Fully perfect elastic collision

AI Thread Summary
The discussion centers on deriving the formula for perfectly elastic collisions, emphasizing the conservation of kinetic energy and momentum. The user presents their derivation steps but expresses concern that it may be overly simplified, as a friend suggested a more complex approach. Another participant shares their experience of simplifying algebraic solutions to enhance problem-solving efficiency. The conversation also touches on the validity of the equations when considering scenarios where velocities remain unchanged, clarifying that the focus is on cases where collisions do occur. The exchange highlights the balance between thoroughness and simplicity in mathematical derivations.
madah12
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Homework Statement


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.

Homework 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

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.
 
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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
 
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...?
 
madah12 said:
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.
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*}<br /> m_1 v_1 + m_2 u_1 &amp;= m_1 v_2 + m_2 u_2 \\<br /> v_1 - u_1 &amp;= -v_2 + u_2<br /> \end{align*}<br />

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 &amp; m_2 \\ v_1-u_1 &amp; 1 \end{vmatrix}}{\begin{vmatrix}m_1 &amp; m_2 \\ -1 &amp; 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.
madah12 said:
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...?
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.
 
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