CELE Engineering Mechanics — Dynamics: KinematicsExam Answer Templates

Examiner tip

The single discriminating phrase is 'without forces.' Include it and the mark is secured.

Model answer

Kinematics is the branch of dynamics that describes the motion of particles and rigid bodies in terms of position, displacement, velocity, and acceleration, without consideration of the forces or masses causing the motion.

Answer structure

• Single sentence: Name the branch, state what it describes (position, velocity, acceleration), and state what it excludes (forces/causes). [1 mark]

Scoring breakdown

Common mark deductions

• Confusing kinematics with kinetics (kinetics involves forces; kinematics does not).
• Defining dynamics instead of kinematics.
• Correct concept but omitting the 'no forces' distinction.

Key phrases to include

• describes motion
• position, velocity, acceleration
• without consideration of forces
• branch of dynamics

Examiner tip

Memorize these three equations as a set. In an exam, writing them takes under 10 seconds and secures the mark.

Model answer

The three equations of motion for constant acceleration are:
(1) v = v₀ + at
(2) s = v₀t + ½at²
(3) v² = v₀² + 2as
where v₀ = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement.

Answer structure

• List all three equations clearly numbered. [1 mark]
• (Optional, no extra mark): Define the symbols if space allows.

Scoring breakdown

Common mark deductions

• Writing s = vt (only true for constant velocity, not constant acceleration).
• Forgetting the ½ coefficient in the second equation.
• Writing v² = v₀ + 2as (missing the squared on v₀).

Key phrases to include

• v = v₀ + at
• s = v₀t + ½at²
• v² = v₀² + 2as
• constant acceleration

Examiner tip

A common board-exam trap: the problem gives N in rpm. Always convert first, before substituting into ω₀ + αt.

Model answer

ω = 2πN / 60
where ω is in rad/s and N is in revolutions per minute (rpm).
Alternatively: ω = 2πf, where f = N/60 is the frequency in Hz.

Answer structure

• State the conversion formula clearly. [1 mark]

Scoring breakdown

Common mark deductions

• Writing ω = 2πN (omitting the division by 60 — valid only when N is in rev/s).
• Inverting the formula as N = 2πω/60.

Key phrases to include

• ω = 2πN/60
• rad/s
• rpm

Examiner tip

For a differentiate question, use a two-column format or two clearly labeled paragraphs. Do not combine the two definitions into one sentence.

Model answer

Velocity (v) is the time rate of change of displacement: v = ds/dt. It is a vector quantity expressed in m/s and indicates how fast a particle moves and in what direction.

Acceleration (a) is the time rate of change of velocity: a = dv/dt = v(dv/ds). It is also a vector quantity expressed in m/s² and indicates how rapidly the velocity changes with time. A negative acceleration (deceleration) means the particle is slowing down in the positive direction.

Answer structure

• Line 1–2: Define velocity with formula v = ds/dt and unit. [1 mark]
• Line 3–4: Define acceleration with formula a = dv/dt and unit, and note the sign significance. [1 mark]

Scoring breakdown

Common mark deductions

• Confusing displacement with distance (displacement is a vector; distance is scalar).
• Stating units but omitting the derivative definition.
• Writing 'acceleration is speed divided by time' — this is only valid for constant acceleration from rest.

Key phrases to include

• v = ds/dt
• a = dv/dt
• vector quantity
• m/s and m/s²
• rate of change

Examiner tip

Free-fall problems from rest are the most common 2-mark kinematics item. Both equations are essentially one-step substitutions — write them out fully anyway to protect your method marks.

Model answer

Given: v₀ = 0 (dropped from rest), t = 3.5 s, g = 9.81 m/s² (downward positive)

(a) Height of bridge:
s = v₀t + ½gt²
s = 0(3.5) + ½(9.81)(3.5)²
s = ½(9.81)(12.25)
s = 60.1 m

(b) Speed at impact:
v = v₀ + gt
v = 0 + 9.81(3.5)
v = 34.3 m/s

Answer structure

• State given data and sign convention. [Method — implicit mark]
• Apply s = v₀t + ½gt² for part (a); correct answer 60.1 m. [1 mark]
• Apply v = v₀ + gt for part (b); correct answer 34.3 m/s. [1 mark]

Scoring breakdown

Common mark deductions

• Using g = 10 m/s² without the problem specifying it — use 9.81 unless told otherwise.
• Forgetting the ½ in the displacement equation.
• Reporting velocity in m/s² instead of m/s.

Key phrases to include

• s = v₀t + ½gt²
• v = v₀ + gt
• v₀ = 0
• g = 9.81 m/s²
• dropped from rest

Examiner tip

The examiners specifically test if you can distinguish sin2θ (range) from sin²θ (height). A mnemonic: R is a wider trajectory → use the wider angle (2θ).

Model answer

(a) Range: R = v₀² sin2θ / g
(b) Maximum height: H = v₀² sin²θ / (2g)
(c) Time of flight: t = 2v₀ sinθ / g

All three apply only when launch and landing elevations are equal. Maximum range occurs at θ = 45°.

Answer structure

• State all three formulas correctly labeled (a), (b), (c). [1.5 marks — or 0.5 each]
• Note the condition (same-elevation launch/landing) and the 45° max-range result. [0.5 mark]

Scoring breakdown

Common mark deductions

• The most common error: writing sin²θ for the range formula instead of sin2θ.
• Omitting the factor of 2 in the time-of-flight formula.
• Forgetting the 2g denominator in the height formula (writing g instead).

Key phrases to include

• R = v₀² sin2θ / g
• H = v₀² sin²θ / (2g)
• t = 2v₀ sinθ / g
• sin2θ = 2sinθ cosθ
• maximum range at θ = 45°

Examiner tip

When given v₀, v, and s (but not t), the v² equation is always the most direct. Never start with the quadratic s = v₀t + ½at² when t is unknown.

Model answer

Given: v₀ = 10 m/s, v = 30 m/s, s = 400 m
Sign convention: direction of motion is positive.

(a) Acceleration — use v² = v₀² + 2as:
(30)² = (10)² + 2a(400)
900 = 100 + 800a
800a = 800
a = 1.0 m/s²

(b) Time — use v = v₀ + at:
30 = 10 + 1.0(t)
t = 20 s

Check: s = v₀t + ½at² = 10(20) + ½(1.0)(20²) = 200 + 200 = 400 m ✔

Answer structure

• State all given data and sign convention. [0.5 mark — method]
• Correctly apply v² = v₀² + 2as and solve for a = 1.0 m/s². [1 mark]
• Correctly apply v = v₀ + at and solve for t = 20 s. [1 mark]
• Verify with check equation. [0.5 mark — bonus or method mark]

Scoring breakdown

Common mark deductions

• Using s = v₀t + ½at² as the first equation when t is unknown — creates two unknowns simultaneously.
• Arithmetic error in expanding 30² − 10² = 900 − 100 = 800 (some students write 700).
• Reporting a in m/s instead of m/s².

Key phrases to include

• v² = v₀² + 2as
• v = v₀ + at
• a = 1.0 m/s²
• t = 20 s
• sign convention

Examiner tip

Rotational kinematics problems almost always require two conversions before you even touch the kinematic equation. Build the habit: convert rpm → rad/s, convert rev → rad, then apply the formula.

Model answer

Given: N₀ = 600 rpm, N = 0 rpm, θ = 40 rev

Step 1 — Convert initial angular velocity:
ω₀ = 2πN₀/60 = 2π(600)/60 = 62.83 rad/s
ω = 0 rad/s

Step 2 — Convert revolutions to radians:
θ = 40 rev × 2π rad/rev = 251.3 rad

Step 3 — Apply ω² = ω₀² + 2αθ:
0 = (62.83)² + 2α(251.3)
0 = 3947.6 + 502.6α
α = −3947.6 / 502.6
α = −7.85 rad/s²

The magnitude of angular deceleration is 7.85 rad/s².

Answer structure

• Convert N₀ from rpm to rad/s. [1 mark]
• Convert θ from revolutions to radians. [0.5 mark]
• Apply ω² = ω₀² + 2αθ and solve for α = −7.85 rad/s². [1 mark]
• State the magnitude and interpret the negative sign as deceleration. [0.5 mark]

Scoring breakdown

Common mark deductions

• Forgetting to convert rpm to rad/s — using 600 directly in the formula.
• Forgetting to convert revolutions to radians — using 40 directly.
• Omitting the negative sign and failing to state this implies deceleration.

Key phrases to include

• ω₀ = 2πN/60
• θ = 40 × 2π = 251.3 rad
• ω² = ω₀² + 2αθ
• α = −7.85 rad/s²
• deceleration

Examiner tip

When a is given as a function of t, the constant-acceleration equations are invalid. Immediately switch to integration. The marker is looking for the integral sign, not v = v₀ + at.

Model answer

Given: a = 6t − 4 m/s², v₀ = 2 m/s, s₀ = 0, t = 3 s

(a) Velocity at t = 3 s:
v = v₀ + ∫₀ᵗ a dt
v = 2 + ∫₀³ (6t − 4) dt
v = 2 + [3t² − 4t]₀³
v = 2 + [3(9) − 4(3) − 0]
v = 2 + [27 − 12]
v = 2 + 15
v = 17 m/s

(b) Position at t = 3 s:
s = s₀ + ∫₀ᵗ v dt
s = 0 + ∫₀³ (2 + 3t² − 4t) dt
s = [2t + t³ − 2t²]₀³
s = 2(3) + (3)³ − 2(3)²
s = 6 + 27 − 18
s = 15 m

Answer structure

• Recognize variable acceleration requires integration; state v = v₀ + ∫a dt. [0.5 mark]
• Correctly integrate a = 6t − 4 and substitute limits to get v = 17 m/s. [1 mark]
• Correctly integrate v to get s = 15 m. [1 mark]
• Show the definite integral substitution clearly. [0.5 mark]

Scoring breakdown

Common mark deductions

• Using the constant-acceleration trio (v = v₀ + at) when a is variable — this is a fundamental conceptual error.
• Forgetting to add v₀ = 2 m/s as the constant of integration for velocity.
• Integrating a to get velocity but forgetting to then integrate v to get position.

Key phrases to include

• v = v₀ + ∫a dt
• s = s₀ + ∫v dt
• variable acceleration
• constants of integration
• definite integral

Examiner tip

A 5-mark projectile problem awards marks at each sub-step. Even if you compute R incorrectly, you can still earn marks for correct H and t formulas. Never skip the component decomposition step — it earns an explicit mark.

Model answer

Given: v₀ = 30 m/s, θ = 40°, g = 9.81 m/s², level surface (launch height = landing height)

Horizontal component: v₀ₓ = v₀ cosθ = 30 cos40° = 30(0.7660) = 22.98 m/s
Vertical component: v₀ᵧ = v₀ sinθ = 30 sin40° = 30(0.6428) = 19.28 m/s

(a) Range R:
R = v₀² sin2θ / g
R = (30)² sin(80°) / 9.81
R = 900(0.9848) / 9.81
R = 886.3 / 9.81
R ≈ 90.3 m

(b) Maximum height H:
H = v₀² sin²θ / (2g)
H = (30)² (sin40°)² / (2 × 9.81)
H = 900(0.6428)² / 19.62
H = 900(0.4132) / 19.62
H = 371.9 / 19.62
H ≈ 18.95 m

(c) Time of flight t:
t = 2v₀ sinθ / g
t = 2(30)(0.6428) / 9.81
t = 38.57 / 9.81
t ≈ 3.93 s

Verification: x = v₀ₓ × t = 22.98 × 3.93 = 90.3 m ✔ (matches range R)

Answer structure

• State given data; decompose v₀ into horizontal and vertical components. [1 mark]
• Apply R = v₀² sin2θ / g; correct answer ≈ 90.3 m. [1 mark]
• Apply H = v₀² sin²θ / (2g); correct answer ≈ 18.95 m. [1 mark]
• Apply t = 2v₀ sinθ / g; correct answer ≈ 3.93 s. [1 mark]
• Verification using x = v₀ₓ × t confirming R; and neat labeled diagram. [1 mark]

Scoring breakdown

Common mark deductions

• Using sin2θ = sin²θ — the most frequent error in projectile problems.
• Not decomposing v₀ into components at the start — forgoing the method mark.
• Using g = 9.81 in the formula for H but forgetting to double it in the denominator (2g vs g).
• Forgetting that the range formula is valid ONLY for equal-elevation launch and landing — marks may be deducted if this is not acknowledged.
• Premature rounding of trigonometric values (use at least 4 decimal places mid-solution).

Key phrases to include

• v₀ₓ = v₀ cosθ
• v₀ᵧ = v₀ sinθ
• R = v₀² sin2θ / g
• H = v₀² sin²θ / (2g)
• t = 2v₀ sinθ / g
• horizontal and vertical motions are independent

Examiner tip

When the problem says 'verify,' the verification is a separate awarded step — not optional. Write it as a clearly labeled 'Verification:' block. The examiner is literally looking for that word and the matching check equation.

Model answer

Given: v₀ = 20 m/s, v = 0 m/s, a = −4 m/s² (deceleration)
Sign convention: direction of initial motion is positive.

(a) Stopping time — use v = v₀ + at:
0 = 20 + (−4)t
4t = 20
t = 5 s

(b) Stopping distance — use s = v₀t + ½at²:
s = 20(5) + ½(−4)(5)²
s = 100 + ½(−4)(25)
s = 100 − 50
s = 50 m

Verification using v² = v₀² + 2as:
0² = (20)² + 2(−4)(s)
0 = 400 − 8s
8s = 400
s = 50 m ✔

Conclusion: The train stops after 5 s over a distance of 50 m.

Answer structure

• State all given data; declare sign convention (a = −4 m/s²). [0.5 mark]
• Apply v = v₀ + at; solve for t = 5 s with unit. [1 mark]
• Apply s = v₀t + ½at²; solve for s = 50 m with unit. [1.5 marks]
• Apply verification equation v² = v₀² + 2as; confirm s = 50 m with tick. [1.5 marks]
• State conclusion sentence. [0.5 mark]

Scoring breakdown

Common mark deductions

• Using a = +4 m/s² (positive) throughout — leads to s = 200 m, which is physically wrong (the car would be accelerating, not stopping).
• Missing the negative sign in ½at²: writing +50 instead of −50, giving s = 150 m.
• Skipping the verification step when the question explicitly requests it — costs 1–1.5 marks.
• Not writing units (m, s, m/s²) beside each intermediate result.

Key phrases to include

• a = −4 m/s² (deceleration)
• v = v₀ + at
• s = v₀t + ½at²
• v² = v₀² + 2as
• t = 5 s
• s = 50 m
• sign convention

Examiner tip

Rotational kinematics mirrors rectilinear kinematics exactly. If you can solve a rectilinear problem, you can solve a rotational one — just replace s→θ, v→ω, a→α. The only extra step is unit conversion.

Model answer

Given: ω₀ = 0 (starts from rest), N = 300 rpm, t = 10 s

Step 1 — Convert final speed to rad/s:
ω = 2πN/60 = 2π(300)/60 = 10π = 31.42 rad/s

(a) Angular acceleration:
α = (ω − ω₀)/t = (31.42 − 0)/10 = 3.14 rad/s²

(b) Total angle turned — use θ = ω₀t + ½αt²:
θ = 0(10) + ½(3.14)(10)²
θ = ½(3.14)(100)
θ = 157 rad

(c) Number of revolutions:
n = θ/(2π) = 157/(2π) = 157/6.283 = 25.0 revolutions

Verification: θ = (ω₀ + ω)/2 × t = (0 + 31.42)/2 × 10 = 15.71 × 10 = 157.1 rad ✔

Answer structure

• Convert ω from rpm to rad/s: ω = 2πN/60 = 31.42 rad/s. [1 mark]
• Apply α = (ω − ω₀)/t; compute α = 3.14 rad/s². [1 mark]
• Apply θ = ω₀t + ½αt²; compute θ = 157 rad. [1.5 marks]
• Convert radians to revolutions: n = θ/2π = 25 rev. [1 mark]
• Verification with average-velocity method. [0.5 mark]

Scoring breakdown

Common mark deductions

• Substituting N = 300 directly into α = (ω − ω₀)/t without converting to rad/s.
• Using degrees instead of radians throughout.
• Forgetting to divide by 2π at the end to get revolutions from radians.
• Arithmetic error: computing ½ × 3.14 × 100 as 314 instead of 157.

Key phrases to include

• ω = 2πN/60
• α = (ω − ω₀)/t
• θ = ω₀t + ½αt²
• n = θ/2π
• 31.42 rad/s
• 157 rad
• 25 revolutions

Examiner tip

Whenever a wheel problem specifies constant speed, state explicitly that α = 0, hence aₜ = 0. This shows the examiner you understand the difference between tangential and centripetal acceleration.

Model answer

Given: r = 0.5 m, N = 120 rpm (constant speed → α = 0)

(a) Angular velocity:
ω = 2πN/60 = 2π(120)/60 = 4π = 12.57 rad/s

(b) Linear (peripheral) speed at the rim:
v = rω = 0.5 × 12.57 = 6.28 m/s

(c) Centripetal (normal) acceleration:
aₙ = v²/r = (6.28)²/0.5 = 39.48/0.5 = 78.96 m/s²
(or equivalently: aₙ = rω² = 0.5 × (12.57)² = 0.5 × 158.0 = 79.0 m/s²)

Note: Since speed is constant, the tangential acceleration aₜ = rα = 0. The total acceleration equals the centripetal acceleration: a = 79.0 m/s² directed radially inward.

Answer structure

• Convert N = 120 rpm to ω = 12.57 rad/s. [1 mark]
• Apply v = rω to get v = 6.28 m/s. [1 mark]
• Apply aₙ = v²/r or rω² to get aₙ ≈ 79.0 m/s². [1 mark]

Scoring breakdown

Common mark deductions

• Confusing centripetal acceleration with tangential acceleration.
• Using r in cm instead of m — always work in SI (meters).
• Omitting the direction of aₙ (it must be stated as 'directed toward the center' or 'radially inward').

Key phrases to include

• ω = 2πN/60
• v = rω
• aₙ = v²/r = rω²
• centripetal acceleration
• radially inward
• aₜ = 0 (constant speed)

Examiner tip

This is a semi-derivation question. The mark is not just for stating '45°' — it is for the chain of reasoning: formula → maximize sin2θ → 2θ = 90° → θ = 45°. Show every link.

Model answer

The range of a projectile (launched and landing at the same elevation) is:

R = v₀² sin2θ / g

For a fixed launch speed v₀ and fixed g, R is maximized when sin2θ is maximum. The sine function achieves its maximum value of 1.0 when its argument equals 90°:

sin2θ = 1 → 2θ = 90° → θ = 45°

Therefore, the maximum range is:
Rₘₐₓ = v₀² / g (at θ = 45°)

Physically, at 45° the launch velocity is optimally shared between the horizontal component (governs how far the projectile travels) and the vertical component (governs how long it stays airborne). Any angle above or below 45° sacrifices one component at the expense of the other, reducing R. Complementary angles (e.g., 30° and 60°) give the same range but different flight times and heights.

Answer structure

• State the range formula R = v₀² sin2θ / g. [1 mark]
• Show that R is maximized when sin2θ = 1, i.e., 2θ = 90°, θ = 45°. [1 mark]
• State Rₘₐₓ = v₀²/g and provide a physical explanation. [1 mark]

Scoring breakdown

Common mark deductions

• Stating θ = 45° without the mathematical justification — earns only 1 of 3 marks.
• Confusing 'maximum range' with 'maximum height' (max height occurs at θ = 90°, not 45°).
• Writing sin²θ instead of sin2θ in the range formula.

Key phrases to include

• R = v₀² sin2θ / g
• sin2θ is maximized at 2θ = 90°
• θ = 45°
• Rₘₐₓ = v₀² / g
• complementary angles give equal range

Examiner tip

In variable-motion problems, 'at rest' means v = 0. But a = 0 means constant velocity. Do not confuse them. A particle at rest with a ≠ 0 is a turning point — always interpret the sign of a when v = 0.

Model answer

Given: s = 2t³ − 9t² + 12t m

Step 1 — Velocity (differentiate s with respect to t):
v = ds/dt = 6t² − 18t + 12 m/s

Step 2 — At rest: v = 0:
6t² − 18t + 12 = 0
Divide by 6: t² − 3t + 2 = 0
Factor: (t − 1)(t − 2) = 0
t = 1 s and t = 2 s

Step 3 — Acceleration (differentiate v):
a = dv/dt = 12t − 18 m/s²

Step 4 — Evaluate a at each rest instant:
At t = 1 s: a = 12(1) − 18 = −6 m/s²
v = 0, a < 0 → particle was moving in +s direction and is now beginning to move in −s direction (decelerating / changing direction)

At t = 2 s: a = 12(2) − 18 = +6 m/s²
v = 0, a > 0 → particle was moving in −s direction and is now beginning to move in +s direction (accelerating from rest in +s direction)

Conclusion: The particle is momentarily at rest at t = 1 s and t = 2 s. At t = 1 s it is decelerating (a = −6 m/s²); at t = 2 s it is accelerating (a = +6 m/s²). Both represent turning points.

Answer structure

• Differentiate s to obtain v = 6t² − 18t + 12. [1 mark]
• Set v = 0 and solve the quadratic; t = 1 s and t = 2 s. [1.5 marks]
• Differentiate v to obtain a = 12t − 18. [1 mark]
• Evaluate a at t = 1 s (= −6 m/s²) and t = 2 s (= +6 m/s²) and interpret. [1.5 marks]

Scoring breakdown

Common mark deductions

• Integrating s instead of differentiating it to find v.
• Finding t = 1 and t = 2 but failing to compute and interpret the acceleration at those instants.
• Stating 'decelerating' without checking the sign of a relative to the sign of v.

Key phrases to include

• v = ds/dt
• a = dv/dt
• set v = 0
• turning point
• decelerating at t = 1 s
• accelerating at t = 2 s

Do

• State the formula or definition in one concise, technically correct sentence.
• Include units (m/s, rad/s, m/s²) even for 1-mark items.
• Use standard engineering notation (v₀, ω, α, θ).
• Answer immediately and move on — never spend more than 90 seconds.

Don't

• Do not write lengthy paragraphs for a 1-mark item.
• Do not derive the formula when only stating it is required.
• Do not leave blank — even a partially correct answer may earn the mark.

Do

• Label each part clearly: (a) and (b).
• State the formula used before substituting numbers.
• Include units for every intermediate and final result.
• Write a brief 'Check' if time permits (takes 20 seconds and can save a mark).

Don't

• Do not combine two steps into one line — the marker needs to see each step.
• Do not round aggressively mid-solution; round only at the final answer.
• Do not assume the marker will fill in missing arithmetic — show it all.

Do

• Write 'Given:', 'Find:', and 'Solution:' headers to organize the answer.
• Number each step.
• State the sign convention explicitly.
• Draw and label a sketch for projectile and rotation problems.
• State your conclusion in words at the end.

Don't

• Do not skip the unit conversion step (rpm → rad/s, rev → rad) — it is usually a dedicated mark.
• Do not write the answer only without showing the working path.
• Do not use the constant-acceleration equations when acceleration is variable.

Do

• Spend 1–2 minutes planning the solution path before writing.
• Use headed sections: 'Given Data', 'Sign Convention', 'Step 1 – Convert Units', 'Step 2 – Find…', 'Verification', 'Conclusion'.
• Draw a fully labeled free-body or trajectory diagram.
• Include a Verification step using an alternative equation.
• Box the final answer with correct units.
• Write a one-sentence engineering conclusion: 'The projectile lands 90.3 m from the launch point.'

Don't

• Do not attempt the problem in one continuous paragraph — examiners cannot identify where to award marks.
• Do not leave out the verification step when there are 5 marks at stake.
• Do not skip stating units — a correct number without units earns zero for that mark.
• Do not use approximate values of sin/cos in your head — write out the trig evaluation.