Representations, Derivations and Experimental Design: Exam Skills Guide (Physics C: Mechanics)

How to use this guide

This guide is for Physics C: Mechanics students who know the physics but lose free-response marks on diagrams, derivations and graphs. It covers Units 1 to 7. Try each worked example before reading the strong answer.

Every question here is original Marlbridge practice, not past exam material. The marks shown are a suggested Marlbridge rubric to help you judge your own work. They are not official scoring.

The free-response section at a glance

Section II has 4 questions in 95 minutes, worth 50% of the exam. You read them in the Bluebook app and write answers by hand in a paper booklet. A four-function, scientific or graphing calculator is allowed. The order is fixed:

Question	Type	Points	Suggested time	What it rewards
1	Mathematical Routines (MR)	10	20–25 min	Deriving and calculating, helped by a diagram or sketch
2	Translation Between Representations (TBR)	12	25–30 min	Diagram, derivation and graph of one scenario, then linking them
3	Experimental Design and Analysis (LAB)	10	25–30 min	A workable procedure, then a linear graph of given data
4	Qualitative/Quantitative Translation (QQT)	8	15–20 min	A justified claim linked to a derivation

Creating representations (Science Practice 1) is tested only in this section, not in multiple choice.

Representations that earn marks

Free-body diagrams. Follow these rules unless the question says otherwise.

• Draw the object as a dot. For torque problems, draw its outline and put each force where it acts.
• Draw one arrow per force, starting on the object and pointing away from it.
• Label each force so its source is clear: F_g (Earth on block), F_N (floor on block), T (rope on block).
• Show only real forces on the object: no “ma”, net-force or velocity arrow, no components.
• If asked for relative lengths, make equal forces equal and bigger forces longer.

Motion graphs. Read the slope and the area.

Graph	Slope gives	Area under it gives
x against t	velocity v	—
v against t	acceleration a	displacement Δx
a against t	—	change in velocity Δv
F against t or F against x	—	impulse J, or work W

A sketch needs no exact values, but its features must be right: starting value, sign, zeros, curvature and any asymptote.

Energy representations. For a bar chart, name the system first, draw one bar per kind of energy at the start and end, and one for work by external forces. The bars must balance. On a U(x) graph, F_x = −dU/dx (the force points downhill), turning points are where the total-energy line meets U(x), and equilibrium is where the slope is zero.

Derivations: the routine markers look for

A “derive” answer has four visible parts.

1. Start from a fundamental principle in general form (ΣF = ma, W_net = ΔK, τ = Iα, I = ∫r² dm) or a reference-sheet equation.
2. Substitute this scenario’s forces or quantities, with signs matching your axis.
3. Solve, showing each step.
4. Finish with the requested quantity alone, in the allowed symbols only.

Three calculus routines come up again and again.

Routine	Starting point	Key step	Common trap
Differential equation from the second law	m dv/dt = ΣF(v), or m d²x/dt² = −kx	Separate variables with matching limits, or recognise d²x/dt² = −ω²x	Using constant-acceleration equations
Integrating a variable force	W = ∫F(x) dx, J = ∫F(t) dt	Integrate between the start and end values	Multiplying the force by the distance or time
Rotational inertia by integration	I = ∫r² dm	Write dm = λ dx (rod) or dm = σ·2πr dr (disk), with all of dm at one distance r	Using a uniform-object formula for a non-uniform object

Full methods are in Topic 2.9, Topic 3.2 and Topic 5.4.

Check every result.

• Units. Added terms share units; the answer has the units of the quantity asked for.
• Limiting cases. With no friction, a light pulley or equal masses, does the answer reduce to a result you know?
• Bounds and trends. Is the answer between two simple cases? Does it grow when it should?

If a question asks whether an answer is reasonable, a check like this is the evidence it wants.

Translating between representations

One scenario can be described in words, a diagram, an equation and a graph. TBR and QQT questions often ask whether two of your answers agree, so decide which feature links them.

From	To	What must match
Free-body diagram	Second-law equation	One term per arrow, same direction and sign
Equation for v(t)	Sketch of v against t	Starting value, sign, zeros, asymptote, curvature
Energy bar chart	Energy equation	One term per bar; totals balance
Words (“speeds up ever more slowly”)	v–t graph	Rising, concave-down curve

A strong “do they agree?” answer names both representations, the shared feature and the physics linking them.

Designing an experiment and linearising data

The design part. A sound procedure:

• changes one variable, measures one, and says what is kept the same;
• names realistic equipment (metre stick, calipers, balance, photogate, motion sensor, force sensor) and what each measures;
• uses at least five values over a wide range, repeating and averaging to reduce uncertainty;
• says what to plot and what the slope or intercept will tell you.

Linearising. Rearrange the theory into y = (slope)x + (intercept), where y and x are measured or calculated. The slope then contains the quantity you want.

Theory	Plot (y against x)	Slope	Expected intercept
T = 2π√(m/k) for a mass on a spring	T² against m	4π²/k	0
v = v₀e^(−t/τ) for coasting with drag	ln v against t	−1/τ	ln v₀

Label axes with units and use most of the grid. Draw one best-fit line; never join the dots. Take the slope from two far-apart points on the line, not from data points. If the line misses the expected intercept, say so and suggest a reason.

Worked example 1: a nonlinear bumper (Unit 3)

Question. A cart of mass m moves at speed v₀ along a level, frictionless track towards a rubber bumper. When the bumper is compressed by x, it exerts a force of magnitude αx² on the cart, where α is a positive constant. (a) Derive an expression for the greatest compression x_max in terms of m, v₀ and α. (b) Calculate x_max for m = 0.80 kg, v₀ = 1.5 m/s and α = 400 N/m².

Weak answer. “½mv₀² = ½αx², so x = v₀√(m/α) = 0.067 m.”

What it misses. It treats the bumper as a Hooke’s-law spring and starts from a remembered formula, not a principle. Units expose it: α is in N/m², so ½αx² is in newtons, not joules.

Strong answer.

1. Work-energy theorem for the cart: W_net = ΔK.
2. Take +x into the bumper. The bumper force is F_x = −αx², so W = ∫₀^x_max (−αx²) dx = −αx_max³/3.
3. ΔK = 0 − ½mv₀². So −αx_max³/3 = −½mv₀², giving x_max = (3mv₀²/(2α))^(1/3).
4. Check. mv₀²/α is in (kg·m²/s²) ÷ (N/m²) = m³, so x_max is in metres. v₀ → 0 gives x_max → 0; a stiffer bumper gives less compression.
5. (b) x_max = (3 × 0.80 × 1.5² ÷ 800)^(1/3) = (0.00675)^(1/3) = 0.19 m.

Suggested Marlbridge rubric	Point
Starts from the work-energy theorem or energy conservation, with work as an integral	1
Correct integral of the variable force with correct limits	1
Correct expression for x_max	1
Correct value with unit	1

Worked example 2: a non-uniform rod (Unit 5)

Question. A thin rod of length L and mass M is pivoted at one end, x = 0. Its linear mass density is greatest at the pivot and falls steadily to zero at the free end: λ = λ₀(1 − x/L). (a) Derive its rotational inertia about the pivot in terms of M and L. (b) Show that your answer is reasonable.

Weak answer. “I = ⅓ML² for a rod about its end.”

What it misses. That formula is for a uniform rod. There is no integral, no link between λ₀ and M, and no check.

Strong answer.

1. I = ∫r² dm, with r = x and dm = λ dx = λ₀(1 − x/L) dx.
2. Mass: M = ∫₀ᴸ λ₀(1 − x/L) dx = λ₀L/2, so λ₀ = 2M/L.
3. I = ∫₀ᴸ x²λ₀(1 − x/L) dx = λ₀(L³/3 − L³/4) = λ₀L³/12 = ⅙ML².
4. Check. Less mass is far from the pivot than in a uniform rod, so I should be below ⅓ML². The centre of mass is at L/3, and I about the pivot cannot be less than M(L/3)² = ML²/9. 1/9 < ⅙ < ⅓, so the answer is reasonable. Units: kg·m².

Suggested Marlbridge rubric	Point
Sets up I = ∫x²λ dx with the given λ	1
Relates λ₀ to M by integrating λ	1
Correct result ⅙ML²	1
Justifies with a valid bound or comparison	1

Worked example 3: friction from a tilted board (Unit 2)

Question. Students want to find the coefficient of kinetic friction μ_k between a wooden block and a board. (a) Describe a procedure that changes the angle θ of the board. (b) They release the block from rest at five angles and measure its acceleration a down the board. Use the data to find μ_k.

θ (°)	25	30	35	40	45
a (m/s²)	1.52	2.32	3.24	4.01	4.87

Weak answer. “Tilt the board until the block slides; then μ = tan θ. Using 45°, μ = 1.0.”

What it misses. μ = tan θ holds only at constant velocity; this block speeds up. It ignores the measured a, names no equipment or repeats, and uses no graph.

Strong answer.

(a) Vary θ over five values from 25° to 45°, finding sin θ = height ÷ length of the board with a metre stick. Keep the block, surface and release point the same. A motion sensor at the top gives v against t; its slope is a. Repeat three times at each angle and average.

(b) Along the board: mg sin θ − μ_k mg cos θ = ma. Divide by m cos θ: a/cos θ = g tan θ − μ_k g. Plot a/cos θ against tan θ; the slope should be g and the intercept is −μ_k g.

• tan θ: 0.47, 0.58, 0.70, 0.84, 1.00; a/cos θ: 1.68, 2.68, 3.96, 5.23, 6.89 m/s².
• Best-fit slope 9.8 m/s²; intercept −2.9 m/s².
• μ_k = −intercept ÷ slope = 2.9 ÷ 9.8 = 0.30.

The slope matching g supports the model (Topic 2.7).

Suggested Marlbridge rubric	Point
Variable changed, variable measured and controls named	1
Equipment that can actually measure θ and a; repeats over a range	2
Linear plot of a/cos θ against tan θ with a best-fit line	2
Uses the intercept and slope to find μ_k	1

Worked example 4: one oscillator, four representations (Unit 7)

Question. A 0.50 kg block on a frictionless horizontal surface is attached to a spring with k = 50 N/m. It is pulled to x = +0.10 m and released from rest at t = 0. (a) Draw the free-body diagram at t = 0. (b) Starting from Newton’s second law, derive the differential equation for x and find ω. (c) Sketch v and a against t for one period. (d) Justify that your answers to (a) and (c) agree.

Weak answer. A spring arrow plus a “force of motion” arrow; a v–t graph starting at its maximum; “they agree because the spring pulls it back”.

What it misses. There is no force of motion. The block starts from rest, so v starts at zero. The justification names no feature of either representation.

Strong answer.

1. (a) Three arrows: F_s (spring on block) in −x, of size kA = 5.0 N; F_g (Earth on block) down; F_N (surface on block) up, as long as F_g.
2. (b) ΣF_x = ma_x gives −kx = m d²x/dt², so d²x/dt² = −(k/m)x: SHM with ω = √(k/m) = 10 rad/s, T = 0.63 s and x = 0.10 cos(10t).
3. (c) v = −1.0 sin(10t) m/s: starts at zero, goes negative first, reaching −1.0 m/s at T/4 = 0.16 s. a = −10 cos(10t) m/s²: starts at −10 m/s², zero at T/4, +10 m/s² at T/2.
4. (d) The diagram’s only horizontal force gives a net force of −5.0 N; ÷ 0.50 kg gives −10 m/s², the starting value of my a–t graph. With v = 0 at release and a negative, the v–t graph must start at zero and head down, as drawn.

Suggested Marlbridge rubric	Point
Correct, labelled forces and nothing extra	1
Differential equation from the second law; correct ω	2
v–t starts at zero and goes negative; a–t starts at its most negative value	2
Justification links the net force to the graph’s starting value	1

Common mistakes

• Adding components, a net-force arrow or an “ma” arrow to a free-body diagram.
• Using constant-acceleration equations when the force depends on position, velocity or time.
• Multiplying a variable force by a distance, or using a uniform-object formula for a non-uniform object.
• Leaving numbers or unlisted symbols in a “derive” answer.
• Sketching a graph whose starting value or curvature contradicts your own equation.
• Joining data points dot-to-dot, or taking the slope from data points instead of the best-fit line.
• Writing a procedure that changes two things at once, or measures something no school lab could measure.
• Justifying by repeating an equation, without naming the physics.

Practise it

Task 1 (Unit 1). A cart has velocity v = 6t − 3t² (m/s, t in s) and is at x = 0 when t = 0. (a) Find x(t). (b) Describe the shape of the x–t graph from 0 to 3.0 s. (c) Find the displacement and the distance travelled in that time.

Model answer

(a) Integrate: x = 3t² − t³. (b) v = 0 at t = 2.0 s, so x peaks at 4.0 m there. a = 6 − 6t, so the graph is concave up until t = 1.0 s, then concave down. At 3.0 s it is back at x = 0, falling steeply (v = −9.0 m/s). (c) Displacement 0; distance 4.0 + 4.0 = 8.0 m.

Task 2 (Unit 2). A crate of mass m slides on a rough floor (coefficient μ_k), pushed by a handle with force P at angle θ below the horizontal. List the forces on its free-body diagram, derive its acceleration, and check your result with two limiting cases.

Model answer

Forces: P along the handle, F_g down, F_N up, f_k backward. F_N = mg + P sin θ (not mg), and P cos θ − μ_kF_N = ma, so a = [P cos θ − μ_k(mg + P sin θ)]/m. Checks: μ_k = 0 gives P cos θ/m; θ = 0 gives (P − μ_kmg)/m. Units: N/kg = m/s².

Task 3 (Unit 4). A bat pushes a 0.15 kg ball, at rest on a stand, with force F = 4F₀t(τ − t)/τ² for 0 ≤ t ≤ τ. Derive the impulse, then calculate the ball’s speed for F₀ = 600 N and τ = 0.010 s. Check your answer with the average force.

Model answer

J = ∫₀^τ 4F₀t(τ − t)/τ² dt = (4F₀/τ²)(τ³/2 − τ³/3) = ⅔F₀τ = 4.0 N·s. v = J/m = 27 m/s. Average force J/τ = ⅔F₀ = 400 N is below the 600 N peak, as it must be.

Task 4 (Units 1, 2, 6, 7). What would you plot to get a straight line, and how do you get the target from the slope? (a) g from a pendulum: T = 2π√(ℓ/g). (b) Earth’s mass from satellite orbits: T² = 4π²r³/(GM). (c) A drag constant k from the terminal speeds of same-shaped objects of different mass: v_T = mg/k. (d) g from a ball launched horizontally at fixed speed v from heights H, landing d away: d = v√(2H/g).

Model answer

(a) T² against ℓ; g = 4π²/slope. (b) T² against r³; M = 4π²/(G × slope). (c) v_T against m; k = g/slope. (d) d² against H; g = 2v²/slope.

Task 5 (Unit 5). For two hanging masses m₁ < m₂ over a pulley of radius R and rotational inertia I, a student writes a = (m₂ − m₁)g/(m₁ + m₂ + IR²). Use units and limiting cases to judge this, and correct it.

Model answer

IR² is in kg·m⁴ and cannot be added to a mass, so the result is wrong. From m₂g − T₂ = m₂a, T₁ − m₁g = m₁a and (T₂ − T₁)R = I(a/R): a = (m₂ − m₁)g/(m₁ + m₂ + I/R²). Checks: I/R² is in kg; I → 0 gives the light-pulley result; m₁ = m₂ gives a = 0; larger I gives smaller a.

Where to practise next

• Topic 1.3 study guide: Representing Motion
• Topic 2.2 study guide: Forces and Free-Body Diagrams
• Topic 2.9 practice: Resistive Forces
• Topic 3.2 practice: Work
• Topic 5.4 practice: Rotational Inertia
• Topic 7.2 study guide: Frequency and Period of SHM
• Unit 7 review: Oscillations