Impulse and Momentum

This page defines impulse and describes its relationship to momentum.

The Main Idea

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]\displaystyle{ \vec{J} = \int \vec{F}_{net}dt }[/math]. (Recall that for constant force, this integral reduces to [math]\displaystyle{ \vec{J} = \vec{F} \Delta t }[/math].)Impulse is represented by the letter [math]\displaystyle{ \vec{J} }[/math]. The most commonly used metric unit for impulse is the Newton*second.

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]\displaystyle{ \Delta \vec{p} = \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 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.

A Mathematical Model

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

[math]\displaystyle{ \vec{F}_{net} = \frac{d\vec{p}}{dt} }[/math]

can be arranged to [math]\displaystyle{ d\vec{p} = \vec{F}_{net}dt }[/math].

Integrating both sides yields [math]\displaystyle{ \int d\vec{p} = \int \vec{F}_{net}dt }[/math]

which simplifies to [math]\displaystyle{ \Delta \vec{p} = \int \vec{F}_{net}dt = \vec{J} }[/math], which may also be written as [math]\displaystyle{ \vec{p}_f = \vec{p}_i + \vec{J} }[/math].

In the above derivation, [math]\displaystyle{ \vec{F}_{net} }[/math] is the instantaneous force acting on the system.

According to calculus,

[math]\displaystyle{ \int \vec{F(t)}dt = \vec{F}_{avg} \Delta t }[/math], where [math]\displaystyle{ \vec{F}_{avg} }[/math] is the time average of the force over the time interval [math]\displaystyle{ \Delta t }[/math]. When the force is constant, the average force is equal to that constant force, so the impulse [math]\displaystyle{ \vec{J} }[/math] exerted by constant force [math]\displaystyle{ \vec{F} }[/math] is [math]\displaystyle{ \vec{F} \Delta t }[/math].

A Computational Model

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.

The following is an example of a line of vPython responsible for updating the momentum of a particle according to the impulse-momentum theorem:

p = p + fnet*deltat

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:

https://trinket.io/glowscript/ce43925647

For more information, see iterative prediction.

Examples

1. (Simple)

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?

[math]\displaystyle{ \vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i = \lt 6, 0, 0\gt - \lt 0, 8, 0\gt = \lt 6, -8, 0\gt . }[/math] N*s

The impulse is the product of the average force and the time interval over which it was applied.

[math]\displaystyle{ \vec{F} = \frac{\vec{J}}{\Delta t} = \lt 6, -8, 0\gt / .02 = \lt 300, -400, 0\gt }[/math] N

[math]\displaystyle{ |\vec{F}| = \sqrt{300^2 + (-400)^2 + 0^2} = 500 }[/math] N

Note that at any given moment, the force acting on the particle may have been something other than <300, -400, 0>; this is simply the average force.

2. (Simple)

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?

Answer: <16,0,11>kg*m/s

Explanation:

[math]\displaystyle{ ∆\vec{p} = \vec{F}_{net} * {∆t} }[/math] since force is constant

[math]\displaystyle{ \vec{p}_{final} - \vec{p}_{initial} }[/math] = [math]\displaystyle{ \vec{F}_{net} * {∆t} }[/math]

[math]\displaystyle{ \vec{p}_{final} }[/math] - <16,0,5>kg*m/s = <0,0,2000>N * (3e-3)s

[math]\displaystyle{ \vec{p}_{final} }[/math] = <16,0,11>kg*m/s

Note: Since the force has no x or y components, these components of the puck's momentum remain constant.

3. (Simple)

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?

Answer: <90.8,106,80>m/s

Explanation:

[math]\displaystyle{ ∆\vec{p} = \vec{F}_{net} * {∆t} }[/math] since force is constant

[math]\displaystyle{ \vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t} }[/math]

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

[math]\displaystyle{ m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t} }[/math]

(5kg * <114,94,112>m/s) - (5kg * [math]\displaystyle{ \vec{v}_{initial} }[/math]) = <29,-15,40>N * 4s

[math]\displaystyle{ \vec{v}_{initial} }[/math] = (<29,-15,40>N * 4s - 5kg * <114,94,112>m/s)/5kg = <90.8,106,80>m/s

4. (Intermediate)

This problem combines the impulse-momentum theorem with a number of kinematic equations, some of which take advantage of the constant acceleration.

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]\displaystyle{ {\vec{v}=\lt 0.1,4.0,0.3\gt m/s} }[/math] Ignore air resistance.

• Initial Momentum?

• Rock's momentum after 0.25 seconds?

• Calculate the average velocity of the rock from just after it leaves your hand to 0.25 seconds later.

• 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.

• Initial Momentum?

[math]\displaystyle{ {m=0.5, \vec{v} = \lt 0.1,4,0.3\gt } }[/math]
[math]\displaystyle{ {\vec{p} = m\vec{v} = \lt 0.05,2,0.15\gt kg*m/s} }[/math]

• Rock's momentum after 0.25 seconds?

[math]\displaystyle{ {\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.)

[math]\displaystyle{ {\vec{p}_f = \lt 0.05,2,0.15\gt + \lt 0,(-9.8)(0.5),0\gt *0.25 = \lt 0.05,2,0.15\gt + \lt 0,-1.225,0\gt } }[/math]

[math]\displaystyle{ {= \lt 0.05,0.775,0.15\gt kgm/s} }[/math]

• Calculate the average velocity of the rock from just after it leaves your hand to 0.25 seconds later.

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

[math]\displaystyle{ {\vec{v}_f=(\vec{p}_f/m)=(1/0.5)*\lt 0.05,0.775,0.15\gt } }[/math]

[math]\displaystyle{ {=\lt 0.1,1.55,0.3\gt m/s} }[/math]

[math]\displaystyle{ {\vec{v}_{avg}=(\vec{v}_i+\vec{v}_f)/2 = (0.5)*[\lt 0.1,4,0.3\gt + \lt 0.1,1.55,0.3\gt ]= (0.5)\lt 0.2,5.55,0.6\gt } }[/math]

[math]\displaystyle{ {=\lt 0.1,2.775,0.3\gt m/s} }[/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.

[math]\displaystyle{ {\vec{r}_i=\lt 0,1.2,0\gt m} }[/math]
[math]\displaystyle{ {\vec{r}_f=\vec{r}_i+\vec{v}_{avg}(\Delta{t})= \lt 0,1.2,0\gt + \lt 0.1,2.775,0.3\gt *0.25} }[/math]

[math]\displaystyle{ {= \lt 0,1.2,0\gt + \lt 0.025,0.694,0.075\gt } }[/math]

[math]\displaystyle{ {=\lt 0.025,1.894,0.075\gt m} }[/math]

5. (Intermediate)

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. Make the somewhat unrealistic assumption that the force, and therefore the acceleration, was constant, which is necessary to answer the questions below.

(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)?

Here, we use kinematic equations to solve for average velocity!

(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?)

(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?

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]\displaystyle{ \vec{F(t)} }[/math] acts on the particle, where [math]\displaystyle{ \vec{F(t)} = \lt t \sin (t^2), 0, 0\gt }[/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]\displaystyle{ \vec{F} }[/math]:

[math]\displaystyle{ \vec{F}_{net} = \lt t \sin (t^2), -80, 0\gt }[/math] N.

Let us find the impulse exerted on the particle:

[math]\displaystyle{ \vec{J} = \int_0^2 \vec{F}_{net} dt }[/math] N*s

[math]\displaystyle{ = \int_0^2 \lt t \sin (t^2), -80, 0\gt dt }[/math] N*s

[math]\displaystyle{ = [\lt - \frac{1}{2} \cos (t^2), -80t, 0\gt ]_0^2 }[/math] N*s (Note: u-substitution was used to integrate [math]\displaystyle{ t \sin (t^2) }[/math], setting u equal to [math]\displaystyle{ t^2 }[/math].)

= <[math]\displaystyle{ - \frac{1}{2} \cos 4 + \frac{1}{2}, -160, 0\gt }[/math] N*s

[math]\displaystyle{ \vec{p}_f = \vec{p}_f + \vec{J} }[/math] so [math]\displaystyle{ \vec{p}_f = \lt - \frac{1}{2} \cos 4 + \frac{1}{2}, -160, 0\gt }[/math] kg*m/s

[math]\displaystyle{ \vec{v}_f = \frac{\vec{p}_f}{m} = \lt - \frac{1}{16} \cos 4 + \frac{1}{16}, -20, 0\gt = \lt .0216, -20, 0\gt }[/math] m/s

Connectedness

Scenario 1

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.

Scenario 2

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. 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, allowing the force acting on the driver to be lower while achieving the same impulse, which minimizes risk of injury.

See also

External links

1. http://acme.highpoint.edu/~atitus/phy221/lecture-notes/2-2-momentum-principle.pdf

References

[1] William, Harris. "How Newton's Laws of Motion Work" 29 July 2008. HowStuffWorks.com. <https://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm> 11 April, 2018.

[2] 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.

[3] Fenton, Flavio. "Momentum and Second Newton's Law." August-September Lectures. 2017.

[4] “Momentum Conservation Principle.” The Physics Classroom, <www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle.> 13 April, 2018.

[5] 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.