Elastic Collisions
Edit by Aidan Looman Summer 2025
The Main Idea
While the term "elastic" 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.
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.
In an elastic collision:
- Objects bounce off each other.
- There is no net loss of kinetic energy.
- There is no lasting deformation or generation of heat within the bodies involved.
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.
Although perfect elastic collisions don'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'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't high enough to cause excitation.
Conditions and Analysis for Elastic Collision
- Total kinetic energy before and after the collision is the same.
- Kinetic energy temporarily converts to elastic potential energy and back.
- Occurs frequently at the atomic or molecular level.
- Relative speed of approach equals relative speed of separation.
Here's a video walkthrough of a basic elastic collision: https://www.youtube.com/watch?v=V4vzNk4qppw
Nuclear Collisions
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.
Types of Nuclear Collisions
Elastic: No internal energy change; kinetic energy is conserved. Inelastic: Part of kinetic energy excites the nucleus or is emitted as radiation. Absorption: The incoming particle is captured, sometimes causing decay or fission. Fission: A heavy nucleus splits into smaller nuclei, releasing energy and particles.
Alpha Particles
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.
Scattering Efficiency Formula
The fraction of energy transferred from an alpha particle (mass m) to a target nucleus (mass M) during an elastic collision is given by: [math]\displaystyle{ (A-1)^2/(A+1)^2 }[/math] Where [math]\displaystyle{ A=M/m }[/math]. This expression helps quantify energy loss per collision based on the mass ratio. This formula is useful in applications like neutron moderation, radiation shielding, and analysis of nuclear reactions.
A Mathematical Model
Now lets take what we know about elastic collisions and talk about it in mathematical terms. There are three main mathematical concepts surrounding elastic collisions:
1. [math]\displaystyle{ K_f = K_i }[/math]
2. [math]\displaystyle{ \Delta E_{int} = 0 }[/math]
[math]\displaystyle{ \Delta E_{sys} + \Delta E_{surr} = 0 \\ }[/math] [math]\displaystyle{ \Delta E_{surr} = 0 }[/math] , so [math]\displaystyle{ \Delta E_{sys} = 0 }[/math] [math]\displaystyle{ E_{final} = E_{initial} \\ }[/math]
3. [math]\displaystyle{ \vec{p}_f = \vec{p}_i }[/math]
[math]\displaystyle{ \Delta \vec{p}_{sys} + \Delta \vec{p}_{surr} = 0 \\ }[/math] [math]\displaystyle{ \Delta \vec{p}_{surr} = 0 }[/math] , so [math]\displaystyle{ \Delta \vec{p}_{sys} = 0 }[/math] [math]\displaystyle{ \vec{p}_{final} = \vec{p}_{initial} \\ }[/math]
[math]\displaystyle{ m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + + m_2*v_{2}) }[/math]
4. [math]\displaystyle{ \vec{KE}_f = \vec{KE}_i }[/math]
[math]\displaystyle{ 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} }[/math]
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 [math]\displaystyle{ v_{COM} = \frac{m_{1}v_{1,i}+m_{2}v_{2,i}}{m_{1}+m_{2}} }[/math]. The elastic collision formula is [math]\displaystyle{ |v_{1,i}-v_{COM}| = |v_{1,f}-v_{COM}| }[/math].
Computational Models
The trinket model linked demonstrates an elastic collision between two spheres.
Elastic Collision Glowscript Model
The video below demonstrates a basic car crash elastic collision using MATLAB. Take note of the changes in displacement and the time.
Elastic Collision MATLAB model
Examples
Reminders
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.
Simple
Question
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?
Solution
1. After choosing a correct system, surroundings, and initial and final states, we can apply the momentum principle mentioned above to get a relationship between the initial and final momentums of the two carts. 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.
Momentum Principle: [math]\displaystyle{ \vec{p}_{final} = \vec{p}_{initial} + \vec{F}_{net}\Delta t }[/math]
And it turns into: [math]\displaystyle{ \vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{1f} }[/math] since during the collision, the [math]\displaystyle{ F_{net} }[/math] is negligible.
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.
The Energy Principle: [math]\displaystyle{ \Delta E_{sys} = W + Q }[/math]
We then get rid of the work, heat transfer and internal energies:
[math]\displaystyle{ K_{1f} + \Delta E_{1int} + K_{2f} + \Delta E_{2int} = K_{1i} + K_{2i} + W + Q \\ }[/math] [math]\displaystyle{ \Delta E_{1int} = \Delta E_{2int} = W = Q = 0 \\ }[/math] [math]\displaystyle{ K_{1f} + K_{2f} = K_{1i} + K_{2i} }[/math]
The reason the internal energies are directly crossed out is because we can put them to one side since [math]\displaystyle{ E_{final} = E_{initial} }[/math] and therefore [math]\displaystyle{ E_{final} -
E_{initial} = 0 }[/math].
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.
Kinetic Energy Definition: [math]\displaystyle{ K = \frac{p^2}{2m} }[/math]
Finished Result: [math]\displaystyle{ \frac{p^2_{1f}}{2m} + \frac{p^2_{2f}}{2m} = \frac{p^2_{1i}}{2m} + \frac{p^2_{2i}}{2m} }[/math]
Question
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.
Solution
Solving for the center of mass velocity:
[math]\displaystyle{ v_{com} = \frac{10(2) + 2(-9)}{10 + 2} = 1 m/s }[/math]
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.
[math]\displaystyle{ |v_{1i} - v_{com}| = |v_{1f} - v_{com}| }[/math]
[math]\displaystyle{ |2 m/s - 1 m/s| = |v_{1f} - 1 m/s| }[/math]
[math]\displaystyle{ 1 m/s = v_{1f} - 1 m/s }[/math]
[math]\displaystyle{ v_{1f} = 0 m/s }[/math]
Question
Jay and Sarah are best friends. Since they’re best friends, they both weigh [math]\displaystyle{ 55 kg }[/math]. Jay hadn’t seen Sarah in a long time, so when she saw her she ran to her with a velocity of [math]\displaystyle{ \lt 4,0,0\gt m/s }[/math]. 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 [math]\displaystyle{ \lt 7,0,0\gt m/s }[/math]. What was Jay’s final velocity?
Solution
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.
Conceptual Question
A 1 kg ball moving at 3 m/s collides head-on with a stationary 1 kg ball. What are the final velocities?
Solution
Equal mass head-on elastic collision: Ball 1 transfers all velocity to Ball 2. Final velocities: Ball 1 = 0 m/s, Ball 2 = 3 m/s.
Conceptual Question
Two ice pucks of equal mass move toward each other at 2 m/s. What is the outcome?
Solution
Both have equal and opposite momenta. After elastic collision, they reverse directions at the same speed: ±2 m/s.
Middling
Question
When far apart, the momentum of a proton is [math]\displaystyle{ \lt 5.2 * 10^{−21}, 0, 0\gt kg · m/s }[/math] 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 [math]\displaystyle{ \lt 1.75, .82, 0\gt m/s }[/math]. At that instant, what is the momentum of the other proton? HINT: The mass of a proton is [math]\displaystyle{ 1.7 *10^{-21} kg }[/math].
Solution
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 [math]\displaystyle{ p = mv }[/math] 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.
Problem
A 5 kg cart at 3 m/s collides with a 2 kg cart at -2 m/s. Find final velocities.
Solution
Find center of mass velocity: [math]\displaystyle{ v_{cm}={(5*3)+(2*(-2))}/7 = {15-4}/7 = 1.57m/s }[/math] Use transformation: [math]\displaystyle{ v_{1f}=2v_{cm} - v_{1i} = 2(1.57) - 3 =0.14m/s }[/math] [math]\displaystyle{ v_{2f}=2v_{cm} - v_{2i} = 2(1.57) - (-2) = 5.14m/s }[/math]
Difficult
Problem
Two smooth spheres collide elastically on a frictionless horizontal surface. Sphere A has a mass of [math]\displaystyle{ m_A = 2kg }[/math] and an initial velocity of [math]\displaystyle{ {v_A} = \lt 4,0,0\gt m/s }[/math]. Sphere B has a mass of [math]\displaystyle{ m_B = 3kg }[/math] 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.
Solution
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart1.jpg
file:///C:/Users/Aidan/Documents/Physics/DifficultProblemPart2.jpg
Newton's Cradle One idea that represents the concept of elastic collisions well is the tool of Newton's Cradle. Newton's Cradle is a fixture in many physics classrooms and inspires awe in the seemingly perfect motion that it undertakes. However, this isn't magic- it's due to to the conservation of momentum in a (in this case , virtually) elastic collision.
Think about it: Let'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?
Since momentum is conserved we can write:
[math]\displaystyle{ p_f = p_i = m *v_i = m*v_f }[/math]
This means that [math]\displaystyle{ v_i = v_f }[/math].
Using the Kinetic Energy equation, we get [math]\displaystyle{ KE = 0.5*m*v_i^2 - 0.5*m*v_i^2 = 0 }[/math] thus ensuring that our collision is elastic.
Now let's see if two balls can leave the other side when one ball is struck.
Again, let's assume that momentum is conserved in our system:
[math]\displaystyle{ p_f = p_i = m *v_i = m*v_f + m*v_f }[/math].
This means that [math]\displaystyle{ v_f = v_i/2 }[/math].
Using the Kinetic Energy equation, we get [math]\displaystyle{ KE = (0.5*m*(v_i/2)^2 + 0.5*m*(v_i/2)^2) - 0.5*m*v_i^2 }[/math].
However this will not equate to 0 meaning that our collision will not preserve elasticity.
Linked here is an interactive visualization of Newton's Cradle made by B. Philhour in 2017. [1]
Connectedness
So how are collisions connected to the real world? Collisions are all around us!
One main example is cars. Lots of car companies will purposely test and wreck cars to test collisions! Data is then sent to places like the Insurance Institute for Highway Safety where we can learn about car agility and more.
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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.
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't feel any vibration when you strike the ball.
Elastic collisions also happen between particles. The Rutherford Scattering experiment mentioned below is a good example.
Here's a video to help you understand how elastic collisions connect to the real world: https://youtu.be/WE0sum7t5XE
History
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.
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 'scattering' or the particles shot through a tin gold foil. Had the collisions been inelastic, the particles would not have been able to bounce back.
Collision Theory
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.
See also
Collisions for a general understanding of all collisions.
Inelastic Collisions to contrast them from elastic collisions.
[2] to see how nuclear collisions show us important physics and chemical principals.
Further reading
http://www.britannica.com/science/elastic-collision
External links
Different Types of Elastic Collisions
References
Main Idea:
Matter and Interactions, 4th Edition
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section4.rhtml
http://blogs.bu.edu/ggarber/archive/bua-physics/collisions-and-conservation-of-momentum/
Nuclear Collisions:
Masterson, R.E. (2017). Nuclear Engineering Fundamentals: A Practical Perspective (1st ed.). CRC Press. https://doi.org/10.1201/9781315156781
History:
Matter and Interactions, 4th Edition
https://en.wikipedia.org/wiki/Elastic_collision
https://www.youtube.com/watch?v=5pZj0u_XMbc
Newton's Cradle Code:
https://trinket.io/glowscript/fd3363c9c0
Connectedness:
Additionally used references from see also.