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.
Full screen applet or
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