<?xml version="1.0"?>
<feed xmlns="http://www.w3.org/2005/Atom" xml:lang="en">
	<id>http://www.physicsbook.gatech.edu/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=Ebarden3</id>
	<title>Physics Book - User contributions [en]</title>
	<link rel="self" type="application/atom+xml" href="http://www.physicsbook.gatech.edu/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=Ebarden3"/>
	<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/Special:Contributions/Ebarden3"/>
	<updated>2026-05-12T10:06:12Z</updated>
	<subtitle>User contributions</subtitle>
	<generator>MediaWiki 1.42.7</generator>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inductors&amp;diff=48157</id>
		<title>Inductors</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inductors&amp;diff=48157"/>
		<updated>2026-04-27T03:48:45Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: Edited by Ella Barden Spring 2026&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Page Claimed by Ella Barden Spring 2026&lt;br /&gt;
&lt;br /&gt;
==Inductors - Conceptual Overview==&lt;br /&gt;
&lt;br /&gt;
An inductor&#039;s broad purpose, within circuits, is to resist changes in current. Upon examining the structure and characteristics of inductors, we are able to develop mathematical and conceptual arguments that explain this behavior. Inductors typically consist of a conductor, usually wire, wrapped in a coil.&lt;br /&gt;
&lt;br /&gt;
The conductor that makes up the inductor will have a current running through it, and as this non-changing current runs, a magnetic field will be created within the center of the coil. Therefore, as the current begins to change, the respective flux through a ring of the coil will change as well. As we know, the ratio of a change in flux to the respective change in time will provide us with the potential difference resulting from the change in current. By multiplying this potential difference by the number of loops in the circuit, we are able to obtain the coil&#039;s overall potential difference due to the single change in current.&lt;br /&gt;
&lt;br /&gt;
This phenomenon is tied to the concept of magnetic energy storage. Unlike a capacitor that stores energy in an electric field between plates, an inductor stores energy within its magnetic field. When the current flows, the magnetic field builds up, and when the current attempts to drop, that stored magnetic energy is released back into the circuit to keep the electrons moving.&lt;br /&gt;
&lt;br /&gt;
Shutterstock&lt;br /&gt;
Explore&lt;br /&gt;
&lt;br /&gt;
The essential directional behavior of the induced emf is to be opposite the coil&#039;s magnetic field when the current increases and in line with the coil&#039;s magnetic field when the current decreases. We can make this assumption because if the behaviors were reversed, any change in current would result in an infinite increase or decrease. This phenomenon occurs because of the direction of the non-Coulomb (NC) electric field&#039;s direction in relation to the Coulomb electric field&#039;s direction (direction of current). When the current through the inductor increases, the NC electric field curls in the opposite direction of the current, but when the current through the inductor decreases, the NC electric field curls in the same direction as the current.&lt;br /&gt;
&lt;br /&gt;
This is fundamentally an expression of Lenz’s Law, which states that the direction of an induced current is such that it creates a magnetic field opposing the change in the original magnetic flux. You can think of it as a form of electrical inertia. Just as a heavy object resists a change in its state of motion, an inductor resists a change in the state of the current flowing through it.&lt;br /&gt;
&lt;br /&gt;
The mathematical characterization of this phenomenon will be described in greater detail below.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model and Explanation===&lt;br /&gt;
&lt;br /&gt;
As we established earlier, an inductor&#039;s emf has roots in ratio of change in flux to change in time. We most often look at solenoids when discussing inductors, so the explanations and diagrams below will be solenoid specific though the concepts can be applied to coils with any shape - this is the case as flux is dependent on area.&lt;br /&gt;
&lt;br /&gt;
The derivation below begins by providing the formula for the magnetic field within a solenoid and continues by applying this magnetic field (and subsequent changes due to change in current) to the equation for flux given the solenoid is comprised of &amp;quot;N&amp;quot; rings with radius &amp;quot;R&amp;quot;. After determining the emf of a single loop, we multiply by &amp;quot;N&amp;quot; to obtain the emf of the entire solenoid. Finally, in order to simplify our expression and provide a segway to the exploration of non-solenoid inductor shapes, we represent the proportion of total emf to change in current by the variable &amp;quot;L&amp;quot; whose units are henries (H).&lt;br /&gt;
&lt;br /&gt;
The value of L is determined by the physical geometry of the component. For a standard solenoid, the inductance is calculated by the permeability of free space multiplied by the square of the number of turns, the cross sectional area, and then divided by the length of the solenoid. This highlights why adding a high permeability core, such as iron, inside the coil significantly boosts the inductance because it concentrates the magnetic flux.&lt;br /&gt;
&lt;br /&gt;
[[File:Screen_Shot_2016-11-27_at_9.52.28_PM.png]]&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
The following sequence is a visualization of inductors in a simple circuit&lt;br /&gt;
&lt;br /&gt;
[[File:physics gif 1.gif]]&lt;br /&gt;
&lt;br /&gt;
Note, inductance and resistance have inverse effects on current. Resistance dissipates energy as heat while inductance stores it temporarily. This relationship is often characterized in RL circuits by a time constant, which is the ratio of inductance to resistance. This constant tells us how quickly the current will reach its maximum value after a switch is closed.&lt;br /&gt;
&lt;br /&gt;
[[File:physics gif 2.gif]]&lt;br /&gt;
&lt;br /&gt;
Here&#039;s one final gif to show how inductors are crucial in the design of power supplies. See how the force is created?&lt;br /&gt;
&lt;br /&gt;
[[File:physics gif 3.gif]]&lt;br /&gt;
&lt;br /&gt;
(original simulation created by Eugene Khutoryansky on Youtube)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
What is the magnitude of the induced emf of one loop of a 300 turn solenoid that has a radius of 5 cm and length of 2 m when the current running through the solenoid decreases by 1 Amp/second?&lt;br /&gt;
&lt;br /&gt;
We must first realize that this question only requests that we provide the emf of a single loop. Therefore, we must ensure that we only use a single &amp;quot;N&amp;quot; term rather than &amp;quot;N^2&amp;quot;.&lt;br /&gt;
Once we simplify our overall emf equation so that it applies to a single loop, we can plug in our values and solve from there.&lt;br /&gt;
&lt;br /&gt;
[[File:Screen_Shot_2016-11-27_at_10.15.27_PM.png]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A 5 meter solenoid with 500 turns of copper wire creates an emf with a magnitude of 20V when current increases by 10 Amps/second.&lt;br /&gt;
What is the radius of the coil?&lt;br /&gt;
&lt;br /&gt;
We begin with our simplified expression and begin algebraic rearrangements to create an expression that provides us with a value for &amp;quot;R&amp;quot;. After this, we simply plug in the values we are given, and compute our radius.&lt;br /&gt;
&lt;br /&gt;
[[File:Screen_Shot_2016-11-27_at_10.03.10_PM.png]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
An infinite straight wire with current I is next to a solenoid of 10 square loops with dimensions L. The side of the solenoid closest to the wire is at a distance &amp;quot;s&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
This problem requires us to integrate the magnetic field of the infinite wire over the area of the square loops. Since the magnetic field of a wire decreases as the inverse of the distance, the flux is not uniform across the square loops. We would set up an integral from s to s plus L to find the total flux through one loop and then multiply by the number of turns.&lt;br /&gt;
&lt;br /&gt;
[[File:Screen_Shot_2016-11-27_at_11.27.25_PM.png]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&#039;&#039;How is this topic connected to something that you are interested in?&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
I find the concept of inductance incredibly interesting because it serves as an almost natural buffer to any deviation from &amp;quot;equilibrium&amp;quot;. This intrigues me because I find biology and chemistry very interesting, and within those two subjects, we often discuss homeostasis and chemical equilibrium and the natural buffers that exist in order to counteract movements away from the state of equilibrium. It is fascinating how the laws of physics provide a mechanism to maintain stability in a system just like biological systems maintain a constant internal environment.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Is there an interesting industrial application?&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Inductors have a vast range of industrial applications ranging from transformers that help manage the power supply to massive cities or energy storage for personal computers. The defining characteristics of inductors also enable them to be used at traffic lights in order to gauge traffic flow at intersections. Living in Atlanta, we all know the horrors of rush hour, so I find this industrial application incredibly interesting. These sensors work as inductive loops buried in the road. When a large metal object like a car sits on top of them, the inductance of the loop changes, which signals the computer to change the light.&lt;br /&gt;
&lt;br /&gt;
For more information on how inductors play a role in traffic control, please visit this link from HowStuffWorks: http://auto.howstuffworks.com/car-driving-safety/safety-regulatory-devices/question234.htm&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;How is it connected to your major?&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The concept of inductors does not directly relate to my major as I am a Business major. However, I am taking Pre-Medical Courses, and as mentioned earlier in this section, the way in which inductors provide a natural buffer within circuits is mirrored through various biology and chemistry concepts. For example, in the human body, the pH of our blood is kept stable by buffer systems that resist change, much like an inductor resists a change in current.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Joseph Henry is the father of inductance. Although Joseph Henry&#039;s work on self-inductance may not hold the same place in common knowledge as Sir Isaac Newton&#039;s Laws, Henry had an enormous impact on history during the 19th century. The American scientist discovered self-inductance when performing experiments with electromagnets, and although much credit for this work is given to the Englishman Michael Faraday, who published his work first, many suggest that Henry discovered these concepts first. Because of Henry&#039;s importance in this field, the units of inductance take his name: henry (H).&lt;br /&gt;
&lt;br /&gt;
Henry also greatly impacted scientific advancement within America as he was the first Smithsonian Secretary. His dedication to the dissemination of knowledge helped establish a foundation for American research that persists today.&lt;br /&gt;
&lt;br /&gt;
The discovery of inductors quickly led to the discovery of something super cool: transformers. The simplest version of a transformer is just two inductors next to each other not touching because the magnetic field of each inductor affects the other. This discovery kicked off years of experimentation. Eventually, William Stanley of George Westinghouse created the first commercially viable transformer in the 1880s. This launched George Westinghouse into the public consciousness and they quickly became a worldwide infrastructure powerhouse--even powering the World’s Columbian Exposition in Chicago in 1893.&lt;br /&gt;
&lt;br /&gt;
This period was known as the War of Currents, where Westinghouse’s alternating current (AC) system, which relied heavily on transformers and inductors, eventually triumphed over Thomas Edison’s direct current (DC) system for long distance power transmission.&lt;br /&gt;
&lt;br /&gt;
In a greater, historical perspective, the introduction of inductors into complex circuits provided the basis for the evolution of modern power grids and electric technology as we know it today. The supply of the massive amounts of electricity that today&#039;s society consumes would be incredibly unstable without transformers, whose properties are made possible by inductors.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://powerelectronics.com/passive-components/one-powerful-decade-update-inductors-provide-important-functions-power-electronic&lt;br /&gt;
&lt;br /&gt;
https://www.lifewire.com/applications-of-inductors-818816&lt;br /&gt;
&lt;br /&gt;
http://electronics.howstuffworks.com/inductor3.htm&lt;br /&gt;
&lt;br /&gt;
http://auto.howstuffworks.com/car-driving-safety/safety-regulatory-devices/question234.htm&lt;br /&gt;
&lt;br /&gt;
https://www.aps.org/programs/outreach/history/historicsites/henry.cfm&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
Matter &amp;amp; Interactions, Volume 2 4th Edition (Section 22.6)&lt;br /&gt;
&lt;br /&gt;
https://www.quora.com/What-is-the-function-of-inductors-and-capacitors&lt;br /&gt;
&lt;br /&gt;
http://electronics.howstuffworks.com/inductor.htm&lt;br /&gt;
&lt;br /&gt;
http://www.edisontechcenter.org/JosephHenry.html&lt;br /&gt;
&lt;br /&gt;
http://siarchives.si.edu/history/joseph-henry&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=ukBFPrXiKWA&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47523</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47523"/>
		<updated>2025-12-01T04:06:22Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve momentum, internal energy, and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision, so kinetic energy=0! &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions occur perfectly only in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
===Inelastic vs. Elastic Collisions===&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: Momentum conserved, kinetic energy conserved (deltaK equals 0), there is no energy transformation, energy is not lost (as always)! Ex= subatomic particles, Billiard balls.&lt;br /&gt;
&lt;br /&gt;
Inelastic collisions: Momentum is conserved, kinetic energy is not conserved (deltaK does not equal 0), energy is transformed but not lost! Ex= Car crashes, clay balls, ballistic pendulum.&lt;br /&gt;
&lt;br /&gt;
Here is a great video describing the difference between elastic and inelastic collisions in simple terms: https://www.youtube.com/watch?v=M2xnGcaaAi4 &lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level, not much in our macroscopic world, there are approximations at best.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
&lt;br /&gt;
Utilizes the conservation of momentum equation, very important for this sector of physics: m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is extremely important and very common especially in the Nuclear field. Nuclear collisions utilize the subatomic particles that have elastic collisions! Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
In elastic collisions on an atomic level: The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
==How to solve elastic collision problems in general==&lt;br /&gt;
&lt;br /&gt;
Here is a video of a simple problem that shows the main concepts described below: https://www.youtube.com/watch?v=CFbo_nBdBco&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Write the conservation of momentum equation: Pi1 + Pi2 = Pf1 + Pf2 ---&amp;gt; P=mv ----&amp;gt; m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
&lt;br /&gt;
2. Separate the problem into vector components x,y, and z&lt;br /&gt;
&lt;br /&gt;
3. Write the conservation of kinetic energy equation: KE= (1/2_mv^2). -----&amp;gt; (1/2)m1(vi1)^2 + (1/2)m2(vi2)^2 = (1/2)m1(vf1)^2 + (1/2)m2(vf2)^2&lt;br /&gt;
&lt;br /&gt;
4. Utilize these two equations for a system of equations problem and solve for variables&lt;br /&gt;
&lt;br /&gt;
5. In the end of the equation, the initial and final sides should equal each other, because kinetic energy and momentum are conserved! &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
&lt;br /&gt;
Here are computational models to help visualize perfectly elastic collisions that occur in computational models, but not in real life due to heat loss, etc.&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
- Be sure to show all steps in your solution and include diagrams whenever possible. &lt;br /&gt;
- 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.&lt;br /&gt;
- It is easiest to solve problems symbolically to find the variable the problem is asking for, then to substitute the real values in at the end of the process.&lt;br /&gt;
- It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple examples===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Mid difficulty example===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult example===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. This is an example of inelastic collisions in the real world.&lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Another example of approximated elastic 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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
The history of specifically elastic collisions started by Isaac Newton formulating the laws of motion and classical mechanics. Later on, James Clerk Maxwell studied gas behavior with kinetic energy and found the molecular interactions participating in elastic collisions with no transformation of energy in a system and kinetic energy=0. In the 20th century, quantum mechanics was developed, which provided a further look into elastic particle interactions! We can now contribute approximations to macroscopic situations, while knowing perfectly elastic collisions only happen in the microscopic world not seen by the human eye, but which can be discovered by physics!&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
https://www.ai-futureschool.com/en/physics/understanding-elastic-collisions-in-physics.php&lt;br /&gt;
&lt;br /&gt;
How to solve elastic problems in general:&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/physics/pages/8-3-elastic-and-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47517</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47517"/>
		<updated>2025-12-01T03:55:32Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve momentum, internal energy, and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision, so kinetic energy=0! &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions occur perfectly only in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
===Inelastic vs. Elastic Collisions===&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: Momentum conserved, kinetic energy conserved (deltaK equals 0), there is no energy transformation, energy is not lost (as always)! Ex= subatomic particles, Billiard balls.&lt;br /&gt;
&lt;br /&gt;
Inelastic collisions: Momentum is conserved, kinetic energy is not conserved (deltaK does not equal 0), energy is transformed but not lost! Ex= Car crashes, clay balls, ballistic pendulum.&lt;br /&gt;
&lt;br /&gt;
Here is a great video describing simply the difference between elastic and inelastic collisions: https://www.youtube.com/watch?v=M2xnGcaaAi4 &lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
&lt;br /&gt;
Utilizes the conservation of momentum equation, very important for this sector of physics: m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
In elastic collisions on an atomic level: The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
===How to solve elastic collision problems in general===&lt;br /&gt;
&lt;br /&gt;
Here is a video of a simple problem that shows the main concepts described below: https://www.youtube.com/watch?v=CFbo_nBdBco&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Write the conservation of momentum equation: Pi1 + Pi2 = Pf1 + Pf2 ---&amp;gt; P=mv ----&amp;gt; m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
&lt;br /&gt;
2. Separate the problem into vector components x,y, and z&lt;br /&gt;
&lt;br /&gt;
3. Write the conservation of kinetic energy equation: KE= (1/2_mv^2). -----&amp;gt; (1/2)m1(vi1)^2 + (1/2)m2(vi2)^2 = (1/2)m1(vf1)^2 + (1/2)m2(vf2)^2&lt;br /&gt;
&lt;br /&gt;
4. Utilize these two equations for a system of equations problem and solve for variables&lt;br /&gt;
&lt;br /&gt;
5. In the end of the equation, the initial and final sides should equal each other, because kinetic energy and momentum are conserved! &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Mid difficulty===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. This is an example of inelastic collisions in the real world.&lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Another example of approximated elastic 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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
The history of specifically elastic collisions started by Isaac Newton formulating the laws of motion and classical mechanics. Later on, James Clerk Maxwell studied gas behavior with kinetic energy and found the molecular interactions participating in elastic collisions with no transformation of energy in a system and kinetic energy=0. In the 20th century, quantum mechanics was developed, which provided a further look into elastic particle interactions! We can now contribute approximations to macroscopic situations, while knowing perfectly elastic collisions only happen in the microscopic world not seen by the human eye, but which can be discovered by physics!&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
https://www.ai-futureschool.com/en/physics/understanding-elastic-collisions-in-physics.php&lt;br /&gt;
&lt;br /&gt;
How to solve elastic problems in general:&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/physics/pages/8-3-elastic-and-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47510</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47510"/>
		<updated>2025-12-01T03:46:57Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve momentum, internal energy, and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision, so kinetic energy=0! &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions occur perfectly only in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
===Inelastic vs. Elastic Collisions===&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: Momentum conserved, kinetic energy conserved (deltaK equals 0), there is no energy transformation, energy is not lost (as always)! Ex= subatomic particles, Billiard balls.&lt;br /&gt;
&lt;br /&gt;
Inelastic collisions: Momentum is conserved, kinetic energy is not conserved (deltaK does not equal 0), energy is transformed but not lost! Ex= Car crashes, clay balls, ballistic pendulum.&lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
&lt;br /&gt;
Utilizes the conservation of momentum equation, very important for this sector of physics: m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
===How to solve elastic collision problems in general===&lt;br /&gt;
&lt;br /&gt;
Here is a video of a simple problem that shows the main concepts described below: https://www.youtube.com/watch?v=CFbo_nBdBco&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Write the conservation of momentum equation: Pi1 + Pi2 = Pf1 + Pf2 ---&amp;gt; P=mv ----&amp;gt; m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
&lt;br /&gt;
2. Separate the problem into vector components x,y, and z&lt;br /&gt;
&lt;br /&gt;
3. Write the conservation of kinetic energy equation: KE= (1/2_mv^2). -----&amp;gt; (1/2)m1(vi1)^2 + (1/2)m2(vi2)^2 = (1/2)m1(vf1)^2 + (1/2)m2(vf2)^2&lt;br /&gt;
&lt;br /&gt;
4. Utilize these two equations for a system of equations problem and solve for variables&lt;br /&gt;
&lt;br /&gt;
5. In the end of the equation, the initial and final sides should equal each other, because kinetic energy and momentum are conserved! &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Mid difficulty===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. This is an example of inelastic collisions in the real world.&lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Another example of approximated elastic 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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
The history of specifically elastic collisions started by Isaac Newton formulating the laws of motion and classical mechanics. Later on, James Clerk Maxwell studied gas behavior with kinetic energy and found the molecular interactions participating in elastic collisions with no transformation of energy in a system and kinetic energy=0. In the 20th century, quantum mechanics was developed, which provided a further look into elastic particle interactions! We can now contribute approximations to macroscopic situations, while knowing perfectly elastic collisions only happen in the microscopic world not seen by the human eye, but which can be discovered by physics!&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
https://www.ai-futureschool.com/en/physics/understanding-elastic-collisions-in-physics.php&lt;br /&gt;
&lt;br /&gt;
How to solve elastic problems in general:&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/physics/pages/8-3-elastic-and-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47499</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47499"/>
		<updated>2025-12-01T03:24:11Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve momentum, internal energy, and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision, so kinetic energy=0! &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions occur perfectly only in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
===Inelastic vs. Elastic Collisions===&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: Momentum conserved, kinetic energy conserved (deltaK equals 0), there is no energy transformation, energy is not lost (as always)! Ex= subatomic particles, Billiard balls.&lt;br /&gt;
&lt;br /&gt;
Inelastic collisions: Momentum is conserved, kinetic energy is not conserved (deltaK does not equal 0), energy is transformed but not lost! Ex= Car crashes, clay balls, ballistic pendulum.&lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
&lt;br /&gt;
Utilizes the conservation of momentum equation, very important for this sector of physics: m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
===How to solve elastic collision problems in general===&lt;br /&gt;
&lt;br /&gt;
Here is a video of a simple problem that shows the main concepts described below: https://www.youtube.com/watch?v=CFbo_nBdBco&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Write the conservation of momentum equation: Pi1 + Pi2 = Pf1 + Pf2 ---&amp;gt; P=mv ----&amp;gt; m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
&lt;br /&gt;
2. Separate the problem into vector components x,y, and z&lt;br /&gt;
&lt;br /&gt;
3. Write the conservation of kinetic energy equation: KE= (1/2_mv^2). -----&amp;gt; (1/2)m1(vi1)^2 + (1/2)m2(vi2)^2 = (1/2)m1(vf1)^2 + (1/2)m2(vf2)^2&lt;br /&gt;
&lt;br /&gt;
4. Utilize these two equations for a system of equations problem and solve for variables&lt;br /&gt;
&lt;br /&gt;
5. In the end of the equation, the initial and final sides should equal each other, because kinetic energy and momentum are conserved! &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Mid difficulty===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
The history of specifically elastic collisions started by Isaac Newton formulating the laws of motion and classical mechanics. Later on, James Clerk Maxwell studied gas behavior with kinetic energy and found the molecular interactions participating in elastic collisions with no transformation of energy in a system and kinetic energy=0. In the 20th century, quantum mechanics was developed, which provided a further look into elastic particle interactions! We can now contribute approximations to macroscopic situations, while knowing perfectly elastic collisions only happen in the microscopic world not seen by the human eye, but which can be discovered by physics!&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
https://www.ai-futureschool.com/en/physics/understanding-elastic-collisions-in-physics.php&lt;br /&gt;
&lt;br /&gt;
How to solve elastic problems in general:&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/physics/pages/8-3-elastic-and-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47497</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47497"/>
		<updated>2025-12-01T03:14:24Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: /* Middling */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve internal energy and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision. &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions often occur in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
===Inelastic vs. Elastic Collisions===&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: Momentum conserved, kinetic energy conserved (deltaK equals 0), there is no energy transformation, energy is not lost (as always)! Ex= subatomic particles, Billiard balls.&lt;br /&gt;
&lt;br /&gt;
Inelastic collisions: Momentum is conserved, kinetic energy is not conserved (deltaK does not equal 0), energy is transformed but not lost! Ex= Car crashes, clay balls, ballistic pendulum.&lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
===How to solve elastic collision problems in general===&lt;br /&gt;
&lt;br /&gt;
Here is a video of a simple problem that shows the main concepts described below: https://www.youtube.com/watch?v=CFbo_nBdBco&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Write the conservation of momentum equation: Pi1 + Pi2 = Pf1 + Pf2 ---&amp;gt; P=mv ----&amp;gt; m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
&lt;br /&gt;
2. Separate the problem into vector components x,y, and z&lt;br /&gt;
&lt;br /&gt;
3. Write the conservation of kinetic energy equation: KE= (1/2_mv^2). -----&amp;gt; (1/2)m1(vi1)^2 + (1/2)m2(vi2)^2 = (1/2)m1(vf1)^2 + (1/2)m2(vf2)^2&lt;br /&gt;
&lt;br /&gt;
4. Utilize these two equations for a system of equations problem and solve for variables&lt;br /&gt;
&lt;br /&gt;
5. In the end of the equation, the initial and final sides should equal each other, because kinetic energy and momentum are conserved! &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Mid difficulty===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
How to solve elastic problems in general:&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/physics/pages/8-3-elastic-and-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47496</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47496"/>
		<updated>2025-12-01T03:09:15Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve internal energy and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision. &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions often occur in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
===Inelastic vs. Elastic Collisions===&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: Momentum conserved, kinetic energy conserved (deltaK equals 0), there is no energy transformation, energy is not lost (as always)! Ex= subatomic particles, Billiard balls.&lt;br /&gt;
&lt;br /&gt;
Inelastic collisions: Momentum is conserved, kinetic energy is not conserved (deltaK does not equal 0), energy is transformed but not lost! Ex= Car crashes, clay balls, ballistic pendulum.&lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
===How to solve elastic collision problems in general===&lt;br /&gt;
&lt;br /&gt;
Here is a video of a simple problem that shows the main concepts described below: https://www.youtube.com/watch?v=CFbo_nBdBco&lt;br /&gt;
&lt;br /&gt;
Steps:&lt;br /&gt;
&lt;br /&gt;
1. Write the conservation of momentum equation: Pi1 + Pi2 = Pf1 + Pf2 ---&amp;gt; P=mv ----&amp;gt; m1vi1 + m2vi2 = m1vf1 + m2vf2&lt;br /&gt;
&lt;br /&gt;
2. Separate the problem into vector components x,y, and z&lt;br /&gt;
&lt;br /&gt;
3. Write the conservation of kinetic energy equation: KE= (1/2_mv^2). -----&amp;gt; (1/2)m1(vi1)^2 + (1/2)m2(vi2)^2 = (1/2)m1(vf1)^2 + (1/2)m2(vf2)^2&lt;br /&gt;
&lt;br /&gt;
4. Utilize these two equations for a system of equations problem and solve for variables&lt;br /&gt;
&lt;br /&gt;
5. In the end of the equation, the initial and final sides should equal each other, because kinetic energy and momentum are conserved! &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
How to solve elastic problems in general:&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/physics/pages/8-3-elastic-and-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47494</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47494"/>
		<updated>2025-12-01T02:57:05Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve internal energy and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision. &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions often occur in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
==Inelastic vs. Elastic Collisions==&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: Momentum conserved, kinetic energy conserved (deltaK equals 0), there is no energy transformation, energy is not lost (as always)! Ex= subatomic particles, Billiard balls.&lt;br /&gt;
&lt;br /&gt;
Inelastic collisions: Momentum is conserved, kinetic energy is not conserved (deltaK does not equal 0), energy is transformed but not lost! Ex= Car crashes, clay balls, ballistic pendulum.&lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47493</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47493"/>
		<updated>2025-12-01T02:53:46Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve internal energy and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision. &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
Overall in elastic collisions-&amp;gt; Energy (kinetic energy and momentum) is conserved. The change in kinetic energy (deltaK)=0. The system doesn&#039;t change its shape, temperature, etc! This never really happens in real life for macroscopic objects, it only happens perfectly on atomic levels, but pool balls colliding is a good approximation of what happens! &lt;br /&gt;
&lt;br /&gt;
These idealized interactions often occur in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
==Inelastic vs. Elastic Collisions==&lt;br /&gt;
&lt;br /&gt;
Elastic collisions: &lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
Inelastic vs. Elastic Collisions:&lt;br /&gt;
&lt;br /&gt;
https://kingofthecurve.org/blog/elastic-vs-inelastic-collisions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47492</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47492"/>
		<updated>2025-12-01T02:48:16Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve internal energy and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision. &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
These idealized interactions often occur in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
==Inelastic vs. Elastic Collisions==&lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47491</id>
		<title>Elastic Collisions</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Elastic_Collisions&amp;diff=47491"/>
		<updated>2025-12-01T02:47:07Z</updated>

		<summary type="html">&lt;p&gt;Ebarden3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Edit by Ella Barden Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
While the term &amp;quot;elastic&amp;quot; may evoke rubber bands or bubble gum, in physics it specifically refers to collisions that conserve internal energy and kinetic energy. Elastic collisions are interactions between two or more objects where no kinetic energy is lost during the collision. &lt;br /&gt;
&lt;br /&gt;
[[File:Elastic collision.svg|Elastic collision|right]]&lt;br /&gt;
&lt;br /&gt;
A collision is a short-duration, high-force interaction between two or more objects where their motion changes dramatically over a very small amount of time. In physics, an elastic collision is defined as one where both momentum and kinetic energy are conserved. Unlike inelastic collisions, no kinetic energy is converted into internal energy forms such as heat, deformation, sound, or vibrations.&lt;br /&gt;
&lt;br /&gt;
In an elastic collision:&lt;br /&gt;
&lt;br /&gt;
*Objects bounce off each other.&lt;br /&gt;
&lt;br /&gt;
*There is no net loss of kinetic energy.&lt;br /&gt;
&lt;br /&gt;
*There is no lasting deformation or generation of heat within the bodies involved.&lt;br /&gt;
&lt;br /&gt;
These idealized interactions often occur in microscopic systems, such as between gas molecules, atoms, or subatomic particles. On a macroscopic level, real-world collisions (such as between billiard balls or on air tracks) can approximate elastic collisions if friction and deformation are minimized. Elastic collisions are particularly important in thermodynamics and statistical mechanics, where they help explain gas pressure and temperature as emergent properties from molecular motion.&lt;br /&gt;
&lt;br /&gt;
Although perfect elastic collisions don&#039;t exist on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation. on a macroscopic scale due to inevitable energy loss (e.g., heat), they can occur in atomic or subatomic interactions where quantized energy levels are involved, provided the collision energy isn&#039;t high enough to cause excitation.&lt;br /&gt;
&lt;br /&gt;
====Conditions and Analysis for Elastic Collision====&lt;br /&gt;
*Total kinetic energy before and after the collision is the same.&lt;br /&gt;
*Kinetic energy temporarily converts to elastic potential energy and back.&lt;br /&gt;
*Occurs frequently at the atomic or molecular level.&lt;br /&gt;
*Relative speed of approach equals relative speed of separation.&lt;br /&gt;
                &lt;br /&gt;
Here&#039;s a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw&lt;br /&gt;
&lt;br /&gt;
===Nuclear Collisions===&lt;br /&gt;
Although elastic collisions are not a common thing you see, on a nuclear scale it is exetremely important and very common especially in the Nuclear field. Nuclear collisions are interactions between atomic nuclei or between a nucleus and a subatomic particle (such as a neutron or alpha particle). These collisions can be elastic, inelastic, or involve absorption. In these events, extremely high energy levels may result in radiation, decay, or even nuclear fission.&lt;br /&gt;
&lt;br /&gt;
====Types of Nuclear Collisions====&lt;br /&gt;
Elastic: No internal energy change; kinetic energy is conserved.&lt;br /&gt;
Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation.&lt;br /&gt;
Absorption: The incoming particle is captured, sometimes causing decay or fission.&lt;br /&gt;
Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.&lt;br /&gt;
&lt;br /&gt;
====Alpha Particles====&lt;br /&gt;
Alpha particles (α) consist of two protons and two neutrons. They are emitted in radioactive decay, are highly ionizing, and can be used in both scattering and absorption experiments. Due to their mass, alpha particles follow relatively straight paths and transfer substantial energy in collisions.&lt;br /&gt;
&lt;br /&gt;
====Scattering Efficiency Formula====&lt;br /&gt;
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: &amp;lt;math&amp;gt; (A-1)^2/(A+1)^2 &amp;lt;/math&amp;gt; Where &amp;lt;math&amp;gt; A=M/m &amp;lt;/math&amp;gt;. This expression helps quantify energy loss per collision based on the mass ratio.&lt;br /&gt;
This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
&lt;br /&gt;
1. &amp;lt;math&amp;gt; K_f = K_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt; \Delta E_{int} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{sys} + \Delta E_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta E_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; E_{final} = E_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3. &amp;lt;math&amp;gt; \vec{p}_f = \vec{p}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta \vec{p}_{surr} = 0 &amp;lt;/math&amp;gt; , so &amp;lt;math&amp;gt; \Delta \vec{p}_{sys} = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} \\&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4. &amp;lt;math&amp;gt; \vec{KE}_f = \vec{KE}_i &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1/2{m_1*v_{1i}^2} + 1/2{m_2*v_{2i}^2} = 1/2{m_1*v_{1f}^2} + + 1/2{m_2*v_{2f}^2} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There is another (easier) method of solving problems involving elastic collisions. This involves the use of the center of mass velocity, which can be calculated as &amp;lt;math&amp;gt; v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} &amp;lt;/math&amp;gt;. The elastic collision formula is &amp;lt;math&amp;gt;|v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}|&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Computational Models===&lt;br /&gt;
[[File:Collision carts elastic.gif|Collision carts elastic|right]]&lt;br /&gt;
The trinket model linked demonstrates an elastic collision between two spheres.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/glowscript/f3bc1dd31f Elastic Collision Glowscript Model]&lt;br /&gt;
&lt;br /&gt;
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time. &lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=AfCfCS6O2VM Elastic Collision MATLAB model]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&#039;&#039;Reminders&#039;&#039;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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. It would be helpful to also write the main principles at the side to remind yourself of how you found it. If you still get stuck, try finding similar examples to that problem and look at the solutions to get on the right track.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Example1good.jpg|300px|thumb|right]]&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship&lt;br /&gt;
between the initial and final momentums of the two carts.  Usually, when we want to consider the system, we will consider the two colliding objects as the system and the rest as the surroundings. The initial state would be before the collision, and the final state would be after.&lt;br /&gt;
&lt;br /&gt;
Momentum Principle: &amp;lt;math&amp;gt; \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And it turns into: &amp;lt;math&amp;gt; \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} &amp;lt;/math&amp;gt; since during the collision, the &amp;lt;math&amp;gt; F_{net} &amp;lt;/math&amp;gt; is negligible.&lt;br /&gt;
&lt;br /&gt;
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 that the reaction happens so quickly that neither work nor heat transfer is done by the surroundings, the equation can be simplified to include just initial and final kinetic energies. &lt;br /&gt;
&lt;br /&gt;
The Energy Principle: &amp;lt;math&amp;gt; \Delta E_{sys} = W + Q &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We then get rid of the work, heat transfer and internal energies: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q  \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; K_{1f} + K_{2f} = K_{1i} + K_{2i} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The reason the internal energies are directly crossed out is because we can put them to one side since &amp;lt;math&amp;gt; E_{final} = E_{initial} &amp;lt;/math&amp;gt; and therefore &amp;lt;math&amp;gt; E_{final} -&lt;br /&gt;
 E_{initial} = 0 &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Kinetic Energy Definition: &amp;lt;math&amp;gt; K = \frac{p^2}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Finished Result: &amp;lt;math&amp;gt; \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 10 kg mass travelling 2 m/s collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocity of the 10 kg mass.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Solving for the center of mass velocity:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1  m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
By definition, the difference between the initial and center of mass velocities and the difference between the final and center of mass velocities are the same.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |v_{1i} - v_{com}| = |v_{1f} - v_{com}| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 1 m/s = v_{1f} - 1 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f} = 0 m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Jay and Sarah are best friends. Since they’re best friends, they both weigh &amp;lt;math&amp;gt; 55 kg &amp;lt;/math&amp;gt;. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of &amp;lt;math&amp;gt;&amp;lt;4,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. 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 &amp;lt;math&amp;gt;&amp;lt;7,0,0&amp;gt; m/s&amp;lt;/math&amp;gt;. What was Jay’s final velocity?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1.0.png|400px|Problem 1]]&lt;br /&gt;
&lt;br /&gt;
In a lot of simple elastic collision problems, the momentum principle is all you need to solve them. Most problems will give you 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Equal mass head-on elastic collision:&lt;br /&gt;
Ball 1 transfers all velocity to Ball 2.&lt;br /&gt;
Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Conceptual Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Both have equal and opposite momenta.&lt;br /&gt;
After elastic collision, they reverse directions at the same speed: ±2 m/s.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When far apart, the momentum of a proton is &amp;lt;math&amp;gt;&amp;lt;5.2 * 10^{−21}, 0, 0&amp;gt; kg · m/s&amp;lt;/math&amp;gt; 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 &amp;lt;math&amp;gt;&amp;lt;1.75, .82, 0&amp;gt; m/s&amp;lt;/math&amp;gt;. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is &amp;lt;math&amp;gt;1.7 *10^{-21} kg&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2.0.png|300px]]&lt;br /&gt;
&lt;br /&gt;
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 &amp;lt;math&amp;gt;p = mv&amp;lt;/math&amp;gt; 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.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Find center of mass velocity:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{cm}={(5*3)+(2*(-2))}/7     =     {15-4}/7  =   1.57m/s   &amp;lt;/math&amp;gt;&lt;br /&gt;
Use transformation:&lt;br /&gt;
&amp;lt;math&amp;gt; v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of &amp;lt;math&amp;gt; m_A = 2kg &amp;lt;/math&amp;gt; and an initial velocity of &amp;lt;math&amp;gt; {v_A} = &amp;lt;4,0,0&amp;gt; m/s &amp;lt;/math&amp;gt;. Sphere B has a mass of &amp;lt;math&amp;gt; m_B = 3kg &amp;lt;/math&amp;gt; and is initially at rest. After the collision, sphere A moves at an angle of 30 degrees above the x-axis, and sphere B moves at 45 degrees below the x-axis. We are asked to find the final speeds of both spheres using conservation of momentum and kinetic energy.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg&lt;br /&gt;
&lt;br /&gt;
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Newton&#039;s Cradle&#039;&#039;&#039;&lt;br /&gt;
One idea that represents the concept of elastic collisions well is the tool of Newton&#039;s Cradle. Newton&#039;s Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn&#039;t magic- it&#039;s due to to the conservation of momentum in a (in this case , virtually) elastic collision.&lt;br /&gt;
&lt;br /&gt;
Think about it: Let&#039;s assume that all 5 balls have the same mass and have elastic collisions. A person lifts one ball to the side and allows it to strike the rest of the balls. The result is another ball leaving the other side at the same speed. Why does this happen?&lt;br /&gt;
Since momentum is conserved we can write:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f&amp;lt;/math&amp;gt;&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_i = v_f&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0&amp;lt;/math&amp;gt; thus ensuring that our collision is elastic.&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Now let&#039;s see if two balls can leave the other side when one ball is struck.&lt;br /&gt;
Again, let&#039;s assume that momentum is conserved in our system:&lt;br /&gt;
&amp;lt;math&amp;gt; p_f = p_i = m *v_i = m*v_f + m*v_f &amp;lt;/math&amp;gt;.&lt;br /&gt;
This means that &amp;lt;math&amp;gt;v_f = v_i/2&amp;lt;/math&amp;gt;.&lt;br /&gt;
Using the Kinetic Energy equation, we get &amp;lt;math&amp;gt;KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 &amp;lt;/math&amp;gt;.&lt;br /&gt;
However this will not equate to 0 meaning that our collision will not preserve elasticity.&lt;br /&gt;
&lt;br /&gt;
Linked here is an interactive visualization of Newton&#039;s Cradle made by B. Philhour in 2017.&lt;br /&gt;
[https://www.glowscript.org/#/user/nikhelkrishna/folder/MyPrograms/program/newton]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
So how are collisions connected to the real world? Collisions are all around us! &lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
&lt;br /&gt;
[[File:Corvette Crash Tester (3695903512).jpg|Corvette Crash Tester (3695903512)|500px]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. We consider this a head on collision of equal masses. &lt;br /&gt;
&lt;br /&gt;
[[File:Billiards_and_snookers.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
A third 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&#039;t feel any vibration when you strike the ball.&lt;br /&gt;
&lt;br /&gt;
[[File:Marcus_Thames_Tigers_2007.jpg|500px]]&lt;br /&gt;
&lt;br /&gt;
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.&lt;br /&gt;
&lt;br /&gt;
Here&#039;s a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Collisions are certainly not a new concept in the world. Ever since the beginning of time things have been colliding and reacting in different ways. It was experiments done by scientists like Newton and Rutherford that started to characterize collisions into categories and apply fundamental principles of physics to the reactions. What we observed above was the Newtonian way.&lt;br /&gt;
&lt;br /&gt;
The history of collisions originates from the Rutherford Scattering experiment. It consisted of shooting alpha particles through a thin gold foil. As Rutherford studied the scattering of alpha particles through metal foils, he first noticed a collision with a single massive positive particle. Although the alpha particles did not hit the nucleus of the gold atoms, they did interact with each other and therefore can be considered as a collision. Since the interaction did not excite the gold atoms, fortunately enough, it was an elastic collision. This lead to the conclusion that the positive charge of the mass was concentrated in the center, the atomic nucleus! The plum pudding model (where the positive and negative charges were stuck within the atom like plums in a pudding), that had been around was disproved. When further research was done, they measured the angle of the &#039;scattering&#039; or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.&lt;br /&gt;
&lt;br /&gt;
==Collision Theory==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Collision Theory is also a concept that has been used to explain collisions but in chemistry. Collision theory in chemistry allows us to predict how fast reactions will happen. For a reaction to work particles need to hit each other in the right way with enough energy. Successful collisions occur, bonds break and new bonds form. More reactants or increased temperatures lead to more collisions this was found by Max Trautz and William Lewis in 1916 and 1918 and describes a historic way how collisions aid in science.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/Collisions Collisions] for a general understanding of all collisions.&lt;br /&gt;
&lt;br /&gt;
[http://physicsbook.gatech.edu/inelastic_collisions Inelastic Collisions] to contrast them from elastic collisions.&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsbook.gatech.edu/Rutherford_Experiment_and_Atomic_Collisions] to see how nuclear collisions show us important physics and chemical principals.&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://www.britannica.com/science/elastic-collision&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/mmedia/momentum/trece.cfm Elastic Collision Example]&lt;br /&gt;
&lt;br /&gt;
[http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html Different Types of Elastic Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://www.boundless.com/physics/textbooks/boundless-physics-textbook/linear-momentum-and-collisions-7/collisions-70/elastic-collisions-in-one-dimension-298-6220/ A Lesson on Collisions]&lt;br /&gt;
&lt;br /&gt;
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html Collision Simulation]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Main Idea:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml&lt;br /&gt;
&lt;br /&gt;
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Nuclear Collisions:&lt;br /&gt;
&lt;br /&gt;
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781&lt;br /&gt;
&lt;br /&gt;
History:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Elastic_collision&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=5pZj0u_XMbc&lt;br /&gt;
&lt;br /&gt;
Newton&#039;s Cradle Code:&lt;br /&gt;
&lt;br /&gt;
https://trinket.io/glowscript/fd3363c9c0&lt;br /&gt;
&lt;br /&gt;
Connectedness:&lt;br /&gt;
&lt;br /&gt;
http://www.iihs.org/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Additionally used references from &#039;&#039;see also&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Collisions]]&lt;/div&gt;</summary>
		<author><name>Ebarden3</name></author>
	</entry>
</feed>