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).

$J = \left[8t - t^2\right]_0^4 = 32 - 16$

$\boxed{J = 16 \text{ N·s}}$

(b) Final velocity:

Using impulse-momentum theorem: $J = \Delta p = m(v_f - v_0)$

$16 = 2.0(v_f - 0)$

$\boxed{v_f = 8.0 \text{ m/s}}$

(c) Work done:

First find position. Using F = ma: $a(t) = \frac{F}{m} = \frac{8 - 2t}{2} = 4 - t$

$v(t) = \int_0^t (4 - t') \, dt' = 4t - \frac{t^2}{2}$

$x(t) = \int_0^t \left(4t' - \frac{t'^2}{2}\right) dt' = 2t^2 - \frac{t^3}{6}$

Distance at t = 4: $x(4) = 2(16) - \frac{64}{6} = 32 - 10.67 = 21.33 \text{ m}$

Work: $W = \int_0^4 F(t) \cdot v(t) \, dt = \int_0^4 (8-2t)\left(4t - \frac{t^2}{2}\right) dt$

Or using work-energy theorem: $W = \Delta KE = \frac{1}{2}mv_f^2 - 0 = \frac{1}{2}(2.0)(8.0)^2$

$\boxed{W = 64 \text{ J}}$