direction be positive. Then:

[math]\displaystyle{ v_i = 40 m/s }[/math]

[math]\displaystyle{ v_f = -50 m/s }[/math]

Find the change in momentum:

[math]\displaystyle{ \Delta p = m(v_f - v_i) }[/math]

[math]\displaystyle{ \Delta p = 0.15(-50 - 40) }[/math]

[math]\displaystyle{ \Delta p = 0.15(-90) }[/math]

[math]\displaystyle{ \Delta p = -13.5 kg \cdot m/s }[/math]

Now use:

[math]\displaystyle{ F_{avg} = \frac{\Delta p}{\Delta t} }[/math]

[math]\displaystyle{ F_{avg} = \frac{-13.5}{0.01} }[/math]

[math]\displaystyle{ F_{avg} = -1350 N }[/math]

The average force is:

[math]\displaystyle{ \boxed{1350N} }[/math]

The negative sign means the force acts opposite the ball's original direction.

Common Mistakes

  • Forgetting that momentum and impulse are vectors.
  • Forgetting to include direction when velocity changes.
  • Confusing momentum with impulse.
  • Using force instead of net force.
  • Forgetting to multiply force by time.
  • Thinking a bigger force always means a bigger impulse, even though time also matters.
  • Forgetting that [math]\displaystyle{ N \cdot s }[/math] and [math]\displaystyle{ kg \cdot m/s }[/math] are equivalent units.

Real-World Applications

Airbags

Airbags reduce injury during car crashes by increasing the time it takes for a person's momentum to change. The person still needs to be brought to rest, so the change in momentum is the same. However, because the stopping time is longer, the average force on the person is smaller.

Sports

Impulse is important in sports such as baseball, tennis, soccer, and football. When a bat, racket, or foot stays in contact with a ball for a longer time, the impulse can increase. This changes the ball's momentum and can make it leave with a greater speed.

Landing from a Jump

When someone lands from a jump, bending the knees increases the stopping time. This lowers the average force on the person's legs. Landing with locked knees creates a shorter stopping time and a larger force.

Computational Model

In a computer simulation, momentum can be updated step by step using the momentum principle: 

p = p + Fnet*deltat

This means that during each small time interval, the object's momentum changes because of the net force acting on it.

A simple GlowScript simulation for this topic could show a ball moving forward while a force acts on it for a short time. The simulation could display the ball's changing momentum after the impulse is applied.

Practice Questions

  1. A 3 kg object moves with velocity [math]\displaystyle{ \langle 2,0,0 \rangle m/s }[/math]. What is its momentum?
  2. A force of [math]\displaystyle{ \langle 10,0,0 \rangle N }[/math] acts on an object for 4 seconds. What impulse is delivered?
  3. A cart has initial momentum [math]\displaystyle{ \langle 5,0,0 \rangle kg \cdot m/s }[/math]. An impulse of [math]\displaystyle{ \langle -2,0,0 \rangle N \cdot s }[/math] acts on it. What is its final momentum?

See Also

References

  • OpenStax. College Physics 2e, Momentum and Impulse.
  • The Physics Classroom. Momentum and Impulse Connection.
  • HyperPhysics. Impulse and Momentum.
  • Khan Academy. Impulse and Momentum.
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