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Author Topic: Elastic Collision (1D)  (Read 12035 times)
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Fu-Kwun Hwang
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on: February 19, 2010, 07:19:56 pm » posted from:Taipei,T\'ai-pei,Taiwan

This is a discussion of elastic collision in one dimension.
Before collision: two particles with mass and velocity as m_1,\vec{v_1} and m_2,\vec{v_1}
After collision: the velocity have been changed to \vec{v_1}' and \vec{v_2}'

Assume there is no external force or the interval is very short, then
total linear momentum is conserved: i.e. m_1\vec{v_1}+m_2\vec{v_2}=m_1\vec{v_1}'+m_2\vec{v_2}'
 so m_1(\vec{v_1}-\vec{v_1}')+m_2(\vec{v_2}-\vec{v_2}')=0

For elastic collision, the total energy is also conserved. \frac{1}{2}m_1 v_1^2+\frac{1}{2}m_2 v_2^2==\frac{1}{2}m_1 v_1^{'2}+\frac{1}{2}m_2 v_2^{'2}
It can be re-write as \frac{1}{2}m_1(v_1^2-v_1^{'2})= -\frac{1}{2}m_2(v_2^2-v_2^{'2})
It is the same as m_1(v_1-v_1')(v_1+v_1')= -m_2 (v_2-v_2')(v_2+v_2')
Since m_1(v_1-v_1')=-m_2 (v_2-v_2'), so (v_1+v_1')=(v_2+v_2') or v_1-v_2=v_2'-v_1'
The result is
v'_1= \frac{m_1-m_2}{m_1+m_2} v_1 +\frac{2m_2}{m_1+m_2}v_2=V_{cm}+\frac{m_2}{m_1+m_2}(v_2-v_1)=2V_{cm}-v_1
and v'_2=\frac{2m_1}{m_1+m_2}v_1+\frac{m_2-m_1}{m_2+m_1}v_2=V_{cm}+\frac{m_1}{m_1+m_2}(v_1-v_2)=2V_{cm}-v_2
where V_{cm}=\frac{m_1V_1+m_2V_2}{m_1+m_2}
 and V_{cm}=\frac{v_1+v_1'}{2}=\frac{v_2+v_2'}{2} or v_1+v_1'=v_2+v_2'=2V_{cm}

It means that from the coordinate of center of mass: V_{cm}=0, it reduced to
v_1'=-v_1 and v_2'=-v_2

Define \rho=\frac{m_2}{m_1}, the above equations can be re-write as
\frac{v'_1}{v_1}=\frac{1-\rho}{1+\rho}+ \frac{2\rho}{1+\rho}\frac{v_2}{v_1}=2\frac{V_{cm}}{v_1}-1
\frac{v'_2}{v_1}=\frac{2}{1+\rho}-\frac{1-\rho}{1+\rho}\frac{v_2}{v_1}=2\frac{V_{cm}}{v_1}-\frac{v_2}{v_1}

The following simulation plot the above two functions.
The X-axis is \frac{v_2}{v_1}, it range from Vscale*xmin to 1. (There is no collision if v_2>v_1)
The blue curve is \frac{v_1^'}{v_1} and red curve is \frac{v_2^'}{v_1}
You can change the ratio of \frac{m_2}{m_1} with slider.

The default value is \frac{m_2}{m_1}=1, so
 v_2^'= v_1 , so \frac{v_2^'}{v_1}=1 is a horizontal line
 v_1^'=v_2 , so v_2 is a straight line with slope 1 (function of \frac{v_2}{v_1})

Special case:
if m_1<<m_2, v_1'=-v_1+2v_2 and v_2'=v_2 ,-*-
 if v_2=0, then v_1'=-v_1 and v_2'=0
   e.g. a ball hit the wall, it will biunced back with almost the same speed (but in oppositive direction).
if m_1>>m_2, v_1'=v_1 and v_2' =2 v_1-v_2,
 if v_2=0, then v_1'=v_1 and v_2'=2 v_1,
   e.g. a speedy car hit you while you stand still, you will be kicked by twice the velocity of the car.

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* collision1Dplot.gif (8.53 KB, 525x410 - viewed 411 times.)
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Fu-Kwun Hwang
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Reply #1 on: June 18, 2011, 07:45:23 pm » posted from:Taipei,T'ai-pei,Taiwan

The following is 1D collision using EJS event

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Reply #2 on: June 18, 2011, 09:56:15 pm » posted from:Taipei,T'ai-pei,Taiwan

Here is 2D collision processed with EJS event.
You can lean how to use EJS event to process collision.
Download the jar file, double click to run it, then
right click mouse and select "open ejs model" to view how it was created with EJS.
(You need to download EJS and installed it -- i.e. unzip downloaded ZIP file.)

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License
  • Please feel free to post your ideas about how to use the simulation for better teaching and learning.
  • Post questions to be asked to help students to think, to explore.
  • Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
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Reply #3 on: April 13, 2013, 04:54:31 pm » posted from:,,Satellite Provider

You are welcomed to check out  Elastic 1D collision inquiry
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