Representations, Derivations and Experimental Design: Exam Skills Guide (Physics 1)

How to use this guide

This guide is for students in the algebra-based Physics 1 course who know the physics but lose marks on diagrams, derivations, explanations and lab questions. It covers the whole course.

Every example and task here is original Marlbridge practice, not past exam material. The marks shown are a suggested Marlbridge rubric, not official scoring.

The free-response section at a glance

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

Question	Type	Points	Suggested time	Main demands
1	Mathematical Routines (MR)	10	20–25 min	Representation, derivation, calculation, justified claim
2	Translation Between Representations (TBR)	12	25–30 min	Diagram, derived equation, graph, then consistency or prediction
3	Experimental Design and Analysis (LAB)	10	25–30 min	Procedure, then plotting and analysing given data
4	Qualitative/Quantitative Translation (QQT)	8	15–20 min	Reasoned claim, derivation, link between them

Practice 1 (creating representations) is tested only in this section, never in multiple choice.

Representations that earn marks

A representation scores when it is complete and agrees with the situation.

Representation	Rules that earn credit	Common way to lose it
Free-body diagram	A dot for the system; one straight arrow per force, starting on the dot; each labelled with its type and what exerts it; longer arrows for bigger forces	Adding “ma”, a net force or velocity; drawing components; missing a force
Position–time graph	Gradient = velocity. Curving upwards = acceleration in the + direction.	Sharp corners where velocity changes smoothly
Velocity–time graph	Gradient = acceleration; area = displacement. A straight line means constant acceleration.	Treating the line’s height as position
Energy bar chart	Name the system and the zero of U_g; one bar per energy store at each moment; a zero value is a line on the zero level; total constant if no outside work	No ΔE_thermal bar when friction acts inside the system

Deriving symbolically, then checking

“Derive” means start from a law or an equation-sheet relationship and show each algebraic step. Four steps:

1. Write the starting principle in symbols, such as ΣF_x = ma_x or K_i + U_i = K_f + U_f + ΔE_thermal.
2. Substitute from your diagram. Each arrow on your free-body diagram becomes a term with a sign.
3. Solve for the target using only the symbols allowed. No numbers yet.
4. Check the result before moving on.

Check	How to do it	What a pass looks like
Units	Replace each symbol with its SI unit and simplify	Both sides have the same unit
Limiting case	Set one quantity to an extreme (0, 90°, very large)	The result matches what you know for that simple case
Direction of change	Increase one variable in your head	The answer changes the way physics says it should

Translating between representations

When a question shows one situation in several forms, use a shared feature as a bridge:

From	To	Bridge to look for
Free-body diagram	Velocity–time graph	Net force direction gives the sign of the gradient
Velocity–time graph	Position–time graph	Height of the v–t line gives the gradient of the x–t curve
Energy bar chart	Equation	Each bar becomes one term; bars before = bars after
Equation	Graph	Which variable is on each axis, and whether the relationship is linear, squared or inverse

To justify agreement, name the shared feature and show it matches in both. “They agree” alone earns nothing.

The paragraph argument

The QQT question, and parts of the MR and TBR questions, ask for a claim backed by reasoning beyond an equation. Use this structure:

1. Claim. One clear sentence: greater than, less than, equal, increases, decreases.
2. Governing law. Name it and say why it applies (“it floats, so it is in equilibrium”).
3. Chain. Two or three linked steps, each saying what changes and why.
4. Link back. Show that any equation you derived predicts the same.

Designing and analysing an experiment

A sound procedure:

• changes one independent variable over at least five values and measures one dependent variable;
• keeps everything else fixed, and says so;
• names realistic lab equipment (metre stick, stopwatch, motion or force sensor, balance);
• repeats and averages, or uses a sensor;
• says how the data will be analysed, usually with a straight-line graph.

Linearising turns a curved relationship into a straight line, so the gradient gives a constant:

Relationship	Plot vertical vs horizontal	Gradient equals
d = ½at²	d vs t²	a ÷ 2
v² = 2gh	v² vs h	2g
T = 2π√(m ÷ k)	T² vs m	4π² ÷ k
F = mv² ÷ r (fixed m, r)	F vs v²	m ÷ r

Take the gradient from two far-apart points on your best-fit line and give its units.

Worked example 1: Deriving a pulling force (Unit 2)

Question. A student drags a 15 kg box across a carpet at constant speed, pulling a strap at angle θ above the horizontal. The coefficient of kinetic friction is μ_k. (a) Draw a free-body diagram. (b) Derive the tension T in terms of m, g, θ and μ_k. (c) Evaluate it for θ = 30° and μ_k = 0.40, and check your expression.

Weak answer. “Forces: weight, normal, friction, tension. Friction = μmg, so T cos θ = μmg and T = μmg ÷ cos θ = 68 N.”

What it misses. No labelled diagram. The strap pulls partly upward, so F_N is less than mg, and friction is overestimated. No check.

Strong answer.

(a) Four arrows from a dot: F_g (Earth on box) down; F_N (carpet on box) up, shorter than F_g; F_f (carpet on box) backwards; F_T (strap on box) at θ above the horizontal, forwards.

(b) Vertical: F_N + T sin θ − mg = 0, so F_N = mg − T sin θ. Horizontal, constant velocity: T cos θ − μ_k(mg − T sin θ) = 0. So T = μ_k mg ÷ (cos θ + μ_k sin θ).

(c) T = 0.40 × 147 ÷ (0.866 + 0.200) = 55 N (55.2 N), with F_N = 119 N.

Checks: units are N, since μ_k and trig terms have none. θ = 0 gives T = μ_k mg = 59 N, the horizontal-pull result. μ_k = 0 gives T = 0: no friction, no force needed at constant speed.

Point	What earns it (suggested Marlbridge rubric)
1	Four labelled forces, F_N shorter than F_g, no extra arrows
1	F_N = mg − T sin θ from the vertical equation
1	Correct expression for T
1	55 N
1	A valid limiting-case or units check, stated

Worked example 2: Energy bar charts and a derivation (Unit 3)

Question. A 2.0 kg block is released from rest at height H = 1.5 m on a ramp inclined at 35°, with μ_k = 0.20. (a) Draw energy bar charts for the top (A) and bottom (B). (b) Derive the speed at the bottom in terms of g, H, θ and μ_k. (c) Show that your chart and equation agree.

Weak answer. “At A it has potential energy, at B kinetic. mgH = ½mv², so v = √(2gH) = 5.4 m/s.” Bar charts show one U_g bar at A and an equal K bar at B.

What it misses. No system named. Friction turns some energy into thermal energy, so the K bar must be shorter than the U_g bar, and the speed is too large.

Strong answer. (a) System: block + Earth + ramp, with U_g = 0 at the bottom.

(b) mgH = ½mv² + ΔE_th. Friction is μ_k mg cos θ over a ramp length H ÷ sin θ, so ΔE_th = μ_k mgH ÷ tan θ. The mass cancels: v = √[2gH(1 − μ_k ÷ tan θ)] = √[2(9.8)(1.5)(1 − 0.286)] = 4.6 m/s (4.58 m/s).

(c) The bracket, 0.714, is the fraction of the 29.4 J that becomes kinetic: 0.714 × 29.4 J = 21.0 J, the K bar. Checks: μ_k = 0 gives √(2gH); μ_k = tan θ gives v = 0, as friction then balances gravity along the slope.

Point	What earns it (suggested Marlbridge rubric)
1	System named; U_g only at A; zeros shown as lines
1	K and ΔE_th at B, with K + ΔE_th = U_g at A
1	Energy equation including friction’s thermal energy
1	Correct expression and 4.6 m/s
1	Consistency stated using a shared feature (bar heights match terms)

Worked example 3: The paragraph argument (Unit 8)

Question. A garden sprayer is a wide, closed tank of water. A pump raises the air above the water to a gauge pressure P_g. Water leaves through a nozzle at the same height as the water surface. (a) A student says that doubling P_g doubles the speed of the jet. Is the student right? Justify your claim without equations. (b) Derive the jet speed v in terms of P_g and the water density ρ, and evaluate it for P_g = 2.0 × 10⁵ Pa. (c) Explain whether (a) and (b) agree.

Weak answer. “More pressure pushes the water harder, so doubling the pressure doubles the speed. P = ½ρv², so v = 20 m/s.”

What it misses. The claim is wrong, and “pushes harder” is not a reason: no law, no chain, and no part (c). Its own equation contradicts its claim.

Strong answer.

(a) No: the speed rises by less than double. Energy is conserved as the water flows (Bernoulli’s equation). The pressure does work on each cubic metre of water pushed out, and that work becomes kinetic energy. Doubling P_g doubles the kinetic energy of each cubic metre. Kinetic energy depends on the square of the speed, so the speed grows by a factor less than 2.

(b) Bernoulli from the surface (pressure P_atm + P_g, speed close to 0 in a wide tank) to the nozzle (pressure P_atm, same height): P_g = ½ρv², so v = √(2P_g ÷ ρ) = √(2 × 2.0 × 10⁵ ÷ 1000) = 20 m/s.

(c) They agree: P_g sits under a square root, so doubling it multiplies v by √2 ≈ 1.4, giving 28 m/s, not 40 m/s. Limiting case: P_g = 0 gives v = 0.

Point	What earns it (suggested Marlbridge rubric)
1	Correct claim: the speed less than doubles
1	Energy conservation (Bernoulli) named as the governing law
1	Chain: double pressure, double kinetic energy per volume, speed up by a factor under 2
1	Correct derivation and 20 m/s
1	Agreement explained using the square root (factor √2)

Worked example 4: Design and linearise (Unit 1)

Question. (a) Describe a procedure to find the acceleration of a cart released from rest on a fixed incline. (b) A group records the data below. Choose what to plot so the graph is a straight line, and find the acceleration.

d (m)	0.20	0.40	0.60	0.80	1.00
t (s)	0.70	0.97	1.21	1.37	1.55

Weak answer. “Roll the cart down and time it. Plot d against t. Using the last row, a = 2d ÷ t² = 0.83 m/s².”

What it misses. One trial, no variable changed. d against t is a curve, and one data point ignores the rest.

Strong answer.

(a) Keep the angle and cart fixed. Release from rest at marked distances d = 0.20 m to 1.00 m (metre stick) from a fixed finish line. Time each run with a stopwatch, three trials averaged per distance, or use a motion sensor.

(b) From rest, d = ½at², so plot d (vertical) against t² (horizontal). The t² values are 0.49, 0.94, 1.46, 1.88 and 2.40 s², close to a straight line through the origin. Two points on the best-fit line are (0.48 s², 0.20 m) and (2.15 s², 0.90 m). Gradient = 0.70 ÷ 1.67 = 0.42 m/s². Since the gradient = a ÷ 2, a = 0.84 m/s².

Point	What earns it (suggested Marlbridge rubric)
1	One independent variable (d), one measured (t), the rest fixed
1	Realistic equipment, with repeats or a sensor
1	d against t² (or equivalent linear pair)
1	Gradient from the best-fit line, with units
1	a = 2 × gradient = 0.84 m/s²

Common mistakes

• Drawing components, “ma” or a “net force” arrow on a free-body diagram.
• Assuming F_N = mg when another force has a vertical component or the object accelerates vertically.
• Putting numbers in at the start of a “derive” part, so the expression cannot be checked.
• Energy bar charts with no system named, or with no thermal bar when friction acts.
• Saying two representations “agree” without naming the feature that matches.
• Designing an experiment that changes two variables at once, or uses one trial.
• Taking the gradient from two data points instead of the best-fit line, or forgetting its units.

Practise it

Task 1 (Unit 1). A cart’s velocity rises steadily from 0 to 3.0 m/s in 2.0 s, then stays at 3.0 m/s for 3.0 s. Describe the position–time graph and find the total displacement.

Model answer

From 0 to 2.0 s the x–t graph curves upward from a zero gradient. From 2.0 s to 5.0 s it is a straight line of gradient 3.0 m/s, joining smoothly with no corner. Displacement = area under v–t = ½(2.0)(3.0) + (3.0)(3.0) = 12 m.

Task 2 (Unit 4). A 0.40 kg ball at rest is struck. The force rises linearly from 0 to 600 N in 2.0 ms, then falls linearly to 0 at 4.0 ms. Find the impulse, the ball’s final speed and the average force.

Model answer

Impulse = area of the triangle = ½(600)(0.0040) = 1.2 N·s. Δp = J, so v = 1.2 ÷ 0.40 = 3.0 m/s. Average force = 1.2 ÷ 0.0040 = 300 N, half the peak, as a triangle should give.

Task 3 (Unit 5). A uniform chest lid of mass M and width L is hinged along its back edge. You hold it open at angle θ above the horizontal by lifting straight up on its front edge. Derive the force you need and check it.

Model answer

Torques about the hinge. Lever arms are horizontal distances: L cos θ for your lift, (L ÷ 2) cos θ for the weight. So F L cos θ − Mg(L ÷ 2) cos θ = 0, and F = Mg ÷ 2, whatever the angle. Units: N. Check: θ = 0 gives a horizontal lid held at one end, which needs half its weight at each end.

Task 4 (Unit 6). A flywheel (I = 0.50 kg·m²) spins at 40 rad/s and drives a small generator that lights a 5.0 W lamp. Assume all of the flywheel’s kinetic energy reaches the lamp at that steady rate. Claim how the lighting time changes if the flywheel starts at 80 rad/s instead, justify, then derive the time and check your claim.

Model answer

The time becomes four times as long. The lamp uses the stored kinetic energy at the same rate either way. Rotational kinetic energy depends on the square of ω, so doubling ω stores four times the energy. Derivation: t = K ÷ P = Iω₀² ÷ (2P) = 0.50 × 40² ÷ (2 × 5.0) = 80 s. At 80 rad/s, t = 320 s, agreeing with the claim. Units: kg·m²/s² ÷ W = J ÷ (J/s) = s.

Where to practise next

• Representing Motion study guide: motion graphs.
• Forces and Free-Body Diagrams practice: diagrams.
• Conservation of Energy study guide: bar charts.
• Rotational Equilibrium study guide: torque derivations.
• Fluids and Conservation Laws study guide: Bernoulli arguments.
• Unit 2 review and Unit 3 review for mixed practice.