Work is defined as the vector dot product: $W=\vec{F}\cdot\vec{s}=|\vec{F}||\vec{s}|\cos\theta$. The $\cos\theta$ factor extracts the force component along displacement.

Derivation steps: ① Decompose $\vec{F}$ into $F_{\parallel}=F\cos\theta$ (parallel to displacement) and $F_{\perp}=F\sin\theta$ (perpendicular); ② Only $F_{\parallel}$ contributes to work; $F_{\perp}$ does zero work; ③ Thus $W=F_{\parallel}\cdot s=Fs\cos\theta$.

Extension to variable forces: $W=\int_{s_1}^{s_2}\vec{F}\cdot d\vec{s}$ (area under $F$-$s$ graph).

Derivation steps:
① Start from Newton's 2nd law: $F=ma$
② Multiply both sides by displacement $s$: $Fs=mas$
③ Substitute the kinematic equation $v_f^2-v_i^2=2as$, so $as=\frac{v_f^2-v_i^2}{2}$
④ $W=Fs=m\cdot\frac{v_f^2-v_i^2}{2}=\frac12mv_f^2-\frac12mv_i^2$

Generalization: The theorem holds even for variable forces and curved paths — $W$ is simply the total net work. This makes it more powerful than Newton's laws + kinematics alone.

Derivation of ME Conservation:
① From Work-Energy Theorem: $W_{\text{net}}=\Delta E_k=E_{k2}-E_{k1}$
② If only gravity does work: $W_{\text{net}}=W_g=mgh_1-mgh_2=-\Delta E_p$
③ Therefore $-\Delta E_p=\Delta E_k$ → $\Delta E_k+\Delta E_p=0$ → $E_{k1}+E_{p1}=E_{k2}+E_{p2}$

Derivation of Elastic PE: Spring force $F=kx$ varies linearly; average force $\bar{F}=\frac12kx$; work $W=\bar{F}\cdot x=\frac12kx^2$. Or integrate: $E_p=\int_0^x kx\,dx=\frac12kx^2$.

Derivation of $P=Fv$: $P=\frac{W}{t}=\frac{Fs}{t}=F\cdot\frac{s}{t}=Fv$. Under constant power, force and speed are inversely proportional — key to understanding engine startup problems.

Two startup modes: ① Constant power: $P=Fv$ fixed, smaller $v$ → larger $F$ (higher $a$), as $v$↑, $F$↓, $a$↓, when $F=f$→$a=0$ at $v_{\max}=P/f$; ② Constant acceleration: $F$ fixed, $P=Fv$ increases with $v$, switching to constant power mode after reaching rated power.

Derivation steps:
① From Newton's 2nd Law: $\vec{F}=m\vec{a}=m\frac{\Delta\vec{v}}{\Delta t}$
② Multiply both sides by $\Delta t$: $\vec{F}\Delta t=m\Delta\vec{v}$
③ Define impulse $\vec{I}=\vec{F}\Delta t$ and momentum $\vec{p}=m\vec{v}$
④ $\vec{I}=m\vec{v}_f-m\vec{v}_i=\Delta\vec{p}$

Comparison with Work-Energy: The Impulse-Momentum theorem handles time-related quantities (collision time, force duration); Work-Energy handles space-related quantities (braking distance, height). Both derive from $F=ma$ and complement each other.

Deriving Momentum Conservation from Newton's 3rd Law:
① The interaction forces satisfy: $\vec{F}_{12}=-\vec{F}_{21}$ (action-reaction pair)
② Apply Impulse-Momentum to object 1: $\vec{F}_{21}\Delta t=m_1\vec{v}_1'-m_1\vec{v}_1$
③ Apply to object 2: $\vec{F}_{12}\Delta t=m_2\vec{v}_2'-m_2\vec{v}_2$
④ Add: $0=(m_1\vec{v}_1'+m_2\vec{v}_2')-(m_1\vec{v}_1+m_2\vec{v}_2)$
⑤ Hence $m_1\vec{v}_1+m_2\vec{v}_2=m_1\vec{v}_1'+m_2\vec{v}_2'$

Elastic collision velocity formulas: Solve the system of momentum conservation + KE conservation to obtain $v_1'=\frac{(m_1-m_2)v_1+2m_2v_2}{m_1+m_2}$, $v_2'=\frac{(m_2-m_1)v_2+2m_1v_1}{m_1+m_2}$.

$E_k=\frac{p^2}{2m}$ derivation: From $p=mv$ → $v=p/m$, substitute into $E_k=\frac12mv^2$ → $E_k=\frac12m(p/m)^2=\frac{p^2}{2m}$.

Geometric derivation of $a=v^2/r$:
① Object moves A→B, velocity changes from $\vec{v}_1$ to $\vec{v}_2$, same magnitude but direction changes by $\Delta\theta$
② Magnitude of velocity change: $|\Delta\vec{v}|\approx v\Delta\theta$ (exact as $\Delta t\to0$)
③ $\Delta\theta=\omega\Delta t=\frac{v}{r}\Delta t$
④ $a=\frac{|\Delta\vec{v}|}{\Delta t}=\frac{v\cdot(v/r)\Delta t}{\Delta t}=\frac{v^2}{r}$
⑤ $\Delta\vec{v}$ direction is toward center → acceleration is centripetal

Vertical circle critical speed derivation:
Top: $mg+T=mv^2/r$. $T\ge0$ (string can't push) → $mg\le mv^2/r$ → $v\ge\sqrt{gr}$. Bottom speed from energy: $\frac12mv_{\text{b}}^2=\frac12mv_{\text{t}}^2+mg\cdot2r$ → $v_{\text{b}}=\sqrt{v_{\text{t}}^2+4gr}$. With $v_{\text{t}}=\sqrt{gr}$, $v_{\text{b}}=\sqrt{5gr}$.

Golden substitution $GM=gR^2$ derivation:
① Object at surface: $F=GMm/R^2$
② Ignoring Earth's rotation, this equals weight $mg$: $mg=GMm/R^2$
③ Cancel $m$: $g=GM/R^2$ → $GM=gR^2$
Usage: replaces the hard-to-measure $GM$ with easily measured $g$ and $R$.

Orbital speed $v=\sqrt{GM/r}$ derivation:
① Gravity = centripetal: $GMm/r^2=mv^2/r$
② Cancel $m$, multiply by $r$: $GM/r=v^2$
③ $v=\sqrt{GM/r}$ (or using golden substitution: $v=\sqrt{gR^2/r}$)

Kepler's Third Law $T^2\propto r^3$ (circular orbit):
① $v=2\pi r/T$
② Substitute into $v^2=GM/r$: $(2\pi r/T)^2=GM/r$
③ $4\pi^2 r^2/T^2=GM/r$ → $\frac{r^3}{T^2}=\frac{GM}{4\pi^2}=\text{constant}$

Escape velocity $v_2=\sqrt{2GM/R}$ derivation:
① Launch from surface, at infinity KE+PE=0 (just escapes): $\frac12mv^2-\frac{GMm}{R}=0$
② $\frac12mv^2=\frac{GMm}{R}$ → $v=\sqrt{2GM/R}=\sqrt{2gR}$
③ Note: $v_2=\sqrt{2}v_1$, escape speed is $\sqrt{2}$ times the orbital speed.

Satellite total energy $E=-\frac{GMm}{2r}$ derivation:
① $E=E_k+E_p=\frac12mv^2-\frac{GMm}{r}$
② For circular orbit $v^2=GM/r$ → $E_k=\frac{GMm}{2r}$
③ $E=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}$
④ Hence $E_k=|E|$, $E_p=2E$, i.e., $E_k=-\frac12E_p$ (virial theorem for gravity).