Choques en 1 dimensión
Inelastic collision
(270)\[m_1 v_{1i} + m_2 v_{2i}=m_1 v_{1f} + m_2 v_{2f}\]
(271)\[m_1 v_{1i} + m_2 v_{2i}=m_1 v_{1f} + m_2 v_{2f}=(m_1+m_2)\vec{v}_f \rightarrow \vec{v}_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1+m_2}\]
Elastic collision
(272)\[m_1 v_{1i} + m_2 v_{2i}=m_1 v_{1f} + m_2 v_{2f}\]
(273)\[\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2=\frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \rightarrow m_1 (v_{1i}-v_{1f})(v_{1i}+v_{1f}) = m_2 (v_{2f}-v_{2i})(v_{2f}+v_{2i})\]
(274)\[v_{1i}+v_{1f}=v_{2f}+v_{2i} \rightarrow v_{1i}-v_{2i}=-(v_{1f}-v_{2f})\]
(275)\[\begin{split}v_{1i}-v_{2i}=-(v_{1f}-v_{2f}) \longrightarrow \left\{\begin{aligned}v_{1f}=v_{2f}+v_{2i}-v_{1i}\\v_{2f}=v_{1f}+v_{1i}-v_{2i}\end{aligned}\right.\end{split}\]
(276)\[v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}\]
(277)\[v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}+\frac{m_2-m_1}{m_1+m_2}v_{2i}\]
Particles with the same mass
(278)\[v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}=v_{2i}\]
(279)\[v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}+\frac{m_2-m_1}{m_1+m_2}v_{2i}=v_{1i}\]
A particle is much heavier than the other
(280)\[v_{1f}=\frac{m-M}{m+M}v_{1i}+\frac{2M}{m+M}v_{2i}\approx-\frac{M}{M}v_{1i}+\frac{2M}{M}v_{2i}=-v_{1i}+2v_{2i}\]
(281)\[v_{2f}=\frac{2m}{m+M}v_{1i}+\frac{M-m}{m+M}v_{2i}=v_{1i}\approx 0 v_{1i}+\frac{M}{M}v_{2i}=v_{2i}\]
Magnitudes escalares
I am writing an equation inline \(x=-i\hbar\psi=\hat{h}\psi\).
If the equation is by itself,
(282)\[-i\hbar\psi=\hat{h}\psi\]
I am going to add a figure
:::{admonition,warning} This is also Markdown
This text is standard Markdown
:::
:::{admonition,note} This is also Markdown
This text is standard Markdown
:::
:::{admonition,tip} This is also Markdown
This text is standard Markdown
:::
Magnitudes vectoriales
There are many ways to write content in Jupyter Book. This short section
covers a few tips for how to do so.
I am going to cite a reference [HdHPK14]
Now I am going to cite section escalares Sec. Magnitudes escalares
The Schrödinger equation is Eq. (195)
I am citing the figure: Fig. 102
Unidades
Problemas y ejemplos resueltos
I can start solving like this
(283)\[x=-2\pi\]
Some text needs to go between sidebars
(284)\[y=-log(e)\]
And at the end