# Fully perfect elastic collision

1. Aug 4, 2010

1. The problem statement, all variables and given/known data
ok so I was trying to figure out how to derive the formula for true elastic collision and I came up with the following but I want to make sure I didn't make a mistake.

2. Relevant equations
(K1+K2)before=(K1+K2)after
1/2 m1 v1^2 + 1/2 m2 u1^2 = 1/2 m1 v2^2 + 1/2 m2 u2^2

m1v1 + m2u1=m1v2 + m2 u2

3. The attempt at a solution
(1/2 m1 v1^2 + 1/2 m2 u1^2 = 1/2 m1 v2^2 + 1/2 m2 u2^2)*2

m1v1^2+m2 u1^2=m1v2^2+m2 u2^2

m1(v1^2-v2^2)=m2(u2^2-u1^2) *(1)

m1v1 +m2u1=m1v2 + m2 u2

m1(v1-v2)=m2(u2-u1) *(2)
(1)/(2)
=
v1+v2=u1+u2
u2=v1+v2-u1

m1v1 + m2u1=m1v2 + m2 (v1+v2-u1)

m1v1 + m2u1=m1v2 + m2 v1+m2 v2 - m2 u1

2m2u1 +m1v1 -m2v1 =v2 (m1+m2)

v2= (2m2u1 +v1(m1-m2)) / (m1+m2)

EDIT: the point of this is that my friend told me that the derivation will take more than 3 pages so I am thinking there must be something incorrect here or I ate some lines lol.

2. Aug 4, 2010

### ehild

This is a very nice and simple way to derive the new velocities in a perfectly elastic collision. Just a note: The equations 1 and 2 allow the possibility that the velocities do not change, so nothing happens (v1=v2, u1=u2 is also a possible solution). Congratulation if you found this out by yourself!

ehild

3. Aug 4, 2010

ok but isn't it possible in an elastic collision that after the collision the objects velocities don't change? so the equation is still valid right...?

4. Aug 4, 2010

### vela

Staff Emeritus
Heh, reminds me of when I was an undergrad. My friends would turn in three or four pages of algebra per problem while I turned in one. Often after I first solved a problem, I'd go back and try figure out how to reduce it to the fewest number of steps so I could follow the logic. Doing this helped develop better algebra skills, plus I learned tricks and techniques that allowed me to solve problems more efficiently later on. (Plus when I was writing up a problem to turn in, I was lazy and wanted to write down just enough so the grader could follow what I was doing.)

Once you reach the point where you have the two equations

\begin{align*} m_1 v_1 + m_2 u_1 &= m_1 v_2 + m_2 u_2 \\ v_1 - u_1 &= -v_2 + u_2 \end{align*}

you can solve them quickly using Cramer's rule. For example, for v2, you'd get

$$v_2 = \frac{\begin{vmatrix}m_1 v_1 + m_2 u_1 & m_2 \\ v_1-u_1 & 1 \end{vmatrix}}{\begin{vmatrix}m_1 & m_2 \\ -1 & 1\end{vmatrix}} = \frac{m_1 v_1 + m_2 u_1 - m_2(v_1-u_1)}{m_1 + m_2} = \frac{(m_1-m_2)v_1 + 2 m_2 u_2}{m_1+m_2}$$

It doesn't make much of a difference for this problem, but occasionally, it can help you avoid some tedious algebra.
If the objects collide, their velocities will change, and when you divided equation (1) by equation (2), you assumed this, otherwise you'd be dividing by 0.

The original equations, however, admit the possibility that the objects don't collide and their velocities remain unchanged, but that's not the case you're interested in anyway.