Module 16 Synthesis & Exam Technique
Every Module 16 mechanics problem follows the same three steps: recognise the form of acceleration (a function of $t$, of $x$, or of $v$); set up the differential equation using $a = dv/dt$, $a = v\,dv/dx$, or $a = d^2x/dt^2$; and solve with the calculus toolkit from Module 15. This final lesson distils that workflow, gives you an exam-day decision tree, and runs three closing examples that span the breadth of M16 — SHM, resisted motion, and circular motion.
The three identities at the heart of Module 16: $a = \dfrac{dv}{dt}$, $a = v\dfrac{dv}{dx}$, $a = \dfrac{d^2x}{dt^2}$. Before checking — for each form of $a$ below, decide which identity gives a separable DE that you would actually try first: (i) $a = -g$; (ii) $a = -\omega^2 x$; (iii) $a = -kv$; (iv) $a = -kv^2$ and we want $v$ as a function of $x$.
Every Module 16 question can be unlocked by the same disciplined sequence: recognise the form of $a$ (function of $t$, $x$, or $v$?), choose the matching identity for the differential equation, then solve carefully and apply initial conditions. Half the marks in a long mechanics question come from explicitly stating which form of $a$ is being used and writing the DE before integrating.
The recognise-set up-solve workflow: (1) read the problem and state $a$ as a function of $t$, $x$, or $v$; (2) write the DE — $\frac{dv}{dt}$ for "what is $v(t)$?", $v\frac{dv}{dx}$ for "what is $v(x)$?", $\frac{d^2x}{dt^2}$ for SHM/equilibrium analyses; (3) separate, integrate, apply ICs, check units and limits.
$a(t)$: integrate w.r.t. $t$ · $a(x)$: use $v\frac{dv}{dx}$ · $a(v)$: choose $\frac{dv}{dt}$ for $v(t)$ or $v\frac{dv}{dx}$ for $v(x)$
Key facts
- Three identities: $a = dv/dt$, $a = v\,dv/dx$, $a = d^2x/dt^2$
- SHM: $\ddot x = -\omega^2 (x - c)$ ⇒ $x = c + A\cos(\omega t + \phi)$, period $2\pi/\omega$
- Resisted motion: $m\ddot x = -mg - mkv^n$ (upward) or $m\ddot x = mg - mkv^n$ (downward)
- Circular motion (uniform): centripetal $a = v^2/r$ towards centre
Concepts
- Why the form of $a$ dictates which DE to write
- Why SHM emerges whenever the restoring force is linear in displacement
- Why terminal velocity is found by setting $a = 0$ in the resisted-motion DE
Skills
- Diagnose $a$ as a function of $t$, $x$, or $v$ and pick the correct DE
- Solve SHM, resisted motion, and circular problems within a 15-minute exam slot
- Apply initial conditions cleanly and sanity-check answers at limits
Almost every Module 16 problem can be tackled in three numbered steps.
- Recognise. Read the problem and write $a$ explicitly. Is it a function of $t$ (e.g., $a = -g$, $a = \sin t$), of $x$ (e.g., $a = -\omega^2 x$), or of $v$ (e.g., $a = -kv$, $a = -g - kv^2$)?
- Set up the DE. Pick the identity that gives the simplest separable equation:
- $a$ depends on $t$: integrate $a = \dfrac{dv}{dt}$ directly.
- $a$ depends on $x$: use $a = v\dfrac{dv}{dx}$, separate $v\,dv = a\,dx$.
- $a$ depends on $v$ and you want $v(t)$: use $\dfrac{dv}{dt} = a(v)$, separate $\dfrac{dv}{a(v)} = dt$.
- $a$ depends on $v$ and you want $v(x)$: use $v\dfrac{dv}{dx} = a(v)$, separate $\dfrac{v\,dv}{a(v)} = dx$.
- Solve and apply ICs. Integrate. State the IC in words. Substitute to find the constant. Simplify. Check: does the answer behave sensibly as $t \to \infty$ or at endpoints?
Step 1: $a$ as function of $t$, $x$, or $v$? · Step 2: pick the DE identity matching the form of $a$ and the target variable · Step 3: separate, integrate, IC, simplify, sanity-check · The four mark-gates: form of $a$ stated · DE written · IC stated · clean integration
Pause — copy the M16 three-step workflow (recognise form of $a$, write and separate DE, IC + simplify), the four mark-gates, and the sanity-check instruction into your book.
Quick check: A particle has $a = -kv$ (with $k > 0$) and starts from $x = 0$ with $v = v_0$. You want to find $v$ as a function of $x$. Which DE form gives the cleanest separation?
We just saw the M16 workflow: identify what $a$ depends on, select the matching DE identity, separate and integrate, apply ICs, and sanity-check — with four mark-gates (form of $a$, DE written, IC stated, clean integration). That raises a question: what does the HSC marker look for in an exam response specifically? This card answers it → always state $a$ first, write the chosen DE form ($dv/dt$ or $v\,dv/dx$), state the IC in words before substituting, find terminal velocity by setting $a = 0$, and close with a one-line sanity check.
Long mechanics questions (typically 5–8 marks) in HSC Extension 2 reward structure as much as algebra. A reliable layout:
- Line 1 — state $a$. "$a = $ \ldots, a function of $t$/$x$/$v$".
- Line 2 — choose the DE. "Since $a$ depends on $x$, use $a = v\,dv/dx$."
- Lines 3-5 — separate & integrate. Show the separated form and the integrated form before substituting ICs.
- Line 6 — apply ICs. "When $t = 0$, $x = x_0$ and $v = v_0$ ⇒ $C = \ldots$".
- Line 7 — answer. Substitute back; state the final relation clearly.
- Line 8 — sanity check. One sentence: "As $t \to \infty$, $v \to v_T$, as expected."
For circular motion, replace step 2 by "Newton II radially: $F_{\text{net}} = mv^2/r$" and step 3 by an FBD or component equation.
Always state $a$ before choosing the DE · $v(t)$ ⇒ $a = dv/dt$; $v(x)$ ⇒ $a = v\,dv/dx$ · Write IC in words before substituting · Terminal velocity: set $a = 0$ and solve for $v$ · One-line sanity check at the end of every long question
Pause — copy the five HSC habits for Module 16: state $a$ before choosing DE; $v(t) \Rightarrow dv/dt$, $v(x) \Rightarrow v\,dv/dx$; state IC in words first; terminal velocity via $a=0$; one-line sanity check at the end into your book.
Did you get this? True or false: to find the terminal velocity of a body falling under gravity with resistance $kv$, you set $\ddot x = 0$ in the equation of motion and solve for $v$.
Worked examples · 3 in a row, reveal as you go
A particle moves so that $\ddot x = -4(x - 3)$. At $t = 0$ it is at $x = 7$ with $\dot x = 0$. Find $x$ as a function of $t$, state the amplitude and period, and find the speed when $x = 3$.
A body of unit mass falls from rest under gravity and a resistive force $kv$ per unit mass ($k > 0$). So $\ddot x = g - kv$ (taking downward positive). Find $v$ as a function of $t$, and find the terminal velocity.
A car of mass $m$ rounds a banked frictionless curve of radius $r$ at the designed speed $v$. The bank angle is $\theta$. Derive $\tan\theta = v^2/(rg)$, then find the normal reaction $N$ from the road in terms of $m$, $g$, and $\theta$.
Fill the gap: A body of unit mass has $\ddot x = g - kv$. Its terminal velocity is $v_T =$ , found by setting in the equation of motion.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for SHM of the form $\ddot x = -\omega^2 x$, the speed at the centre is $A\omega$, where $A$ is the amplitude.
Activities · practice with the ideas
A particle satisfies $\ddot x = -9(x - 2)$ with $x(0) = 5$ and $\dot x(0) = 0$. Write $x(t)$, state the period and amplitude, and the maximum speed.
A particle has $a = -kv$ (with $k > 0$) and at $t = 0$ has $v = v_0$. Find $v$ as a function of $t$ and as a function of $x$ (taking $x = 0$ at $t = 0$).
A body of unit mass falls from rest with $\ddot x = g - kv^2$. Find $v$ as a function of $x$ (target: $v(x)$, not $v(t)$).
A particle moves in a horizontal circle of radius $r = 0.4$ m on a string of length $\ell$, the string making $40^\circ$ with the vertical. Find $\ell$ and the speed.
Diagnose, for each form of $a$, which DE identity you would use and write the separated form: (a) $a = -kx$; (b) $a = -kv$ for $v(t)$; (c) $a = -kv^2$ for $v(x)$; (d) $a = g - kv$ for $v(t)$.
Odd one out: Three of these are sensible first moves for a Module 16 mechanics question. Which one is NOT?
Earlier you diagnosed several forms of $a$ and chose the matching DE identity.
The whole of Module 16 is built on the three identities $a = dv/dt = v\,dv/dx = d^2x/dt^2$, applied with discipline: state the form of $a$, choose the right DE, separate, integrate, apply ICs, check. Whether the problem is SHM, resisted motion, or circular motion, the workflow is the same — only the specific algebra changes. That habit is what carries marks in the HSC.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A particle moves so that $\ddot x = -16(x - 1)$. State the centre of motion, the angular frequency $\omega$, and the period. (2 marks)
Q2. A body of unit mass moves under $\ddot x = -g - kv^2$ (rising; $k > 0$) and is projected upward from $x = 0$ with speed $u$. Find $v$ as a function of $x$ (i.e., on the way up). (3 marks)
Q3. A particle of mass $m$ on a light string of length $\ell$ moves as a conical pendulum at angle $\theta$ to the vertical with period $T$. Show that $T = 2\pi\sqrt{\ell\cos\theta/g}$ and explain in one sentence why this is independent of $m$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\omega = 3$, centre $c = 2$. With $x(0) = 5$, $\dot x(0) = 0$: $A = 3$, $\phi = 0$. $x(t) = 2 + 3\cos(3t)$. Period $2\pi/3$. Amplitude $3$. Max speed $A\omega = 9$.
2. $v(t)$: $dv/dt = -kv \Rightarrow \ln(v/v_0) = -kt \Rightarrow v = v_0 e^{-kt}$. $v(x)$: $v\,dv/dx = -kv \Rightarrow dv = -k\,dx \Rightarrow v = v_0 - kx$ (valid until $v = 0$).
3. $v\,dv/dx = g - kv^2$. Let $u = g - kv^2$, $du = -2kv\,dv$. So $-du/(2k) = u\,dx/v$ — easier: $v\,dv/(g - kv^2) = dx$, integrate: $-(1/(2k)) \ln|g - kv^2| = x + C$. IC $v = 0$ at $x = 0$: $C = -(1/(2k))\ln g$. Result: $g - kv^2 = g e^{-2kx}$, i.e., $v^2 = (g/k)(1 - e^{-2kx})$.
4. $\ell = r/\sin 40^\circ = 0.4/\sin 40^\circ \approx 0.622$ m. $v^2 = rg\tan 40^\circ = 0.4(9.8)(0.839) \approx 3.29$, so $v \approx 1.81$ m s$^{-1}$.
5. (a) $v\,dv = -kx\,dx$. (b) $dv/v = -k\,dt$. (c) $dv/v = -k\,dx$. (d) $dv/(g - kv) = dt$.
Q1 (2 marks): Centre $x = 1$; $\omega = 4$ [1]; period $T = 2\pi/4 = \pi/2$ [1].
Q2 (3 marks): $a$ depends on $v$, want $v(x)$, so use $v\,dv/dx = -(g + kv^2)$ [1]. Separate and integrate: $(1/(2k))\ln(g + kv^2) = -x + C$ [1]. IC $v = u$ at $x = 0$: $v^2 = \big((g + ku^2)e^{-2kx} - g\big)/k$ [1].
Q3 (3 marks): Vertical $T_s\cos\theta = mg$ [1]; radial $T_s\sin\theta = m\omega^2\ell\sin\theta$, hence $\omega^2 = g/(\ell\cos\theta)$ [1]; period $= 2\pi/\omega = 2\pi\sqrt{\ell\cos\theta/g}$; mass cancels because both force equations are proportional to $m$ [1].
Six timed questions covering SHM, resisted motion, and circular motion — the breadth of Module 16. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick mechanics questions. Lighter alternative to the boss.
End of course — mark complete
This lesson completes Module 16 and the entire Mathematics Extension 2 course. From induction and complex numbers through proof, integration, vectors, and mechanics — you now have the full Extension 2 toolkit. Tick the box when you've finished practice and review, and take a moment to look back at how far you've come.