Momentum with respect to external Forces: Difference between revisions
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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. | 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=== | ===A Mathematical Model=== | ||
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{| | {| | ||
|- | |- | ||
| | |<small> </small> | ||
Acceleration is defined as:<br> | Acceleration is defined as:<br> | ||
<math>{\vec{a} = \frac{\vec{v}_{final} - \vec{v}_{intial}}{\Delta{t}} = \frac{d\vec{v}}{dt}}</math><br> | <math>{\vec{a} = \frac{\vec{v}_{final} - \vec{v}_{intial}}{\Delta{t}} = \frac{d\vec{v}}{dt}}</math><br> | ||
| | |<small> </small> | ||
:Momentum and change in momentum are defined as: | :Momentum and change in momentum are defined as: | ||
:<math>{p = mv}</math> and <math>{\Delta{p} = \frac{d\vec{p}}{dt} = m*\frac{d\vec{v}}{dt}}</math> | :<math>{p = mv}</math> and <math>{\Delta{p} = \frac{d\vec{p}}{dt} = m*\frac{d\vec{v}}{dt}}</math> | ||
|} | |} | ||
This yields this alternate equation to consider scenarios | This yields this alternate equation to consider scenarios:<br> | ||
<big><math>{\vec{F}_{net} = m\vec{a}}</math></big> | :<big><math>{\vec{F}_{net} = m\vec{a}}</math></big><br> | ||
:'''F<sub>net</sub>''' is the net force | |||
:'''m''' is the mass of the object in question | |||
:'''a''' is the acceleration of the object in question<br><br> | |||
The superposition principle states that the net force is the resultant of the sum of all the component forces. This | 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.<br> | ||
:<math>{\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + ... + \vec{F}_n}</math><br> | :<math>{\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + ... + \vec{F}_n}</math><br> | ||
:'''F<sub>net</sub>''' is the net force | :'''F<sub>net</sub>''' is the net force | ||
:'''F<sub>n</sub>''' is a component force in the system | :'''F<sub>n</sub>''' is a component force in the system | ||
:'''n''' is the possible number of component forces in the system | :'''n''' is the possible number of component forces in the system | ||
This is very important for solving problems concerning net force. | |||
===A Computational Model=== | ===A Computational Model=== | ||
==Examples== | ==Examples== | ||
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 v=<0.1,4.0,0.3> m/s Ignore air resistance. | 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. | ||
* 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? | |||
<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> | |||
:<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. | |||
m | <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> | |||
:<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{r}_i=<0,1.2,0> m}</math><br> | |||
<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> | |||
===Simple=== | |||
===Middling=== | |||
===Difficult=== | |||
==Connectedness== | ==Connectedness== | ||
Revision as of 21:48, 18 April 2016
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
- Fnet is the net force from the surroundings.
The first equation can be further rearranged.
|
Acceleration is defined as: |
|
This yields this alternate equation to consider scenarios:
- [math]\displaystyle{ {\vec{F}_{net} = m\vec{a}} }[/math]
- Fnet 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]
- Fnet is the net force
- Fn 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.
A Computational Model
Examples
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]
Simple
Middling
Difficult
Connectedness
This topic is the basis behind calculating most forms of linear movement with simple 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.
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 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.