Elastic Collision and conservation of momentum

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Homework Help Overview

The discussion revolves around an elastic collision between two objects, focusing on demonstrating that the speed of approach equals the speed of separation. Participants reference the conservation of momentum and kinetic energy as key principles in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevant equations for momentum and kinetic energy conservation, with some attempting to manipulate these equations to show the relationship between speeds. Questions arise about the conditions for perfect elasticity and the implications for energy conservation.

Discussion Status

The conversation includes attempts to derive relationships between the variables involved in the collision. Some participants provide insights into the mathematical manipulation of the equations, while others express uncertainty about the application of these principles. There is no explicit consensus, but productive lines of reasoning are being explored.

Contextual Notes

Participants note the importance of distinguishing between perfectly elastic collisions and other types, which affects the conservation of kinetic energy. There is also mention of the complexity introduced in three-dimensional scenarios.

futb0l
In an elastic collision between two objects. Show that the speed of approach is equal to the speed of separation.

I know you have two equations, the conservation of momentum and the conservation of kinetic energy, but I wouldn't know how to solve the equations.
 
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You have the two equations right?

M1Vi1+M2Vi2=M1fV1f+M2fV2f...sorry no subscripts
and
(1/2M1Vi1^2)+(1/2M2Vi2^2)=(1/2M1Vf1^2)=(1/2M2Vf2^2)

All you have to do is take the mass of your objects, substitute them in the correct places and then do the same thing for velocities. Just replace the variables with the numbers. Its basically plug and chug the whole thing out with the info you have.

You'll know if you've done the work properly if both sides of the equation equal on another. The only thing you need to ask yourself is: Is this a perfectly elastic collison?
Because then the equation for the conservation of Kinetic Energy comes in. But if its not then you don't have to worry about the extra equation. Kinetic energy isn't conserved in an elastic collision. Only in Perfectly elastic collisions.

MissKitty
 
futb0l said:
In an elastic collision between two objects. Show that the speed of approach is equal to the speed of separation.

I know you have two equations, the conservation of momentum and the conservation of kinetic energy, but I wouldn't know how to solve the equations.

The total momentum is conserved:
[tex]m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2[/tex]

which can be rewritten:
(1)[tex]m_1(v_1 - v'_1) = m_2(v'_2 - v_2)[/tex]

Since energy is conserved:
[tex]m_1v_1^2 + m_2v_2^2 = m_1v'_1^2 + m_2v'_2^2[/tex]

which can be rewritten:
[tex]m_1v_1^2 - m_1v'_1^2 = m_2v'_2^2 + m_2v_2^2[/tex] or

(2)[tex]m_1(v_1 - v'_1)(v_1 + v'_1) = m_2(v'_2 - v_2)(v'_2 + v_2)[/tex]

so divide (2) by (1)

[tex]v_1 + v'_1 = v'_2 + v_2[/tex] or

[tex]v_1 - v_2 = -(v'_1 - v'_2)[/tex]

This is all one dimensional, of course. The math gets a little cumbersome for 3 dimensions, but you get the idea.

AM
 
Last edited:
Ooo.. Thanks.
I remembered doing this before a while ago.
 

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