Physics Book - User contributions [en]

Length and Stiffness of an Interatomic Bond

2017-11-29T02:04:25Z

Nicoletansey:

'''Claimed by Andy Stevens Spring 2017'''
Claimed by Nicole Tansey Fall 2017

This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid object is made up of a large amount of atoms that are held together by chemical bonds in a lattice-like structure as seen in ''Figure 1''. Each atom is connected to its neighboring atoms by these chemical bonds. One way to approximate this solid object model is to imagine the atoms as balls and the chemical bonds as springs-- a ball and spring model! The relaxed length of the microscopic spring between two atoms is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

''Figure 1''
[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

With this information, we can determine the stiffness of an interatomic bond. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms, which is just the distance of the two atoms' radii added together since we use the space-filling model. To find the radius, we divide the diameter of a single atom in half. So, the center-to-center distance is essentially equal to the diameter of a single atom since in the space-filling model there is no space between each atom. If we can calculate the length of the interatomic bond (the diameter of a single atom), we can use this data to find the stiffness of the interatomic bond, modeled as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

== Mathematical Model ==
'''Formulas to Know'''

In many problems, you will be given the density of the solid that you are being questioned on. It is important to remember that <math>{density} = \frac{mass}{volume}</math>
Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<div align ="center">

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

</div>

''b. What is the diameter of a single zinc atom?''

<div align ="center">

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

</div>

''c. How many zinc atoms make up one side of the penny?''

<div align ="center">

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

</div>

''d. Calculate the interatomic spring stiffness for zinc.''

<div align ="center">

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

</div>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire
<div align ="center">

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

</div>

'''Step 2'''
Cross-Sectional Area of Wire

<div align ="center">

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

</div>

'''Step 3'''
Cross-Sectional Area of One Atom

<div align ="center">

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

</div>

'''Step 4'''
Number of Chains

<div align ="center">

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

</div>

'''Step 5'''
Number of Bonds in One Chain

<div align ="center">

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

</div>

'''Step 6'''
Stiffness of One Interatomic Spring

<div align ="center">

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

</div>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's Modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn

Length and Stiffness of an Interatomic Bond

2017-11-29T01:45:14Z

Nicoletansey:

'''Claimed by Andy Stevens Spring 2017'''
Claimed by Nicole Tansey Fall 2017

This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid object is made up of a large amount of atoms that are held together by chemical bonds in a lattice-like structure as seen in ''Figure 1''. Each atom is connected to its neighboring atoms by these chemical bonds. One way to approximate this solid object model is to imagine the atoms as balls and the chemical bonds as springs-- a ball and spring model! The relaxed length of the microscopic spring between two atoms is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

''Figure 1''
[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

With this information, we can determine the stiffness of an interatomic bond. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms, which is just the distance of the two atoms' radii added together since we use the space-filling model. To find the radius, we divide the diameter of a single atom in half. So, the center-to-center distance is essentially equal to the diameter of a single atom since in the space-filling model there is no space between each atom. If we can calculate the length of the interatomic bond (the diameter of a single atom), we can use this data to find the stiffness of the interatomic bond, modeled as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

===Formulas to know===

Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<div align ="center">

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

</div>

''b. What is the diameter of a single zinc atom?''

<div align ="center">

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

</div>

''c. How many zinc atoms make up one side of the penny?''

<div align ="center">

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

</div>

''d. Calculate the interatomic spring stiffness for zinc.''

<div align ="center">

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

</div>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire
<div align ="center">

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

</div>

'''Step 2'''
Cross-Sectional Area of Wire

<div align ="center">

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

</div>

'''Step 3'''
Cross-Sectional Area of One Atom

<div align ="center">

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

</div>

'''Step 4'''
Number of Chains

<div align ="center">

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

</div>

'''Step 5'''
Number of Bonds in One Chain

<div align ="center">

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

</div>

'''Step 6'''
Stiffness of One Interatomic Spring

<div align ="center">

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

</div>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's Modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn

Length and Stiffness of an Interatomic Bond

2017-11-29T01:39:24Z

Nicoletansey:

'''Claimed by Andy Stevens Spring 2017'''
Claimed by Nicole Tansey Fall 2017

This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid object is made up of a large amount of atoms that are held together by chemical bonds in a lattice-like structure as seen in ''Figure 1''. Each atom is connected to its neighboring atoms by these chemical bonds. One way to approximate this solid object model is to imagine the atoms as balls and the chemical bonds as springs-- a ball and spring model! The relaxed length of the microscopic spring between two atoms is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

''Figure 1''
[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

We can determine the stiffness of an interatomic bond by considering it as a spring. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms. The diameter of an atom is the space-filling model of a solid. To find the radius, we divide the diameter in half. If we can calculate the length of the interatomic bond and the diameter of a single atom, we can use this data to find the stiffness of the interatomic bond, considered as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

===Formulas to know===

Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<div align ="center">

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

</div>

''b. What is the diameter of a single zinc atom?''

<div align ="center">

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

</div>

''c. How many zinc atoms make up one side of the penny?''

<div align ="center">

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

</div>

''d. Calculate the interatomic spring stiffness for zinc.''

<div align ="center">

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

</div>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire
<div align ="center">

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

</div>

'''Step 2'''
Cross-Sectional Area of Wire

<div align ="center">

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

</div>

'''Step 3'''
Cross-Sectional Area of One Atom

<div align ="center">

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

</div>

'''Step 4'''
Number of Chains

<div align ="center">

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

</div>

'''Step 5'''
Number of Bonds in One Chain

<div align ="center">

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

</div>

'''Step 6'''
Stiffness of One Interatomic Spring

<div align ="center">

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

</div>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's Modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn

Maximally Inelastic Collision

2017-11-29T01:18:42Z

Nicoletansey:

Maximally Inelastic Collisions between multiple objects.
claimed by dwinegarden3 (spring 2017)

==The Main Idea==

A collision is a brief interaction between large forces. Collisions can include such phenomena as a ball bouncing off a wall, two people running into each other on their way to class, or atoms bouncing around within a star. There are two primary types of collisions: elastic and inelastic. Elastic collisions occur when the total kinetic energy of the system is conserved while inelastic collisions occur when the objects' kinetic energies are not conserved. Maximally inelastic collisions are a subset of inelastic collisions in which the objects in the system collide and stick to form one object. The new object has a new velocity and the mass of the object equals the total mass of all the objects that have become one. An example of a maximally inelastic collision is a car crash or stellar collision, when two stars collide and form a new star.

[[File:DW4.jpg]]

===A Mathematical Model===

Maximally inelastic collisions are first and foremost momentum problems, so modeling such a collision usually begins with the fundamental momentum principle, modeled in the equation:
::<math>{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}{Δt}</math>
where '''p''' is the momentum of the system, '''F''' is the net force from the surroundings, '''Δt''' is the change in time for the process.

Because the momentum principle states that the change in momentum, also called impulse, within a system in which the objects stick is not changed, the time of interaction is negligible, so <math>{Δt} ≈ {0}</math>. By substituting '''Δt''' with zero and rewriting the equation yields:
::<math>{ΔP_{system}} = {0}</math> where we can break the change in momentum of the system to it's initial and final components to get
::<math>{P_{final}} - {P_{initial}} = {0}</math> which becomes
::<math>{P_{final}} = {P_{initial}}</math>.

If we substitute the mass and velocity of multiple objects into the above equation, we see that:
::<math>m_1 v_1 + m_2 v_2 + ... m_n v_n = M V </math>
where '''M''' is sum of the masses of all collided objects and '''V''' is the final velocity of the amalgamated object
::<math> V=\frac{\sum {m v}}{M}</math>

In maximally inelastic collisions, kinetic energy is not conserved between the initial and final states of the objects. However, total energy is conserved. This is possible because some of the energy is lost to heat and changes in shape, accounted for in the change of internal energy called '''ΔU'''. Thus, we can us the energy principle:
::<math>{E} = {Q} + {W}</math> where '''E''' is the total energy of the system, '''Q''' is the change in thermal energy, and '''W''' is the work done.
If we input the various types of energy in for the total energy such as kinetic, potential, and internal we get
:: <math>{ΔE_k}+{ΔE_p}+{ΔU}= {Q} + {W} </math> where '''ΔE_k''' is the change in kinetic energy of the system,'''ΔE_p''' is the change in the potential energy of the system, and '''ΔU''' is the change in internal energy.

Since we assume that the objects all end up as one combined object with no significant loss of mass, we can also assume that the work done is negligible and the process is adiabatic, or a process that produces negligible heat, and the change in overall potential energy of the system is negligible. With these assumptions in place, the equation above simplifies to
:: <math> {ΔE_k} + {ΔU} = 0 </math> which we can break into the initial and final components to get
:: <math> {ΔU} = -({K_{final}} - ({K_{initial,1}} + {K_{initial,2}} +... {K_{initial,n}})) </math>
Kinetic Energy for objects that have a velocity significantly smaller than the speed of light is defined as <math> \frac{1}{2}{mv^2} </math>, so putting in the values for mass and velocity we get that:
:: <math> {ΔU} = \frac{1}{2}{(v_1 + v_2 +... v_n)^2}{(M)} - (\frac{1}{2}{m_1 (v_1)^2} + \frac{1}{2}{m_2 (v_2)^2}+...\frac{1}{2}{m_n (v_n)^2)} </math>
However, when the kinetic energy of an object is close to the speed of light, relativistic equations are required. In this case, the kinetic energy is equal instead to
:: <math> {K} = \frac{mc^2}{\sqrt{1-(\frac{v}{c})^2}} - mc^2 </math>
where '''m''' is the mass of the object, '''V''' is the velocity of the object, '''c''' is the speed of light (~3e8 m/s). The first element on the left of the above equation accounts for all of the energy of an object and the second element removes the rest energy of the system from the total energy, leaving the relativistic kinetic energy. When this is put into the overall equation with the known values, we get
:: <math> {ΔU} = (\frac{Mc^2}{\sqrt{1-(\frac{V}{c})^2}} - Mc^2) - (\sum_{n=1}^N \frac{m_N c^2}{\sqrt{1-(\frac{v_N}{c})^2}} - m_N c^2)</math>

===A Computational Model===
[https://trinket.io/glowscript/02316a1ea9 Maximally Inelastic Collisions using Glowscript]

==Examples==
===Simple===
'''Problem:'''
Greco wants to eat a very large flan, but Fenton and Gumbart both have small flans, so they decide to combine the flans. Fenton throws a 9 kg mass flan at a velocity of 14 m/s which strikes Gumbart's flan that weighs 5 kg mass with a velocity of -5 m/s head-on, and the two flans stick together to make an ultra mega flan for Greco to eat. At what speed will the giant flan have? Greco can only catch objects that are flying at 5 m/s, will he catch it or will it go past him? Assume negligible air resistance.

'''Solution:'''
Use conservation of momentum to solve for the final velocity : <math>m_1 v_1 + m_2 v_2 = \left( m_1 + m_2 \right) v \,</math>.
Solving for the final velocity we get the equation: <math> v=\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>.
Plugging in the numbers yields us a final velocity of 7.214 m/s, therefore Greco will be unable to catch the flan.

===Middling===
'''Problem:'''
A .1 kg bullet is launched at 100 m/s a stationary block that weighs 5kg. The bullet embeds itself into the block. When someone goes to retrieve the bullet and remove it from the block, they notice the block is slightly warmer than it was before. By how much did the block's thermal energy change? The specific heat of the block and bullet is 6 J/(g*C) so by how many degrees did the block warm up by?

'''Solution:'''
Using the conservation of momentum principle, we can find the final velocity of the object: <math>m_1 v_1 = \left( m_1 + m_2 \right) v \,</math> which is 1.961 m/s.
Now we can use the conservation of Energy principle to solve for the thermal energy: <math> {ΔE_k} + {ΔU} = 0. </math> Then we break the change of kinetic energy into it's initial and final conditions to get :
<math> {ΔU} = -({K_{final}} - ({K_{initial,bullet}})) </math>
Plugging in the values we get :
<math> {ΔU} = \frac{1}{2}{(v_{final})^2}{(m_{final})} - (\frac{1}{2}{m_{bullet} (v_{bullet})^2}) </math>.
Entering the values from the word problem, our final answer is 490J.

To find the change in temperature, we need to use the heat equation : <math> Q = mCΔT </math>
Plugging in the given information of specific heat, thermal energy that we solved for, and the mass of the final object, we get an increase of 16 degrees.

===Difficult===

[[File: 11-091-playground.jpg |border|right]]
'''Problem:'''
Sally (5 years old) is at the playground and decides that she wants to go on the ride that spins really fast in circles. The playground ride consists of a currently stationary disk of mass M = 45 kg and radius R = 2.4 m mounted on a low-friction axle. Sally weighs about 18 kg. She is super excited to play and runs at speed of 2.3 m/s on a line tangential to the disk and jumps onto the outer edge of the disk. She initially decided to just run around and spin the wheel, but later she decides that she wants to jump on and let the wheel keep going. What is the change of energy of the system?

'''Solution:'''
This problem is difficult because it requires the use of the three big equations of classical physics 1.
First we need to use the conservation of angular momentum and calculate the initial angular momentum which is <math> L = RPsinθ </math>.
In this case, the momentum is 60 <math> m^2 kg/s </math>.
Next we need to find the angular velocity of the ride with Sally. Using the equation: <math> L_i = Iω + RPsinθ </math> where we can use the relationship that <math> v = ωr </math> to find the more useful equation : <math> L_i = Iω + RmwR </math>. Since this is a wheel, the moment of inertia will be <math>.5MR^2</math>. Solving for ω gives us .639 radians.

Now we can solve for the change in energies using the formula : <math> {ΔE_k} + {ΔU} = 0 </math>. However in this case we need to break the kinetic energies into rotational and translational and then their respective final and initial parts. So we get the rather large equation : <math> K_{rot,f} + K_{trans,f} + ΔU = K_{rot,i} + K_{trans,i} </math> where we can convert <math> K_{rot} </math> to <math>.5Iω^2
</math>. Using the conservation of linear momentum theorem, we can find Sally's final velocity with the equation : <math> v=\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>. Plugging in values, we find that her final velocity is 1.02m/s. Now we can plug all our numbers into the previous equation and solve for the change in energy which is 373J.

==Connectedness==

This topic is of interest to me as I am very interested in astronomy and a very important Maximally Inelastic Collision in space is the collision of asteroids. Because asteroids usually stick together after a collision their collisions can be modeled with the equations of Maximally Inelastic Collisions. Maximally Inelastic Collisions are also important to NASA, where I want to work; NASA is planning to use the Asteroid Redirect Mission, or the ARM program, in order to keep asteroids from hitting the Earth and must use the Maximally Inelastic Collision equations in order to plot the path of the asteroids after collision to ensure that they do not hit the Earth in their new flight path. Most other celestial collisions are generally modeled as inelastic collisions because they stick together. When two stars collide they absorb each other and form one star, when they don't explode. Because this forms one massive sphere with evenly distributed mass this collision is an almost ideal Maximally Inelastic Collision.

[[File:NASAARM.jpg]]

==History==

The idea of maximally inelastic collisions is part of the conservation of linear momentum which is implied by Newton's Laws. Most objects tend to bounce off each other creating the idea of collision. Most objects that are visibly colliding tend to lose energy through a variety of ways but theoretically objects could collide and not lose kinetic energy. This created the two types of collision inelastic and elastic. Elastic defines collisions that have no change in kinetic energy, these are usually microscopic such as Rutherford Scattering. All other objects tend to be inelastic since some energy in lost between the objects and so their respective kinetic energy changes. However, some objects can stick together and therefore combine their masses. This is a special case of inelastic collisions and tend to be a small portion of collisions that actually occur in real life, but some examples seen are when car's collide or when a sticky substance stays on the object it is throw at.

[[File:DW2.gif]]

== See also ==

There are several other topics within this wiki that can give you more information on collisions as a whole including
[[Inelastic Collisions]] and [[Elastic Collisions]]. These topics along with the basic fundamentals such as [[Kinetic Energy]] and [[ Momentum Principle]] that we used to derive the equations will show how exactly maximally inelastic collisions are shown by classical physics.
===Further reading===
These extensive resources cover this topic in more depth
* A physics resource written by experts for an expert audience [https://en.wikipedia.org/wiki/Portal:Physics Physics Portal]
* A wiki book on modern physics [https://en.wikibooks.org/wiki/Modern_Physics Modern Physics Wiki]
* The MIT open courseware for intro physics [http://ocw.mit.edu/resources/res-8-002-a-wikitextbook-for-introductory-mechanics-fall-2009/index.htm MITOCW Wiki]
* An online concept map of intro physics [http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html HyperPhysics]
* Interactive physics simulations [https://phet.colorado.edu/en/simulations/category/physics PhET]
* OpenStax algebra based intro physics textbook [https://openstaxcollege.org/textbooks/college-physics College Physics]
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]
* A resource guide compiled by the [http://www.aapt.org/ AAPT] for educators [http://www.compadre.org/ ComPADRE]

===External links===

These are some internet articles that can show more animations and pictures to help understand this concept
*[http://www.ux1.eiu.edu/~cfadd/1150/07Mom/Inelastic.html Inelastic Collisions]
*[http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes:examples:maximally_inelastic_collision_of_two_identical_carts Carts]

==References==

The primary reference I used was the textbook : Matter and Interactions 4th edition.
Full Citation:
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. Hoboken, NJ: Wiley, 2011. Print.

http://www.howitworksdaily.com/wp-content/uploads/2016/02/818x818xAsteroid-Redirect-Mission-1024x1024.jpg.pagespeed.ic.ywLacxEeka.jpg

http://www.kshitij-iitjee.com/Study/Physics/Part1/Chapter9/33.jpg

[[Category:Collisions]]

Length and Stiffness of an Interatomic Bond

2017-11-29T01:17:24Z

Nicoletansey:

'''Claimed by Andy Stevens Spring 2017'''
Claimed by Nicole Tansey Fall 2017
This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid objects is made up of a bunch of tiny balls (atoms) that are held together by springs (chemical bonds). The relaxed legth of the little spring between two atoms (the interatomic bond) is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

We can determine the stiffness of an interatomic bond by considering it as a spring. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms. The diameter of an atom is the space-filling model of a solid. To find the radius, we divide the diameter in half. If we can calculate the length of the interatomic bond and the diameter of a single atom, we can use this data to find the stiffness of the interatomic bond, considered as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

===Formulas to know===

Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<div align ="center">

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

</div>

''b. What is the diameter of a single zinc atom?''

<div align ="center">

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

</div>

''c. How many zinc atoms make up one side of the penny?''

<div align ="center">

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

</div>

''d. Calculate the interatomic spring stiffness for zinc.''

<div align ="center">

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

</div>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire
<div align ="center">

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

</div>

'''Step 2'''
Cross-Sectional Area of Wire

<div align ="center">

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

</div>

'''Step 3'''
Cross-Sectional Area of One Atom

<div align ="center">

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

</div>

'''Step 4'''
Number of Chains

<div align ="center">

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

</div>

'''Step 5'''
Number of Bonds in One Chain

<div align ="center">

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

</div>

'''Step 6'''
Stiffness of One Interatomic Spring

<div align ="center">

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

</div>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's Modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn

Maximally Inelastic Collision

2017-11-28T20:49:26Z

Nicoletansey:

Maximally Inelastic Collisions between multiple objects.
claimed by dwinegarden3 (spring 2017)
Claimed by Nicole Tansey (Fall 2017)
==The Main Idea==

A collision is a brief interaction between large forces. Collisions can include such phenomena as a ball bouncing off a wall, two people running into each other on their way to class, or atoms bouncing around within a star. There are two primary types of collisions: elastic and inelastic. Elastic collisions occur when the total kinetic energy of the system is conserved while inelastic collisions occur when the objects' kinetic energies are not conserved. Maximally inelastic collisions are a subset of inelastic collisions in which the objects in the system collide and stick to form one object. The new object has a new velocity and the mass of the object equals the total mass of all the objects that have become one. An example of a maximally inelastic collision is a car crash or stellar collision, when two stars collide and form a new star.

[[File:DW4.jpg]]

===A Mathematical Model===

Maximally inelastic collisions are first and foremost momentum problems, so modeling such a collision usually begins with the fundamental momentum principle, modeled in the equation:
::<math>{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}{Δt}</math>
where '''p''' is the momentum of the system, '''F''' is the net force from the surroundings, '''Δt''' is the change in time for the process.

Because the momentum principle states that the change in momentum, also called impulse, within a system in which the objects stick is not changed, the time of interaction is negligible, so <math>{Δt} ≈ {0}</math>. By substituting '''Δt''' with zero and rewriting the equation yields:
::<math>{ΔP_{system}} = {0}</math> where we can break the change in momentum of the system to it's initial and final components to get
::<math>{P_{final}} - {P_{initial}} = {0}</math> which becomes
::<math>{P_{final}} = {P_{initial}}</math>.

If we substitute the mass and velocity of multiple objects into the above equation, we see that:
::<math>m_1 v_1 + m_2 v_2 + ... m_n v_n = M V </math>
where '''M''' is sum of the masses of all collided objects and '''V''' is the final velocity of the amalgamated object
::<math> V=\frac{\sum {m v}}{M}</math>

In maximally inelastic collisions, kinetic energy is not conserved between the initial and final states of the objects. However, total energy is conserved. This is possible because some of the energy is lost to heat and changes in shape, accounted for in the change of internal energy called '''ΔU'''. Thus, we can us the energy principle:
::<math>{E} = {Q} + {W}</math> where '''E''' is the total energy of the system, '''Q''' is the change in thermal energy, and '''W''' is the work done.
If we input the various types of energy in for the total energy such as kinetic, potential, and internal we get
:: <math>{ΔE_k}+{ΔE_p}+{ΔU}= {Q} + {W} </math> where '''ΔE_k''' is the change in kinetic energy of the system,'''ΔE_p''' is the change in the potential energy of the system, and '''ΔU''' is the change in internal energy.

Since we assume that the objects all end up as one combined object with no significant loss of mass, we can also assume that the work done is negligible and the process is adiabatic, or a process that produces negligible heat, and the change in overall potential energy of the system is negligible. With these assumptions in place, the equation above simplifies to
:: <math> {ΔE_k} + {ΔU} = 0 </math> which we can break into the initial and final components to get
:: <math> {ΔU} = -({K_{final}} - ({K_{initial,1}} + {K_{initial,2}} +... {K_{initial,n}})) </math>
Kinetic Energy for objects that have a velocity significantly smaller than the speed of light is defined as <math> \frac{1}{2}{mv^2} </math>, so putting in the values for mass and velocity we get that:
:: <math> {ΔU} = \frac{1}{2}{(v_1 + v_2 +... v_n)^2}{(M)} - (\frac{1}{2}{m_1 (v_1)^2} + \frac{1}{2}{m_2 (v_2)^2}+...\frac{1}{2}{m_n (v_n)^2)} </math>
However, when the kinetic energy of an object is close to the speed of light, relativistic equations are required. In this case, the kinetic energy is equal instead to
:: <math> {K} = \frac{mc^2}{\sqrt{1-(\frac{v}{c})^2}} - mc^2 </math>
where '''m''' is the mass of the object, '''V''' is the velocity of the object, '''c''' is the speed of light (~3e8 m/s). The first element on the left of the above equation accounts for all of the energy of an object and the second element removes the rest energy of the system from the total energy, leaving the relativistic kinetic energy. When this is put into the overall equation with the known values, we get
:: <math> {ΔU} = (\frac{Mc^2}{\sqrt{1-(\frac{V}{c})^2}} - Mc^2) - (\sum_{n=1}^N \frac{m_N c^2}{\sqrt{1-(\frac{v_N}{c})^2}} - m_N c^2)</math>

===A Computational Model===
[https://trinket.io/glowscript/02316a1ea9 Maximally Inelastic Collisions using Glowscript]

==Examples==
===Simple===
'''Problem:'''
Greco wants to eat a very large flan, but Fenton and Gumbart both have small flans, so they decide to combine the flans. Fenton throws a 9 kg mass flan at a velocity of 14 m/s which strikes Gumbart's flan that weighs 5 kg mass with a velocity of -5 m/s head-on, and the two flans stick together to make an ultra mega flan for Greco to eat. At what speed will the giant flan have? Greco can only catch objects that are flying at 5 m/s, will he catch it or will it go past him? Assume negligible air resistance.

'''Solution:'''
Use conservation of momentum to solve for the final velocity : <math>m_1 v_1 + m_2 v_2 = \left( m_1 + m_2 \right) v \,</math>.
Solving for the final velocity we get the equation: <math> v=\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>.
Plugging in the numbers yields us a final velocity of 7.214 m/s, therefore Greco will be unable to catch the flan.

===Middling===
'''Problem:'''
A .1 kg bullet is launched at 100 m/s a stationary block that weighs 5kg. The bullet embeds itself into the block. When someone goes to retrieve the bullet and remove it from the block, they notice the block is slightly warmer than it was before. By how much did the block's thermal energy change? The specific heat of the block and bullet is 6 J/(g*C) so by how many degrees did the block warm up by?

'''Solution:'''
Using the conservation of momentum principle, we can find the final velocity of the object: <math>m_1 v_1 = \left( m_1 + m_2 \right) v \,</math> which is 1.961 m/s.
Now we can use the conservation of Energy principle to solve for the thermal energy: <math> {ΔE_k} + {ΔU} = 0. </math> Then we break the change of kinetic energy into it's initial and final conditions to get :
<math> {ΔU} = -({K_{final}} - ({K_{initial,bullet}})) </math>
Plugging in the values we get :
<math> {ΔU} = \frac{1}{2}{(v_{final})^2}{(m_{final})} - (\frac{1}{2}{m_{bullet} (v_{bullet})^2}) </math>.
Entering the values from the word problem, our final answer is 490J.

To find the change in temperature, we need to use the heat equation : <math> Q = mCΔT </math>
Plugging in the given information of specific heat, thermal energy that we solved for, and the mass of the final object, we get an increase of 16 degrees.

===Difficult===

[[File: 11-091-playground.jpg |border|right]]
'''Problem:'''
Sally (5 years old) is at the playground and decides that she wants to go on the ride that spins really fast in circles. The playground ride consists of a currently stationary disk of mass M = 45 kg and radius R = 2.4 m mounted on a low-friction axle. Sally weighs about 18 kg. She is super excited to play and runs at speed of 2.3 m/s on a line tangential to the disk and jumps onto the outer edge of the disk. She initially decided to just run around and spin the wheel, but later she decides that she wants to jump on and let the wheel keep going. What is the change of energy of the system?

'''Solution:'''
This problem is difficult because it requires the use of the three big equations of classical physics 1.
First we need to use the conservation of angular momentum and calculate the initial angular momentum which is <math> L = RPsinθ </math>.
In this case, the momentum is 60 <math> m^2 kg/s </math>.
Next we need to find the angular velocity of the ride with Sally. Using the equation: <math> L_i = Iω + RPsinθ </math> where we can use the relationship that <math> v = ωr </math> to find the more useful equation : <math> L_i = Iω + RmwR </math>. Since this is a wheel, the moment of inertia will be <math>.5MR^2</math>. Solving for ω gives us .639 radians.

Now we can solve for the change in energies using the formula : <math> {ΔE_k} + {ΔU} = 0 </math>. However in this case we need to break the kinetic energies into rotational and translational and then their respective final and initial parts. So we get the rather large equation : <math> K_{rot,f} + K_{trans,f} + ΔU = K_{rot,i} + K_{trans,i} </math> where we can convert <math> K_{rot} </math> to <math>.5Iω^2
</math>. Using the conservation of linear momentum theorem, we can find Sally's final velocity with the equation : <math> v=\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>. Plugging in values, we find that her final velocity is 1.02m/s. Now we can plug all our numbers into the previous equation and solve for the change in energy which is 373J.

==Connectedness==

This topic is of interest to me as I am very interested in astronomy and a very important Maximally Inelastic Collision in space is the collision of asteroids. Because asteroids usually stick together after a collision their collisions can be modeled with the equations of Maximally Inelastic Collisions. Maximally Inelastic Collisions are also important to NASA, where I want to work; NASA is planning to use the Asteroid Redirect Mission, or the ARM program, in order to keep asteroids from hitting the Earth and must use the Maximally Inelastic Collision equations in order to plot the path of the asteroids after collision to ensure that they do not hit the Earth in their new flight path. Most other celestial collisions are generally modeled as inelastic collisions because they stick together. When two stars collide they absorb each other and form one star, when they don't explode. Because this forms one massive sphere with evenly distributed mass this collision is an almost ideal Maximally Inelastic Collision.

[[File:NASAARM.jpg]]

==History==

The idea of maximally inelastic collisions is part of the conservation of linear momentum which is implied by Newton's Laws. Most objects tend to bounce off each other creating the idea of collision. Most objects that are visibly colliding tend to lose energy through a variety of ways but theoretically objects could collide and not lose kinetic energy. This created the two types of collision inelastic and elastic. Elastic defines collisions that have no change in kinetic energy, these are usually microscopic such as Rutherford Scattering. All other objects tend to be inelastic since some energy in lost between the objects and so their respective kinetic energy changes. However, some objects can stick together and therefore combine their masses. This is a special case of inelastic collisions and tend to be a small portion of collisions that actually occur in real life, but some examples seen are when car's collide or when a sticky substance stays on the object it is throw at.

[[File:DW2.gif]]

== See also ==

There are several other topics within this wiki that can give you more information on collisions as a whole including
[[Inelastic Collisions]] and [[Elastic Collisions]]. These topics along with the basic fundamentals such as [[Kinetic Energy]] and [[ Momentum Principle]] that we used to derive the equations will show how exactly maximally inelastic collisions are shown by classical physics.
===Further reading===
These extensive resources cover this topic in more depth
* A physics resource written by experts for an expert audience [https://en.wikipedia.org/wiki/Portal:Physics Physics Portal]
* A wiki book on modern physics [https://en.wikibooks.org/wiki/Modern_Physics Modern Physics Wiki]
* The MIT open courseware for intro physics [http://ocw.mit.edu/resources/res-8-002-a-wikitextbook-for-introductory-mechanics-fall-2009/index.htm MITOCW Wiki]
* An online concept map of intro physics [http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html HyperPhysics]
* Interactive physics simulations [https://phet.colorado.edu/en/simulations/category/physics PhET]
* OpenStax algebra based intro physics textbook [https://openstaxcollege.org/textbooks/college-physics College Physics]
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]
* A resource guide compiled by the [http://www.aapt.org/ AAPT] for educators [http://www.compadre.org/ ComPADRE]

===External links===

These are some internet articles that can show more animations and pictures to help understand this concept
*[http://www.ux1.eiu.edu/~cfadd/1150/07Mom/Inelastic.html Inelastic Collisions]
*[http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes:examples:maximally_inelastic_collision_of_two_identical_carts Carts]

==References==

The primary reference I used was the textbook : Matter and Interactions 4th edition.
Full Citation:
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. Hoboken, NJ: Wiley, 2011. Print.

http://www.howitworksdaily.com/wp-content/uploads/2016/02/818x818xAsteroid-Redirect-Mission-1024x1024.jpg.pagespeed.ic.ywLacxEeka.jpg

http://www.kshitij-iitjee.com/Study/Physics/Part1/Chapter9/33.jpg

[[Category:Collisions]]