Introduction to Newton's Laws of Motion
Welcome to one of the most fundamental chapters in your AS 1 Physics course: Newton's Laws of Motion. Everything that moves, stops, speeds up, or changes direction in our universe behaves according to three simple, elegant principles first published by Sir Isaac Newton in 1687.
In this module, you will learn how forces cause objects to accelerate, why a car moving at \(70\text{ m s}^{-1}\) can have a total resultant force of zero, what really happens to your weight in an accelerating lift, and how skydivers reach terminal velocity. Don't worry if dynamics problems have felt overwhelming in the past; we will break down every concept step-by-step using clear free-body diagrams and straightforward algebra.
1. Newton's First Law of Motion (The Law of Inertia)
Formal CCEA Definition
Newton's First Law of Motion: A body will remain at rest or continue to move in a straight line at a constant speed (constant velocity) unless acted upon by a resultant (unbalanced/net external) force.
Understanding the Concept
Newton's First Law tells us about equilibrium. An object does not need a resultant force to keep moving; it only needs a resultant force to change how it is moving.
If the vector sum of all forces acting on an object is zero (\(\Sigma F = 0\)), the object is in equilibrium:
\(\Sigma F = 0 \implies a = 0\text{ m s}^{-2}\)
This leads to two possible states of motion:
1. Static Equilibrium: The object is stationary (\(v = 0\text{ m s}^{-1}\)) and remains at rest.
2. Dynamic Equilibrium: The object moves at a constant speed in a straight line (constant velocity, \(v = \text{constant}\)).
Everyday Example: Cruising on a Straight Road
Imagine a car travelling along a straight, flat motorway at a steady speed of \(30\text{ m s}^{-1}\). The engine produces a forward driving force of \(1500\text{ N}\). Because the car is moving at a constant velocity, Newton's First Law guarantees that the resultant force must be zero. Therefore, the total resistive forces (air drag and rolling friction) acting backwards must equal exactly \(1500\text{ N}\).
Quick Review: Key Takeaway
Key Takeaway: Constant velocity means zero acceleration, which strictly means zero resultant force (\(F_{\text{net}} = 0\)). Never assume that high speed requires a net forward force!
2. The SI Unit of Force: The Newton
Definition of the Newton
The standard SI unit of force is the newton (\(\text{N}\)).
Formal Definition: One newton is defined as the unbalanced (resultant) force that produces an acceleration of \(1\text{ m s}^{-2}\) when applied to a mass of \(1\text{ kg}\).
SI Base Units of the Newton
By applying \(F = ma\), we can express the newton in SI base units:
\(\text{Force} = \text{mass} \times \text{acceleration}\)
\(1\text{ N} = 1\text{ kg} \times 1\text{ m s}^{-2} = 1\text{ kg m s}^{-2}\)
3. Newton's Second Law of Motion
Formal CCEA Definitions
CCEA assesses Newton's Second Law in two equivalent ways depending on whether mass is constant:
1. The Fundamental Statement (Momentum Formulation):
The rate of change of momentum of an object is directly proportional to the resultant force acting on it and occurs in the direction of that force.
Mathematically: \(F_{\text{net}} \propto \frac{\Delta p}{\Delta t}\)
2. The Constant Mass Formulation:
The acceleration of an object is directly proportional to the resultant force acting on it, inversely proportional to its mass, and acts in the direction of the unbalanced force.
Mathematically: \(a = \frac{F_{\text{net}}}{m} \implies F_{\text{net}} = ma\)
The Resultant Force Equation
When solving mechanics problems, you must always look at the resultant (net) force along the axis of motion:
\(F_{\text{net}} = \Sigma F = ma\)
\((\Sigma F_{\text{in direction of motion}} - \Sigma F_{\text{opposing motion}}) = ma\)
Step-by-Step Problem Solving Strategy
Step 1: Draw a clear free-body diagram showing every individual force acting on the body.
Step 2: Choose a positive direction (usually the direction of acceleration).
Step 3: Write down the equation: \((\text{Forces forwards}) - (\text{Forces backwards}) = ma\).
Step 4: Substitute known values using \(g = 9.81\text{ m s}^{-2}\) for gravity and solve for the unknown.
Quick Review: Key Takeaway
Key Takeaway: In any \(F = ma\) calculation, \(F\) is never just "any force"—it is always the unbalanced (resultant) force acting on the mass.
4. Newton's Third Law of Motion
Formal CCEA Definition
Newton's Third Law of Motion: If Body \(A\) exerts a force on Body \(B\), then Body \(B\) exerts an equal and opposite force on Body \(A\).
The Four Essential Conditions for a Newton III Pair
Two forces only form a true Newton's Third Law action-reaction pair if they satisfy all four of the following rules:
1. Equal Magnitude: Both forces have the exact same size (\(F_A = F_B\)).
2. Opposite Direction: The two forces act in exactly opposite directions.
3. Same Physical Nature: Both forces must be of the same type (e.g., both are gravitational, both are normal contact/electrostatic, or both are frictional).
4. Act on Different Bodies: Force 1 acts on Body \(B\); Force 2 acts on Body \(A\).
The Classic Exam Trap: Book on a Table
Consider a book resting stationary on a horizontal table. Two forces act on the book:
1. The downward gravitational pull of the Earth on the book (Weight, \(W = mg\)).
2. The upward normal contact reaction force of the table on the book (\(R\)).
Question: Do \(W\) and \(R\) form a Newton's Third Law pair?
Answer: NO! Even though \(W = R\) in magnitude and they point in opposite directions:
• Both forces act on the same object (the book).
• They are different types of forces (\(W\) is gravitational, while \(R\) is an electrostatic contact force between atoms).
• This is an example of Newton's First Law equilibrium (\(\Sigma F = R - W = 0\)), NOT Newton's Third Law.
The True Newton III Pairs:
• For the book's weight (Earth pulls book down): The pair is the gravitational force of the book pulling the Earth upwards.
• For the normal contact force (table pushes book up): The pair is the downward contact force of the book pushing the table down.
5. Core Application Scenarios in CCEA AS 1
A. Apparent Weight in Lifts (Elevators)
When a person of mass \(m\) stands on a set of bathroom scales inside a lift, the scales do not directly measure weight; they measure the normal contact reaction force (\(R\)) pushing upwards on the person's feet.
Scenario 1: Lift at Rest or Moving at Constant Velocity
Acceleration is zero (\(a = 0\text{ m s}^{-2}\)):
\(R - mg = m(0) \implies R = mg\)
Scale reading equals true weight.
Scenario 2: Lift Accelerating Upwards (or Decelerating Downwards)
The lift accelerates upwards with acceleration \(a\). Taking upwards as positive:
\(R - mg = ma \implies R = m(g + a)\)
The scale reading increases. You feel heavier than normal.
Scenario 3: Lift Accelerating Downwards (or Decelerating Upwards)
The lift accelerates downwards with acceleration \(a\). Taking downwards as positive:
\(mg - R = ma \implies R = m(g - a)\)
The scale reading decreases. You feel lighter than normal.
B. Connected Bodies and Towing Systems
When two objects are connected by a light, inextensible string (or towbar), both objects share the same magnitude of acceleration (\(a\)), and the tension (\(T\)) throughout the string is uniform.
Example: A Car Towing a Trailer
A car of mass \(m_1 = 1200\text{ kg}\) tows a trailer of mass \(m_2 = 400\text{ kg}\). The engine provides a driving force of \(3200\text{ N}\). Total resistive forces are \(400\text{ N}\) on the car and \(200\text{ N}\) on the trailer.
Step 1: Treat the whole system as a single combined mass to find acceleration \(a\):
\(m_{\text{total}} = m_1 + m_2 = 1200 + 400 = 1600\text{ kg}\)
\(F_{\text{net, total}} = F_{\text{drive}} - R_{\text{total}} = 3200 - (400 + 200) = 2600\text{ N}\)
\(a = \frac{F_{\text{net, total}}}{m_{\text{total}}} = \frac{2600}{1600} = 1.625\text{ m s}^{-2}\)
Step 2: Apply \(F_{\text{net}} = ma\) to the trailer alone to find Tension (\(T\)):
\(T - R_{\text{trailer}} = m_2 a\)
\(T - 200 = 400 \times 1.625\)
\(T - 200 = 650 \implies T = 850\text{ N}\)
C. Free Fall and Terminal Velocity
When an object falls through a fluid (such as air), it experiences a downward gravitational force (weight, \(W = mg\)) and an upward resistive force (drag / air resistance, \(F_D\)). Drag increases as speed increases (\(F_D \propto v\) or \(v^2\)).
Phase 1: Instant of Release (\(t = 0\text{ s}\), \(v = 0\text{ m s}^{-1}\))
• Air resistance is zero: \(F_D = 0\text{ N}\).
• Resultant force: \(F_{\text{net}} = mg - 0 = mg\).
• Acceleration is maximum: \(a = g = 9.81\text{ m s}^{-2}\).
Phase 2: Falling and Accelerating (\(v\) increasing)
• As velocity increases, drag \(F_D\) increases.
• Resultant force decreases: \(F_{\text{net}} = mg - F_D\).
• Acceleration decreases (\(a < 9.81\text{ m s}^{-2}\)), but the object is still speeding up.
Phase 3: Terminal Velocity Reached (\(v = v_t\))
• The upward drag force grows until it equals the downward weight: \(F_D = mg\).
• Resultant force is zero: \(F_{\text{net}} = mg - F_D = 0\text{ N}\).
• Acceleration becomes zero: \(a = 0\text{ m s}^{-2}\).
• By Newton's First Law, the object continues falling at a constant maximum velocity called terminal velocity.
6. Summary of Common Exam Pitfalls
1. Forgetting "Resultant" in Newton's Second Law: Never write "force equals mass times acceleration" without stating that the force is the resultant or unbalanced force.
2. Mixing up Newton I and Newton III: Remember that Newton III pairs must act on two different objects and be of the same type.
3. Sign Errors in Lift Problems: If a lift accelerates upwards, the scale reads \(R = m(g + a)\). If it accelerates downwards, the scale reads \(R = m(g - a)\).
4. Incorrect Gravity Value: Always use \(g = 9.81\text{ m s}^{-2}\) from the CCEA AS 1 Data and Formulae Sheet (never \(9.8\text{ m s}^{-2}\) or \(10\text{ m s}^{-2}\)).
5. Unit Formats: Always write units in negative index format (e.g., \(\text{m s}^{-2}\), \(\text{kg m s}^{-2}\)) as required by CCEA marking schemes.