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]. 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]. 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.

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].

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 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

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]

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]

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

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.

Middling

This problem combines the impulse-momentum theorem with a number of kinematic equations that 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?

[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]

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)} }[/math] = <[math]\displaystyle{ t \sin (t^2) }[/math], 0, 0>. 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} }[/math] = <[math]\displaystyle{ t \sin (t^2) }[/math], -80, 0>.

Let us find the impulse exerted on the particle:

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

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

[math]\displaystyle{ = [\lt - \frac{1}{2} \cos (t^2), -80t, 0\gt ]_0^2 }[/math] (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} }[/math], -160, 0>

[math]\displaystyle{ \vec{p}_f = \vec{p}_f + \vec{J} }[/math] so [math]\displaystyle{ \vec{p}_f }[/math] = <[math]\displaystyle{ - \frac{1}{2} \cos 4 + \frac{1}{2} }[/math], -160, 0>

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