Ultimate Guide to Modelling Motion Without Resistance

Introduction

Before diving into the complexities of air resistance, it is crucial to master the mechanics of idealised environments. Modelling Motion Without Resistance allows us to focus purely on the core interactions of gravity, applied forces, and the normal force. In HSC Mathematics Extension 2, this topic serves as the critical transition from 1D kinematics to resolving multi-force vector systems like inclined planes and simple pulleys.

Executive Summary

This comprehensive guide covers the essential techniques for modelling unresisted motion:

Equations of Motion: Deriving and applying kinematics formulas for constant and variable acceleration.
Smooth Inclined Planes: Resolving gravitational and normal forces into parallel and perpendicular components to determine acceleration along a slope.
Pulley Systems: Analysing connected particles where one body moves horizontally or on an incline while the other hangs vertically, utilizing a single smooth pulley.

What is this about?

When we ignore air resistance (drag) and friction (smooth surfaces), we are left with a simplified but powerful mathematical playground. In this topic, you will learn how to use vectors to break down forces that act at angles. If a block slides down a ramp, gravity pulls it straight down, but the ramp forces it to move diagonally. How much of gravity is actually accelerating the block? By resolving forces, you can answer this algebraically and build the differential equations needed to track the block's velocity and position over time.

Main Content

1. Constant and Variable Acceleration

If a particle travels in a straight line without resistance under a constant force, its acceleration $a$ is constant. From this, we derive the standard kinematic equations:

$v = u + at$
$x = ut + \frac{1}{2}at^2$
$v^2 = u^2 + 2ax$

However, if the force is variable (e.g., a rocket losing mass, or a magnetic field changing strength), you cannot use these formulas. You must return to the calculus definitions ( $a = \frac{dv}{dt}$ or $a = v\frac{dv}{dx}$ ) and integrate from scratch, applying Newton's Second Law ( $F = ma$ ).

2. Smooth Inclined Planes

Consider a mass $m$ resting on a smooth (frictionless) plane inclined at an angle $\theta$ to the horizontal. There are two primary forces acting on it:

Weight ( $mg$ ): Acting straight down towards the center of the Earth.
Normal Force ( $N$ ): Acting perpendicular (at $90^\circ$ ) to the surface of the plane.

To analyze the motion, we tilt our coordinate system so the $x$ -axis is parallel to the plane and the $y$ -axis is perpendicular to it. We resolve the Weight vector into two components:

Perpendicular to the plane: $mg \cos \theta$
Parallel to the plane (downwards): $mg \sin \theta$

Since the mass doesn't fly off the ramp or sink into it, the net force perpendicular is zero: $N - mg \cos \theta = 0 \implies N = mg \cos \theta$

The net force parallel to the plane causes acceleration: $F_{\text{net}} = mg \sin \theta$ Using $F = ma$ : $ma = mg \sin \theta \implies a = g \sin \theta$

The mass accelerates down the plane at a constant rate of $g \sin \theta$ .

3. Connected Particles (Pulley Systems)

A common extension is connecting two masses with a light, inextensible string over a smooth pulley. Consider Mass 1 ( $m_1$ ) lying on a smooth horizontal table, connected via string over a pulley at the edge to Mass 2 ( $m_2$ ) hanging vertically.

Because the string is inextensible, both masses move with the same acceleration ( $a$ ). Because the pulley is smooth, the tension ( $T$ ) in the string is constant throughout.

We write an equation of motion for each mass separately: For the hanging mass ( $m_2$ ): It moves downwards under gravity, but tension pulls it up. $m_2 g - T = m_2 a$ (Equation 1)

For the mass on the table ( $m_1$ ): The only horizontal force acting on it is the tension $T$ . $T = m_1 a$ (Equation 2)

We can solve these simultaneously by substituting Equation 2 into Equation 1: $m_2 g - (m_1 a) = m_2 a$ $m_2 g = a(m_1 + m_2)$ $a = \frac{m_2 g}{m_1 + m_2}$

Simple Worked Example

Question: A particle of mass $3 \text{ kg}$ rests on a smooth plane inclined at $30^\circ$ to the horizontal. It is connected by a light inextensible string passing over a smooth pulley at the top of the plane to a particle of mass $5 \text{ kg}$ which hangs freely vertically. The system is released from rest. Find the acceleration of the system and the tension in the string. (Let $g = 9.8 \text{ m/s}^2$ ).

Solution: Let the tension in the string be $T$ and the common acceleration be $a$ . Since the hanging mass ( $5 \text{ kg}$ ) is heavier than the component of gravity acting on the $3 \text{ kg}$ mass down the plane, the $5 \text{ kg}$ mass will accelerate downwards, and the $3 \text{ kg}$ mass will accelerate up the plane.

Step 1: Equation for the hanging mass ( $5 \text{ kg}$ ) Forces acting downwards: Weight = $5g$ Forces acting upwards: Tension = $T$ Equation: $5g - T = 5a$ (Eq 1)

Step 2: Equation for the mass on the incline ( $3 \text{ kg}$ ) Forces acting up the plane (direction of motion): Tension = $T$ Forces acting down the plane: Component of weight = $3g \sin(30^\circ)$ Equation: $T - 3g \sin(30^\circ) = 3a$ Since $\sin(30^\circ) = 0.5$ : $T - 1.5g = 3a$ (Eq 2)

Step 3: Solve simultaneously Add (Eq 1) and (Eq 2) to eliminate $T$ : $(5g - T) + (T - 1.5g) = 5a + 3a$ $3.5g = 8a$ $a = \frac{3.5 \times 9.8}{8}$ $a = \frac{34.3}{8} = 4.2875 \text{ m/s}^2$

Step 4: Find Tension $T$ Substitute $a$ back into (Eq 1): $5(9.8) - T = 5(4.2875)$ $49 - T = 21.4375$ $T = 49 - 21.4375 = 27.5625 \text{ Newtons}$ .

mini-FAQ page

Q: Do I need to memorize the formula $a = g \sin \theta$ for an inclined plane? A: No, and you shouldn't! In the HSC, you are generally expected to draw a diagram, state the forces, and derive the acceleration from Newton's Second Law. Memorizing the final result won't help if the question adds an external pushing force.

Q: What does "light, inextensible string" mean? A: "Light" means the string has zero mass, so we don't need to factor its weight into our $F=ma$ equations. "Inextensible" means it doesn't stretch like a rubber band, which guarantees that both connected masses share the exact same velocity and acceleration.

Common mistakes to avoid

Using $mg \cos \theta$ for the parallel force: A classic trigonometry error. When resolving gravity on an inclined plane where $\theta$ is the angle with the horizontal, the component pulling down the slope is always $mg \sin \theta$ , and the component pressing into the slope is $mg \cos \theta$ .
Treating the whole system as one mass: While there are shortcuts that treat connected particles as a single mass ( $F_{\text{net}} = (m_1 + m_2)a$ ), this method completely hides the Tension ( $T$ ). It is much safer to write separate equations for each mass and solve them simultaneously.

Practice on Vu's Maths Hub

Modelling motion requires you to accurately draw force diagrams and resolve vectors under exam pressure. Build your confidence with our resources:

Practice drawing force diagrams and solving connected particle problems in the HSC Mechanics Booklet.
Review the basic trigonometry required for resolving vectors in the HSC Trigonometry Booklet.
Tackle challenging exam-style variations in the HSC Last Resorts Booklet.

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Want to master Mechanics and lock in top marks for HSC Mathematics Extension 2? Visit Vu's Maths Hub for in-depth booklets, detailed video explanations, and expert advice to help you ace your exams!