Worked Solutions
Mechanics — Worked Solutions (HSC Maths Extension 2)
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Worked examples for HSC Maths Extension 2 mechanics. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
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Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
In mechanics, start by writing the equation of motion clearly, then choose the right form of acceleration — $\ddot{x}$, $v\dfrac{dv}{dx}$ or $\dfrac{dv}{dt}$ — for the variables in the question.
Example 1 — Simple harmonic motion
Question
A particle moves in simple harmonic motion so that $\ddot{x} = -9x$, where $x$ is the displacement in metres from the origin and $t$ is in seconds. The amplitude of the motion is $5$ m. Find the period of the motion and the maximum speed of the particle.
Solution
Compare $\ddot{x} = -9x$ with the SHM standard form $\ddot{x} = -n^2 x$. So $n^2 = 9$, giving $n = 3$.
Period: $T = \dfrac{2\pi}{n} = \dfrac{2\pi}{3}$ seconds.
Maximum speed in SHM occurs at the centre and equals $na$, where $a$ is the amplitude: $v_{\max} = na = 3 \times 5 = 15$ m/s.
Memorise $\ddot{x} = -n^2 x$, $T = \tfrac{2\pi}{n}$, and $v_{\max} = na$ — they fall straight out once you read off $n$.
The equation $\ddot{x} = -9x$ has the tell-tale shape of simple harmonic motion: acceleration proportional to displacement and pointing back toward the centre. The standard form is $\ddot{x} = -n^2 x$, so matching the coefficients, $n^2 = 9$ and therefore $n = 3$ (we take the positive value — $n$ is an angular frequency).
The period is the time for one full oscillation, $T = \dfrac{2\pi}{n} = \dfrac{2\pi}{3}$ seconds.
For the maximum speed, think about where the particle is moving fastest — at the centre, $x = 0$, where all the energy is kinetic. The standard result is $v_{\max} = na$ with $a$ the amplitude. So $v_{\max} = 3 \times 5 = 15$ m/s.
If you ever forget $v_{\max} = na$, it comes from $v^2 = n^2(a^2 - x^2)$ evaluated at $x = 0$, which gives $v^2 = n^2 a^2$.
Match to standard SHM form $\ddot{x} = -n^2 x$.
- $n^2 = 9 \Rightarrow n = 3$
- Period: $T = \dfrac{2\pi}{n} = \dfrac{2\pi}{3}$ s
- $v_{\max} = na = 3 \times 5 = 15$ m/s
Where the marks go
- 1 mark: Identifies $n^2 = 9$, so $n = 3$
- 1 mark: Correct period $T = \frac{2\pi}{3}$ s
- 1 mark: Uses $v_{\max} = na$ at the centre
- 1 mark: Correct maximum speed $15$ m/s
Key idea
For $\ddot{x} = -n^2 x$ the motion is SHM with period $T = \dfrac{2\pi}{n}$ and maximum speed $na$ at the centre, from $v^2 = n^2(a^2 - x^2)$.
Example 2 — Resisted motion (terminal velocity)
Question
A particle of mass $2$ kg falls vertically from rest. It experiences gravity and a resistive force of magnitude $0.5v$ newtons, where $v$ is the speed in m/s. Taking $g = 10$ m/s$^2$ and downward as positive, write the equation of motion and find the terminal velocity.
Solution
Newton's second law downward: $m\ddot{x} = mg - 0.5v$. With $m = 2$ and $g = 10$:
$2\ddot{x} = 20 - 0.5v$, i.e. $\ddot{x} = 10 - 0.25v$.
Terminal velocity is reached when acceleration is zero — the particle stops speeding up. Set $\ddot{x} = 0$:
$0 = 10 - 0.25v \Rightarrow v = 40$ m/s.
Terminal velocity is always just "acceleration = 0" — don't try to solve the differential equation if all they want is the limiting speed.
Let's build the equation of motion with Newton's second law, $m\ddot{x} = F_{\text{net}}$. Two forces act: gravity pulling down, $mg$, and resistance opposing the motion (so upward as the particle falls), $0.5v$. Taking down as positive, $m\ddot{x} = mg - 0.5v$.
Substituting $m = 2$ and $g = 10$: $2\ddot{x} = 20 - 0.5v$, and dividing by 2, $\ddot{x} = 10 - 0.25v$.
Now, what is terminal velocity? As the particle speeds up, the resistive force grows until it balances gravity; at that point there's no net force, so no acceleration, and the speed levels off. That means we set $\ddot{x} = 0$:
$0 = 10 - 0.25v$, which gives $v = \dfrac{10}{0.25} = 40$ m/s. It makes sense that this is a fixed limiting speed — beyond it, resistance would exceed gravity and slow the particle back down.
Equation of motion from $m\ddot{x} = mg - 0.5v$.
- $2\ddot{x} = 20 - 0.5v \Rightarrow \ddot{x} = 10 - 0.25v$
- Terminal velocity: $\ddot{x} = 0$
- $0 = 10 - 0.25v \Rightarrow v = 40$ m/s
Where the marks go
- 1 mark: Correct equation of motion $2\ddot{x} = 20 - 0.5v$ (or equivalent)
- 1 mark: Sets acceleration to zero for terminal velocity
- 1 mark: Correct terminal velocity $v = 40$ m/s
Key idea
Resisted motion uses $m\ddot{x} = mg - (\text{resistance})$; terminal velocity is the speed at which acceleration is zero, found by setting $\ddot{x} = 0$.