Momentum with respect to external Forces

Editing claimed by estaniforth3

Claimed by vkt3

The Main Idea

Momentum in an open system is fundamentally different from that within a closed system. Because an open system allows for the existence of external forces, momentum is no longer conserved. The sum of the individual elements' respective momentums, the net momentum, is not guaranteed to equal zero. However, this is not as much of an obstacle as one might think. Because the net momentum is zero in the absence of external force, the change in momentum is directly related to the net external force. This was established in Newton's Second Law of Motion or Momentum Principle: in a system of constant mass, the change in momentum over time is equal to the net force.

A Mathematical Model

An equation expressing this idea is:

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

p is the momentum of the system

F_net is the net force from the surroundings.

The first equation can be further rearranged.

Acceleration is defined as: [math]\displaystyle{ {\vec{a} = \frac{\vec{v}_{final} - \vec{v}_{intial}}{\Delta{t}} = \frac{d\vec{v}}{dt}} }[/math]

Momentum and change in momentum are defined as: [math]\displaystyle{ {p = mv} }[/math] and [math]\displaystyle{ {\Delta{p} = \frac{d\vec{p}}{dt} = m*\frac{d\vec{v}}{dt}} }[/math]

This yields this alternate equation to consider scenarios:

[math]\displaystyle{ {\vec{F}_{net} = m\vec{a}} }[/math]

F_net is the net force

m is the mass of the object in question

a is the acceleration of the object in question

The superposition principle states that the net force is the resultant of the sum of all the component forces. This results in the following equation.

[math]\displaystyle{ {\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + ... + \vec{F}_n} }[/math]

F_net is the net force

F_n is a component force in the system

n is the possible number of component forces in the system

This is very important for solving problems concerning net force.

Examples

Simple

A molecule traveling to the right at 5 m/s splits into two equal parts. What are the two parts' new velocity?

• Because there are no external forces, the initial and final momentum are the same.

[math]\displaystyle{ {\vec{p}_{inital} = \vec{p}_{final}} }[/math]
[math]\displaystyle{ {M\vec{v}_{intial} = m\vec{v}_{final} + m\vec{v}_{final}} }[/math]
[math]\displaystyle{ {M = m + m = 2m} }[/math] so [math]\displaystyle{ {m = \frac{M}{2}} }[/math]
[math]\displaystyle{ {\vec{v}_{final} = \frac{\vec{v}_{intial}}{2}} }[/math]

[math]\displaystyle{ {=\lt 2.5,0,0\gt m/s} }[/math]

Difficult

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]

[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]

Connectedness

This topic is the basis behind calculating and predicting movement due to external forces. While a simple formula, this equation is a powerful tool as that it can include any number of forces acting on a system, and show the change in momentum of an object. It holds a particular interest to me because it can be used for simulation and animation of almost anything, so long as you put in the work to include every force acting in the scenario.

This equation can be used to calculate the simple movements of objects in the vacuum of space with respect to the magnitude of forces acting upon the system.

History

The Momentum Principle was established by Isaac Newton as Newton's Second Law of Motion. It was born from Newton's First Law which states that an object at rest will remain at rest, and an object in motion will remain in motion unless acted upon by an external force, the Principle of Inertia.