content(warnings): Add W8-W11, X4-X5 — collision, inertia, and torque cross-refs

concepts/mechanics/u4/m4-4-collisions.tex

@@ -27,6 +27,12 @@ A collision is \emph{perfectly inelastic} if the objects stick together after th
 \]
 Every perfectly inelastic collision is inelastic.}
 
+\wc{Momentum is conserved in every collision type}{Total momentum of an isolated system is conserved in elastic, inelastic, \emph{and} perfectly inelastic collisions. The word ``elastic'' only tells you whether kinetic energy is also conserved. A common exam error is to drop the momentum equation for inelastic collisions --- \emph{never} do this.}
+
+\wc{``Momentum is lost'' in a collision}{Momentum of the \emph{system} is never lost in any isolated collision. What is ``lost'' in an inelastic collision is \emph{kinetic energy}, which is converted to thermal energy, sound, deformation, etc. Momentum conservation holds regardless of how ``bouncy'' the collision is.}
+
+Collisions connect momentum conservation (Unit 4) with energy conservation (Unit 3): elastic collisions conserve both, inelastic conserve only momentum. See also the coefficient of restitution in Section 4.5 for explosive separation.
+
 \nt{Do not decide whether momentum is conserved by asking whether the collision is elastic. Those are different ideas. Momentum conservation depends on the net external impulse on the chosen system. If the system is isolated during the collision, then total momentum is conserved for elastic, inelastic, and perfectly inelastic collisions alike. Kinetic energy supplies an \emph{extra} condition only in the elastic case. In two-dimensional AP problems, conserve momentum separately in the $x$- and $y$-directions.}
 
 \ex{Illustrative example}{On a frictionless track, cart 1 has mass $m_1=0.40\,\mathrm{kg}$ and initial velocity

concepts/mechanics/u5/m5-3-torque.tex

@@ -30,6 +30,8 @@ For rotation about a chosen fixed axis with unit vector $\hat{k}$,
 \]
 so the scalar $\tau$ is positive or negative according to the declared sign convention.}
 
+Torque about a fixed axis connects directly to angular acceleration through Newton's second law for rotation (Section 5.6), just as net force connects to linear acceleration through Newton's second law (Section 2.1).
+
 \pf{Why $rF\sin\phi$ equals $F\ell$}{From the cross-product magnitude formula,
 \[
 |\vec{\tau}|=|\vec{r}\times \vec{F}|=rF\sin\phi.

concepts/mechanics/u5/m5-4-moment-of-inertia.tex

@@ -16,6 +16,8 @@ I=mr_\perp^2.
 \]
 The SI unit of moment of inertia is $\mathrm{kg\cdot m^2}$. Because the distance to the axis is squared, mass farther from the axis contributes much more strongly to $I$.}
 
+\wc{Moment of inertia does not depend on speed or rotation}{The moment of inertia $I$ is a geometric property of the mass distribution about an axis. It depends only on the object's mass, shape, and axis position --- \emph{not} on angular velocity, angular acceleration, or torque. A spinning top and a stationary top have the same $I$.}
+
 \thm{Key fixed-axis relations and axis dependence}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $I$ denote the moment of inertia about that same axis. Then
 \[
 \tau_{\mathrm{net}}=I\alpha,

concepts/mechanics/u5/m5-6-rotation-dynamics.tex

@@ -13,6 +13,8 @@ The net torque about the axis is the sum of the signed torques:
 
 Let $I$ denote the moment of inertia of the body about that same axis, and let $\alpha$ denote the signed angular acceleration. The quantity $I$ measures the rotational inertia of the body: for the same net torque, a larger $I$ gives a smaller $\alpha$. Thus $I$ plays the rotational role that mass plays in translational motion.}
 
+\wc{Heavier objects do not fall faster (in vacuum)}{In the absence of air resistance, all objects fall with the same acceleration $g$, regardless of mass. The gravitational force is larger on a heavier object ($F_g=mg$), but so is the object's inertia ($F=ma$), and the $m$ cancels: $a=g$. Air resistance is what makes feathers fall slower in real life.}
+
 \thm{Newton's second law for fixed-axis rotation}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $I$ denote the moment of inertia about that axis, let $\alpha$ denote the signed angular acceleration, and let $\sum \tau$ denote the net external torque about the axis using the declared sign convention. Then
 \[
 \sum \tau = I\alpha.