Component form: $F_x = m\frac{dv_x}{dt}, \quad F_y = m\frac{dv_y}{dt}, \quad F_z = m\frac{dv_z}{dt}$

Impulse-Momentum Theorem

From Newton's second law: $\vec{F} = \frac{d\vec{p}}{dt}$

where momentum $\vec{p} = m\vec{v}$.

Integrating both sides: $\int_{t_1}^{t_2} \vec{F} \, dt = \int_{t_1}^{t_2} \frac{d\vec{p}}{dt} \, dt = \vec{p}_2 - \vec{p}_1 = \Delta \vec{p}$

Impulse: $\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \Delta \vec{p}$

Variable Force Example

If $F(t) = F_0\cos(\omega t)$ for time $0$ to $T = \pi/(2\omega)$:

$J = \int_0^T F_0\cos(\omega t) \, dt = \frac{F_0}{\omega}\sin(\omega t)\Big|_0^T = \frac{F_0}{\omega}$

Variable Mass Systems

For systems where mass changes with time (rockets):

$\vec{F}_{ext} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}$

Rocket equation: $m\frac{dv}{dt} = -v_{rel}\frac{dm}{dt} + F_{ext}$

where $v_{rel}$ is exhaust velocity relative to rocket.

For no external forces and integrating: $\Delta v = v_{rel}\ln\frac{m_0}{m_f}$

(Tsiolkovsky rocket equation)

Drag Forces

Linear drag: $F_d = -bv$

$m\frac{dv}{dt} = -bv$

Solution: $v(t) = v_0e^{-bt/m}$

Quadratic drag: $F_d = -cv^2$

$m\frac{dv}{dt} = -cv^2$

Separating variables: $\int_{v_0}^v \frac{dv'}{v'^2} = -\frac{c}{m}\int_0^t dt'$

$v(t) = \frac{v_0}{1 + \frac{cv_0t}{m}}$

Falling with Air Resistance

Equation of motion: $m\frac{dv}{dt} = mg - bv$

Terminal velocity: $v_t = \frac{mg}{b}$

Rewrite as: $\frac{dv}{dt} = g - \frac{b}{m}v = \frac{b}{m}(v_t - v)$

Solution: $v(t) = v_t(1 - e^{-bt/m})$

Forces in Polar Coordinates

Newton's second law in polar coordinates:

Radial direction: $F_r = m(\ddot{r} - r\dot{\theta}^2)$

Tangential direction: $F_{\theta} = m(r\ddot{\theta} + 2\dot{r}\dot{\theta})$

Example: Central Force

For motion under central force (depends only on $r$):

$F_{\theta} = 0$, so: $r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0$

$\frac{d}{dt}(r^2\dot{\theta}) = 0$

This means $r^2\dot{\theta} = L/m$ is constant (angular momentum conservation).

$F_{thrust} = -(2000)(-10)$

$\boxed{F_{thrust} = 20,000 \text{ N} = 20 \text{ kN}}$

(Constant as long as burn rate is constant)

(b) Acceleration as function of time:

Mass at time t: $m(t) = m_0 + \left(\frac{dm}{dt}\right)t = 1000 - 10t$

Newton's second law: $F = ma$ $20000 = (1000 - 10t)a$

$a(t) = \frac{20000}{1000 - 10t} = \frac{2000}{100 - t}$

$\boxed{a(t) = \frac{2000}{100 - t} \text{ m/s}^2}$

(c) Velocity after 30 s:

Using Tsiolkovsky rocket equation: $\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)$

At t = 30 s: $m_f = 1000 - 10(30) = 700 \text{ kg}$

$\Delta v = 2000 \ln\left(\frac{1000}{700}\right) = 2000 \ln(1.429)$

$\Delta v = 2000(0.357)$

$\boxed{v = 714 \text{ m/s}}$

Alternative (integrate acceleration): $v = \int_0^{30} a(t) \, dt = \int_0^{30} \frac{2000}{100-t} dt$

$v = -2000\ln(100-t)\Big|_0^{30} = -2000[\ln(70) - \ln(100)]$

$v = -2000\ln(0.7) = 2000(0.357) = 714 \text{ m/s}$ ✓

$v_t = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(5.0)(9.8)}{0.25}}$

$v_t = \sqrt{196}$

$\boxed{v_t = 14 \text{ m/s}}$

(b) Differential equation:

Newton's second law (taking down as positive): $m\frac{dv}{dt} = mg - bv^2$

$\boxed{\frac{dv}{dt} = g - \frac{b}{m}v^2}$

Or: $\frac{dv}{dt} = g\left(1 - \frac{v^2}{v_t^2}\right)$

(c) Time to reach 90% of v_t:

$\frac{dv}{g(1 - v^2/v_t^2)} = dt$

Using partial fractions or substitution u = v/v_t:

$\int_0^{0.9v_t} \frac{dv}{g(1 - v^2/v_t^2)} = \int_0^t dt$

Result (hyperbolic tangent): $\frac{v_t}{g}\tanh^{-1}\left(\frac{v}{v_t}\right) = t$

At v = 0.9v_t: $t = \frac{v_t}{g}\tanh^{-1}(0.9)$

$t = \frac{14}{9.8}\tanh^{-1}(0.9) = (1.43)(1.472)$

$\boxed{t = 2.1 \text{ s}}$

Note: $\tanh^{-1}(0.9) = \frac{1}{2}\ln\left(\frac{1+0.9}{1-0.9}\right) = \frac{1}{2}\ln(19) = 1.472$