Fully perfect elastic collision

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Homework Help Overview

The discussion revolves around deriving the formulas for true elastic collisions in physics, specifically focusing on the conservation of kinetic energy and momentum. Participants are examining the mathematical relationships that govern the velocities of colliding objects.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the equations governing elastic collisions, expressing concern over the complexity of their derivation. Some participants question whether the derived equations account for scenarios where the velocities of the objects do not change after the collision.

Discussion Status

Participants are actively engaging with the derivation process, with some offering insights into the implications of the equations presented. There is a recognition of the potential for simpler methods, such as using Cramer's rule, to solve the equations efficiently. However, the discussion does not reach a consensus on the correctness of the original poster's derivation.

Contextual Notes

There is an ongoing debate about the assumptions made regarding the velocities of the objects post-collision, particularly whether they can remain unchanged in an elastic collision scenario.

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

[tex]\begin{align*}<br /> m_1 v_1 + m_2 u_1 &= m_1 v_2 + m_2 u_2 \\<br /> v_1 - u_1 &= -v_2 + u_2<br /> \end{align*}[/tex]

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

[tex]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}[/tex]

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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