Unit 2: Linear motion and waves

QLDPhysicsSyllabus dot point

How is linear motion analysed using Newton's laws?

Linear motion (displacement, velocity, acceleration, suvat equations), Newton's three laws, free-body diagrams, momentum $p = mv$, impulse $J = F \Delta t$, work, energy, power

A focused answer to the QCE Physics Unit 2 subject-matter point on linear motion. Kinematics (suvat), Newton's three laws, momentum and impulse, work, kinetic and potential energy, conservation of energy, and power; foundation for Unit 3 Newtonian motion in 2D.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answer

Have a quick question? Jump to the Q&A page

What this dot point is asking

QCAA wants Year 11 students to apply Newton's laws, momentum, energy and the SUVAT equations to linear motion problems.

SUVAT equations

For uniformly accelerated motion:

  • IMATH_0
  • IMATH_1
  • IMATH_2
  • IMATH_3

Where uu initial velocity, vv final, aa acceleration, ss displacement, tt time.

Newton's laws

1st law (inertia). Object at rest stays at rest; in motion stays in motion at constant velocity, unless acted on by net external force.

2nd law. F=maF = ma. Net force equals mass times acceleration.

3rd law. Action-reaction pair: equal and opposite forces on different bodies.

Free-body diagrams

Show all forces on the object (gravity, normal force, friction, tension, applied). Sum vectorially, set equal to mama.

Momentum and impulse

p=mvp = mv (kg m/s). Vector.

Impulse J=FΔt=ΔpJ = F\Delta t = \Delta p.

Conservation: in isolated system, total momentum conserved through collisions.

Energy

Work W=FdcosθW = Fd\cos\theta.

Kinetic KE=12mv2KE = \frac{1}{2}mv^2.

Gravitational PE =mgh= mgh.

Conservation of mechanical energy: KE+PE=KE + PE = constant (no friction).

With friction: KEi+PEi=KEf+PEf+ElostKE_i + PE_i = KE_f + PE_f + E_{\text{lost}}.

Power

P=W/t=FvP = W/t = Fv (W = J/s).

Common errors

Substituting before differentiating (in derivative-like manipulations). Not relevant here but a related issue.

Confusing distance and displacement. Distance is path; displacement is net change.

Sign errors in free fall. Choose positive direction, stick to it.

Inelastic vs elastic. Momentum always conserved. Kinetic energy only in elastic.

In one sentence

Linear motion is analysed using suvat equations (v=u+atv = u + at, s=ut+12at2s = ut + \frac{1}{2}at^2, v2=u2+2asv^2 = u^2 + 2as) and Newton's three laws (inertia, F=maF = ma, action-reaction); momentum p=mvp = mv is conserved in collisions (elastic also conserves KEKE, inelastic does not); work, kinetic energy (12mv2\frac{1}{2}mv^2), potential energy (mghmgh) and conservation of energy govern energy transfers.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksA $1500$ kg car at $20$ m/s collides with a stationary $1000$ kg car. After collision they move together. (a) Velocity after? (b) Kinetic energy lost?
Show worked answer →

(a) Conservation of momentum: 1500×20+0=2500×v1500 \times 20 + 0 = 2500 \times v. v=12v = 12 m/s.

(b) KE before: 12(1500)(400)=300,000\frac{1}{2}(1500)(400) = 300,000 J. KE after: 12(2500)(144)=180,000\frac{1}{2}(2500)(144) = 180,000 J. Lost: 120,000120,000 J.

Inelastic collision: momentum conserved, KE not.

Markers reward momentum equation, KE calculation, and elastic/inelastic distinction.

Related dot points