Elastric Collisions: Shortcut to Solving

[bfr]

I given to objects of masses m1 and m2 and initial velocities v1 and v2, I know how to find out out v1' and v2', their velocities after the collision by settings up simultaneous equations. However, that takes a long time to solve, and there is a shortcut that my teacher told my class that I wrote down on a sheet of paper but then lost. I tried searching Google and haven't gotten any luck yet. I remember the shortcut was something like:

v1'=2((m1v1+m2v2)/(m1+m2))-v1
v2'=2((m1v1+m2v2)/(m1+m2))-v2

But I'm not sure. Would anybody know what the shortcut is?

 

[awvvu]

I just remember this equation for elastic collisions:

$$v_1 + v_1' = v_2 + v_2'$$
or
$$v_1 - v_2 = -(v_1' - v_2')$$

Then it's fairly simple to plug this into the conservation of momentum equation to find $$v_1'$$ or $$v_2'$$. (solve for $$v_1'$$ and plug into get $$v_2'$$ and vice versa.)

This doesn't look anything like what you remember though.

 

[rock.freak667]

I just remember this equation for elastic collisions:

$$v_1 + v_1' = v_2 + v_2'$$
or
$$v_1 - v_2 = -(v_1' - v_2')$$

If I remember correctly that comes from solving:
$\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

AND

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

 

[awvvu]

If I remember correctly that comes from solving:
$\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

AND

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Yeah, energy and momentum is conserved for elastic collisions.