Module 5: Advanced Mechanics — Worked Solutions (HSC Physics)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our HSC Physics course →
Worked examples for HSC Physics Module 5: Advanced 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.
In short: full step-by-step worked solutions for HSC Physics — Module 5: Advanced Mechanics. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
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.
Take $g = 9.8\ \text{m s}^{-2}$ and the universal gravitational constant $G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ unless stated otherwise.
Example 1 — Projectile motion
Question
A ball is launched from ground level at $25\ \text{m s}^{-1}$ at an angle of $30^\circ$ above the horizontal. Take $g = 9.8\ \text{m s}^{-2}$ and ignore air resistance. Calculate the maximum height reached by the ball and its total horizontal range.
Solution
Split the launch velocity into components first. $u_x = 25\cos 30^\circ = 21.65\ \text{m s}^{-1}$ and $u_y = 25\sin 30^\circ = 12.5\ \text{m s}^{-1}$.
Maximum height is where $v_y = 0$. Use $v_y^2 = u_y^2 - 2gH$:
$$H = \frac{u_y^2}{2g} = \frac{12.5^2}{2(9.8)} = 7.97\ \text{m}$$
For range, get the time of flight. By symmetry the ball returns to the ground when $u_y t = \tfrac{1}{2}gt^2$, so $t = \dfrac{2u_y}{g} = \dfrac{2(12.5)}{9.8} = 2.551\ \text{s}$.
$$R = u_x t = 21.65 \times 2.551 = 55.2\ \text{m}$$
Maximum height $\approx 7.97\ \text{m}$, range $\approx 55.2\ \text{m}$. Resolve the velocity before anything else — every projectile question collapses once you have the components.
The trick with projectiles is that the horizontal and vertical motions are completely independent, so we treat them separately. First resolve the launch velocity:
$u_x = 25\cos 30^\circ = 21.65\ \text{m s}^{-1}$ (constant), and $u_y = 25\sin 30^\circ = 12.5\ \text{m s}^{-1}$ (slowed by gravity).
At the very top of the flight the ball is momentarily not rising, so $v_y = 0$. Using $v_y^2 = u_y^2 - 2gH$ and setting $v_y = 0$:
$$H = \frac{u_y^2}{2g} = \frac{12.5^2}{2(9.8)} = 7.97\ \text{m}$$
For the range we need how long the ball is in the air. It goes up and comes back to the same height, so the time of flight is twice the time to the top: $t = \dfrac{2u_y}{g} = \dfrac{2(12.5)}{9.8} = 2.551\ \text{s}$.
Because horizontal velocity never changes, the range is simply $R = u_x t = 21.65 \times 2.551 = 55.2\ \text{m}$.
So the ball reaches about $7.97\ \text{m}$ high and lands $55.2\ \text{m}$ away. The reason this works is gravity only ever acts vertically, so the horizontal motion just coasts.
Resolve, then treat axes separately.
- $u_x = 25\cos 30^\circ = 21.65\ \text{m s}^{-1}$
- $u_y = 25\sin 30^\circ = 12.5\ \text{m s}^{-1}$
Max height ($v_y = 0$):
- $H = \dfrac{u_y^2}{2g} = \dfrac{12.5^2}{19.6} = 7.97\ \text{m}$
Range:
- $t = \dfrac{2u_y}{g} = \dfrac{25}{9.8} = 2.551\ \text{s}$
- $R = u_x t = 21.65 \times 2.551 = 55.2\ \text{m}$
$H \approx 7.97\ \text{m}$, $R \approx 55.2\ \text{m}$.
Where the marks go
- 1 mark: Resolves launch velocity into $u_x = 21.65\ \text{m s}^{-1}$ and $u_y = 12.5\ \text{m s}^{-1}$
- 1 mark: Correct maximum height $H = 7.97\ \text{m}$
- 1 mark: Correct time of flight $t = 2.55\ \text{s}$
- 1 mark: Correct range $R = 55.2\ \text{m}$ with units
Key idea
Horizontal and vertical motion are independent: vertical uses $v_y^2 = u_y^2 - 2gH$ for height, and horizontal range is $R = u_x t$ where $t$ comes from the vertical motion.
Example 2 — Circular motion and orbits
Question
A satellite orbits the Earth in a circular orbit of radius $7.0 \times 10^6\ \text{m}$ from the Earth's centre. The mass of the Earth is $5.97 \times 10^{24}\ \text{kg}$ and $G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$. Calculate the orbital speed of the satellite and its period.
Solution
Gravity supplies the centripetal force, so set them equal: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$. The satellite mass cancels.
$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{7.0\times10^6}}$$
$$v = \sqrt{5.689\times10^{7}} = 7543\ \text{m s}^{-1} \approx 7.54\times10^3\ \text{m s}^{-1}$$
Period is the circumference over the speed:
$$T = \frac{2\pi r}{v} = \frac{2\pi (7.0\times10^6)}{7543} = 5831\ \text{s} \approx 5.83\times10^3\ \text{s}$$
That's about 97 minutes. Always cancel the satellite mass straight away — it never matters for orbital speed.
An orbiting satellite is in free fall — the only force on it is gravity, and that force is exactly what keeps it turning in a circle. So we equate gravitational force with the required centripetal force:
$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$
The satellite's own mass $m$ appears on both sides and cancels, which is why all satellites at a given radius orbit at the same speed. Rearranging gives:
$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{7.0\times10^6}} = 7543\ \text{m s}^{-1}$$
To find the period, remember the satellite travels one full circumference $2\pi r$ in one period at this steady speed:
$$T = \frac{2\pi r}{v} = \frac{2\pi(7.0\times10^6)}{7543} = 5.83\times10^3\ \text{s}$$
So it moves at roughly $7.54\ \text{km s}^{-1}$ and circles the Earth in about $5.83\times10^3\ \text{s}$ (just over an hour and a half). The physics insight is that gravity is the centripetal force here — nothing else.
Gravity = centripetal force; $m$ cancels.
- $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow v = \sqrt{\dfrac{GM}{r}}$
- $v = \sqrt{\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{7.0\times10^6}} = 7543\ \text{m s}^{-1}$
- $v \approx 7.54\times10^3\ \text{m s}^{-1}$
Period:
- $T = \dfrac{2\pi r}{v} = \dfrac{2\pi(7.0\times10^6)}{7543} = 5.83\times10^3\ \text{s}$
Where the marks go
- 1 mark: Equates gravitational and centripetal force and cancels satellite mass
- 1 mark: Correct orbital speed $v = 7.54\times10^3\ \text{m s}^{-1}$
- 1 mark: Uses $T = \dfrac{2\pi r}{v}$ correctly
- 1 mark: Correct period $T = 5.83\times10^3\ \text{s}$ with units
Key idea
For a circular orbit, gravity provides the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ gives $v = \sqrt{\dfrac{GM}{r}}$, and the period follows from $T = \dfrac{2\pi r}{v}$.
Frequently asked questions
Step-by-step solutions to exam-style questions on Advanced Mechanics in HSC Physics, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Physics syllabus for Advanced Mechanics, using the methods and notation expected in the exam.