Momentum and Energy¶
Syllabus mapping
PY-11-01 — Fundamentals of Mechanics. Covers: work done by a force; conservative vs non-conservative forces; gravitational potential energy and kinetic energy; conservation of mechanical energy; momentum; conservation of momentum; elastic and inelastic collisions.
Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.
You need to know: how to calculate work, kinetic energy and gravitational potential energy, when mechanical energy is conserved, and how to apply conservation of momentum to collisions — including telling elastic and inelastic collisions apart.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Algebra — rearranging multi-step equations
- Forces and Motion — Newton's second law (momentum builds directly on this)
Core¶
Work¶
Work is done by the component of a force along the direction of displacement — a force perpendicular to motion (like the normal force on a horizontal surface) does zero work.
Worked example
A \(50\ \text{N}\) force is applied at \(40^\circ\) above horizontal to drag a box \(3\ \text{m}\) across the floor. Find the work done by the applied force.
Conservative vs non-conservative forces¶
| Conservative | Non-conservative | |
|---|---|---|
| Examples | Gravity, spring force | Friction, air resistance |
| Work done over a closed path | Zero | Not zero |
| Energy | Recoverable, stored as potential energy | Dissipated (usually as heat/sound), not recoverable as mechanical energy |
This is why gravity gets a potential energy (height above the ground stores recoverable energy) but friction doesn't — energy lost to friction is gone from the mechanical system, not stored anywhere retrievable.
Gravitational potential energy and kinetic energy¶
Worked example
A \(0.5\ \text{kg}\) ball is dropped from rest and falls \(2\ \text{m}\). Find its speed just before landing, using energy.
All the lost PE converts to KE (no air resistance):
Cross-check with SUVAT: \(v^2 = u^2+2as = 0 + 2(9.8)(2)\), giving the same \(v \approx 6.26\ \text{m s}^{-1}\) — the energy method and the kinematics method must always agree for the same problem.
Conservation of mechanical energy¶
When only conservative forces do work, total mechanical energy is constant:
If friction or another non-conservative force is present, some mechanical energy converts to other forms and this equation no longer balances on its own — the "missing" energy went somewhere (usually heat), it isn't destroyed.
Momentum¶
Momentum is a vector — direction matters, same as velocity.
Conservation of momentum¶
Total momentum of an isolated system (no external net force) is conserved — true for every collision, elastic or not.
Worked example
A \(2\ \text{kg}\) trolley moving at \(3\ \text{m s}^{-1}\) collides with a stationary \(1\ \text{kg}\) trolley, and they stick together. Find their combined velocity after the collision.
Elastic vs inelastic collisions¶
Momentum is conserved in every collision. What changes is kinetic energy:
- Elastic: kinetic energy is also conserved — \(\sum \tfrac{1}{2}mv^2_{before} = \sum \tfrac{1}{2}mv^2_{after}\). No energy converts to heat/sound/deformation.
- Inelastic: kinetic energy is not conserved — some converts to heat, sound, or permanent deformation, even though momentum still is.
- Perfectly inelastic: the extreme case — the objects stick together and move with a common final velocity (as in the worked example above).
Worked example
Using the trolley collision above: is it elastic?
KE dropped from \(9\ \text{J}\) to \(6\ \text{J}\), so this collision is inelastic (in fact, perfectly inelastic, since the trolleys stuck together) — \(3\ \text{J}\) converted to other forms.
Practical¶
Suggested prac: conservation of momentum in collisions — trolleys with a spring-loaded or Velcro coupling on a track, measuring velocities before and after (light gates) to confirm total momentum is conserved, and to compare elastic vs perfectly-inelastic setups. TODO — link this school's actual prac instructions once written.
Advanced
2D collisions — momentum conservation still holds, but must be applied component-wise: total \(x\)-momentum before equals total \(x\)-momentum after, and separately for \(y\), using the same component method from Quantities of Motion.
Explosions are the reverse of a perfectly inelastic collision — a single object separates into pieces, and total momentum is still conserved (if it started at rest, the pieces' momenta must sum to zero, meaning they fly apart in exactly the directions and speeds needed to balance).
Extension
Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Newton's second law in its general form is \(\vec{F}_{net} = \dfrac{d\vec{p}}{dt}\) (flagged already in the Extension box on Forces and Motion). Integrating both sides over a time interval gives the impulse–momentum theorem:
which for a constant force reduces to \(\Delta \vec{p} = \vec{F}_{net}\Delta t\) — useful whenever a force varies over the course of a collision (e.g. a real car crash, where the force isn't constant) and you only care about the net effect, not the instant-by-instant force. Worth a depth study modelling a specific variable-force collision (crumple zones, a bouncing ball) using the area under a force–time graph to find impulse directly, without ever needing an explicit \(F(t)\) formula.
Video/visual resources¶
- 🖥️ PhET Simulation — Collision Lab — TODO: confirm current PhET link (flagged in
project.mdSection 7 as a strong fit for this topic) - 🎥 Khan Academy — TODO: source a momentum/collisions explainer
- 🎥 Physics High — TODO: check for a NSW-syllabus-aligned momentum video
Check yourself¶
-
A \(60\ \text{kg}\) skateboarder at the top of a \(2\ \text{m}\) high ramp rolls down (frictionless). Find their speed at the bottom using energy conservation.
Answer
\(mg\Delta h = \tfrac{1}{2}mv^2 \implies v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.8 \times 2} \approx 6.26\ \text{m s}^{-1}\) (mass cancels, same as free fall)
-
A \(1500\ \text{kg}\) car moving at \(10\ \text{m s}^{-1}\) collides head-on with a stationary \(1000\ \text{kg}\) car, and the two lock together. Find their common velocity after the collision.
Answer
\(m_1v_1 = (m_1+m_2)v_f \implies v_f = \dfrac{(1500)(10)}{2500} = 6\ \text{m s}^{-1}\)
-
Two \(1\ \text{kg}\) trolleys collide: before, one moves at \(4\ \text{m s}^{-1}\) and the other is stationary; after, the first is stationary and the second moves at \(4\ \text{m s}^{-1}\). Is this collision elastic?
Answer
\(KE_{before} = \tfrac{1}{2}(1)(4)^2 = 8\ \text{J}\), \(KE_{after} = \tfrac{1}{2}(1)(4)^2 = 8\ \text{J}\) — kinetic energy is conserved, so yes, this collision is elastic.