Trap 1: Rod vs. String in Vertical Circular Motion

The Trap: "Minimum speed at the top for a ball in a vertical circle" — applying the string result ($\sqrt{gR}$) to a rigid rod.

Correct Logic:

• String: Cannot push → T ≥ 0 → minimum T = 0 → $v_{top}$ = $\sqrt{gR}$
• Rod: CAN push → v can be zero at top (rod provides centripetal force) → $v_{top}$ = 0

Distractor options often give "$\sqrt{gR}$ for both" (wrong) or "$\sqrt{5gR}$ and $\sqrt{4gR}$" which are the bottom speeds, not top speeds.

Trap 2: KE in Perfectly Inelastic Collision

The Trap: Assuming KE = 0 after perfectly inelastic collision (thinking "all KE is lost").

Correct Logic: Momentum is conserved. If $p_{total}$ ≠ 0, then $v_f$ ≠ 0, so $KE_{after}$ ≠ 0. Only 50% of KE is lost (for equal masses).

Trap 3: Forgetting Friction in Work-Energy Theorem

The Trap: Treating only the applied force's work as $W_{net}$.

Correct Logic: $W_{net}$ = $W_{all}$ forces = $W_{applied}$ + $W_{friction}$ + $W_{gravity}$ + $W_{normal}$. On rough surfaces, $W_{friction}$ = −μₖNd is always negative and must be included.

Trap 4: Work by Normal Force

The Trap: Assuming normal force always does zero work.

Correct Logic: On a horizontal surface, $W_N$ = 0 (N vertical, d horizontal). But on an inclined surface, if normal force has a component along displacement, it can do work. However, N is always perpendicular to the surface and displacement is along the surface, so $W_N$ = 0 always on any incline. (This trap is the reverse — students sometimes think N does work on an incline.)

Trap 5: KE vs Momentum After Doubling Speed

The Trap: "Doubling speed doubles KE" — wrong.

Correct: Doubling speed → KE quadruples (KE ∝ $v^{2}$). Doubling speed → momentum doubles ($p \propto v$).