Dynamics: Kinetics - Memory Anchors
Memory techniques can multiply your recall rate by 3–5×. The brain encodes vivid, emotional, and story-based information far more durably than dry text. For PRC board exam preparation, where you must recall dozens of formulas and decision rules under timed pressure, anchoring each concept to a striking image, story, or acronym means the formula 'pops up' automatically when you see a problem — no blanking, no panic. This set covers every key concept in Dynamics: Kinetics using mnemonics, analogies, micro-stories, visual associations, and rhymes — with Filipino cultural hooks to make them stick even harder.
Memory anchors
Choosing the correct kinetics method: Newton's 2nd Law vs Work-Energy vs Impulse-Momentum
Memory aid
Use the acronym FAD-TV: (F)orce + (A)cceleration → use ΣF = ma; (D)istance + (T)wo speeds → use Work-Energy; (T)ime + (V)elocity change → use Impulse-Momentum. Think of a FAD (fashion trend) on TV — you see a celebrity (the given info) and immediately know which show (method) they belong to.
Why it works
Acronyms create a lookup table in memory. The TV metaphor adds a visual hook. On exam day, scanning the given data automatically triggers FAD-TV.
Example usage
Problem says: 'A 1500 kg car goes from 0 to 20 m/s in 8 s — find the force.' Time (T) and velocity change (V) are given → Impulse-Momentum. FAD-TV instantly points you there.
Recall trigger
What info is given? → FAD-TV
Newton's Second Law: ΣF = ma
Memory aid
Imagine pushing a jeepney (F) loaded with passengers (m) — the heavier it is, the slower it moves (less a). The jeepney driver (Σ = net, total effort) pushes all forces together. ΣF = ma: Sum-of-Forces = mass × acceleration. The jeepney never lies — more mass, less acceleration for the same push.
Why it works
The jeepney is a universally familiar Filipino vehicle. The push-mass-acceleration relationship is physically intuitive and the analogy is personally relatable.
Example usage
A 50 kg crate, 200 N push, μ = 0.2: Draw FBD, friction = 0.2×50×9.81 = 98.1 N, net F = 200−98.1 = 101.9 N, a = 101.9/50 = 2.04 m/s². The jeepney image reminds you to include friction as resistance.
Recall trigger
Picture a heavily loaded jeepney being pushed
D'Alembert's Principle: Add inertia force −ma to treat dynamics as statics
Memory aid
Juan is riding a tricycle accelerating forward. He feels pushed backward into his seat — that backward 'ghost force' is the inertia force −ma. D'Alembert says: 'Pretend that ghost force is real, add it to your FBD, and now the body is in equilibrium — treat it like a statics problem!' Juan's backward lean = the inertia force. Once he adds it, everything balances (ΣF = 0).
Why it works
The felt sensation of being pushed back in an accelerating vehicle is a universal physical experience. The micro-story creates an embodied memory.
Example usage
Same crate problem: Add −ma = −50a pointing left. Now ΣF = 0: 200 − 98.1 − 50a = 0 → a = 2.04 m/s². Same answer, treated as statics — useful when a problem says 'use D'Alembert.'
Recall trigger
Lean-back feeling in an accelerating tricycle
Work-Energy Principle: U₁→₂ = ½mv₂² − ½mv₁²
Memory aid
Work is like money deposited into a bank account called Kinetic Energy. Every peso (joule) of work you put in increases your KE balance. If friction takes some money out (negative work), your final balance is less. The principle: Net deposit = Final balance − Initial balance. U_net = KE₂ − KE₁.
Why it works
Money and bank accounts are concrete, emotionally familiar concepts. The deposit/withdrawal framing maps perfectly onto positive/negative work.
Example usage
100 N force over 5 m on a 10 kg block: Deposit = 500 J; Friction withdrawal = 73.6 J; Net deposit = 426.4 J = ½(10)v² → v = 9.24 m/s. ATM reminder: net deposit goes to KE.
Recall trigger
ATM transaction — depositing and withdrawing energy
Friction always does NEGATIVE work
Memory aid
Rhyme: 'Friction is a thief in the night — it steals your energy, left and right. Always subtract, never add — friction work is always bad!' On an exam: friction work = −μNd, every single time.
Why it works
Rhymes encode information in both semantic and phonological memory — two storage pathways. The 'thief' metaphor also reinforces the concept emotionally.
Example usage
Any problem with a surface: U_friction = −μNd. If μ = 0.15, m = 10 kg, d = 5 m: U_f = −0.15(98.1)(5) = −73.6 J. Always negative — the thief always steals.
Recall trigger
Thief stealing energy
Work of gravity: U = ±Wh (positive going DOWN)
Memory aid
Visualize a coconut falling from a tall coconut tree. As it falls DOWN, gravity does POSITIVE work — gravity is 'helping' the coconut move in the direction it pulls. If you carry the coconut UP, you work AGAINST gravity — negative work by gravity. Down = positive; Up = negative. The falling coconut image: DOWN = POSITIVE.
Why it works
The coconut falling is a familiar tropical image. The direction (down = positive) is reinforced by the physical sensation of gravity pulling things downward naturally.
Example usage
A 5 kg ball falls 10 m: U_gravity = +Wh = 5(9.81)(10) = 490.5 J (positive, falling down). Ball thrown upward 10 m: U_gravity = −490.5 J (gravity opposes upward motion).
Recall trigger
Coconut falling from a tree
Impulse-Momentum: ΣF·Δt = mv₂ − mv₁
Memory aid
Impulse is like a long jeepney horn blast. The longer and louder (force × time) the honk, the more it changes a pedestrian's walking speed (momentum). A quick tap (small Δt) barely changes anything. A sustained blast (large Δt) makes people jump and change direction dramatically. F·Δt = Δ(mv): the sustained honk equals the momentum change.
Why it works
Sound duration × intensity → effect is a natural intuitive analogy. The jeepney horn is a distinctly Filipino sensory memory.
Example usage
Car 1500 kg, 0 to 20 m/s in 8 s: F = mΔv/Δt = 1500(20)/8 = 3750 N. The impulse (F×8) = momentum change (1500×20). Hear the honk, write the formula.
Recall trigger
Jeepney horn blast duration
Conservation of Momentum: no external impulse → m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Memory aid
Two paluwagan (communal savings) members, Nena (m₁) and Berto (m₂), bump into each other on a slippery floor at the paluwagan office. Nobody from outside pushes them. Their total money (momentum) before the bump equals their total money after — the paluwagan pot is conserved! Internal reshuffling (the collision) doesn't change the total. No outsider (external force) = total momentum conserved.
Why it works
The paluwagan is a deeply familiar Filipino financial tradition. 'Total pot stays the same unless someone from outside adds or takes' maps perfectly onto momentum conservation.
Example usage
2000 kg truck at 15 m/s + 1000 kg stationary car: (2000)(15) + (1000)(0) = (3000)v → v = 10 m/s. The paluwagan of momentum: total before = total after, no outsider involved.
Recall trigger
Paluwagan pot stays the same
Coefficient of Restitution: e = (v₂ − v₁)/(u₁ − u₂) — separation over approach
Memory aid
Remember 'e = SOBA': Separation Over Before (i.e., Approach). The numerator is the relative velocity of SEPARATION (after impact), the denominator is the relative velocity of APPROACH (before impact). SOBA also sounds like 'sober' — stay SOBER and keep your signs consistent on a single axis!
Why it works
The SOBA acronym encodes both the fraction structure (S/A) and the warning about sign consistency. The 'sober' reminder addresses the #1 sign error on board exams.
Example usage
Ball A at 5 m/s hits ball B at 1 m/s (same direction): approach = 5−1 = 4 m/s. If e = 0.8: separation = 0.8×4 = 3.2 m/s → v_B − v_A = 3.2. Combine with momentum to solve. Remember: SOBA = e definition.
Recall trigger
SOBA — Separation Over Before Approach
e = 1 (perfectly elastic) vs e = 0 (perfectly plastic)
Memory aid
e = 1: Think of a super bouncy rubber ball (like the ones sold at 5-peso machines in SM). It bounces back perfectly — KE is conserved, bodies separate. e = 0: Think of two lumps of sticky rice (puto/kakanin) that smash together and stick — they move as one body after impact, KE is lost. Rubber ball = elastic (e=1); sticky rice = plastic (e=0).
Why it works
Rubber ball and sticky rice are vivid, tactile images with Filipino cultural resonance. The contrast (bounces vs sticks) directly encodes the two extremes.
Example usage
e = 0 problem: bodies move together after collision → use momentum conservation alone (one unknown: common velocity v). e = 1: use both momentum conservation and KE conservation to get two equations.
Recall trigger
Rubber ball vs. sticky kakanin
Normal and Tangential components in curvilinear motion: ΣFₙ = mv²/r, ΣFt = mat
Memory aid
Imagine riding a motorcycle on a banked curve (like NLEX). The centripetal (normal) force pulls you toward the center of the curve — that's ΣFₙ = mv²/r. The tangential force is what makes you speed up or slow down along the curve — ΣFt = mat. N is for 'Nobody falls toward center' (ΣFₙ keeps you on the curve); T is for 'Throttle/Brake' (ΣFt changes your speed along the road).
Why it works
Philippine expressway driving is a real engineering context. The N=center/T=along split is reinforced by Nobody/Throttle memory hooks.
Example usage
Car at 15 m/s on r = 50 m curve: ΣFₙ = mv²/r → friction force N = m(225/50) = 4.5m. For minimum μ: μmg = mv²/r → μ = v²/(rg) = 225/(50×9.81) = 0.459.
Recall trigger
Motorcycle on NLEX curve — center vs. throttle
Mass (kg) vs Weight (N) — critical distinction in ΣF = ma
Memory aid
Mnemonic: 'M is for Mass — measured in Manila in kilograms on a scale (it never changes). W is for Weight — what gravity does to you (W = mg, changes on the Moon).' On every board exam: if the problem gives weight in N, divide by g to get m in kg before using ΣF = ma. 'Manila kilograms = Mass; gravity gives Weight.'
Why it works
Attaching Manila (a familiar city) to mass creates a geographic memory peg. The Moon contrast reinforces why weight is not mass.
Example usage
Problem: '500 N crate pushed by 200 N force.' m = 500/9.81 = 50.97 kg. ΣF = ma: always use kg for m. Never substitute 500 directly for m — Manila scale rule.
Recall trigger
Scale in Manila = mass in kg
Work of a spring: U = ½k(x₁² − x₂²)
Memory aid
Aling Rosa has a spring sling-shot (tirador). When she stretches it (x₁ large), it has stored energy. When she releases it toward x₂ = 0, the spring DOES work on the stone — positive work. The formula: Work done BY the spring = ½k(x₁² − x₂²). Notice: it's x₁² MINUS x₂² — the spring does positive work as it returns toward natural length (x decreases). Don't confuse this with stored PE = ½kx², which is the energy still left in the spring.
Why it works
The tirador (slingshot) is a familiar Filipino childhood toy. The story of energy stored then released maps directly onto the formula structure.
Example usage
Spring k = 500 N/m stretched from x₁ = 0.1 m to x₂ = 0.05 m: U_spring = ½(500)(0.1² − 0.05²) = ½(500)(0.01 − 0.0025) = 1.875 J (positive: spring contracts toward natural length).
Recall trigger
Aling Rosa's tirador releasing stored spring energy
Energy Conservation (conservative forces only): KE₁ + PE₁ = KE₂ + PE₂
Memory aid
Imagine a children's seesaw: when one side (KE) goes up, the other (PE) goes down — but the total weight on the seesaw (total mechanical energy) stays the same, as long as nobody (friction, external force) pushes from outside. KE + PE = constant, like the seesaw balance. Only conservative forces (gravity, springs) — no friction allowed on the seesaw!
Why it works
The seesaw is a universal playground image with an inherent balance/conservation metaphor. The 'nobody pushes' condition maps onto 'no non-conservative forces.'
Example usage
Ball released from 10 m height, find impact speed: KE₁ = 0, PE₁ = mgh = m(9.81)(10). KE₂ = ½mv², PE₂ = 0. Seesaw balance: m(98.1) = ½mv² → v = √(196.2) = 14.0 m/s.
Recall trigger
Balanced seesaw — KE and PE trading places
Banking angle formula derived from ΣFₙ = mv²/r
Memory aid
Picture the Skyway or SLEX ramps — they are banked (tilted) precisely so your car stays on the curve without friction. The banking angle θ satisfies tan θ = v²/(rg). Visualize the ramp: the steeper the bank (bigger θ), the faster you can go. Remember: tan of the tilt = v²/(rg). The Skyway tilt = tan = v-squared over r-g.
Why it works
Philippine expressway ramps are a real, visible engineering application. Linking a formula to a daily commute creates an episodic memory anchor.
Example usage
Road curve r = 100 m, design speed v = 20 m/s: tan θ = (20²)/(100 × 9.81) = 400/981 = 0.408 → θ = 22.2°. Skyway ramp: the tilt angle for that speed.
Recall trigger
Skyway/SLEX banked ramp tilt angle
Sign convention in impact problems — use ONE consistent positive direction
Memory aid
Mnemonic: 'ONE ARROW, ONE BOSS.' Draw ONE arrow (positive direction) before you write any velocity. Every velocity — u₁, u₂, v₁, v₂ — gets a sign based on that one arrow. Never flip signs mid-solution. The positive direction is the BOSS; everyone follows the boss. Head-on collisions (opposite initial velocities) are where students forget to make u₂ negative.
Why it works
Authority figures are memorable. The 'one boss' rule creates a clear procedural memory: establish the direction first, never change it.
Example usage
Ball A (5 m/s right, +) hits Ball B (3 m/s LEFT, −). ONE BOSS: right is positive. u₁ = +5, u₂ = −3. Do NOT use 3 as positive for B. Momentum: m₁(5) + m₂(−3) = m₁v₁ + m₂v₂. The boss prevents sign errors.
Recall trigger
One boss arrow — draw it before writing any velocity
Centripetal acceleration aₙ = v²/r (for circular/curved path)
Memory aid
Rhyme: 'v-squared over r — that's how fast you curve and star! Normal force points to center near — that's the centripetal, crystal clear!' aₙ = v²/r: velocity squared divided by radius. The inward direction is the normal direction — always pointing toward the center of curvature.
Why it works
The rhythm and rhyme make the formula easy to recall under exam pressure. The 'center near' phrase reinforces the direction of the normal acceleration.
Example usage
Car at 15 m/s on r = 50 m: aₙ = 15²/50 = 225/50 = 4.5 m/s² (directed toward center). ΣFₙ = maₙ = m(4.5). For minimum friction: μmg = m(4.5) → μ = 4.5/9.81 = 0.459.
Recall trigger
v-squared over r rhyme
Work = F·d·cosθ (component of force along displacement)
Memory aid
Imagine pulling a kariton (push cart) with a rope at angle θ above the horizontal. Only the HORIZONTAL component (F cosθ) actually moves the cart forward. The vertical component just lifts (does no work on horizontal displacement). Work = F·d·cosθ — only the component IN the direction of motion counts. The useless vertical pull = the sin component = zero work on horizontal ground.
Why it works
The kariton/push cart is a universally recognized Filipino street vendor image. The physical experience of angled pulling is relatable and the cosθ rule follows naturally.
Example usage
Force 100 N at 30° above horizontal, distance 5 m: U = 100(5)cos30° = 500(0.866) = 433 J. If the force were vertical (θ=90°): U = 500·cos90° = 0 J — no work on horizontal motion.
Recall trigger
Kariton being pulled at an angle — only the horizontal pull counts
Perfectly plastic collision: bodies stick together, use only momentum conservation
Memory aid
Engineer Dino drives his old 2000 kg MMDA truck at 15 m/s and rear-ends a parked 1000 kg barangay vehicle. They crumple and move together (e = 0 — sticky kakanin collision). You only need ONE equation: total momentum before = combined momentum after. (2000)(15) = (3000)v → v = 10 m/s. No KE conservation needed — energy is lost to the crumple.
Why it works
The MMDA/barangay vehicle scenario is a vivid local context. The crumple image reinforces energy loss. The single-equation procedure is memorable through the story.
Example usage
Always check: do bodies stick together? → e = 0, one equation, common final velocity v. (m₁u₁ + m₂u₂)/(m₁+m₂) = v.
Recall trigger
MMDA truck crumple collision — one equation only
Normal force on a vertical circular loop (top vs bottom)
Memory aid
Picture a rollercoaster loop (like at Star City). At the BOTTOM: both N (up) and mg (down) act — net upward = centripetal → N − mg = mv²/r → N = m(g + v²/r). At the TOP: both N and mg point DOWN toward center → N + mg = mv²/r → N = m(v²/r − g). For minimum speed at top: N ≥ 0 → v_min = √(rg). Star City loop: bottom = N is BIG (you feel heavy); top = N is SMALL (you feel light or weightless).
Why it works
The rollercoaster physical sensation (heavy at bottom, light at top) is universally experienced and emotionally memorable. The equations follow from the felt sensations.
Example usage
Loop r = 10 m, m = 70 kg, v = 12 m/s at bottom: N = 70(9.81 + 144/10) = 70(9.81+14.4) = 70(24.21) = 1694.7 N (much greater than weight 686.7 N). At top: N = 70(14.4−9.81) = 70(4.59) = 321.3 N.
Recall trigger
Star City rollercoaster — heavy bottom, light top
Revision game
I am the only tool you need when a problem gives you force, time, and wants velocity — but never asks about distance. What method am I?
Answer
Impulse-Momentum Principle (ΣF·Δt = mΔv)
Memory link
A7 — Jeepney horn blast analogy: force × time = momentum change
Two cars collide and crumple together, moving as one. What is the coefficient of restitution, and how many equations do you need?
Answer
e = 0 (perfectly plastic); only ONE equation (momentum conservation) since they move together
Memory link
A10 — Sticky kakanin analogy; A19 — MMDA truck crumple micro-story
I look like a statics problem, but I'm actually dynamics in disguise. I add a fake backward force equal to −ma to make everything 'balance.' Who am I?
Answer
D'Alembert's Principle
Memory link
A3 — Juan leaning back in the accelerating tricycle micro-story
Nena pushes a 20 kg crate with 80 N on a floor where μ = 0.2. Before solving, I insist you write my equation as ΣF = ma. What force must you include that opposes motion?
Answer
Friction force = μN = 0.2 × (20 × 9.81) = 39.24 N (opposing direction of motion, subtract from net force)
Memory link
A2 — Jeepney push analogy; A5 — Friction is the thief rhyme
I am the ratio of relative velocity of separation to relative velocity of approach. My acronym sounds like something you stay in a road trip. What am I?
Answer
Coefficient of Restitution e — recalled by acronym SOBA (Separation Over Before-Approach)
Memory link
A9 — SOBA mnemonic
A 5 kg ball is released from 8 m height onto a floor. Using the seesaw analogy, what is its speed just before impact? (g = 9.81 m/s²)
Answer
v = √(2gh) = √(2 × 9.81 × 8) = √157.0 = 12.53 m/s (KE₂ = PE₁ seesaw balance, no friction)
Memory link
A14 — Seesaw KE/PE balance analogy
A car drives around a curve of radius 80 m at 18 m/s. I am the equation governing the forces in the direction pointing toward the center. Write me.
Answer
ΣFₙ = mv²/r = m(18²/80) = m(324/80) = 4.05m N (centripetal direction)
Memory link
A11 — Motorcycle on NLEX curve analogy; A17 — v²/r rhyme
In a work-energy problem, I am always subtracted — never added. I equal negative μNd. Who am I, and what rhyme helps you remember?
Answer
Friction work (U_friction = −μNd) — Rhyme: 'Friction is a thief in the night — it steals your energy, left and right. Always subtract, never add — friction work is always bad!'
Memory link
A5 — Friction thief rhyme
Formula mnemonics
ΣF = ma
Mnemonic
FAce MA — put your FACE on MA (mass × acceleration). ΣF = m × a. Sum of Forces = mass times acceleration. 'Face the mass times acceleration!'
When to use
When forces AND acceleration are the unknowns, OR when forces are given and acceleration is needed. Time and distance are NOT required.
What each part means
ΣF = net force in Newtons (N); m = mass in kilograms (kg); a = acceleration in m/s². Note: m is NOT weight — divide W by g first if given in N.
U₁→₂ = ½mv₂² − ½mv₁²
Mnemonic
NET WORK = FINAL KE minus INITIAL KE. 'Net Work is the NET of your KE bank — final balance minus starting balance.' U = KE₂ − KE₁.
When to use
When speed/velocity at a certain position is unknown and distance is given (or vice versa). Time does NOT need to be involved.
What each part means
U₁→₂ = total net work done on the body (J); ½mv₂² = final kinetic energy (J); ½mv₁² = initial kinetic energy (J). Work includes all forces: applied force (+), friction (−μNd), gravity (±Wh), spring (½k(x₁²−x₂²)).
U = Fd cosθ
Mnemonic
U = Fortune·Distance·Cosine of angle = FDC. 'Fortune (F) times Distance (d) times Cosine (C) = your energy fortune.' Only the component along the path counts.
When to use
Computing work done by any constant force that makes angle θ with the direction of motion.
What each part means
F = magnitude of force (N); d = displacement (m); θ = angle between force vector and displacement direction. If θ = 0°: full work. If θ = 90°: zero work.
ΣF·Δt = mv₂ − mv₁
Mnemonic
FLIP = Force-Linear Impulse = momentum change. 'Force FLIPS momentum over time.' ΣF × time = final momentum − initial momentum.
When to use
When force and time are involved (given or unknown). Especially when distance is NOT given.
What each part means
ΣF = average net force (N); Δt = time interval (s); mv₂ = final linear momentum (kg·m/s); mv₁ = initial linear momentum (kg·m/s). Left side = impulse; right side = change in momentum.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Mnemonic
TOTAL BEFORE = TOTAL AFTER: 'Both Bodies Before = Both Bodies After.' (m₁u₁ + m₂u₂) = (m₁v₁ + m₂v₂). Total momentum before impact equals total momentum after — paluwagan rule.
When to use
For ALL collision/impact problems. Must be combined with the restitution equation (e definition) when e is given and both post-collision velocities are unknown.
What each part means
m₁, m₂ = masses (kg); u₁, u₂ = velocities BEFORE impact (m/s, signed on one axis); v₁, v₂ = velocities AFTER impact (m/s, signed). Valid when no external horizontal impulse acts.
e = (v₂ − v₁)/(u₁ − u₂)
Mnemonic
SOBA: Separation Over Before-Approach. 'e = S/A' — e equals Separation (after, numerator) over Approach (before, denominator). Range: 0 ≤ e ≤ 1.
When to use
In all impact problems where e is given or to find e from before/after velocities. Always pair with momentum conservation equation.
What each part means
v₂ − v₁ = relative velocity of separation (body 2 relative to body 1, after impact); u₁ − u₂ = relative velocity of approach (body 1 approaching body 2, before impact). All velocities SIGNED on one axis.
ΣFₙ = mv²/r
Mnemonic
Normal Force Needs v-squared-over-r: 'N needs v²/r for the Curve.' The net force toward the center = mass × centripetal acceleration = mv²/r.
When to use
For any body moving on a curved path: banking problems, looping, pendulum, satellite orbits, car on curve. Always resolve forces into normal (toward center) and tangential (along path) components.
What each part means
ΣFₙ = net force directed toward center of curvature (N); m = mass (kg); v = speed along the path (m/s); r = radius of curvature (m). This force is provided by friction, banking component, rope tension, etc.
U_spring = ½k(x₁² − x₂²)
Mnemonic
Spring work = ½k × DIFFERENCE of squares (first minus second). 'Spring DELIVERS work as it SHRINKS (x₁ > x₂ = positive work delivered).' Stored PE = ½kx² (single x). Work done BY spring = ½k(x₁² − x₂²).
When to use
When a spring is involved and it changes deformation from x₁ to x₂. Add to the work-energy equation alongside other work terms.
What each part means
k = spring stiffness (N/m); x₁ = initial deformation from natural length (m); x₂ = final deformation (m). Positive when spring contracts toward natural length (releases energy).
Quick recall chains
Steps to Solve a Newton's 2nd Law Problem
Recall test
Without looking: what are the 5 steps to solve a ΣF = ma problem? Start with 'D' — Draw the...
Memory chain
The story of DISUM: Engineer Dina DRAWS her FBD in her diary (Draw FBD), then IDENTIFIES all forces like identifying suspects (identify forces), COMPUTES friction like calculating her allowance (compute friction), SUMS all forces in Newton's equation (sum forces), and UNCOVERS the unknown answer (solve unknown). DRAWS → IDENTIFIES → COMPUTES → SUMS → UNCOVERS = DICUS.
Items to remember
- 1. Draw the Free Body Diagram (FBD)
- 2. Identify all forces (applied, weight, normal, friction)
- 3. Compute friction: F_f = μN
- 4. Sum forces along direction of motion: ΣF = ma
- 5. Solve for the unknown (a or F)
Method Selection Flowchart: Which Kinetics Method to Use
Recall test
A problem gives you mass, initial velocity, time, and asks for force. Which method? (Answer: Impulse-Momentum)
Memory chain
FAD-TV-C: Force+Acceleration → Direct Newton. Distance+Speed → Work-Energy. Time+Velocity → Impulse. Collision with e → Two equations. Curved → Normal/Tangential split. Remember: 'FADded TV shows have Curved screens!' F=force, A=accel, D=distance, T=time, V=velocity, C=curved.
Items to remember
- Given: Force + Acceleration (or find a) → Newton's 2nd Law (ΣF = ma)
- Given: Force + Distance, find Speed → Work-Energy
- Given: Force + Time, find Velocity → Impulse-Momentum
- Given: Collision with e → Momentum Conservation + e equation
- Given: Curved path → Resolve into Normal (mv²/r) and Tangential (mat)
Types of Collisions and What to Use
Recall test
What two equations do you use for a partially elastic collision? (Answer: Momentum conservation AND e = separation/approach)
Memory chain
Elastic, Partial, Plastic = EPP: 'Every Problem in Physics has EPP!' E=elastic (bouncy ball, two equations), P=partial (normal collision, two equations with e), P=plastic (sticky kakanin, one equation). Always set the ONE BOSS arrow first.
Items to remember
- e = 1: Perfectly elastic — KE conserved, use momentum + KE conservation
- 0 < e < 1: Partially elastic — use momentum conservation + e equation (two unknowns, two equations)
- e = 0: Perfectly plastic — bodies stick, use momentum conservation only, one unknown v
- Always: define ONE positive direction before assigning signs
All Work Terms in the Work-Energy Equation
Recall test
Name all four types of work terms in the work-energy equation. (Hint: AGSF)
Memory chain
AGSF = Applied, Gravity, Spring, Friction — 'AGgressively Stop Friction!' (A = applied Fd cosθ, G = gravity ±Wh, S = spring ½k(x₁²−x₂²), F = friction −μNd). Sum all four = net work = change in KE.
Items to remember
- Work by applied force: +Fd cosθ
- Work by gravity: +Wh (down) or −Wh (up)
- Work by spring: +½k(x₁² − x₂²)
- Work by friction: −μNd (ALWAYS negative)
- Total U₁→₂ = ΣU = ½mv₂² − ½mv₁²
Signs in Curvilinear (Normal-Tangential) Analysis
Recall test
At the TOP of a vertical loop, which direction do N and mg both point? (Answer: Both toward the center — downward)
Memory chain
Star City rollercoaster memory: CENTER = positive normal direction. THROTTLE = tangential direction. At TOP: both N and mg toward center (ADD). At BOTTOM: N toward center, mg away (SUBTRACT). 'Center is king; throttle is along.'
Items to remember
- Normal direction: TOWARD center = positive
- ΣFₙ = mv²/r (net centripetal force)
- Tangential direction: ALONG path = positive in direction of motion
- ΣFt = mat (net tangential force changes speed)
- At top of loop: N and mg both point DOWN (toward center) — add them
- At bottom of loop: N points UP (toward center), mg points DOWN — subtract
Diagrams
Kinetics Method Selection Decision Tree (FAD-TV-C)
A decision flowchart showing how to select the correct kinetics method based on the given problem data. Follow the arrows from the given information to the correct method and formula. This is the FAD-TV-C system visualized.
Dynamics Kinetics Master Mind Map
A comprehensive mind map of all Kinetics topics covered in this chapter. Start from the center (Kinetics) and branch outward to see how Newton's 2nd Law, Work-Energy, Impulse-Momentum, Impact, and Curvilinear Kinetics are all organized under one framework.
Impact Problem Solution Flowchart — Step by Step
A step-by-step flowchart for solving any impact/collision problem. The critical Step 1 (draw ONE positive direction) prevents the most common board exam sign error. The branching at 'Is e given?' separates plastic (one equation) from elastic/partial (two equations) scenarios.
States of a Body Under Kinetics Analysis
A state diagram showing how a body transitions between states (at rest, accelerating, constant speed, on curve, impact) and which kinetics equations govern each transition. Helps students identify which phase of motion they are analyzing and which equation applies.
Board Exam Distribution: Kinetics Topics
Approximate distribution of kinetics problems in PRC Civil Engineer board examinations. Newton's 2nd Law and Work-Energy each account for roughly 25% of kinetics problems, making them the highest priority study areas. Allocate study time proportionally.