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Easy Java Simulations (2001- ) => Dynamics => Topic started by: Fu-Kwun Hwang on February 19, 2010, 07:19:56 pm

Title: Elastic Collision (1D)
Post by: Fu-Kwun Hwang on February 19, 2010, 07:19:56 pm
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
$\frac{v'_1}{v_1}=\frac{1-\rho}{1+\rho}+ \frac{2\rho}{1+\rho}\frac{v_2}{v_1}=2\frac{V_{cm}}{v_1}-1$

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< 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.

Title: Re: Elastic Collision (1D) with EJS event
Post by: Fu-Kwun Hwang on June 18, 2011, 07:45:23 pm
The following is 1D collision using EJS event

Title: Re: Elastic Collision (2D) processed with EJS event
Post by: Fu-Kwun Hwang on June 18, 2011, 09:56:15 pm
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 (http://www.um.es/fem/EjsWiki/Main/Download) and installed it -- i.e. unzip downloaded ZIP file.)

Title: Re: Elastic Collision (1D)
Post by: Fu-Kwun Hwang on April 13, 2013, 04:54:31 pm
You are welcomed to check out  Elastic 1D collision inquiry (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2709.msg9899;topicseen#msg9899)