Lesson 20: Full AP Exam ReviewAP

This final lesson assembles the whole course into exam form. The AP Physics 1 exam is half multiple choice and half free response, with free-response styles that include multi-part mathematical problems, translation between representations (graphs, diagrams, equations, words), an experimental-design question, and questions demanding a justified written argument. Every one of those formats has appeared in this course's Exam Experience sections; here we organise the physics itself into a retrieval system, then rehearse the two skills that decide borderline grades: choosing the right principle, and chaining principles in multi-stage problems.

The master table: principle, equation, trigger

Choosing the principle: three diagnostic questions

• Does the problem give/ask about time? Time given with forces → impulse or kinematics. Time absent → energy methods.
• Is there a collision or explosion? Momentum conservation first; energy audit second (elastic only if stated or spring-mediated and slow).
• What is the system, and what crosses its boundary? Half of all wrong answers are right calculations on the wrong system. Write the system down before the equations.

Worked example 1 — a three-stage problem (energy → momentum → energy)

Block A (2.0 kg) is released from rest at height 1.25 m on a frictionless ramp. At the bottom it collides with stationary block B (2.0 kg) and they stick together. The pair then runs into a spring (k = 500 N/m) on the level, frictionless surface. Find the spring's maximum compression (g = 10 m/s²). Stage 1 — energy (A alone): v_A = √(2gh) = √(2 × 10 × 1.25) = 5.0 m/s. Stage 2 — momentum (collision): v' = (2.0 × 5.0)/(4.0) = 2.5 m/s. Stage 3 — energy (pair + spring): ½(4.0)(2.5²) = 12.5 J = ½(500)x² → x² = 0.050 → x ≈ 0.22 m.

Stage answers: 5.0 m/s → 2.5 m/s → 0.22 m

Check: energy is not conserved across stage 2 — the sticking collision halved the kinetic energy (25 J → 12.5 J) — and using conservation end-to-end would wrongly give x = 0.32 m. Recognising which law governs each stage is the entire problem.

Worked example 2 — one problem, two methods (and a cross-check)

A 2.0 kg block slides from rest down a 37° incline with μ_k = 0.25, dropping vertically 3.0 m (sin 37° ≈ 0.6, cos 37° ≈ 0.8). Find its speed at the bottom. Method 1 — energy with friction's work: slope length L = 3.0/0.6 = 5.0 m. W_f = −μ_k mg cosθ × L = −0.25 × 20 × 0.8 × 5.0 = −20 J. mgh = 60 J. K_f = 60 − 20 = 40 J → v = √(2 × 40/2.0) = √40 ≈ 6.3 m/s. Method 2 — dynamics then kinematics: a = g(sinθ − μ_k cosθ) = 10(0.6 − 0.20) = 4.0 m/s²; v² = 2aL = 2 × 4.0 × 5.0 = 40 → v ≈ 6.3 m/s. Check: two independent routes, one number. On the exam, when time permits, this cross-check is your insurance policy — and notice Method 1 never needed the acceleration.

Habits of the top band

• Annotate before calculating: system, knowns, target, principle — four short lines that prevent most errors.
• Answer the question asked (a speed is not a velocity; a magnitude is not a direction).
• Exploit proportional reasoning in multiple choice: most ratio questions (Lessons 8, 13–16) fall to scaling arguments in seconds.
• In free response, every point is signposted by a step: a diagram, an equation in symbols, a substitution, a result with units. Symbolic-first also earns method marks when arithmetic slips.
• Never leave a justification blank. A relevant principle stated and applied earns partial credit; a blank earns none.

AP Exam Experience

MCQ 1. A 1.0 kg ball falls freely from rest through 20 m (g = 10 m/s²). Just before landing, its momentum has magnitude:

• (A) 10 kg·m/s
• (B) 20 kg·m/s
• (C) 40 kg·m/s
• (D) 200 kg·m/s

Answer: (B). v = √(2 × 10 × 20) = 20 m/s, so p = mv = 20 kg·m/s. (Energy first, momentum second — the cross-unit chaining this lesson rehearses.)

MCQ 2. A solid disk and a hoop have the same mass and radius. Each is spun with the same angular velocity about its centre. The ratio of their rotational kinetic energies (disk : hoop) is:

• (A) 1 : 2
• (B) 1 : 1
• (C) 2 : 1
• (D) 4 : 1

Answer: (A). K_rot = ½Iω² at equal ω reduces to the ratio of moments of inertia: ½MR² : MR² = 1 : 2.

MCQ 3. A pendulum and a mass–spring system are both taken to the Moon (g ≈ g_Earth/6). Their periods change by factors of:

• (A) pendulum ×√6, spring ×1
• (B) pendulum ×1, spring ×√6
• (C) both ×√6
• (D) both ×1

Answer: (A). T_pendulum = 2π√(L/g) grows by √6 when g falls sixfold; T_spring = 2π√(m/k) contains no g at all. This contrast is a perennial exam question.

FRQ (loop-the-loop, qualitative–quantitative). A small cart of mass 0.50 kg runs on a frictionless track containing a vertical loop of radius R = 0.40 m. The cart is released from rest at height h above the bottom of the loop (g = 10 m/s²). (a) Derive an expression for the minimum release height h_min from which the cart stays on the track at the top of the loop. (b) Calculate h_min. (c) The cart is instead released from h = 1.6 m. Calculate its speed at the top of the loop and the normal force exerted on it there. (d) A student releases the cart from h = 0.90 m and observes it leaves the track before reaching the top. Explain, in terms of forces and the centripetal requirement, why this happens. Scoring guideline. (a) 2 points: at the top, the limiting condition is N = 0, so mg = mv_top²/R → v_top² = gR (1 point). Energy from release to top: mgh_min = mg(2R) + ½m(gR) = (5/2)mgR, so h_min = 2.5R (1 point). (b) 1 point: h_min = 2.5 × 0.40 = 1.0 m. (c) 2 points: energy: v_top² = 2g(h − 2R) = 2 × 10 × (1.6 − 0.80) = 16 → v_top = 4.0 m/s (1 point). At the top: N + mg = mv²/R → N = 0.50 × 16/0.40 − 5.0 = 20 − 5 = 15 N (1 point). (d) 1 point: staying on the track requires mv²/R ≥ mg at the top region so the track must push inward; released from 0.90 m the cart reaches the upper part of the loop with speed too low — the required centripetal force exceeds what gravity plus the (one-sided, push-only) normal force can supply, so the cart follows a projectile path inside the loop instead.

Worked example 3 — the flywheel brake (rotation meets energy)

A flywheel with I = 0.50 kg·m² spins at 20 rad/s. A brake pad applies a constant friction torque of 0.25 N·m. Find (a) the time to stop, (b) the angle turned while stopping, (c) the energy dissipated.

(a) α = τ/I = 0.25/0.50 = 0.50 rad/s² (opposing motion); t = ω/α = 20/0.50 = 40 s
(b) θ = ω_avg × t = 10 × 40 = 400 rad (≈ 64 revolutions)
(c) K = ½Iω² = ½ × 0.50 × 400 = 100 J — all converted to thermal energy at the pad

Check: W = τθ = 0.25 × 400 = 100 J ✓ — the work–energy theorem balances in rotational dress, and the kinematics (b) mirrors the linear stopping problems of Lesson 1 with symbols swapped, one last time.

The final checklist: units, signs, sense

Before submitting any answer, run the ten-second audit that markers wish every student ran. Units: does the answer carry the SI unit of the quantity asked (a force in N, not kg)? Signs/direction: does a velocity or force answer state or imply direction where the question demands it? Sense: is the size physically believable — orbital speeds in km/s, lab accelerations within a few g, periods of hand-sized pendulums near a second? Limits: does your formula reduce correctly if a mass vanishes, a friction coefficient goes to zero, or an angle closes? The asked question: did you answer speed when they asked for speed — not time, not energy? More marks are lost to these five checks than to any gap in knowledge. Close the course with them, and good luck — you have earned it.

Rapid-fire proportionalities (the 60-second drill)

Before the exam, these scaling answers should be reflexive. Doubling the speed on a curve: centripetal force ×4. Doubling a satellite's orbital radius: period ×2.8 (2^1.5), speed ×0.71. Doubling a spring's compression: stored energy ×4. Doubling an oscillator's mass: period ×1.41; doubling its amplitude: period ×1, energy ×4. Halving the distance between two masses: gravitational force ×4. Doubling a string's tension: wave speed ×1.41, pitch (fixed length) ×1.41. Doubling the length of a pendulum: period ×1.41; taking it to the Moon: ×2.45. Each is one line of algebra you have already done in this course; the drill converts them from derivations into recognition.