Game plan 8 · Unit 4
Motion in a line and forces
Diagram before algebra. Then match the acceleration form to what the answer wants.
Reach for it when
You see named forces (weight, normal reaction, an applied force, resistance), “smooth inclined plane”, “equilibrium” or “non-equilibrium”, concurrent forces, “momentum”, straight-line motion with air resistance, or $\ddot x=-\omega^2x$. If you aren’t sure this is the right game plan, start with Game plan 0, the Motion Map.
Part 1
Drawing the force diagram
Not examined
Forces here are always concurrent: they act at a single point. Pulleys and connected bodies are outside the syllabus, so a question will never ask you to link two masses over a pulley. There is no friction anywhere in the course.
Drawing the force diagram
- 1Draw a free-body diagram with exactly three kinds of force: the weight straight down, the normal reaction perpendicular to the plane, and every applied force named in the question. Nothing else.
- 2Choose axes parallel and perpendicular to the plane. Resolve the weight into $mg\sin\theta$ down the slope and $mg\cos\theta$ into it.
- 3Resolve any applied force acting at an angle to the plane into its parallel and perpendicular components.
- 4Apply equilibrium perpendicular to the plane, $\Sigma F_\perp=0$, to find $R$. $R$ is only $mg\cos\theta$ when nothing else acts perpendicular. An applied force with a perpendicular component changes it.
- 5Apply Newton’s second law parallel to the plane: $\Sigma F_\parallel=ma$, or $\Sigma F_\parallel=0$ in equilibrium.
- 6Choose the acceleration form (Game plan 8.2), solve, and state the direction and units in your conclusion.
Try it
Resolve the weight on a smooth plane
The plane is smooth, so nothing acts along the slope except the weight’s component. Changing the mass changes every force, but not the acceleration. This uses $g=9.8$.
Watch for
“Smooth” means frictionless. The surface contributes only the normal reaction.
In the Topic Guides
Part 2
Choosing the acceleration form
Choosing the acceleration form
- 1Ask the two questions from the Motion Map: what is $a$ written in terms of, and does the answer want time or displacement?
- 2$a$ in terms of $t$, or constant: use $a=\frac{dv}{dt}$ and integrate directly.
- 3$a$ in terms of $v$: use $\frac{dv}{dt}$ if the answer wants time, and $v\frac{dv}{dx}$ if it wants displacement.
- 4$a$ in terms of $x$: use $v\frac{dv}{dx}$. Reach for $\frac{d}{dx}\left(\frac12v^2\right)$ only when there is a reason, usually that $v^2$ is itself given in terms of $x$.
- 5Separate, integrate, and apply the conditions. Expect inverse trig from time-based integrals and a natural log from displacement-based ones.
Part 3
Vertical motion with air resistance
Vertical motion with air resistance
- 1Choose a positive direction and write Newton’s second law with the resistance term as given, proportional to $v$ or $v^2$. Its sign opposes the motion, so it flips between the upward and downward phases.
- 2Terminal velocity: set $a=0$ and solve. No integration is needed.
- 3For the full motion, choose the acceleration form (Game plan 8.2), separate and integrate.
- 4Check the behaviour: as $t\to\infty$, $v$ should approach the terminal value. If asked to show $v$ can never exceed it, finish with an inequality (Game plan 7.4).
Remember
Resistance forces oppose the current direction of motion, so the sign flips between the upward and downward phases of vertical motion.
In the Topic Guides
Part 4
Simple harmonic motion
Simple harmonic motion
- 1Recognise $\ddot x=-\omega^2x$, or an equation you can rearrange into it, and read off $\omega$.
- 2The solutions $x=A\sin(\omega t+\alpha)$ or $x=A\cos(\omega t+\beta)$, plus $v=\pm\omega\sqrt{A^2-x^2}$, $T=\frac{2\pi}{\omega}$ and $f=\frac1T$, are all on the sheet.
- 3Use the initial conditions, the position and velocity at $t=0$, to fix the amplitude and phase.
- 4For the speed at a position, use $v=\pm\omega\sqrt{A^2-x^2}$. The maximum speed is at the centre and the maximum acceleration is at the endpoints.
| At $P$ | At $O$ | At $Q$ | |
|---|---|---|---|
| Position | $x=-A$ | $x=0$ | $x=A$ |
| Velocity | $\dot x=0$ | $\dot x=\pm\omega A$ | $\dot x=0$ |
| Acceleration | $\ddot x=\omega^2A$ | $\ddot x=0$ | $\ddot x=-\omega^2A$ |
Speed is greatest at the centre and zero at the endpoints, and acceleration does the opposite. You can answer most SHM sub-questions by reading a position off this line.
In the Topic Guides
Part 5
Momentum
Momentum
- 1$P=mv$. Momentum is a vector, so its direction carries a sign.
- 2Change in momentum: $\Delta P=mv_2-mv_1$.
- 3For a constant force, $F=ma$ with constant $a$. Integrate directly for $v$ and $x$. For a non-constant force $F(t)$, use $a=\frac{F(t)}{m}$ and integrate.
In the Topic Guides
Formulas
Formula panel
Each formula is labelled with how to treat it in the exam.
QCAA language
What the question says, and what it means
| Question type | What you’ll see | What it means |
|---|---|---|
| Forces | “Smooth inclined plane”, “in equilibrium”, named forces | Game plan 8.1. Diagram before algebra |
| Variable acceleration | $a$ given in terms of $v$ or $x$ | Game plan 8.2. Match the form to what’s wanted |
| Air resistance | “Resistance proportional to”, “terminal velocity” | Game plan 8.3 |
| SHM | $\ddot x=-\omega^2x$, “oscillates”, period or amplitude | Game plan 8.4. Everything is on the sheet |
| Momentum | “Momentum”, a mass and a velocity | Game plan 8.5 |
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