Difference between revisions of "Notes:Collision"

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(Created page with "==Conservation of momentum== Yields: * {{M|v_2'\eq -\frac{m_1}{m_2}v_1'+\frac{1}{m_2}(m_1v_1+m_2v_2)}} (written in the form {{M|y\eq m\cdot x+c}}) As this is the equation for...")
 
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==Conservation of energy==
 
==Conservation of energy==
 
We will consider:
 
We will consider:
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* {{M|E_0:\eq \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2}} to be energy before
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* {{M|E_0:\eq \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+I}} to be energy before
−
* {{M|E_1:\eq \frac{1}{2}m_2(v_1')^2+\frac{1}{2}m_2(v_2')^2+W}} to be energy after
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* {{M|E_1:\eq \frac{1}{2}m_1(v_1')^2+\frac{1}{2}m_2(v_2')^2+L}} to be energy after
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** here we can use {{M|W}} (for ''W''ork) to allow us to be able to describe results where we have more energy afterwards then we did before, put in by the explosion of a firework. Or what energy must be taken out to get an observed result
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Notice:
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** It is tempting to consider only {{M|W:\eq -L}} where {{M|L}} is for 'latent' energy, energy which is hidden (and has become heat in collisions which don't conserve mechanical energy) however I use a general {{M|W}} term to allow us to put work in.
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* we consider the "before" energy amount to have some separate amount of work {{M|I}} (for input) which is not a part of the kinetic energy initially, if {{M|I>0}} then this is adding energy to the system (which may be taken out by {{M|L}} of course)
−
** If we do consider {{M|W}} as putting work in we must really define it as {{M|W_{\text{in} } -L}} where we consider the total work in, and what amount was lost due to the imperfect nature of the collisions
+
* In the "after" calculation, {{M|L}} is used to represent ''L''atent energy, which is used here to mean "hidden" or lost energy - say thermal. So the energy in {{M|E_0}} will become kinetic, kinetic and "hidden" (not kinetic). In an [[elastic collision]] where {{M|I\eq 0}} we will have {{M|L\eq 0}}, in an [[inelastic condition]] momentum will be conserved but some energy "lost" - transferred from kinetic to something else.
 +
 
 +
 
 +
Rearranging we see:
 +
* {{MM|(v_s')^2\eq \frac{\frac{1}{2}m_r(v_r')^2+L-\frac{1}{2}m_1v_1^2-\frac{1}{2}m_2v_2^2-I}{\frac{-1}{2}m_s} }} for {{M|s\in\{1,2\} }} and {{M|r\in\{1,2\} }} such that {{M|m\neq r}}
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*: {{MM|\implies\ \ (v_s')^2\eq \frac{\frac{1}{2}m_r(v_r')^2+L-\frac{1}{2}m_sv_s^2-\frac{1}{2}m_rv_r^2-I}{\frac{-1}{2}m_s} }}
 +
*::: {{MM|\eq \frac{I-L+\frac{1}{2}m_r\big(v_r^2-(v_r')^2\big)+\frac{1}{2}m_sv_s^2}{\frac{1}{2}m_s} }}
 +
*::: {{MM|\eq \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2}}
 +
** We arrive at:
 +
*** {{MM|v_s' \eq \pm\sqrt{ \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2 } }}, which gives us a constraint:
 +
**** This requires (for real solutions):
 +
***** {{MM|{ \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2 }\ge 0}} - note we could make this into a quadratic ({{M|\times}} both sides by {{M|m_s^2}} - we'd get one with no constant term - should we?
 +
 
 +
 
 +
==Equations of motion==
 +
* From conservation of momentum:
 +
** {{M|v_2'\eq -\frac{m_1}{m_2}v_1'+\frac{1}{m_2}(m_1v_1+m_2v_2)}} (written in the form {{M|y\eq m\cdot x+c}})
 +
* From conservation of energy:
 +
** {{MM|v_s' \eq \pm\sqrt{ \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2 } }}, for {{M|s,r\in \{1,2\} }} such that {{M|s\neq r}}
 +
*** Note: {{M|I}} is "input work" (if {{M|\ge 0}}) representing an additional quantity of energy that isn't in the initial kinetic energy, and {{M|L}} represents "latent" (literally: hidden) energy, which if {{M|\ge 0}} represents some of the initial energy ending up as heat (and not as kinetic)

Latest revision as of 11:19, 25 September 2018

Conservation of momentum

Yields:

  • [ilmath]v_2'\eq -\frac{m_1}{m_2}v_1'+\frac{1}{m_2}(m_1v_1+m_2v_2)[/ilmath] (written in the form [ilmath]y\eq m\cdot x+c[/ilmath])

As this is the equation for a line (where [ilmath]x\eq v_1'[/ilmath] and [ilmath]y\eq v_2'[/ilmath]) we can see for huge {{|v_1'}} we get huge [ilmath]v_2'[/ilmath] values (but going the other way)

This is because technically if we have 2 particles just next to each other but not moving and suddenly they fly apart in opposite directions - momentum is conserved.

Think for example of a firework, fired vertically there's no momentum in the horizontal plane, hence the relative symmetry of the explosion.

If the firework is misfired and has a "general left-ward trend" shall we say, after it explodes, considering all the pellets as the system, it still has a "left-ward trend", this trend is the [ilmath]\mathbf{+c} [/ilmath] term in the line.

Let us now introduce energy, who's magnitude is based purely on the magnitude of velocity rather than signed like momentum. This will cap extreme situations we get from the huge (and opposite) [ilmath]v_i'[/ilmath]

Conservation of energy

We will consider:

  • [ilmath]E_0:\eq \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+I[/ilmath] to be energy before
  • [ilmath]E_1:\eq \frac{1}{2}m_1(v_1')^2+\frac{1}{2}m_2(v_2')^2+L[/ilmath] to be energy after

Notice:

  • we consider the "before" energy amount to have some separate amount of work [ilmath]I[/ilmath] (for input) which is not a part of the kinetic energy initially, if [ilmath]I>0[/ilmath] then this is adding energy to the system (which may be taken out by [ilmath]L[/ilmath] of course)
  • In the "after" calculation, [ilmath]L[/ilmath] is used to represent Latent energy, which is used here to mean "hidden" or lost energy - say thermal. So the energy in [ilmath]E_0[/ilmath] will become kinetic, kinetic and "hidden" (not kinetic). In an elastic collision where [ilmath]I\eq 0[/ilmath] we will have [ilmath]L\eq 0[/ilmath], in an inelastic condition momentum will be conserved but some energy "lost" - transferred from kinetic to something else.


Rearranging we see:

  • [math](v_s')^2\eq \frac{\frac{1}{2}m_r(v_r')^2+L-\frac{1}{2}m_1v_1^2-\frac{1}{2}m_2v_2^2-I}{\frac{-1}{2}m_s} [/math] for [ilmath]s\in\{1,2\} [/ilmath] and [ilmath]r\in\{1,2\} [/ilmath] such that [ilmath]m\neq r[/ilmath]
    [math]\implies\ \ (v_s')^2\eq \frac{\frac{1}{2}m_r(v_r')^2+L-\frac{1}{2}m_sv_s^2-\frac{1}{2}m_rv_r^2-I}{\frac{-1}{2}m_s} [/math]
    [math]\eq \frac{I-L+\frac{1}{2}m_r\big(v_r^2-(v_r')^2\big)+\frac{1}{2}m_sv_s^2}{\frac{1}{2}m_s} [/math]
    [math]\eq \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2[/math]
    • We arrive at:
      • [math]v_s' \eq \pm\sqrt{ \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2 } [/math], which gives us a constraint:
        • This requires (for real solutions):
          • [math]{ \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2 }\ge 0[/math] - note we could make this into a quadratic ([ilmath]\times[/ilmath] both sides by [ilmath]m_s^2[/ilmath] - we'd get one with no constant term - should we?


Equations of motion

  • From conservation of momentum:
    • [ilmath]v_2'\eq -\frac{m_1}{m_2}v_1'+\frac{1}{m_2}(m_1v_1+m_2v_2)[/ilmath] (written in the form [ilmath]y\eq m\cdot x+c[/ilmath])
  • From conservation of energy:
    • [math]v_s' \eq \pm\sqrt{ \frac{I-L}{2m_s}+\frac{m_r}{m_s}\big(v_r^2-(v_r')^2\big)+v_s^2 } [/math], for [ilmath]s,r\in \{1,2\} [/ilmath] such that [ilmath]s\neq r[/ilmath]
      • Note: [ilmath]I[/ilmath] is "input work" (if [ilmath]\ge 0[/ilmath]) representing an additional quantity of energy that isn't in the initial kinetic energy, and [ilmath]L[/ilmath] represents "latent" (literally: hidden) energy, which if [ilmath]\ge 0[/ilmath] represents some of the initial energy ending up as heat (and not as kinetic)
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