Welcome to Newton's Laws of Motion!
Ever wondered why you jolt forward when a bus suddenly brakes, how rockets launch into space, or why catching a cricket ball stings less when you pull your hands back? The answers to all these questions lie in Newton's Laws of Motion.
These laws are the bedrock of classical mechanics in your CCEA AS-Level Physics course. Don't worry if mechanics has felt intimidating before — we are going to break down every concept step by step with clear explanations, visual analogies, and practical exam tips.
1. Linear Momentum and Impulse
What is Momentum?
Before diving into Newton's laws, we need to understand momentum. In simple terms, momentum is a measure of "how difficult it is to stop a moving object."
Definition: Linear momentum (\( p \)) is the product of an object's mass (\( m \)) and its velocity (\( v \)).
\( p = mv \)
Units: Kilogram metres per second (\( \text{kg}\cdot\text{m}\cdot\text{s}^{-1} \)) or Newton-seconds (\( \text{N}\cdot\text{s} \)).
Important Note: Momentum is a vector quantity. This means direction matters! Always choose a positive direction (e.g., to the right = positive, to the left = negative).
What is Impulse?
When a force acts on an object over a period of time, it changes the object's momentum. This change in momentum is called impulse.
\( \text{Impulse} = F \Delta t = \Delta p = mv - mu \)
Where:
• \( F \) = applied force (\( \text{N} \))
• \( \Delta t \) = time interval during which the force acts (\( \text{s} \))
• \( m \) = mass (\( \text{kg} \))
• \( u \) = initial velocity (\( \text{m}\cdot\text{s}^{-1} \))
• \( v \) = final velocity (\( \text{m}\cdot\text{s}^{-1} \))
Force-Time Graphs
On a graph of Force against Time (\( F \text{ vs } t \)):
• The area under the graph is equal to the Impulse (or the change in momentum, \( \Delta p \)).
Everyday Connection: Car Safety and Sports
Why do cars have crumple zones and airbags? Why do gymnasts land on soft mats?
Rearranging the impulse equation gives: \( F = \frac{\Delta p}{\Delta t} \)
To stop a moving car, a fixed change in momentum (\( \Delta p \)) must occur. By increasing the impact time (\( \Delta t \)) using crumple zones, seatbelts, and airbags, the average force (\( F \)) acting on the passengers is drastically reduced, preventing severe injuries.
Key Takeaway: Momentum is mass in motion (\( p = mv \)). Impulse is the change in momentum (\( \Delta p = F \Delta t \)), and it is represented by the area under a force-time graph.
2. Newton's First Law of Motion
The Law of Inertia
Statement: An object will remain at rest or continue to move with a constant velocity in a straight line unless acted upon by a resultant external force.
Breaking It Down:
1. If the resultant force on an object is zero (\( \Sigma F = 0 \)):
• A stationary object stays stationary.
• A moving object keeps moving at the exact same speed and in the exact same direction.
2. If a resultant force does act (\( \Sigma F \neq 0 \)), the object will accelerate (speed up, slow down, or change direction).
What is Inertia?
Inertia is the natural tendency of an object to resist any change in its state of motion. The greater the mass of an object, the greater its inertia.
Analogy: It is much harder to push-start a massive lorry than a small bicycle because the lorry has vastly more inertia.
Common Mistake to Avoid: Thinking a force is needed to keep an object moving. In everyday life, friction and air resistance slow things down. In deep space, without friction, a kicked football would glide forever at constant velocity without needing any ongoing force!
Key Takeaway: Balanced forces (\( F_{\text{net}} = 0 \)) mean constant velocity (\( a = 0 \)). Unbalanced forces cause acceleration.
3. Newton's Second Law of Motion
The Official Definition
Statement: 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.
\( F \propto \frac{\Delta p}{\Delta t} \)
Deriving \( F = ma \):
Starting with \( F = \frac{\Delta p}{\Delta t} \):
Since \( \Delta p = \Delta(mv) = m\Delta v \) (for a constant mass \( m \)):
\( F = \frac{m \Delta v}{\Delta t} \)
Since acceleration \( a = \frac{\Delta v}{\Delta t} \), this simplifies to the famous formula:
\( F = ma \)
Where:
• \( F \) = resultant (net) force in newtons (\( \text{N} \))
• \( m \) = mass in kilograms (\( \text{kg} \))
• \( a \) = acceleration in metres per second squared (\( \text{m}\cdot\text{s}^{-2} \))
Defining the Newton
Definition of the Newton (\( \text{N} \)): One Newton is the resultant force that provides a mass of \( 1\text{ kg} \) with an acceleration of \( 1\text{ m}\cdot\text{s}^{-2} \).
Weight vs Mass
• Mass (\( m \)): The amount of matter in an object, measured in \( \text{kg} \). It never changes regardless of location.
• Weight (\( W \)): The gravitational force acting on that mass, measured in \( \text{N} \).
\( W = mg \)
Where \( g \) is the acceleration of free fall (\( g \approx 9.81\text{ m}\cdot\text{s}^{-2} \) on Earth).
Applying \( F = ma \) to Common Problem Scenarios
Scenario A: Objects in an Elevator / Lift
Consider a person of mass \( m \) standing on a weighing scale inside a lift. The scale reads the normal contact force \( R \).
1. Lift accelerating upwards at \( a \):
Resultant force is upwards: \( R - W = ma \implies R = mg + ma \)
The person feels heavier!
2. Lift accelerating downwards at \( a \):
Resultant force is downwards: \( W - R = ma \implies R = mg - ma \)
The person feels lighter!
3. Lift moving at constant velocity (\( a = 0 \)):
\( R - W = 0 \implies R = mg \)
The scale reads their normal weight.
Scenario B: Connected Bodies (e.g., Car towing a Trailer)
For a car of mass \( m_{\text{car}} \) pulling a trailer of mass \( m_{\text{trailer}} \) with an engine drive force \( D \) and total friction \( f_{\text{total}} \):
Step 1: Treat the whole system as one single object:
\( F_{\text{res}} = (m_{\text{car}} + m_{\text{trailer}}) \times a \)
\( D - f_{\text{total}} = (m_{\text{car}} + m_{\text{trailer}})a \)
Step 2: Isolate one object (e.g., the trailer) to find the tension (\( T \)) in the tow bar:
\( T - f_{\text{trailer}} = m_{\text{trailer}} a \)
Understanding Terminal Velocity
When an object (like a skydiver) falls through air:
1. At the moment of release (\( t = 0 \)): Speed is zero, so air resistance (drag) is zero. The only force is weight (\( W \)). Acceleration is maximum: \( a = g \).
2. As speed increases: Air resistance increases. The resultant downward force decreases (\( F_{\text{res}} = W - \text{Drag} \)), so acceleration decreases.
3. Terminal Velocity reached: Eventually, air resistance grows until it exactly equals weight (\( \text{Drag} = W \)). The resultant force is zero (\( F_{\text{res}} = 0 \)), and the object falls at a constant maximum velocity called terminal velocity.
Key Takeaway: \( F = ma \) always refers to the resultant (net) force. When forces balance, acceleration becomes zero.
4. Newton's Third Law of Motion
Action and Reaction
Statement: If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.
The 4 Golden Rules of a Newton III Pair:
Two forces form a genuine Newton's Third Law pair if and only if they:
1. Are equal in magnitude.
2. Act in opposite directions.
3. Are of the exact same type (e.g., both gravitational, both normal contact, both electrostatic).
4. Act on two different objects.
The Classic Exam Trap: Book on a Table
Consider a book resting on a horizontal table:
• Force 1: The gravitational pull of the Earth down on the book (\( \text{Weight} \)).
• Force 2: The normal contact push of the table up on the book (\( R \)).
Question: Are these two forces a Newton's Third Law pair?
Answer: NO! Even though they are equal and opposite, they are not a Newton III pair because:
• They act on the same object (the book).
• They are different types of forces (gravitational vs electrostatic/contact).
The actual Newton III pairs are:
• Gravitational: Earth pulls book downwards \( \leftrightarrow \) Book pulls Earth upwards with equal gravitational force.
• Contact: Book pushes table downwards \( \leftrightarrow \) Table pushes book upwards with equal normal contact force.
Key Takeaway: Newton III pairs never cancel each other out because they always act on two completely different objects.
5. Principle of Conservation of Linear Momentum
The Law
Statement: For an isolated system (a system with no external resultant forces), the total linear momentum before a collision or explosion is equal to the total linear momentum after.
\( \text{Total Initial Momentum} = \text{Total Final Momentum} \)
\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
Types of Collisions
1. Elastic Collisions:
• Linear momentum is conserved.
• Total energy is conserved.
• Kinetic energy is conserved (\( E_{\text{k, initial}} = E_{\text{k, final}} \)).
2. Inelastic Collisions:
• Linear momentum is conserved.
• Total energy is conserved.
• Kinetic energy is NOT conserved (some \( E_{\text{k}} \) is transformed into thermal energy, sound, or work done deforming the objects).
• If two objects stick together after collision, it is called a perfectly inelastic collision.
3. Explosions:
• A single stationary object separates into parts.
• Total initial momentum is zero, so total final momentum must also be zero:
\( 0 = m_1 v_1 + m_2 v_2 \implies m_1 v_1 = -m_2 v_2 \)
(One part moves left, the other moves right with equal and opposite momentum).
Worked Example Step-by-Step
Problem: A trolley of mass \( 2.0\text{ kg} \) travelling to the right at \( 4.0\text{ m}\cdot\text{s}^{-1} \) collides head-on with a stationary trolley of mass \( 3.0\text{ kg} \). After the collision, the two trolleys stick together. Find their combined velocity.
Step 1: Define positive direction: Let right be positive (\( + \)).
Step 2: Write down what you know:
\( m_1 = 2.0\text{ kg} \), \( u_1 = +4.0\text{ m}\cdot\text{s}^{-1} \)
\( m_2 = 3.0\text{ kg} \), \( u_2 = 0\text{ m}\cdot\text{s}^{-1} \)
Combined final mass = \( m_1 + m_2 = 2.0 + 3.0 = 5.0\text{ kg} \)
Final velocity = \( v \)
Step 3: Apply conservation of momentum:
\( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \)
\( (2.0 \times 4.0) + (3.0 \times 0) = (5.0) \times v \)
\( 8.0 = 5.0 v \)
Step 4: Solve for \( v \):
\( v = \frac{8.0}{5.0} = 1.6\text{ m}\cdot\text{s}^{-1} \) (to the right)
Key Takeaway: Momentum is always conserved in any collision or explosion, provided no external forces act. Always watch your positive and negative signs for velocities!
Quick Review Summary
• Momentum: \( p = mv \) (vector quantity).
• Impulse: \( F \Delta t = \Delta p = \text{Area under } F\text{-}t \text{ graph} \).
• Newton 1: No resultant force (\( \Sigma F = 0 \)) \( \implies \) constant velocity.
• Newton 2: \( F = \frac{\Delta p}{\Delta t} \), which gives \( F = ma \) when mass is constant.
• Newton 3: Equal and opposite force pairs of the same type acting on two different bodies.
• Conservation of Momentum: Total momentum before = Total momentum after (in a closed system).
• Elastic vs Inelastic: Kinetic energy is conserved only in elastic collisions.