Dynamics Synthesis — Module Capstone
On 12 April 1981, Space Shuttle Columbia (STS-1) re-entered Earth's atmosphere from 27,875 km/h, decelerating to 343 km/h at touchdown — shedding $\Delta KE = \frac{1}{2}(68{,}585)(7{,}716^2 - 95^2) = 2.97\text{ TJ}$ of kinetic energy via atmospheric drag. NASA engineers used all three phases of Module 2 dynamics simultaneously: Phase 1 (drag force and deceleration), Phase 2 (energy dissipation over re-entry), and Phase 3 (impulse and momentum change). This is precisely the multi-phase integration this capstone lesson develops.
Before reading — in one sentence, what is the big difference between a perfectly inelastic collision and an elastic collision?
A crash reconstruction requires finding: the braking friction force, the kinetic energy at impact, and the post-collision velocity. Which phase sequence covers all three?
Core Content
A 1000 kg car brakes to rest from 80 km/h on a rough road. You are asked: what braking force acted, how much kinetic energy was lost, what was the impulse, and how long did the stop take? Each question pulls from a different phase of Module 2. You cannot answer all four with a single formula — you need to recognise which phase each sub-question belongs to and then chain the results together.
Sixteen formulae across three phases. Every HSC Dynamics exam question draws from this set.
| Phase 1 — Forces | ||
|---|---|---|
| $F_{net} = ma$ | Net force from mass and acceleration | Any dynamics problem |
| $W = mg$ | Weight force | All incline/normal problems |
| $N = mg\cos\theta$ | Normal force on incline | Inclined plane |
| $f = \mu N$ | Friction from coefficient and normal | When $\mu$ given |
| $a = g\sin\theta - \mu g\cos\theta$ | Net acceleration on rough incline | Inclined plane synthesis |
| Phase 2 — Energy | ||
| $W = Fs\cos\theta$ | Work done by a force | Force at angle to displacement |
| $W_{net} = \Delta KE$ | Work-energy theorem | Bridge between work and velocity |
| $KE = \frac{1}{2}mv^2$ | Kinetic energy | Any moving object |
| $\Delta U = mg\Delta h$ | Change in gravitational PE | Height change problems |
| $E_{mech} = KE + U$ | Mechanical energy | Conservation check |
| $P = W/t = Fv$ | Power from work/time or force×velocity | Engine or motor problems |
| Phase 3 — Momentum | ||
| $p = mv$ | Momentum | First step of every collision |
| $J = F\Delta t = \Delta p$ | Impulse-momentum theorem | Force×time or change in momentum |
| $F = \Delta p/\Delta t$ | Average force from impulse | Bounce/impact force |
| $\Sigma p_{before} = \Sigma p_{after}$ | Conservation of momentum | All closed-system collisions |
| $\Sigma KE_{before} = \Sigma KE_{after}$ | Elastic collision test | Only when elastic is possible |
Velocity is the common thread linking all three phases: $a \to v$ (Phase 1 kinematics), $v \to KE = \frac{1}{2}mv^2$ (Phase 2 energy), $v \to p = mv$ (Phase 3 momentum); always identify given values and the target quantity first, then solve one phase at a time and carry the result forward.
Pause — copy the highlighted synthesis insight into your book before moving on.
Complete: Work-energy theorem states that the net work done on an object equals its change in ___. This links Phase 2 to Phase 1 via $F_{net} \times s = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2$, and the final ___ can be fed into Phase 3 as $p = mv$.
We just saw the complete Module 2 formula set with velocity as the linking variable across all three phases. That raises a question: how do you put all three phases together in a single multi-part exam question without losing track? This card answers it → a four-step decode strategy applied to a full capstone problem spanning kinematics, energy, and momentum.
A 1200 kg sedan travelling at 25 m/s brakes to rest over 50 m (Phase 1+2), then collides with a stationary 2000 kg truck (Phase 3). After sticking together, contact lasted 0.15 s. Find: (a) friction force during braking, (b) KE at start of braking, (c) post-collision velocity, (d) impulse on the sedan, (e) average force on the sedan.
Four-step decode: (1) list all given values and what is asked; (2) classify each asked quantity by phase — force/acceleration (Ph1), energy/work/power (Ph2), momentum/impulse (Ph3); (3) identify the shared velocity linking the phases; (4) solve the earliest phase first and carry the result forward.
Pause — copy the highlighted four-step decode strategy into your book before moving on.
A ski jumper (80 kg) descends a 30 m high ramp (frictionless). At the bottom she launches horizontally. Which formulae chain gives her launch speed?
Activities
For each scenario, identify the correct two- or three-phase formula chain. Do not solve — just identify the formulae and in what order:
- A box on a rough incline slides down 8 m. Find its speed at the bottom and its final momentum.
- A motor drives a car at constant 30 m/s against 600 N of friction. Find the motor's power output.
- A 2 kg ball dropped from 5 m hits the floor and bounces back to 4 m. Find the impulse and average force (contact time 12 ms).
A 900 kg car accelerates from rest to 20 m/s in 6 s along a 400 m horizontal road. The driving force is 3600 N; rolling friction is 600 N.
- Find net force and verify the acceleration using $F_{net} = ma$.
- Calculate the work done by the engine and by friction over 400 m.
- Show that the net work equals the change in kinetic energy.
- Find the engine power at the moment the car reaches 20 m/s (use $P = Fv$ with driving force).
- Calculate the car's momentum at 20 m/s and the impulse delivered by the engine's net force.
ApplyBand 4(3 marks) 1. A 500 kg crate slides down a frictionless 15 m ramp inclined at 30°. Find: (a) the gravitational PE lost, (b) the speed at the bottom, (c) the crate's momentum at the bottom.
AnalyseBand 5(4 marks) 2. Two trolleys collide and stick: Trolley A (3 kg, 6 m/s east) and Trolley B (2 kg, 4 m/s west). (a) Find the post-collision velocity. (b) Calculate KE lost. (c) If the collision took 0.08 s, find the average force on Trolley A. (d) Explain what happened to the lost KE.
EvaluateBand 6(5 marks) 3. A 1000 kg car decelerates from 30 m/s to rest over 120 m on a flat road. (a) Find the net braking force (using the work-energy theorem). (b) Hence find the stopping time using impulse-momentum. (c) During a separate test the car hits a 1500 kg stationary wall section (the section breaks free). If the car rebounds at 5 m/s, find the wall section's velocity after impact. (d) Is this collision elastic? Justify with calculations.
Show all answers
Short Answer — Model Answers
Q1 (3 marks): (a) Height: $h = 15\sin30° = 7.5\text{ m}$. $\Delta U = mgh = 500 \times 10 \times 7.5 = 37\,500\text{ J}$. (b) $KE = \Delta U = 37\,500\text{ J}$. $v = \sqrt{2 \times 37\,500/500} = \sqrt{150} \approx 12.2\text{ m/s}$. (c) $p = 500 \times 12.2 = 6100\text{ kg m/s}$ down the ramp.
Q2 (4 marks): (a) Define east as positive. $\Sigma p = 3 \times 6 + 2 \times (-4) = 18 - 8 = 10\text{ kg m/s}$. $v_f = 10/(3+2) = 2\text{ m/s east}$. (b) $KE_{before} = \frac{1}{2}(3)(6^2) + \frac{1}{2}(2)(4^2) = 54 + 16 = 70\text{ J}$. $KE_{after} = \frac{1}{2}(5)(2^2) = 10\text{ J}$. $KE_{lost} = 70 - 10 = 60\text{ J}$. (c) $\Delta p_A = 3(2-6) = 3 \times (-4) = -12\text{ kg m/s}$. $F = -12/0.08 = -150\text{ N}$ (westward force on A). Magnitude = 150 N. (d) The 60 J of kinetic energy was converted to thermal energy (heat) from the deformation of the trolleys and internal vibration/sound during the perfectly inelastic collision. Energy is conserved overall but is no longer in mechanical form.
Q3 (5 marks): (a) $\Delta KE = 0 - \frac{1}{2}(1000)(30^2) = -450\,000\text{ J}$. $F_{net} \times s = \Delta KE \Rightarrow F = 450\,000/120 = 3750\text{ N}$ (braking force). (b) $\Delta p = 0 - 1000 \times 30 = -30\,000\text{ kg m/s}$. $t = |\Delta p|/F = 30\,000/3750 = 8\text{ s}$. (c) Define original direction as positive. $\Sigma p_{before} = 1000 \times 30 + 0 = 30\,000\text{ kg m/s}$. $\Sigma p_{after} = 1000 \times (-5) + 1500 \times v_{wall}$. $-5000 + 1500v_{wall} = 30\,000 \Rightarrow v_{wall} = 35\,000/1500 \approx 23.3\text{ m/s}$ (forward). (d) $KE_{before} = \frac{1}{2}(1000)(30^2) = 450\,000\text{ J}$. $KE_{after} = \frac{1}{2}(1000)(5^2) + \frac{1}{2}(1500)(23.3^2) = 12\,500 + 406\,967 \approx 419\,467\text{ J}$. Since $KE_{after} < KE_{before}$, kinetic energy is not conserved — the collision is inelastic (approximately 30.5 kJ was converted to heat/deformation during the impact).
You have completed all 15 lessons of Physics Year 11 Module 2: Dynamics. Across the module you mastered:
- Phase 1 (L01–L05): Newton's Laws, vector resolution, inclined planes, friction, v-t graphs
- Phase 2 (L06–L10): Work, kinetic energy, gravitational PE, power, energy conservation
- Phase 3 (L11–L15): Momentum, impulse, conservation of momentum, elastic/inelastic collisions
These concepts are the foundation for Module 5 (Advanced Mechanics) in Year 12.
Looking back across all 15 lessons: the connecting thread is that velocity links all three phases. Phase 1 tells you how force changes velocity; Phase 2 tells you the energy cost of that change; Phase 3 tells you the impulse required. The Space Shuttle Columbia re-entry on 12 April 1981 — decelerating from 27,875 km/h to 343 km/h and dissipating 2.97 TJ — is the ultimate demonstration: NASA needed all three phases simultaneously. Every real-world dynamics scenario uses this chain.