Problem-solving¶

Syllabus mapping

Working Scientifically outcome — Problem-solving. Covers applying mathematical and conceptual relationships systematically, including in unfamiliar or multi-step contexts, and selecting the appropriate relationship from the data sheet rather than a memorised list of "the formula for this topic."

Exact NESA outcome code TODO — confirm against the official syllabus PDF (Resources) before treating any wording on this page as verbatim NESA text.

You need to know: a reliable process for turning a word problem into a solved answer, and — the harder skill — how to handle a problem that needs two relationships chained together, not just one you can look up directly.

Core¶

A reliable process¶

Every numeric physics problem in this course can be worked through the same way:

List what's given, with units, and what's being asked for.
Identify the relevant relationship(s) — from the HSC data sheet, not memory. NESA expects you to read from it, not recite it (see the syllabus notes referenced in project.md Section 0).
Rearrange algebraically first, before substituting numbers — this catches mistakes far more easily than rearranging with numbers already in place.
Substitute with units, and calculate.
Check the answer: right order of magnitude, right unit, right number of significant figures (see Precision and Uncertainty).

Step 5 is the one students skip under time pressure, and it's the one that catches a genuinely large fraction of careless errors — an answer of \(4500\ \text{m s}^{-1}\) for a trolley's speed should immediately look wrong before you've even checked the working.

Reading the data sheet, not memorising a topic list¶

The exam data sheet groups formulae by relationship, not by "topic," and doesn't tell you which one applies to your specific problem — that's a skill NESA is deliberately testing. Practising with the actual data sheet (not a topic-labelled formula list a textbook gives you) is part of building this skill; a formula you can only recognise when it's under a heading that says "Motion" is a formula you don't actually know yet.

Multi-step problems: chaining relationships¶

The single biggest jump between a "Core" problem and a genuinely Advanced one is whether it needs one relationship or several, chained so the result of one calculation becomes an input to the next.

Worked example — a chained problem

A 2.0 kg block is released from rest at the top of a frictionless incline, 30° from the horizontal, and slides 4.0 m down the slope before reaching the bottom. Find its speed at the bottom.

Step 1 — find the acceleration (see Motion on Inclined Planes):

\[ a = g\sin\theta = 9.8 \times \sin30^\circ = 4.9\ \text{m s}^{-2} \]

Step 2 — use that acceleration in a SUVAT equation (see Motion Relationships) to find speed, with \(u=0\), \(s=4.0\) m:

\[ v^2 = u^2 + 2as = 0 + 2(4.9)(4.0) = 39.2 \]

\[ v = \sqrt{39.2} \approx 6.3\ \text{m s}^{-1} \]

Neither step is hard on its own — Motion on Inclined Planes and Motion Relationships each cover one piece individually. The actual skill being tested is recognising that the incline gives you an acceleration, and that acceleration then plugs straight into a completely different, already-familiar relationship — nothing about the problem told you to do this explicitly.

Sanity-checking with units¶

Checking that units cancel correctly through a rearrangement is a genuinely fast way to catch an algebra mistake, especially under exam time pressure. If a rearrangement for a speed produces units of \(\text{m s}\) instead of \(\text{m s}^{-1}\), something was inverted — worth checking before finishing the calculation, not after.

Advanced

Recognising which relationship applies when the problem doesn't say. The hardest version of a multi-step problem doesn't name the topic at all — it gives you a physical scenario and expects you to figure out which relationships are relevant, sometimes from more than one focus area at once (e.g. a charged object's motion involving both an electric force calculation and SUVAT). There's no shortcut here beyond genuinely understanding what each formula represents physically, not just its symbol pattern — a formula memorised only as "the third one on the data sheet" can't be recognised in an unfamiliar wrapper, but one understood as "force causes acceleration" can be spotted in almost any disguise.

Extension

Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.

📎 Depth study idea

Not every physical situation has a clean, closed-form (SUVAT-style) solution — a falling object with significant air resistance, for instance, doesn't accelerate at a constant rate, so none of the standard equations apply directly. Professional and university physics handles this with numerical methods: breaking time into many small steps, and repeatedly recalculating velocity and position step-by-step (e.g. Euler's method, easily built in a spreadsheet: at each tiny time step, use the current velocity to update position, then use the current forces to update velocity, and repeat). Building a simple step-by-step spreadsheet model of a scenario the syllabus's own equations can't solve directly is a strong, genuinely technical depth study angle — and a real first taste of how physics gets modelled computationally once you're past the equations that have a tidy exact answer.

Video/visual resources¶

🎥 Khan Academy — TODO: source a multi-step problem-solving / worked-examples explainer. Nice-to-have — a genuinely generic "problem-solving strategy" video is less valuable here than the page's own worked multi-step incline+SUVAT example; more useful would be linking a couple of specific multi-step physics worked examples instead of a single generic-strategy video.
🎥 Physics High — TODO: likely to have specific multi-step worked-example videos even if not framed as "problem-solving skill" generically — worth checking their back catalogue for chained/combined-topic problems specifically.

Check yourself¶

A ball is thrown horizontally at \(8.0\ \text{m s}^{-1}\) from a cliff \(20\) m high. Outline (don't fully solve) the sequence of relationships you'd use to find how far from the base of the cliff it lands, and explain why this needs more than one step.

Answer

First, use \(s = \tfrac{1}{2}gt^2\) (vertical motion, starting from rest vertically) to find the time to fall 20 m. Then use that time in \(s = vt\) (horizontal motion, constant velocity, since there's no horizontal acceleration) to find the horizontal distance travelled. It needs more than one step because the vertical and horizontal motions are independent, but they share the same time — the time found from the vertical calculation is the missing piece needed to solve the horizontal one.
A student calculates a car's acceleration and gets an answer of \(250\ \text{m s}^{-2}\). Explain why this result should immediately be treated as suspicious, without redoing the calculation.

Answer

\(250\ \text{m s}^{-2}\) is roughly 25 times the acceleration due to gravity — far beyond what any ordinary car is capable of. The order of magnitude alone is a strong sign of an arithmetic or rearrangement error (a common cause: a value that should have been squared wasn't, or a division was inverted), worth checking before trusting the number.
Rearrange \(F = \dfrac{kq_1q_2}{r^2}\) (Coulomb's Law, see Electrostatics) to make \(r\) the subject, and check that the units work out to metres.

Answer

\[ r = \sqrt{\frac{kq_1q_2}{F}} \]

Units: \(k\) has units \(\text{N m}^2\text{C}^{-2}\), \(q_1q_2\) has units \(\text{C}^2\), and \(F\) has units \(\text{N}\), so inside the square root: \(\dfrac{\text{N m}^2\text{C}^{-2} \times \text{C}^2}{\text{N}} = \text{m}^2\), and taking the square root gives metres — confirming the rearrangement is dimensionally consistent.