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

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

References

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

Impulse and Momentum: Difference between revisions

Latest revision as of 16:42, 27 April 2026

Contents

Impulse and Momentum

Introduction

Main Idea

Key Equations

Units

Conceptual Explanation

Worked Example 1: Finding Final Momentum

Worked Example 2: Finding Average Force

Common Mistakes

Real-World Applications

Airbags

Sports

Landing from a Jump

Computational Model

Practice Questions

See Also

References

Navigation menu

@@ Line 1: / Line 1: @@
-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, represented by the letter <math>\vec{J}</math>, 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>. Recall from calculus that this is equivalent to <math>\vec{J} = \vec{F}_{net, avg} * \Delta t</math>, where <math>\Delta t</math> is the time interval over which the force is exerted and <math>\vec{F}_{net, avg}</math> is the time average of the net force over that time interval. For constant force, the average force is equal to that constant force, so the impulse <math>\vec{J}</math> exerted by constant force <math>\vec{F}</math> is <math>\vec{F} * \Delta t</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}_{system} = \vec{J}</math>. (It is important to note that these are vector quantities, so the impulse determines both the magnitude and direction of the change in momentum of a system.) 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. Often, the system in question consists of a single particle whose motion we want to predict. The impulse-momentum theorem is often used to calculate the final momentum of a particle using its initial momentum, the forces acting on it, and the durations of those forces.
+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.
-[[File:Impulsemomentumtheorem.png]]
+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.
-===A Mathematical Model===
+==Main Idea==
-====Derivation====
+Momentum is a vector quantity that depends on mass and velocity:
-The impulse-momentum theorem is a consequence of the momentum principle (see [[Newton's Second Law: the Momentum Principle]]). Below is its derivation:
+<math>\vec{p} = m\vec{v}</math>
-<math>\vec{F}_{net} = \frac{d\vec{p}}{dt}</math>
+Impulse is also a vector quantity. It is the effect of a net force acting over a time interval:
-can be arranged to <math>d\vec{p} = \vec{F}_{net}dt</math>.
+<math>\vec{J} = \vec{F}_{net,avg}\Delta t</math>
-Integrating both sides yields <math>\int d\vec{p} = \int \vec{F}_{net}dt</math>
+The impulse-momentum theorem says that impulse equals the change in momentum:
-which simplifies to <math>\Delta \vec{p} = \int \vec{F}_{net}dt = \vec{J}</math>, which may also be written as <math>\vec{p}_f = \vec{p}_i + \vec{J}</math>.
+<math>\vec{J} = \Delta \vec{p}</math>
-In the above derivation, <math>\vec{F}_{net}</math> is the <b>instantaneous</b> force acting on the system.
+This can also be written as:
-====Constant Force====
+<math>\vec{p}_f = \vec{p}_i + \vec{F}_{net}\Delta t</math>
-Remember that for constant force, <math>\vec{J}</math> and therefore <math>\Delta \vec{p}</math> are equal to <math>\vec{F} * \Delta t</math>. This form is often used for constant and nearly-constant forces because multiplying force by time is easier than performing a true integral.
+This equation is useful because it lets us predict how an object's momentum changes when a force acts on it.
-====Large Forces With Small Durations====
+==Key Equations==
-Sometimes, an impulse is delivered by a very strong force over a very short duration. An example of such a situation the striking of a baseball with a baseball bat. In these situations, when modeling the motion of the particle on which the impulse is exerted, the impulse is often considered to be instantaneous (that is, it takes place at a single point in time instead of over a time interval). A common consequence of this approximation is that the position of the particle does not change while the impulse is applied. Another consequence is that the magnitude of the force is considered infinite, although the infinitely small duration of the force results in a finite impulse.
+* <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>
-For example, consider a particle with momentum <1, 0, 0> kg*m/s. The particle is moving with constant speed in the +x direction. Then, an impulse of <0, 1, 0>N*s acts on the object. At the end of the impulse, the particle's new momentum is <1, 1, 0> kg*m/s; the particle is now moving diagonally. If the impulse was the product of a weak force over a long duration, the particle will have continued to move during its application, resulting in a curved trajectory like the one on the left. If the impulse was the product of a very strong force over a very short duration, the particle will abruptly change momentum, resulting in a sharp turn as shown on the right.
+Where:
-[[File:Shortdurationimpulse.jpg]]
+* <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
-Note that in both cases, the impulse vector is the same and so the final momentum is the same, but the shape of the trajectory is different because it depends on more than just the impulse vector- it depends on the magnitude and duration of the force.
+==Units==
-===A Computational Model===
+The unit for momentum is:
-In computational simulations of particles using [[Iterative Prediction]], a momentum variable is assigned to each particle. Such simulations 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 so small that the force does not significantly vary over the course of any one time step. By approximating the force as constant over these short intervals, we are allowed to use the formula <math>\Delta \vec{p} = \vec{J} = \vec{F} * \Delta t</math>. This is called the "update momentum" technique.
+<math>kg \cdot m/s</math>
-The following is an example of a line of vPython responsible for updating the momentum of a particle according to the impulse-momentum theorem:
+The unit for impulse is:
-p = p + fnet*deltat
+<math>N \cdot s</math>
-(This line would typically be written within a loop, continually adjusting momentum as the simulation progresses as part of the iterative prediction algorithm.)
+These units are equivalent because:
-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>1N = 1kg \cdot m/s^2</math>
-https://trinket.io/glowscript/ce43925647
+So:
-For more information, see [[Iterative Prediction]].
+<math>N \cdot s = kg \cdot m/s</math>
-==Examples==
+This makes sense because impulse equals change in momentum.
-===1. (Simple)===
+==Conceptual Explanation==
-A 2kg particle is travelling in the +y direction at 4m/s. After an impulse is applied to it, the particle is travelling in the +x direction at 3m/s. The impulse consisted of a force applied for .02s. What was the magnitude of the average force?
+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.
-<math>\vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i = <6, 0, 0> - <0, 8, 0> = <6, -8, 0>.</math> N*s
+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.
-The impulse is the product of the average force and the time interval over which it was applied.
+This is why catching a ball with stiff hands hurts more than catching it while moving your hands backward.
-<math>\vec{F} = \frac{\vec{J}}{\Delta t} = <6, -8, 0> / .02 = <300, -400, 0> </math> N
+==Worked Example 1: Finding Final Momentum==
-<math>|\vec{F}| = \sqrt{300^2 + (-400)^2 + 0^2} = 500</math> N
+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?
-Note that at any given moment, the force acting on the particle may be something other than <300, -400, 0>N; this is simply the <b>average</b> force over the .02s interval. (However, according to the mean value theorem, there must be <b>some</b> point in time within the interval at which the force acting on the particle is the average force.)
+First find the initial momentum:
-===2. (Simple)===
+<math>\vec{p}_i = m\vec{v}</math>
-A hockey puck is sliding along the ice with nearly constant momentum <16,0,5>kg*m/s when it is suddenly struck by a hockey stick with a constant force of <0,0,2000>N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?
+<math>\vec{p}_i = 2\langle 3,0,0 \rangle = \langle 6,0,0 \rangle kg \cdot m/s</math>
-'''Answer: <16,0,11>kg*m/s'''
+Now find the impulse:
-'''Explanation:'''
+<math>\vec{J} = \vec{F}_{net}\Delta t</math>
-<math>∆\vec{p} = \vec{F}_{net} * {∆t}</math> since force is constant
+<math>\vec{J} = \langle 4,0,0 \rangle(2) = \langle 8,0,0 \rangle N \cdot s</math>
-<math>\vec{p}_{final} - \vec{p}_{initial}</math> = <math>\vec{F}_{net}  * {∆t}</math>
+Now use the impulse-momentum theorem:
-<math>\vec{p}_{final}</math> - <16,0,5>kg*m/s = <0,0,2000>N * (3e-3)s
+<math>\vec{p}_f = \vec{p}_i + \vec{J}</math>
-<math>\vec{p}_{final}</math> = <16,0,11>kg*m/s
+<math>\vec{p}_f = \langle 6,0,0 \rangle + \langle 8,0,0 \rangle</math>
-Note: Since the force has no x or y components, these components of the puck's momentum remain constant.
+<math>\vec{p}_f = \langle 14,0,0 \rangle kg \cdot m/s</math>
-===3. (Simple)===
+The final momentum of the cart is:
-In outer space a rock of mass 5kg is acted on by a constant net force <29,-15,40>N during a 4s time interval. At the end of this time interval the rock has a velocity of <114,94,112>m/s. What is the rock's velocity at the beginning of the time interval?
+<math>\boxed{\langle 14,0,0 \rangle kg \cdot m/s}</math>
-'''Answer: <90.8,106,80>m/s'''
+==Worked Example 2: Finding Average Force==
-'''Explanation:'''
+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?
-<math>∆\vec{p} = \vec{F}_{net} * {∆t}</math> since force is constant
+Let the original direction be positive. Then:
-<math>\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}</math>
+<math>v_i = 40 m/s</math>
-<math>\vec{p} = m * \vec{v}</math>
+<math>v_f = -50 m/s</math>
-<math>m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}</math>
+Find the change in momentum:
-(5kg * <114,94,112>m/s) - (5kg * <math>\vec{v}_{initial}</math>) = <29,-15,40>N * 4s
+<math>\Delta p = m(v_f - v_i)</math>
-<math>\vec{v}_{initial}</math> = (<29,-15,40>N * 4s - 5kg * <114,94,112>m/s)/5kg = <90.8,106,80>m/s
+<math>\Delta p = 0.15(-50 - 40)</math>
-===4. (Middling)===
+<math>\Delta p = 0.15(-90)</math>
-This problem combines the impulse-momentum theorem with a number of kinematic equations, some of which take advantage of the constant acceleration due to gravity.
+<math>\Delta p = -13.5 kg \cdot m/s</math>
-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.
+Now use:
-* Initial Momentum?
+<math>F_{avg} = \frac{\Delta p}{\Delta t}</math>
-* Rock's momentum after 0.25 seconds?
+<math>F_{avg} = \frac{-13.5}{0.01}</math>
-* Calculate the average velocity of the rock from just after it leaves your hand to 0.25 seconds later.
+<math>F_{avg} = -1350 N</math>
-* 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.
+The average force is:
+<math>\boxed{1350N}</math>
-* Initial Momentum?
+The negative sign means the force acts opposite the ball's original direction.
-<math>{m=0.5, \vec{v} = <0.1,4,0.3>}</math><br>
+==Common Mistakes==
-<math>{\vec{p} = m\vec{v} = <0.05,2,0.15> kg*m/s}</math><br><br>
-* Rock's momentum after 0.25 seconds?
+* Forgetting that momentum and impulse are vectors.
-<math>{\vec{p}_f=\vec{p}_i+\vec{F}_{net}(\Delta{t})}</math> (Note: this is the step involving the impulse-momentum theorem. We can simply multiply force by time instead of integrating because the force is constant.)
+* 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.
-<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>
+==Real-World Applications==
-:<math>{= <0.05,0.775,0.15> kgm/s}</math><br><br>
-* Calculate the velocity of the rock from just after it leaves your hand to 0.25 seconds later.
+===Airbags===
-<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>
+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.
-:<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.
+===Sports===
-<math>{\vec{r}_i=<0,1.2,0> m}</math><br>
+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.
-<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>
-===5. (Middling)===
+===Landing from a Jump===
-In a crash test, a truck with mass 2200 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.67 m shorter than before. <b>Make the somewhat unrealistic assumption that the force, and therefore the acceleration, is constant. This assumption is necessary to answer some of the questions below because several of the kinematic equations rely on constant acceleration.</b>
+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.
-(a) What is the average speed of the truck during the collision (that is, during the interval between first contact with the wall and coming to a stop)?
+==Computational Model==
-Here, we use kinematic equations to solve for average velocity!
+In a computer simulation, momentum can be updated step by step using the momentum principle:
+<pre>
+p = p + Fnet*deltat
+</pre>
-[[File:Wiki_Image_1.png]]
+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.
-(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)
+==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?
-[[File:Wiki_Image_2.0.png]]
+==See Also==
+* [[Linear Momentum]]
+* [[Newton's Second Law: the Momentum Principle]]
-(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?
+* [[Conservation of Momentum]]
+* [[Collisions]]
-[[File:Wki Image 3.0.png]]
+* [[Iterative Prediction]]
-===6. (Difficult)===
-At t=0, an 8kg particle is released from rest near the surface of the earth. In addition to the gravitational force, a time dependent force <math>\vec{F(t)}</math> acts on the particle, where <math>\vec{F(t)} = < 200 t \sin (t^2), 0, 0> </math> Newtons. Use g = 10 for this problem.
-What is the velocity of the particle at t=2?
-The net force acting on the particle can be found by adding the gravitational force and <math>\vec{F}</math>:
-<math>\vec{F}_{net} = < 200 t \sin (t^2), -80, 0></math> N.
-Let us find the impulse exerted on the particle:
-<math>\vec{J} = \int_0^2 \vec{F}_{net} dt</math> N*s
-<math>= \int_0^2 < t \sin (t^2), -80, 0> dt</math> N*s
-<math>= [< -100 \cos (t^2), -80t, 0>]_0^2 </math> N*s (Note: u-substitution was used to integrate <math>t \sin (t^2)</math>, setting u equal to <math>t^2</math>.)
-= <math> < -100 \cos 4 + 100, -160, 0></math> N*s
-<math>\vec{p}_f = \vec{p}_f + \vec{J}</math> so <math>\vec{p}_f = < -100 \cos 4 + 100, -160, 0></math> kg*m/s
-<math>\vec{v}_f = \frac{\vec{p}_f}{m} = <- \frac{100}{8} \cos 4 + \frac{100}{8}, -20, 0> = <4.32, -20, 0></math> m/s
-==Connectedness==
-===Scenario: follow through===
-Often in baseball and similar sports, batters are encouraged to "follow through" when hitting the ball. This does not increase the force with which the batter hits the ball, which depends on the strength of the batter, but it does increase the time of contact between the bat and the ball. This increases the duration of the force of the hit, which increases the impulse exerted on the ball, and therefore, its final speed at the end of contact.
-===Application: airbag industry===
-When a car comes to a sudden stop, such as in a collision, the driver's body will continue to move forward with its original momentum until it collides with something, such as the dashboard or an airbag, that brings them to rest. Because of the impulse-momentum theorem, for a driver with a given initial momentum, a specific impulse is required to bring the driver to rest. The purpose of airbags is to deliver this impulse over a longer time interval than a dashboard would by deforming. This longer time interval allows the force acting on the driver to be lower while achieving the same impulse, which minimizes risk of injury.
-Similarly, it is better for one's joints if the knees are bent when landing from a jump because the contraction of the legs allows the impulse of landing to last longer and therefore have a smaller force.
-== See also ==
-*[[Linear Momentum]]
-*[[Newton's Second Law: the Momentum Principle]]
-*[[Conservation of Momentum]]
-*[[Iterative Prediction]]
-===Further reading===
-===External links===
-<ol>
-<li>http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection</li>
-<li>https://www.asc.ohio-state.edu/humanic.1/p111_lecture13</li>
-<li>https://study.com/academy/lesson/momentum-and-impulse-definition-theorem-and-examples.html</li>
-<li>https://www.physicsclassroom.com/class/momentum/Lesson-1/Real-World-Applications</li>
-<li>https://www.youtube.com/watch?v=qu_P4lbmV_I</li>
-</ol>
 ==References==
-*Sherwood, Bruce A. "2.1 The Momentum Principle." Matter & Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley & Sons, 2015. 45-50. Print. Modern Mechanics. 8 April, 2018.
+* OpenStax. ''College Physics 2e'', Momentum and Impulse.
+* The Physics Classroom. ''Momentum and Impulse Connection.''
-*Michael Fowler, U. Va. Physics. "Momentum, Work and Energy." Galileo and Einstein: Lecture Index. <http://galileoandeinstein.physics.virginia.edu/lectures/lecturelist.html>14 April, 2018.
+* HyperPhysics. ''Impulse and Momentum.''
+* Khan Academy. ''Impulse and Momentum.''
-[[Category: Momentum]]

Impulse and Momentum: Difference between revisions

Latest revision as of 16:42, 27 April 2026

Impulse and Momentum

Introduction

Main Idea

Key Equations

Units

Conceptual Explanation

Worked Example 1: Finding Final Momentum

Worked Example 2: Finding Average Force

Common Mistakes

Real-World Applications

Airbags

Sports

Landing from a Jump

Computational Model

Practice Questions

See Also

References

Navigation menu

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