### Author Topic: Elastic Collision (1D)  (Read 15209 times)

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Elastic Collision (1D)
« on: February 19, 2010, 08:19:56 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_1vec{v_1}+m_2vec{v_2}=m_1vec{v_1}'+m_2vec{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 \$ho=frac{m_2}{m_1}\$, the above equations can be re-write as
\$frac{v'_1}{v_1}=frac{1-ho}{1+ho}+ frac{2ho}{1+ho}frac{v_2}{v_1}=2frac{V_{cm}}{v_1}-1\$
\$frac{v'_2}{v_1}=frac{2}{1+ho}-frac{1-ho}{1+ho}frac{v_2}{v_1}=2frac{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< 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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#### Fu-Kwun Hwang

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• Posts: 3062
##### Re: Elastic Collision (1D) with EJS event
« Reply #1 on: June 18, 2011, 08:45:23 pm »
The following is 1D collision using EJS event

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#### Fu-Kwun Hwang

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• Posts: 3062
##### Re: Elastic Collision (?D) processed with EJS event
« Reply #2 on: June 18, 2011, 10:56:15 pm »
Here is 2D collision processed with EJS event.
You can lean how to use EJS event to process collision.
right click mouse and select "open ejs model" to view how it was created with EJS.

Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
• 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!