Linear motion and Newton's laws: QCE Physics Unit 2 Year 11

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 motion with uniform (constant) acceleration in a straight line, five quantities are linked by four equations. Each equation omits one of the five variables, so you pick the equation that contains the three knowns and the one unknown.

• $v = u + at$ (omits $s$)
• $s = ut + \frac{1}{2}at^2$ (omits $v$)
• $v^2 = u^2 + 2as$ (omits $t$)
• $s = \frac{1}{2}(u + v)t$ (omits $a$)

Here $u$ is the initial velocity, $v$ the final velocity, $a$ the acceleration, $s$ the displacement and $t$ the time. The equations are valid only while $a$ is constant, so for free fall on Earth take $a = g = 9.8\ \text{m s}^{-2}$ downward and choose a consistent positive direction. Because displacement, velocity and acceleration are all vectors, sign conventions decide the result; a ball thrown upward has positive initial velocity and negative acceleration if up is taken as positive.

Newton's laws

First law (inertia)

A body at rest stays at rest, and a body in motion stays in motion at constant velocity, unless acted on by a net external force. Inertia is the resistance of mass to a change in motion, so a more massive object is harder to accelerate or stop. Seatbelts and headrests are direct applications: when a car stops suddenly, the passenger continues forward at the original velocity until a force restrains them.

Second law

The net force on a body equals the rate of change of its momentum; for constant mass this reduces to $F_{\text{net}} = ma$. The acceleration is in the direction of the net force, is proportional to the net force, and is inversely proportional to mass. This is the central problem-solving equation: resolve all forces, find the net force, then divide by mass.

Third law

When body A exerts a force on body B, body B exerts an equal and opposite force on body A. The two forces are equal in magnitude, opposite in direction, of the same type, and crucially act on different bodies, so they never cancel each other. A swimmer pushes water backward and the water pushes the swimmer forward.

Free-body diagrams

A free-body diagram isolates a single object and shows every force acting on it as a labelled arrow from the object's centre: weight ($mg$ downward), the normal force (perpendicular to the surface), friction (parallel to the surface, opposing relative motion), tension and any applied force. Forces are then summed as vectors. On an inclined plane it is efficient to resolve weight into a component down the slope ($mg\sin\theta$) and a component into the slope ($mg\cos\theta$), so that the net force calculation aligns with the direction of motion.

Momentum and impulse

Momentum is $p = mv$, a vector measured in $\text{kg m s}^{-1}$, pointing in the direction of the velocity. Impulse is the change in momentum produced by a force acting over a time interval:

In an isolated system (no net external force) total momentum is conserved, which is the key tool for analysing collisions and explosions. In an elastic collision both momentum and kinetic energy are conserved; in an inelastic collision momentum is conserved but some kinetic energy is converted to heat, sound and deformation. The impulse relationship explains why crumple zones, airbags and a cricketer's follow-through reduce force: for a fixed change in momentum, extending the contact time $\Delta t$ lowers the average force $F$.

Energy

Work is done when a force moves its point of application through a displacement:

where $\theta$ is the angle between the force and the displacement, so a force perpendicular to the motion does no work. Kinetic energy is $KE = \tfrac{1}{2}mv^2$ and gravitational potential energy near Earth's surface is $PE = mgh$. The work-energy theorem states that the net work done on a body equals its change in kinetic energy.

In the absence of friction, mechanical energy is conserved:

When friction or air resistance acts, the lost mechanical energy appears as heat:

Power

Power is the rate of doing work or transferring energy:

measured in watts ($1\ \text{W} = 1\ \text{J s}^{-1}$). The form $P = Fv$ is useful for vehicles moving at constant speed against a resistive force, where the engine power equals the driving force times the velocity.

Examples in context

Example 1. A Bremer River bridge maintenance gondola hangs from a $2000 \text{ kg}$ counterweight. Treating the rope as massless, Newton's second law on the gondola ($m = 400 \text{ kg}$) gives $T - mg = ma$ where $T$ is tension and $a$ is upward acceleration. When the operator commands $a = 0.5 \text{ m s}^{-2}$, $T = 400(9.8+0.5) = 4120 \text{ N}$. This single Newton's-law calculation, combined with SUVAT to predict travel time over $30 \text{ m}$ of bridge underside, is the kind of design verification a Queensland Transport and Main Roads engineer signs off.

Example 2. A Gladstone port forklift accelerating a $2.0 \text{ tonne}$ container from rest to $3.0 \text{ m s}^{-1}$ over $5.0 \text{ m}$ requires $a = v^2/(2s) = 0.90 \text{ m s}^{-2}$, force $F = ma = 1800 \text{ N}$. Work done is $W = Fs = 9000 \text{ J}$ which equals the container's kinetic energy $\tfrac{1}{2}mv^2 = 9000 \text{ J}$, confirming the work-energy theorem. QCAA EA Unit 2 thematic sets blend SUVAT, $F = ma$ and energy in exactly this way.

Try this

Q1. State Newton's three laws of motion. [3 marks]

• Cue. First (inertia); second ($F = ma$); third (equal and opposite reaction).

Q2. A $1200 \text{ kg}$ car accelerates from $10 \text{ m s}^{-1}$ to $25 \text{ m s}^{-1}$ in $5.0 \text{ s}$. Calculate the acceleration, the net force and the impulse. [4 marks]

• Cue. $a = 3.0 \text{ m s}^{-2}$; $F = 3600 \text{ N}$; $J = \Delta p = 18000 \text{ N s}$.

Q3. A $400 \text{ kg}$ Bremer bridge gondola accelerates upward at $0.5 \text{ m s}^{-2}$ for $4.0 \text{ s}$ from rest. (a) Calculate the rope tension and the displacement. (b) Determine the kinetic energy and check via work-energy theorem. (c) Identify and explain one safety implication of a sudden cable failure. [4+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $T = 4120 \text{ N}$, $s = 4.0 \text{ m}$; (b) $v = 2.0 \text{ m s}^{-1}$, $KE = 800 \text{ J}$, work $= (T-mg)s = 800 \text{ J}$; (c) free-fall acceleration immediately.