Coefficient of Restitution in x and y

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Discussion Overview

The discussion revolves around the application of the coefficient of restitution in analyzing two-body collisions, specifically how to decompose velocities into x and y components. Participants explore whether this approach is valid for both linear and planar collisions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires whether the values in the relative velocity terms of the coefficient of restitution can be expressed in x and y components during a two-body collision.
  • Another participant suggests that for linear collisions, fixing a positive orientation allows for interpreting negative velocities as opposite directions, and this can be extended to planar collisions by analyzing components individually.
  • A participant proposes a specific equation for the coefficient of restitution in the x direction, using the angle of one body’s trajectory relative to the horizontal.
  • There is a request for clarification regarding the notation used for the coefficient of restitution in the context of the proposed equation.
  • One participant clarifies their notation, confirming that they meant the coefficient of restitution when referring to ##e_x##.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and interpretation regarding the application of the coefficient of restitution in different contexts, but no consensus is reached on the validity of the proposed equations or methods.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in applying the coefficient of restitution to components, and there is ambiguity in the definitions and notations used.

unseeingdog
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I am currently studying collisions in high school and my teacher told us that, in order to calculate the direction of each object after a 2-body collision, we could change the values in the relative velocity terms of the equation of the coefficient of restitution to the components in x and y. Is this true? Thanks.
 
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If collisions are on a line, you can fix a positive orientation of the line so you have positive and negative velocities respect the two opposit directions... it is possible to interpret the minus in front of the vector as a velocity in the opposite sense ...
If the collision is in the plane you can always do the same component by component ...
I don't know if I answered ...
Ssnow
 
Ssnow said:
If collisions are on a line, you can fix a positive orientation of the line so you have positive and negative velocities respect the two opposit directions... it is possible to interpret the minus in front of the vector as a velocity in the opposite sense ...
If the collision is in the plane you can always do the same component by component ...
I don't know if I answered ...
Ssnow
So, say, if one of the bodies moves along the x axis, and the other moves with an angle of 120 with respect to the horizontal, one can write ##e_x = (v_2cos(120) - v_1)/(u_1 - u_2cos(120)## ?
 
mmmmh, what is ##e_{x}## ? ... if the ##\vec{v}=(v_{1},v_{2})## is the first vector and ##\vec{u}=(u_{1},u_{2})## the second forming an angle of ##120°## then ##\vec{v}=(v_{1},0)## because is on the ##x## axis and ##\vec{u}=(u\cos{(120)},u\sin{(120)})## where ##u## is the magnitude of ##\vec{u}##. Now you must fix a sign ##\pm## to each component that describes the collision ...
Ssnow
 
Ssnow said:
mmmmh, what is ##e_{x}## ? ... if the ##\vec{v}=(v_{1},v_{2})## is the first vector and ##\vec{u}=(u_{1},u_{2})## the second forming an angle of ##120°## then ##\vec{v}=(v_{1},0)## because is on the ##x## axis and ##\vec{u}=(u\cos{(120)},u\sin{(120)})## where ##u## is the magnitude of ##\vec{u}##. Now you must fix a sign ##\pm## to each component that describes the collision ...
Ssnow
I meant ##e_x## to be the coefficient of restitution. Sorry for not specifying. Anyways, I get it now. Thanks
 

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