CELE Engineering Mechanics — Dynamics: KineticsRevision Notes

Formulas

Exam tips

• Always write ΣF = ma with a declared positive direction. Stick to that convention throughout the solution.
• For an incline problem, set up axes parallel and perpendicular to the incline surface — this eliminates components and simplifies both equations.
• D'Alembert's principle is especially efficient when a problem involves a system of connected bodies (e.g., two blocks connected by a cord over a pulley) — treat the entire system as one free body.
• Board exams frequently give μ and ask for acceleration on an incline: a = g(sin θ – μ cos θ) for a block sliding down. Memorise this shortcut.
• If acceleration comes out negative, the body actually moves in the direction opposite to your assumed positive direction — report the magnitude and state the correct direction.

Key points

• Newton's Second Law: The net force on a particle equals its mass multiplied by its acceleration — ΣF = ma. This is the most fundamental equation in kinetics.
• Always draw a complete Free Body Diagram (FBD) before writing equations. Identify all forces: applied forces, weight (W = mg), normal force N, and friction force f = μN.
• Normal force N is perpendicular to the contact surface. On a horizontal surface, N = mg. On an incline of angle θ, N = mg cos θ.
• Kinetic friction force: f_k = μ_k × N, always opposing the direction of motion. Static friction: f_s ≤ μ_s × N.
• D'Alembert's Principle converts a dynamics problem into a pseudo-static equilibrium by introducing an inertia force (–ma) opposing the acceleration: ΣF – ma = 0. This is useful when you are comfortable with statics equilibrium methods.
• For inclined plane problems: resolve forces parallel (∑F_parallel = ma) and perpendicular (∑F_perp = 0, since no motion perpendicular to surface) to the incline.
• Units check: Force in Newtons (N = kg·m/s²), mass in kg, acceleration in m/s².

Definitions

Common mistakes

• Using weight (W = mg, in Newtons) instead of mass (m, in kg) in ΣF = ma. Remember: ΣF = ma uses mass, not weight.
• Forgetting to resolve the applied force into components parallel and perpendicular to the surface when the force is inclined.
• Assuming N = mg on an inclined plane. Correct value is N = mg cos θ.
• Adding friction as a positive term instead of subtracting it (friction always opposes motion).
• Not redrawing the FBD after identifying all forces — skipping this step is the leading cause of setup errors on board exams.

Formulas

Exam tips

• The work–energy method is FASTEST when only the initial and final states matter and time is not asked. If a board exam problem asks 'find the speed after traveling X metres,' immediately think work–energy.
• List every force on the FBD. Check each one: Does it do work? (Is there a displacement component in the force direction?) Normal force and weight on a horizontal surface — one does work (weight if there is vertical movement), the other does not (normal force, perpendicular to horizontal motion).
• For incline problems with friction: U_net = (mg sin θ – μ mg cos θ)d = mg·d(sin θ – μ cos θ). This combined formula saves time.
• When a problem gives efficiency η of a machine: Useful work output = η × Work input. Power output = η × Power input.
• Memorise: 1 hp = 745.7 W ≈ 0.746 kW. Board exams frequently mix hp and kW in the same problem.

Key points

• The Work–Energy Principle states: the net work done on a body equals the change in its kinetic energy — U_net = ΔKE = ½mv₂² – ½mv₁².
• Work is done by a force only if there is a displacement with a component in the direction of the force: U = F·d·cos θ. A force perpendicular to displacement does zero work (e.g., normal force on horizontal motion).
• Gravity does positive work when the body moves downward (in the direction of gravity): U_gravity = +mgh (downward). Gravity does negative work when the body moves upward: U_gravity = –mgh.
• Spring work: When a spring is compressed or stretched, the work done BY the spring on the body is U_spring = ½k(x₁² – x₂²). Note: this is NOT the stored potential energy — it is the work output of the spring on the attached body.
• Friction does negative work equal to –μNd, regardless of direction of motion. It always removes energy from the system.
• For conservative force systems only (no friction, no damping): total mechanical energy is conserved — KE₁ + PE₁ = KE₂ + PE₂.
• Best method when: (a) unknown is a speed after a given displacement, and (b) time is not needed. If time appears in the problem, switch to impulse–momentum.
• Power = Work/Time = F·v (force times velocity). Units: Watts (W = J/s). 1 hp = 745.7 W.

Definitions

Common mistakes

• Confusing spring work U_spring = ½k(x₁² – x₂²) with spring potential energy PE_spring = ½kx². Work done by the spring and energy stored in the spring are related but not the same quantity in the work–energy equation.
• Forgetting the cosine factor when an applied force is inclined to the direction of motion. Using U = Fd instead of U = Fd cos θ is a frequent error.
• Applying conservation of energy (KE₁ + PE₁ = KE₂ + PE₂) when friction is present. Friction makes energy non-conservative; you must subtract the friction work loss.
• Using height h as the slant distance along an incline instead of the true vertical height. Always use the vertical component for gravity work.
• Setting up the spring work formula with x measured from the equilibrium position of a dynamic system rather than from the spring's natural (unstretched) length.

Formulas

Exam tips

• The impulse–momentum method is FASTEST when a problem involves force and time but NOT distance. Keywords: 'in X seconds,' 'average force during impact,' 'time to stop.'
• For collision problems: ALWAYS set up a single positive direction axis. Write all velocities as signed values (positive in chosen direction, negative in opposite direction). Do NOT work with magnitudes only.
• For a perfectly plastic collision (bodies stick together): use one combined mass (m₁ + m₂) and one common final velocity v_f on the right side.
• For a rebound problem: if a ball hits a wall at v₁ and rebounds at v₂ (opposite direction), the impulse on the ball is m(v₂ – (–v₁)) = m(v₁ + v₂). Board exams test this sign sensitivity.
• Variable force problems: the impulse equals the area under the F–t graph. Triangular loading: J = ½·F_max·Δt. Rectangular: J = F·Δt.

Key points

• The Impulse–Momentum Principle: The impulse of the net force over a time interval equals the change in linear momentum — ΣF·Δt = mv₂ – mv₁.
• Linear momentum p = mv. Units: kg·m/s. Momentum is a vector quantity — direction matters.
• Impulse J = F·Δt. Units: N·s = kg·m/s. A large force over a short time can equal a small force over a long time.
• Best method when: force and time are given or unknown, and distance/displacement is not required.
• Conservation of Momentum: When the net external impulse on a system is zero (no external forces, or very short collision time), total momentum is conserved: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'. This is the basis of collision analysis.
• For a variable force, impulse = area under the F–t diagram (integrate force over time).
• Average force: F_avg = Δp/Δt = m(v₂ – v₁)/Δt. Used when a force varies and only the average is needed.
• Sign convention is critical: choose a positive direction, then assign positive and negative values to velocities accordingly. A body moving left when the positive direction is rightward gets a negative velocity.

Definitions

Common mistakes

• Ignoring signs (directions) of velocities. In impulse–momentum problems, all velocities must be assigned positive or negative values based on a declared positive axis. Using bare magnitudes leads to wrong answers for head-on collision problems.
• Confusing impulse (N·s) with work (J = N·m). They look similar dimensionally but are fundamentally different: one involves time, the other involves distance.
• Applying conservation of momentum without checking whether external forces exist. On a rough horizontal surface, friction is an external horizontal force and momentum is NOT conserved over time (only over the infinitesimally short collision instant).
• Using ΣF·Δt = Δ(mv) but substituting speed (scalar) instead of velocity (vector). A ball bouncing back has a velocity change of v₂ – v₁, where v₁ and v₂ have opposite signs — the impulse is much larger than if you just subtracted magnitudes.

Formulas

Exam tips

• Board exam collision problems ALWAYS require two equations for two unknowns. The two equations are invariably: (1) momentum conservation and (2) the e-equation. Set them up simultaneously.
• For equal masses and e = 1 (elastic): the bodies exchange velocities. v₁ = u₂ and v₂ = u₁. This is a quick shortcut to verify your simultaneous solution.
• For e = 0 (perfectly plastic): single equation — (m₁u₁ + m₂u₂) / (m₁ + m₂) = v_common. No need for the e-equation.
• Declare your positive direction FIRST, then substitute signed velocities into BOTH equations. A body moving left gets a negative value if rightward is positive.
• If the problem says 'the ball rebounds to 3/4 of its original height after dropping from H,' you can find e: h_rebound = e²·H, so e = √(h_rebound/H). This is a derivation of the e-formula for vertical drop–bounce problems.

Key points

• During a collision (impact), two principles govern the outcome: (1) Conservation of Momentum, and (2) Coefficient of Restitution (e) which characterises the elasticity of the collision.
• The coefficient of restitution e is defined as: e = (relative velocity of separation) / (relative velocity of approach) = (v₂ – v₁) / (u₁ – u₂). All velocities are along the line of impact (signed).
• e = 1: Perfectly elastic impact — kinetic energy is conserved, bodies bounce apart. Total KE before = Total KE after.
• e = 0: Perfectly plastic (inelastic) impact — bodies move together after impact. Maximum KE is lost. This reduces to simple momentum conservation with one final velocity.
• 0 < e < 1: Partially elastic — some KE is lost. Most real-world collisions fall here.
• For a central impact (head-on), two simultaneous equations are needed: momentum conservation (gives one equation with two unknowns v₁, v₂) and the e-equation (gives a second equation). Solve simultaneously.
• For impact with a fixed wall or very massive body: the wall's velocity is zero before AND after. The e-equation simplifies to: v_ball = –e·u_ball (the ball rebounds at e times its approach speed, in the opposite direction).
• Kinetic energy lost in an impact: ΔKE_lost = ½m₁u₁² + ½m₂u₂² – ½m₁v₁² – ½m₂v₂². Always positive (energy lost, not gained).

Definitions

Common mistakes

• Putting velocities in the e-formula as magnitudes (speeds) instead of signed values along the line of impact. The formula e = (v₂ – v₁)/(u₁ – u₂) requires all velocities to be signed consistently. Using absolute values gives wrong answers for head-on collisions.
• Mixing up numerator and denominator: the numerator is always the AFTER (separation) velocities minus each other, and the denominator is always the BEFORE (approach) velocities. Inverting this gives e > 1, which is physically impossible.
• Applying conservation of kinetic energy instead of momentum conservation to find post-collision velocities (KE is conserved only for e = 1, and even then you need the momentum equation too).
• For a ball bouncing off a wall, students often forget to set the wall's velocity as zero both before and after, leading to an overcomplicated setup.
• Forgetting that for oblique impacts, only the components along the line of impact are affected by e. Tangential components are unchanged (assuming smooth surfaces).

Formulas

Exam tips

• For any curved-path problem: set up BOTH the normal equation (ΣF_n = mv²/r, pointing inward) and the tangential equation (ΣF_t = m·a_t, along path). Often only one is needed, but write both to identify which applies.
• The condition for 'losing contact' or 'minimum speed' problems is always N = 0. Substitute N = 0 into the normal equation and solve for v_min.
• For banking problems, the key insight: at the ideal bank (no friction), the horizontal component of N provides all centripetal force, and the vertical component of N balances weight. This gives two equations → tan θ = v²/(rg).
• Philippine highway design uses superelevation (banking) rates per DPWH standards. Board exams may ask the ideal banking angle — use tan θ = v²/(rg) directly.
• For a conical pendulum (ball on string moving in a horizontal circle): T cos θ = mg (vertical), T sin θ = mv²/r (horizontal). Period = 2π√(L cos θ / g). Common board exam setup.

Key points

• For a body moving along a curved path, resolve Newton's Second Law into two perpendicular components: tangential (along the path, direction of motion) and normal (perpendicular to the path, pointing toward the centre of curvature).
• Tangential equation: ΣF_t = m·a_t. This governs the change in speed along the curved path. If speed is constant (uniform circular motion), a_t = 0 and ΣF_t = 0.
• Normal (centripetal) equation: ΣF_n = m·a_n = m·v²/r. This governs the change in direction. There must always be a net force toward the centre of curvature to maintain circular motion.
• For a vehicle on a banked curve: the normal component of the resultant provides the centripetal force. Ideal banking angle (no friction needed): tan θ = v²/(rg).
• For a vehicle on a flat (unbanked) curve: friction provides all the centripetal force. Minimum μ to prevent skidding: μ = v²/(rg).
• For a body on the inside of a loop (e.g., roller coaster): at the top of the loop, both weight W and normal force N point toward the centre. Minimum speed at top: v_min = √(rg) (when N = 0).
• For a body swinging on a rope: at the bottom, ΣF_n = N – W = mv²/r, so N = W + mv²/r (normal force exceeds weight — you feel heavier). At the top, N = mv²/r – W.
• Centrifugal force in D'Alembert's sense: the inertia force mv²/r directed outward from the centre — added as a pseudo-force to convert the problem into pseudo-equilibrium.

Definitions

Common mistakes

• Adding centripetal force (mv²/r) as an extra force ON the free body diagram in addition to the actual forces. The centripetal force is the RESULT of summing actual forces, not an additional one.
• Using diameter instead of radius in the formula ΣF_n = mv²/r. Always use the radius r (half the diameter).
• Forgetting that at the top of a vertical loop, both weight AND normal force are directed downward (toward the centre). Students sometimes subtract N from W instead of adding them.
• Confusing the flat-curve friction requirement μ = v²/(rg) with the banking formula tan θ = v²/(rg). Both have the same right-hand side but represent different physical scenarios.
• Not converting speed from km/h to m/s (divide by 3.6) before substituting into v²/r formulas.