Physics Book - User contributions [en]

RC Circuit

2016-11-28T04:15:54Z

Sfischer8: /* References */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:Primary1.jpg]]

The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.
[[File:Primary2.jpg]]

Once you correctly label the diagram you can begin the problem.

1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.

2.Determine the initial current passing through each bulb.

3. Determine the current passing through each bulb for a long time after the switch has been closed.

[[File:Workwork.jpg]]

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I enjoy listening to music and RC circuits are used in speakers and headphones. They allow the current to be regulated and for the music to to played at the right pitch and volume.
[[File:Speakercirc.jpg]]

2. How is it connected to your major?
:: I'm a Mechanical Engineering major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled. I would be designing the machines not wiring the curcuits, but it is important to understand how the RC circuits are used in order to properly design and implement the machines.
3. Is there an interesting industrial application?
:: RC circuits are used as filters for almost all types of electronics. The initial input is taken across the RC circuit and the output is taken across the capacitor. This allows the capacitor in the circuit to be charged to different values and used when the desired one is obtained. It is an easy way to allow multiple energy outputs across a single circuit rather than using a bunch of different circuits.
[[File:Fil42.gif]]

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists. Once his work was recognized and accepted by the scientific community people began to realize the many applications of RC circuits as filters and began to further research and implement them in their own work.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

http://www.animations.physics.unsw.edu.au/jw/RCfilters.html

[[Category:Simple Circuits]]

RC Circuit

2016-11-28T04:15:01Z

Sfischer8: /* History */

RC Circuit

2016-11-28T04:13:24Z

Sfischer8: /* Connectedness */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:Primary1.jpg]]

The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.
[[File:Primary2.jpg]]

Once you correctly label the diagram you can begin the problem.

1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.

2.Determine the initial current passing through each bulb.

3. Determine the current passing through each bulb for a long time after the switch has been closed.

[[File:Workwork.jpg]]

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I enjoy listening to music and RC circuits are used in speakers and headphones. They allow the current to be regulated and for the music to to played at the right pitch and volume.
[[File:Speakercirc.jpg]]

2. How is it connected to your major?
:: I'm a Mechanical Engineering major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled. I would be designing the machines not wiring the curcuits, but it is important to understand how the RC circuits are used in order to properly design and implement the machines.
3. Is there an interesting industrial application?
:: RC circuits are used as filters for almost all types of electronics. The initial input is taken across the RC circuit and the output is taken across the capacitor. This allows the capacitor in the circuit to be charged to different values and used when the desired one is obtained. It is an easy way to allow multiple energy outputs across a single circuit rather than using a bunch of different circuits.
[[File:Fil42.gif]]

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

File:Fil42.gif

2016-11-28T04:13:17Z

Sfischer8:

RC Circuit

2016-11-28T04:03:48Z

Sfischer8: /* Connectedness */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:Primary1.jpg]]

The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.
[[File:Primary2.jpg]]

Once you correctly label the diagram you can begin the problem.

1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.

2.Determine the initial current passing through each bulb.

3. Determine the current passing through each bulb for a long time after the switch has been closed.

[[File:Workwork.jpg]]

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I enjoy listening to music and RC circuits are used in speakers and headphones. They allow the current to be regulated and for the music to to played at the right pitch and volume.
[[File:Speakercirc.jpg]]

2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

File:Speakercirc.jpg

2016-11-28T04:01:43Z

Sfischer8:

RC Circuit

2016-11-28T04:01:24Z

Sfischer8: /* Connectedness */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:Primary1.jpg]]

The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.
[[File:Primary2.jpg]]

Once you correctly label the diagram you can begin the problem.

1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.

2.Determine the initial current passing through each bulb.

3. Determine the current passing through each bulb for a long time after the switch has been closed.

[[File:Workwork.jpg]]

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I enjoy listening to music and RC circuits are used in speakers and headphones. They allow the current to be regulated and for the music to to played at the right pitch and volume.

2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

RC Circuit

2016-11-28T03:58:02Z

Sfischer8: /* Difficult */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:Primary1.jpg]]

The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.
[[File:Primary2.jpg]]

Once you correctly label the diagram you can begin the problem.

1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.

2.Determine the initial current passing through each bulb.

3. Determine the current passing through each bulb for a long time after the switch has been closed.

[[File:Workwork.jpg]]

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

File:Workwork.jpg

2016-11-28T03:57:19Z

Sfischer8:

RC Circuit

2016-11-28T03:46:22Z

Sfischer8: /* Difficult */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:Primary1.jpg]]

The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.
[[File:Primary2.jpg]]

Once you correctly label the diagram you can begin the problem.

1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.

2.Determine the initial current passing through each bulb.

3. Determine the current passing through each bulb for a long time after the switch has been closed.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

RC Circuit

2016-11-28T03:41:32Z

Sfischer8: /* Difficult */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:Primary1.jpg]]

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

File:Primary2.jpg

2016-11-28T03:40:59Z

Sfischer8:

File:Primary1.jpg

2016-11-28T03:40:47Z

Sfischer8:

File:Circuittwo.jpg

2016-11-28T03:34:37Z

Sfischer8:

File:Circuituno.jpg

2016-11-28T03:34:21Z

Sfischer8:

RC Circuit

2016-11-28T03:29:41Z

Sfischer8: /* Difficult */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

RC Circuit

2016-11-28T03:04:15Z

Sfischer8: /* A Mathematical Model */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
Suppose one wished to capture the picture of a bullet (moving at 0.04 m/s ) that was passing through an orange. The duration of the flash is related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was 10.0 Ω? Assume the oragne is a sphere with a diameter of 0.08 m.

Solution:

You know the velocity of the bullet and the distance. You can find the time using Physics I principles such as <math>{time} = {\frac{distance}{velocity}} = {\frac{.08}{.04}} = {.0032 seconds} </math>

the time becomes equal to τ, so:

<math>{C} = {\frac{τ}{v}} = {\frac{0.0032}{10Ω}} = {32μF} </math>

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

File:Outputrc.jpg

2016-11-28T03:03:43Z

Sfischer8:

File:RC output.png

2016-11-28T02:58:53Z

Sfischer8:

RC Circuit

2016-11-28T02:55:41Z

Sfischer8: /* A Mathematical Model */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
Suppose one wished to capture the picture of a bullet (moving at 0.04 m/s ) that was passing through an orange. The duration of the flash is related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was 10.0 Ω? Assume the oragne is a sphere with a diameter of 0.08 m.

Solution:

You know the velocity of the bullet and the distance. You can find the time using Physics I principles such as <math>{time} = {\frac{distance}{velocity}} = {\frac{.08}{.04}} = {.0032 seconds} </math>

the time becomes equal to τ, so:

<math>{C} = {\frac{τ}{v}} = {\frac{0.0032}{10Ω}} = {32μF} </math>

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

File:Rc234.jpg

2016-11-28T02:54:34Z

Sfischer8:

RC Circuit

2016-11-28T02:54:08Z

Sfischer8: /* A Mathematical Model */

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

You know that <math>{I} = {\frac{dQ}{dt}}</math> due to the rate at which charge builds up the positive capactior plate.

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

If you exponentiate both sides, the following equation is achieved.

<math>{\frac{IR}{emf}} = {e^{\frac{-t}{RC}}}</math>

<math>{I} = {\frac{emf}{R}e^{\frac{-t}{RC}}={\frac{dQ}{dt}}}</math>

<math>{dQ} = \int_0^t{\frac{emf}{R}e^{\frac{-t}{RC}}\,\mathrm{d}t}</math>

<math>{Q} = {C(emf)(1-e^{\frac{-t}{RC}})}</math>.

Since <math>{V} = {Q/C}</math>, thus

<math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>.

This is the formula for the change in voltage of a series RC circuit with respect to time.

The RC time constant formula is:

<math>{τ} = {RC}</math>

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
Suppose one wished to capture the picture of a bullet (moving at 0.04 m/s ) that was passing through an orange. The duration of the flash is related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was 10.0 Ω? Assume the oragne is a sphere with a diameter of 0.08 m.

Solution:

You know the velocity of the bullet and the distance. You can find the time using Physics I principles such as <math>{time} = {\frac{distance}{velocity}} = {\frac{.08}{.04}} = {.0032 seconds} </math>

the time becomes equal to τ, so:

<math>{C} = {\frac{τ}{v}} = {\frac{0.0032}{10Ω}} = {32μF} </math>

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

File:Rc123.jpg

2016-11-28T02:52:43Z

Sfischer8:

RC Circuit

2016-11-27T18:37:02Z

Sfischer8:

Claimed by Sean Fischer Fall 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems

<math>{ΔV}_{round trip} = {0} </math>

<math>{ΔV} = {I}{R} </math>

<math>{Q} = {C}{V} </math>

In a loop of a circuit, the change of potential difference has to be zero. The energy equation for the RC Circuit in the figure at the top of the page is:

<math>{ΔV}_{round trip} = emf-RI-Q/C = 0</math>

At the final state of the circuit after the current has dropped to zero and the capacitor is fully charged, <math>{RI} = {0} </math>. The new equation for the final state is:

<math>{V}_{round trip} = emf-Q/C = 0</math>

<math>{Q} = emf*C</math>

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

You know that <math>{I} = {\frac{dQ}{dt}}</math> due to the rate at which charge builds up the positive capactior plate.

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

If you exponentiate both sides, the following equation is achieved.

<math>{\frac{IR}{emf}} = {e^{\frac{-t}{RC}}}</math>

<math>{I} = {\frac{emf}{R}e^{\frac{-t}{RC}}={\frac{dQ}{dt}}}</math>

<math>{dQ} = \int_0^t{\frac{emf}{R}e^{\frac{-t}{RC}}\,\mathrm{d}t}</math>

<math>{Q} = {C(emf)(1-e^{\frac{-t}{RC}})}</math>.

Since <math>{V} = {Q/C}</math>, thus

<math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>.

This is the formula for the change in voltage of a series RC circuit with respect to time.

The RC time constant formula is:

<math>{τ} = {RC}</math>

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
Suppose one wished to capture the picture of a bullet (moving at 0.04 m/s ) that was passing through an orange. The duration of the flash is related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was 10.0 Ω? Assume the oragne is a sphere with a diameter of 0.08 m.

Solution:

You know the velocity of the bullet and the distance. You can find the time using Physics I principles such as <math>{time} = {\frac{distance}{velocity}} = {\frac{.08}{.04}} = {.0032 seconds} </math>

the time becomes equal to τ, so:

<math>{C} = {\frac{τ}{v}} = {\frac{0.0032}{10Ω}} = {32μF} </math>

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

Elastic Collisions

2016-04-12T02:03:03Z

Sfischer8: /* History */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

[[File:Elastischer_stoß2.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible. For a lot of the complicated problems, starting with a diagram and writing down the given information and plugging that into either the momentum or energy principle can help you move forward in the problem. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-21 kg.

[[File:Problem2.0.png|300px]]

A lot of people freak out when they see problems involving atomic particles and electrical forces, but the concept is the same no matter the scale. For this problem we are given masses, velocities, and momentum, but the equation p = mv allows us to easily handle these different values. Once again drawing a diagram helps to understand what is actually happening in the problem, making it easier to put the values in the correct places and solve.

===Difficult===

Okay, let's get a little bit trickier here...

There is a 400 kg train traveling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

[[File:Problem3.0.png|400px]]

This was a difficult problem because of the assumptions we had to make in order to solve it. Because the collision was elastic we were able to disregard the change in internal energy in the energy principle, and since it happened so quickly the work done by the track and heat transfer between the surroundings was negligible. It is tough to tell what assumptions you can make and which ones you can't in problems like these, which is what can make them difficult. On most exams or quizzes the collision problems will be inelastic simply because they are harder, but it is important to understand the fundementals of elastic collisions and how to solve problems involving them.

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

Another example of collisions in real life is billiards. This is one of the most accurate real life examples of elastic collisions. One ball hits another ball at rest, and if done right the first ball stops and transfers nearly all its kinetic energy into kinetic energy in the second ball.

[[File:Pool1.0.jpg]]

Another example is hitting a baseball. Hitting a ball off the end or to close the hands of a bat will cause some vibrational energy from the collision, but if the ball catches the sweet spot the collision is very elastic, and you don't feel any vibration when you strike the ball.

[[File:Hittingabaseball.jpg|500px]]

==History==

Collisions are certainly not a knew concept in the world. Ever since the beginning or time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton or Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions.

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-12T01:59:39Z

Sfischer8: /* The Main Idea */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

[[File:Elastischer_stoß2.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible. For a lot of the complicated problems, starting with a diagram and writing down the given information and plugging that into either the momentum or energy principle can help you move forward in the problem. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-21 kg.

[[File:Problem2.0.png|300px]]

A lot of people freak out when they see problems involving atomic particles and electrical forces, but the concept is the same no matter the scale. For this problem we are given masses, velocities, and momentum, but the equation p = mv allows us to easily handle these different values. Once again drawing a diagram helps to understand what is actually happening in the problem, making it easier to put the values in the correct places and solve.

===Difficult===

Okay, let's get a little bit trickier here...

There is a 400 kg train traveling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

[[File:Problem3.0.png|400px]]

This was a difficult problem because of the assumptions we had to make in order to solve it. Because the collision was elastic we were able to disregard the change in internal energy in the energy principle, and since it happened so quickly the work done by the track and heat transfer between the surroundings was negligible. It is tough to tell what assumptions you can make and which ones you can't in problems like these, which is what can make them difficult. On most exams or quizzes the collision problems will be inelastic simply because they are harder, but it is important to understand the fundementals of elastic collisions and how to solve problems involving them.

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

Another example of collisions in real life is billiards. This is one of the most accurate real life examples of elastic collisions. One ball hits another ball at rest, and if done right the first ball stops and transfers nearly all its kinetic energy into kinetic energy in the second ball.

[[File:Pool1.0.jpg]]

Another example is hitting a baseball. Hitting a ball off the end or to close the hands of a bat will cause some vibrational energy from the collision, but if the ball catches the sweet spot the collision is very elastic, and you don't feel any vibration when you strike the ball.

[[File:Hittingabaseball.jpg|500px]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

File:Elastischer stoß2.gif

2016-04-12T01:58:33Z

Sfischer8:

Elastic Collisions

2016-04-12T01:55:28Z

Sfischer8: /* Connectedness */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible. For a lot of the complicated problems, starting with a diagram and writing down the given information and plugging that into either the momentum or energy principle can help you move forward in the problem. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-21 kg.

[[File:Problem2.0.png|300px]]

A lot of people freak out when they see problems involving atomic particles and electrical forces, but the concept is the same no matter the scale. For this problem we are given masses, velocities, and momentum, but the equation p = mv allows us to easily handle these different values. Once again drawing a diagram helps to understand what is actually happening in the problem, making it easier to put the values in the correct places and solve.

===Difficult===

Okay, let's get a little bit trickier here...

There is a 400 kg train traveling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

[[File:Problem3.0.png|400px]]

This was a difficult problem because of the assumptions we had to make in order to solve it. Because the collision was elastic we were able to disregard the change in internal energy in the energy principle, and since it happened so quickly the work done by the track and heat transfer between the surroundings was negligible. It is tough to tell what assumptions you can make and which ones you can't in problems like these, which is what can make them difficult. On most exams or quizzes the collision problems will be inelastic simply because they are harder, but it is important to understand the fundementals of elastic collisions and how to solve problems involving them.

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

Another example of collisions in real life is billiards. This is one of the most accurate real life examples of elastic collisions. One ball hits another ball at rest, and if done right the first ball stops and transfers nearly all its kinetic energy into kinetic energy in the second ball.

[[File:Pool1.0.jpg]]

Another example is hitting a baseball. Hitting a ball off the end or to close the hands of a bat will cause some vibrational energy from the collision, but if the ball catches the sweet spot the collision is very elastic, and you don't feel any vibration when you strike the ball.

[[File:Hittingabaseball.jpg|500px]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

File:Hittingabaseball.jpg

2016-04-12T01:53:51Z

Sfischer8:

File:Pool1.0.jpg

2016-04-12T01:50:53Z

Sfischer8:

Elastic Collisions

2016-04-12T01:25:42Z

Sfischer8: /* Difficult */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible. For a lot of the complicated problems, starting with a diagram and writing down the given information and plugging that into either the momentum or energy principle can help you move forward in the problem. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-21 kg.

[[File:Problem2.0.png|300px]]

A lot of people freak out when they see problems involving atomic particles and electrical forces, but the concept is the same no matter the scale. For this problem we are given masses, velocities, and momentum, but the equation p = mv allows us to easily handle these different values. Once again drawing a diagram helps to understand what is actually happening in the problem, making it easier to put the values in the correct places and solve.

===Difficult===

Okay, let's get a little bit trickier here...

There is a 400 kg train traveling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

[[File:Problem3.0.png|400px]]

This was a difficult problem because of the assumptions we had to make in order to solve it. Because the collision was elastic we were able to disregard the change in internal energy in the energy principle, and since it happened so quickly the work done by the track and heat transfer between the surroundings was negligible. It is tough to tell what assumptions you can make and which ones you can't in problems like these, which is what can make them difficult. On most exams or quizzes the collision problems will be inelastic simply because they are harder, but it is important to understand the fundementals of elastic collisions and how to solve problems involving them.

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

File:Problem3.0.png

2016-04-12T01:21:33Z

Sfischer8:

Elastic Collisions

2016-04-12T00:57:14Z

Sfischer8: /* Middling */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible. For a lot of the complicated problems, starting with a diagram and writing down the given information and plugging that into either the momentum or energy principle can help you move forward in the problem. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-21 kg.

[[File:Problem2.0.png|300px]]

A lot of people freak out when they see problems involving atomic particles and electrical forces, but the concept is the same no matter the scale. For this problem we are given masses, velocities, and momentum, but the equation p = mv allows us to easily handle these different values. Once again drawing a diagram helps to understand what is actually happening in the problem, making it easier to put the values in the correct places and solve.

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

File:Problem2.0.png

2016-04-12T00:53:43Z

Sfischer8:

Elastic Collisions

2016-04-11T22:25:32Z

Sfischer8: /* Examples */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible. For a lot of the complicated problems, starting with a diagram and writing down the given information and plugging that into either the momentum or energy principle can help you move forward in the problem. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T22:17:52Z

Sfischer8: /* Simple */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T22:17:40Z

Sfischer8: /* Simple */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|thumb|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T22:17:21Z

Sfischer8: /* Simple */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|Problem 1]]

In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will the initial velocities, masses, and one final velocity and you will be asked to solve for a either a final velocity or final momentum. By setting up an equation like the one in this problem, one can easily handle this type of problem.

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T22:14:28Z

Sfischer8: /* Simple */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

[[File:Problem1.0.png|300px|thumb|middle|Problem 1]]

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

File:Problem1.0.png

2016-04-11T22:14:01Z

Sfischer8:

Elastic Collisions

2016-04-11T20:41:10Z

Sfischer8: /* A Mathematical Model */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|350px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

Okay, so first let’s think about what we know.
This is an elastic collision, so as we mentioned before: Pf = Pi.
Breaking that down, we have:

m1v1 + m2v2 = m1v1f + m2v2f, and we are looking for v1f. So let’s rearrange the equation:

v1f= (m1v1 + m2v2) – (m2v2f) / m1

Now, let’s plug in the numbers!

V1f= (55*<4,0,0> + 55*<0,0,0>) – (55*<7,0,0>)/(55)

V1f= <-3,0,0> m/s (Don't forget it's negative because Jay is going in the opposite direction that she started out,not because she's incredibly slow.)

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T20:40:35Z

Sfischer8: /* A Mathematical Model */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

Example Problem

Cart 1, moving in the positive x direction, collides with cart 2 moving in the negative x direction. Both carts have identical masses and the collisions is (nearly) elastic, as it would be if the carts interacted magnetically or repelled each other through soft springs. What are the final momenta of the two carts?

[[File:Example1good.jpg|400px|thumb|right]]

1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship
between the initial and final momentums of the two carts.

2. Next, by applying the energy principle we can gain knowledge about the final and initial kinetic energies. By acknowledging the fact that the change in internal energy is 0 and the fact the reaction happens so quickly that not work or heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies.

3. Once we have simplified the momentum and energy principles, one can use the relationship between kinetic energy and momentum mentioned above to get a relationship involving both energy and momentum. Once this is obtained, the equation can be shifted around to get both the final momentums.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

Okay, so first let’s think about what we know.
This is an elastic collision, so as we mentioned before: Pf = Pi.
Breaking that down, we have:

m1v1 + m2v2 = m1v1f + m2v2f, and we are looking for v1f. So let’s rearrange the equation:

v1f= (m1v1 + m2v2) – (m2v2f) / m1

Now, let’s plug in the numbers!

V1f= (55*<4,0,0> + 55*<0,0,0>) – (55*<7,0,0>)/(55)

V1f= <-3,0,0> m/s (Don't forget it's negative because Jay is going in the opposite direction that she started out,not because she's incredibly slow.)

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T20:27:58Z

Sfischer8: /* A Mathematical Model */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

[[File:Example1good.jpg|200px|thumb|right|Example 1]]

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

Okay, so first let’s think about what we know.
This is an elastic collision, so as we mentioned before: Pf = Pi.
Breaking that down, we have:

m1v1 + m2v2 = m1v1f + m2v2f, and we are looking for v1f. So let’s rearrange the equation:

v1f= (m1v1 + m2v2) – (m2v2f) / m1

Now, let’s plug in the numbers!

V1f= (55*<4,0,0> + 55*<0,0,0>) – (55*<7,0,0>)/(55)

V1f= <-3,0,0> m/s (Don't forget it's negative because Jay is going in the opposite direction that she started out,not because she's incredibly slow.)

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T20:25:24Z

Sfischer8: /* A Mathematical Model */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

[[File:Example1good.jpg]]

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

Okay, so first let’s think about what we know.
This is an elastic collision, so as we mentioned before: Pf = Pi.
Breaking that down, we have:

m1v1 + m2v2 = m1v1f + m2v2f, and we are looking for v1f. So let’s rearrange the equation:

v1f= (m1v1 + m2v2) – (m2v2f) / m1

Now, let’s plug in the numbers!

V1f= (55*<4,0,0> + 55*<0,0,0>) – (55*<7,0,0>)/(55)

V1f= <-3,0,0> m/s (Don't forget it's negative because Jay is going in the opposite direction that she started out,not because she's incredibly slow.)

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

File:Example1good.jpg

2016-04-11T20:24:40Z

Sfischer8:

File:Example1.jpg

2016-04-11T20:20:50Z

Sfischer8: Sfischer8 uploaded a new version of "File:Example1.jpg"

Elastic Collisions

2016-04-11T18:39:24Z

Sfischer8: /* A Mathematical Model */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
----

[[File: CodeCogsEqn copy.gif]]
----

[[File: CodeCogsEqn_copy_2.gif]]

This simple principal can be expanded into:
[[File:elascollisionexpanded.jpg]]

Starting from the left side we have m1 and m2 which are the masses of object 1 and 2 respectively, and v1i and v2i represent their initial velocities respectively. On the right side of the equation, the final kinetic energy, it’s the same equation just different numbers! For the final kinetic energy, unless stuff is breaking or its otherwise stated in the problem, you can assume the masses won’t change BUT the velocities may. Imagine if you toss a pool ball towards another one that is initially at rest. The one that is at rest will move after it is hit, so there is a change in velocity! However, please remember that there is no overall change in kinetic energy. Let’s say the mass of our pool balls are 2 kilograms and I will make up some speeds. Applying the formula from above we get:
[[File:elascollisionexample.jpg]]

As you see the final velocities changed. One ball was initially at rest but it wasn't after the collision because it was struck by a ball that was already moving...But there was no change in kinetic energy because 32=32 yay! Put it in your calculator and see for yourself.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

Okay, so first let’s think about what we know.
This is an elastic collision, so as we mentioned before: Pf = Pi.
Breaking that down, we have:

m1v1 + m2v2 = m1v1f + m2v2f, and we are looking for v1f. So let’s rearrange the equation:

v1f= (m1v1 + m2v2) – (m2v2f) / m1

Now, let’s plug in the numbers!

V1f= (55*<4,0,0> + 55*<0,0,0>) – (55*<7,0,0>)/(55)

V1f= <-3,0,0> m/s (Don't forget it's negative because Jay is going in the opposite direction that she started out,not because she's incredibly slow.)

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

Elastic Collisions

2016-04-11T18:38:37Z

Sfischer8: /* A Mathematical Model */

'''CLAIMED BY SEAN FISCHER'''
==The Main Idea==

So what exactly is an elastic collision? I know you’re thinking, “Oh I know all about elasticity,” but LOL this is not bubble gum or rubber bands guys!

An elastic collision is a collision between two or more objects in which there is no loss in kinetic energy before and after the collision. If we assume that the colliding objects are part of the system and that there is no force from the surroundings, the final kinetic energy is still in the same form as it was initially. To keep it simple, this means that kinetic energy in= kinetic energy out. Remember how your parents always told you what goes in must come out? They were talking about elastic collisions... probably.

Additionally, elastic collisions are a wonderful representation of the conservation of momentum which states that the momentum of an isolated system is constant. For an isolated system undergoing an elastic collision momentum in = momentum out.

And I love playing pool because every time a ball hits another and they bounce off one another I’m witnessing an elastic collision so basically I’m a physicist in the lab. The image below demonstrates the main idea of an elastic collision. Boing!!

It's important to note that with macroscopic systems there are no perfectly elastic collisions because there's always at least a dab of dissipation (like some thermal energy given off), but most are nearly elastic. The only time there are perfect collisions is on a microscopic level when atomic systems with quantized energy collide, but that's only if there is enough energy available to raise the systems to an excited quantum state-- but quantum energy is really a whole other topic. Let's focus on elastic collisions!

[[File: boing.gif]]

Check out this funny video on elastic collisions also! https://www.youtube.com/watch?v=W9EqU1_DXUw

Now check out this more informative video on elastic collisions: https://www.youtube.com/watch?v=V4vzNk4qppw

===A Mathematical Model===

Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
[[File: CodeCogsEqn.gif]]
[[File: CodeCogsEqn copy.gif]]
[[File: CodeCogsEqn_copy_2.gif]]

[[File:kfki.jpg]]
(Initial kinetic energy = final kinetic energy)

This simple principal can be expanded into:
[[File:elascollisionexpanded.jpg]]

Starting from the left side we have m1 and m2 which are the masses of object 1 and 2 respectively, and v1i and v2i represent their initial velocities respectively. On the right side of the equation, the final kinetic energy, it’s the same equation just different numbers! For the final kinetic energy, unless stuff is breaking or its otherwise stated in the problem, you can assume the masses won’t change BUT the velocities may. Imagine if you toss a pool ball towards another one that is initially at rest. The one that is at rest will move after it is hit, so there is a change in velocity! However, please remember that there is no overall change in kinetic energy. Let’s say the mass of our pool balls are 2 kilograms and I will make up some speeds. Applying the formula from above we get:
[[File:elascollisionexample.jpg]]

As you see the final velocities changed. One ball was initially at rest but it wasn't after the collision because it was struck by a ball that was already moving...But there was no change in kinetic energy because 32=32 yay! Put it in your calculator and see for yourself.

===A Computational Model===

The link below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.

[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

Okay let’s start with a simple example like the ones you don't see on the test.

Jay and Sarah are best friends. Since they’re best friends they both weigh 55 kg. Jay hadn’t seen Sarah in a long time so when she saw her she ran to her with a velocity of <4,0,0> m/s. Instead of a hug, they were both too excited and collided and bounced back off of each other, and Sarah flew back with a velocity of <7,0,0> m/s. What was Jay’s final velocity?

Okay, so first let’s think about what we know.
This is an elastic collision, so as we mentioned before: Pf = Pi.
Breaking that down, we have:

m1v1 + m2v2 = m1v1f + m2v2f, and we are looking for v1f. So let’s rearrange the equation:

v1f= (m1v1 + m2v2) – (m2v2f) / m1

Now, let’s plug in the numbers!

V1f= (55*<4,0,0> + 55*<0,0,0>) – (55*<7,0,0>)/(55)

V1f= <-3,0,0> m/s (Don't forget it's negative because Jay is going in the opposite direction that she started out,not because she's incredibly slow.)

===Middling===

When far apart, the momentum of a proton is <5.2 ✕ 10−21, 0, 0> kg · m/s as it approaches another proton that is initially at rest. The two protons repel each other electrically, but they are not close enough to touch. When they are far apart again later, one of the protons now has a velocity of <1.75, .82, 0> m/s. At that instant, what is the momentum of the other proton? HINT: Mass of proton is 1.7 *10^-27 kg.

Alright, don't freak out because this collision is on an atomic level, the idea is still the same!

Lets start out with that principle we mentioned before:

Pf= Pi

If we were given the momentum of both protons, we could both subtract one from the other but we only have one velocity and one momentum... but don't forget the hint! We can calculate that second momentum, let's do that first.

mass of proton*velocity=

1.7e-21 * <1.75, .82, 0> = <2.975e-21, 1.394e-21,0> kg * m/s

Okay so now that we have both momentums, we can just subtract the final from the initial right? Right!

<2.975e-21,1.394e-21,0> - <5.2e-21, 0, 0> = <-2.225e-21, -1.394e-21, 0> kg * m/s

There we go!

===Difficult===

Okay, let's get a little bit trickier here...

There is a 35 kg train travelling at 55 m/s that collides, elastically of course, with a random 2 kg trashcan that's stationary on the tracks. So afterwards, what are the speeds of both the train and the trashcan after the collision?

Let's start out with our handy-dandy momentum in = momentum our principle!

Pf= Pi

m1v1 + m2v2 = m1v1f + m2v2f

Wait, we have two unknowns! So let's plug in numbers:

35*55 + 2*0 = 35*v1f + 2*v2f

1925 = 35*v1f + 2*v2f

Let's multiply this equation by a random number

Let's use the kinetic energy principle now!

1/2mv1^2 = 1/2mv1f^2 + 1/2mv2f^2

==Connectedness==

So how are collisions connected to the real world? Collisions are all around us!

One main example is cars! Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.

[[File:crash-test.jpg]]

==History==

The history of collisions originates from the Rutherford Scattering experiment. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus!

[[File:gold-foil-experiment.jpg]]

== See also ==

http://physicsbook.gatech.edu/Collisions for a general understanding of all collisions.

http://physicsbook.gatech.edu/inelastic_collisions to contrast them from elastic collisions.

===Further reading===

http://www.britannica.com/science/elastic-collision

===External links===

[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm]

[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html]

[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/]

==References==

Main Idea:

''Matter and Interactions, 4th Edition''

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml

http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/

History:

https://en.wikipedia.org/wiki/Elastic_collision

Connectedness:

http://www.iihs.org/

Additionally used references from ''see also''.

File:CodeCogsEqn copy 2.gif

2016-04-11T18:38:04Z

Sfischer8:

File:CodeCogsEqn copy.gif

2016-04-11T18:34:25Z

Sfischer8: