Mechanics Chapter 2 WIP

docs/physics/mechanics/2-projectiles-charged-particles.md

@@ -18,3 +18,63 @@ Note that the result is expected to be of the same order of magnitude as the *Re
 
 ## Section 2.2 - Linear Air Resistance
 
+First, consider the case in which the drag force is negligible. Then, we see that with $\vb{F}_g = \vb{w} = mg$ and $\vb{F}_{drag} = \vb{f} = -b\vb{v}$, then Newton's second law tells us that
+
+$$\vb{F} = m \dot{\vb{v}} = m\vb{g} - b\vb{b}$$
+
+This separates into two equations:
+
+$$\begin{align}
+\dot{v}_x &= -\frac{b}{m}v_x \\
+\dot{v}_y &= g - \frac{b}{m}v_y
+\end{align}$$
+
+These are separable equations and can be trivially solved.
+
+For an object on a surface in which the only motion is in the $x$-direction, only the first of the above equations applies. Then, we can define $k = \frac{b}{m}$, so that $\dot{v_x} = -k v_x$. This can be integrated to see that $v_x(t) = Ae^{-kt}$. We can also define $\tau = \frac{1}{t} = \frac{m}{b}$, so that $v(t) = Ae^{-\frac{t}{\tau}}$
+
+Now, consider the second equation. This is still separable, however, the math is more complicated. Note that when we set $\dot{v}_y = 0$, we are at some maximum (terminal) velocity $v_{ter}$. Solving, we see that $v_{ter} = \frac{mg}{b}$. Then, we can rewrite our equation as
+
+$$m\dot{v}_y = -b(v_y - v_ter)$$
+
+This can be solved to see that $v(y) = v_{ter} + (v_{y0} - v_{ter})e^{-\frac{t}{\tau}}$.
+
+## Section 2.3 - Trajectory and Range in a Linear Medium
+
+From the velocity functions in the previous section, we can integrate to obtain the position functions.
+
+$$\begin{align}
+x(t) &= v_{x0}\tau(1-e^{-\frac{t}{\tau}}) \\
+y(t) &= (v_{y0} + v_{ter})\tau(1-e^{-\frac{t}{\tau}})-v_{ter}t
+\end{align}$$
+
+We can then solve the first equation for $t$ and substitute into the second to see that
+
+$$y(t) = \frac{v_{y_0} + v_{ter}}{v_{x0}} + v_{ter} \tau \ln(1 - \frac{x}{v_{x0}\tau})$$
+
+Note the trajectory has a vertical asymptote at $x = v_{x0} \tau$.
+
+We can find the horizontal range by setting $y(R) = 0$. This results in an equation that contains many exponential terms, so we want to find an approximation. Truncating the Taylor series for $\ln(1 - \varepsilon)$, we see that $R \approx R_{vac}(1 - \frac{4}{3} \frac{v_{y0}}{v_{ter}})$, where $R_{vac}$, or the air resistance assuming no air resistance, is $R_{vac} = \frac{2 v_{x0}v_{y0}}{g}$.
+
+## Section 2.4 - Quadratic Air Resistance
+
+In the case of quadratic air resistance, we see that $\vb{f} = -c v^2 \vu{v}$. Then, we can see that in the horizontal case, $m\frac{dv}{v_x^2} = -c dt$. Thus,
+
+$$v_x(t) = \frac{v_{x0}}{1+cv_{x0}t/m} = \frac{v_{x0}}{1+\tau/t}$$
+
+where $\tau = m/cv_{x0}$. We can see that at $t = \tau$, $v = v_{x0}/2$.
+
+Thus, $x(t) = v_{x0} \tau \ln(1 + t / \tau)$.
+
+For exclusively vertical motion, we first define $v_{ter} = \sqrt{\frac{mg}{c}}$. Then, we see that $\dot{v_y} = g(1 - \frac{v^2}{v_{ter}^2})$. Solving via separation of variables yields
+
+$$v_y(t) = v_{ter}\tanh(\frac{gt}{v_{ter}})$$
+
+Then, we see that
+
+$$y(t) = \frac{v_{ter}^2}{g} \ln(\cosh(\frac{gt}{v_{ter}}))$$
+
+In the case where there is both vertical and horizontal motion, the differential equation no longer has an analytic solution.
+
+## Section 2.5 - Motion of a Charge in a Uniform Magnetic Field
+