# Difference between revisions of "Notes:Collision"

m (→Conservation of energy: Correcting typos and rephrasing) |
m (→Conservation of energy: corrections and finishing equations) |
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Rearranging we see: | Rearranging we see: | ||

− | * {{MM|(v_s')^2\eq \frac{\frac{1}{2}m_r(v_r')^2+ | + | * {{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}} |

− | *: {{MM|\implies\ \ (v_s')^2\eq \frac{\frac{1}{2}m_r(v_r')^2+ | + | *: {{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- | + | *::: {{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? |

## Revision as of 11:10, 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
*L*atent 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?

- This requires (for real solutions):

- [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:

- [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]