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 Author Topic: Elastic Collision (1D)  (Read 13638 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message 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<, $v_1'=-v_1+2v_2$ 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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Fu-Kwun Hwang
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 « Embed this message 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

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
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Fu-Kwun Hwang
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 « Embed this message 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.
right click mouse and select "open ejs model" to view how it was created with EJS.

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
• 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.
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Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
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Fu-Kwun Hwang
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 « Embed this message 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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The softest thing overcomes the hardest thing in the universe. ...Lao Tzu (570-490 BC)