Menu Top


Newton's Second and Third Laws (Basic)

Second Law Of Motion

While the First Law of Motion describes what happens when no unbalanced force acts on an object (it maintains constant velocity), Newton's Second Law quantifies the effect of an unbalanced force on an object's motion. It connects the force acting on an object to the change in its motion.

The Second Law is considered the most fundamental of Newton's laws because it provides a means to calculate the effect of forces. In fact, Newton's First Law can be derived from the Second Law as a special case where the net force is zero.

Statement of Newton's Second Law

Newton's Second Law of Motion states:

"The rate of change of momentum of a body is directly proportional to the applied unbalanced force in the direction of the force."

Let's break this statement down:

Momentum:

The term 'momentum' is crucial here. Momentum ($ \vec{p} $) is defined as the product of an object's mass ($ m $) and its velocity ($ \vec{v} $).

Momentum is a vector quantity, having the same direction as the velocity. Its SI unit is kilogram-metre per second (kg m/s).

$ \vec{p} = m\vec{v} $

Rate of Change of Momentum:

This refers to how quickly the momentum of the object changes over time. Mathematically, the rate of change of momentum is $ \frac{\Delta \vec{p}}{\Delta t} $ for a finite time interval $ \Delta t $, or $ \frac{d\vec{p}}{dt} $ in calculus form for an instantaneous rate of change.

Directly Proportional to Applied Unbalanced Force:

The law states that the rate of change of momentum is proportional to the net external force ($ \vec{F}_{net} $) acting on the object. The change in momentum occurs in the same direction as this net force.

So, we can write: $ \vec{F}_{net} \propto \frac{d\vec{p}}{dt} $

Mathematical Formulation Of Second Law Of Motion

We can write the proportionality as an equation by introducing a constant of proportionality, $ k $:

$ \vec{F}_{net} = k \frac{d\vec{p}}{dt} $

In SI units, the unit of force (Newton) is defined in such a way that the constant $ k $ is equal to 1. Thus, the equation becomes:

$ \vec{F}_{net} = \frac{d\vec{p}}{dt} $

Now, substituting $ \vec{p} = m\vec{v} $ into the equation:

$ \vec{F}_{net} = \frac{d(m\vec{v})}{dt} $

For most common situations at speeds much less than the speed of light, the mass ($ m $) of the object remains constant over time ($ \frac{dm}{dt} = 0 $). In this case, we can take the mass out of the differentiation:

$ \vec{F}_{net} = m \frac{d\vec{v}}{dt} $

The term $ \frac{d\vec{v}}{dt} $ is the rate of change of velocity, which is the definition of acceleration ($ \vec{a} $).

Thus, we arrive at the most commonly used form of Newton's Second Law:

$ \vec{F}_{net} = m\vec{a} $

This equation is profound. It tells us:

Unit of Force: Newton

The SI unit of force is the Newton (N). From the formula $ F = ma $, we can define 1 Newton:

One Newton (1 N) is the amount of force that, when applied to an object with a mass of 1 kilogram (1 kg), produces an acceleration of 1 metre per second squared (1 m/s$^2$) in the direction of the force.

$ 1 \, \text{N} = 1 \, \text{kg} \times 1 \, \text{m/s}^2 $

Examples illustrating the Second Law:

Example 1. A net force of 10 N acts on a trolley of mass 2 kg initially at rest. What is the acceleration of the trolley?

Answer:

Using Newton's Second Law, $ F_{net} = ma $.

Given: $ F_{net} = 10 $ N, $ m = 2 $ kg.

We need to find the acceleration, $ a $.

Rearranging the formula: $ a = \frac{F_{net}}{m} $

Substitute the given values: $ a = \frac{10 \, \text{N}}{2 \, \text{kg}} = 5 \, \text{m/s}^2 $.

The trolley accelerates at $ 5 \, \text{m/s}^2 $ in the direction of the applied net force.

Example 2. What net force is required to accelerate a 1500 kg car from rest to a speed of 20 m/s in 5 seconds on a level road?

Answer:

First, we need to find the acceleration ($ a $) of the car. We can use a kinematic equation: $ v = u + at $.

Given: initial velocity $ u = 0 $ m/s (from rest), final velocity $ v = 20 $ m/s, time $ t = 5 $ s.

Rearranging for $ a $: $ a = \frac{v - u}{t} = \frac{20 \, \text{m/s} - 0 \, \text{m/s}}{5 \, \text{s}} = \frac{20}{5} \, \text{m/s}^2 = 4 \, \text{m/s}^2 $.

Now, use Newton's Second Law: $ F_{net} = ma $.

Given: $ m = 1500 $ kg, $ a = 4 $ m/s$^2$.

Substitute the values: $ F_{net} = (1500 \, \text{kg}) \times (4 \, \text{m/s}^2) = 6000 \, \text{N} $.

A net force of 6000 N is required to produce this acceleration.

Newton's Second Law is fundamental to understanding and solving problems involving forces and motion in classical mechanics.


Third Law Of Motion

Newton's Third Law of Motion describes the interaction between two objects when a force is exerted. It states that forces always occur in pairs.

Statement of Newton's Third Law

Newton's Third Law of Motion states:

"To every action, there is always an equal and opposite reaction."

In simpler terms, for every force that one object exerts on a second object, the second object simultaneously exerts a force of equal magnitude and opposite direction on the first object.

Let's consider two objects, A and B. If object A exerts a force $ \vec{F}_{A \text{ on } B} $ on object B, then object B simultaneously exerts a force $ \vec{F}_{B \text{ on } A} $ on object A. According to the Third Law:

$ \vec{F}_{A \text{ on } B} = - \vec{F}_{B \text{ on } A} $

The negative sign indicates that the forces are in opposite directions. The magnitudes of the forces are equal: $ |\vec{F}_{A \text{ on } B}| = |\vec{F}_{B \text{ on } A}| $.

Key Characteristics of Action-Reaction Pairs

Understanding these points is crucial to correctly applying the Third Law:

Why Action-Reaction Forces Don't Cancel

Because the action and reaction forces act on different objects, they cannot cancel each other out. Forces can only cancel if they act on the same object. For example, when you push a wall (action), the wall pushes back on you (reaction). The force you exert is on the wall, and the force the wall exerts is on you. These two forces act on different objects (the wall and you), so they cannot cancel to produce zero net force on either object.

Examples Illustrating the Third Law

Example 1. When you walk, how does Newton's Third Law explain your motion?

Answer:

When you walk, you push backward on the ground with your feet. This is the action force. According to Newton's Third Law, the ground simultaneously pushes forward on your feet with an equal and opposite reaction force. It is this reaction force from the ground on your feet that propels you forward. If you were to walk on a frictionless surface (like ice), you wouldn't be able to push backward effectively, and thus there would be no significant forward reaction force from the surface, making it difficult to walk.

Example 2. A book is resting on a table. Identify the action-reaction forces related to the force of gravity on the book.

Answer:

The Earth exerts a gravitational force pulling the book downwards. This is an action force (Let's call it $ \vec{F}_{\text{Earth on Book}} $). According to Newton's Third Law, the book simultaneously exerts an equal and opposite gravitational force pulling the Earth upwards ($ \vec{F}_{\text{Book on Earth}} $).

$ \vec{F}_{\text{Earth on Book}} = - \vec{F}_{\text{Book on Earth}} $

Note that the normal force exerted by the table on the book is not the reaction force to the Earth's gravitational pull on the book. The normal force is part of a different action-reaction pair involving the contact between the book and the table (Book pushes down on table, table pushes up on book).

Example 3. Explain the recoil of a gun when a bullet is fired using Newton's Third Law.

Answer:

When the trigger of a gun is pulled, an explosion inside the cartridge propels the bullet forward. The gun exerts a force on the bullet, pushing it out of the barrel (action force). According to Newton's Third Law, the bullet simultaneously exerts an equal and opposite force on the gun, pushing the gun backward. This backward force on the gun is called recoil. Since the gun's mass is much larger than the bullet's mass, the gun's acceleration backward (recoil velocity) is much smaller than the bullet's acceleration forward, but the forces are equal in magnitude.

Example 4. How do rockets move in space where there is no air to push against?

Answer:

Rockets work based on Newton's Third Law. The rocket engine expels hot gases at very high speed downwards (action force on the gases). According to the Third Law, the expelled gases exert an equal and opposite force upwards on the rocket (reaction force). This upward reaction force is the thrust that propels the rocket forward, even in the vacuum of space where there is no air or ground to push against.

Newton's Third Law is essential for understanding interactions between objects and is closely related to the principle of conservation of momentum.


Conservation Of Momentum

The Law of Conservation of Linear Momentum is one of the fundamental principles in physics. It is a direct consequence of Newton's laws of motion, particularly the Third Law.

Definition of Momentum

As introduced earlier, momentum ($ \vec{p} $) of a single object is the product of its mass ($ m $) and velocity ($ \vec{v} $):

$ \vec{p} = m\vec{v} $

For a system consisting of multiple objects, the total momentum of the system is the vector sum of the momenta of all the individual objects.

For a system of $ n $ objects with masses $ m_1, m_2, \ldots, m_n $ and velocities $ \vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n $, the total momentum $ \vec{P}_{total} $ is:

$ \vec{P}_{total} = \vec{p}_1 + \vec{p}_2 + \ldots + \vec{p}_n = m_1\vec{v}_1 + m_2\vec{v}_2 + \ldots + m_n\vec{v}_n $

Statement of the Law of Conservation of Linear Momentum

The Law of Conservation of Linear Momentum states:

"In an isolated system, the total linear momentum of the system remains constant."

Let's define what an isolated system is in this context:

An isolated system is a system of objects upon which no external unbalanced force acts. The forces acting between the objects within the system (internal forces) do not affect the total momentum of the system. Only external forces can change the total momentum of a system.

Derivation from Newton's Laws (for a two-body system)

Consider a system of two objects, A and B, that interact with each other (e.g., collide). Assume this is an isolated system, meaning no external forces are acting on either A or B.

Let $ \vec{F}_{A \text{ on } B} $ be the force exerted by A on B, and $ \vec{F}_{B \text{ on } A} $ be the force exerted by B on A.

According to Newton's Third Law, these forces form an action-reaction pair:

$ \vec{F}_{A \text{ on } B} = - \vec{F}_{B \text{ on } A} $

Now, let's apply Newton's Second Law ($ \vec{F} = \frac{d\vec{p}}{dt} $) to each object:

The force on object B is $ \vec{F}_{A \text{ on } B} $, so the rate of change of momentum of B is:

$ \vec{F}_{A \text{ on } B} = \frac{d\vec{p}_B}{dt} $

The force on object A is $ \vec{F}_{B \text{ on } A} $, so the rate of change of momentum of A is:

$ \vec{F}_{B \text{ on } A} = \frac{d\vec{p}_A}{dt} $

Substitute these into the Third Law equation:

$ \frac{d\vec{p}_B}{dt} = - \frac{d\vec{p}_A}{dt} $

Rearranging the equation, we get:

$ \frac{d\vec{p}_A}{dt} + \frac{d\vec{p}_B}{dt} = 0 $

The sum of the rates of change is equal to the rate of change of the sum:

$ \frac{d}{dt} (\vec{p}_A + \vec{p}_B) = 0 $

The total momentum of the system is $ \vec{P}_{total} = \vec{p}_A + \vec{p}_B $. So, the equation is:

$ \frac{d\vec{P}_{total}}{dt} = 0 $

This means that the rate of change of the total momentum of the system is zero. A quantity whose rate of change is zero is constant.

Therefore, for an isolated system:

$ \vec{P}_{total} = \text{constant} $

The total momentum before the interaction (e.g., collision) is equal to the total momentum after the interaction.

$ \vec{p}_{A, \text{initial}} + \vec{p}_{B, \text{initial}} = \vec{p}_{A, \text{final}} + \vec{p}_{B, \text{final}} $

Or, in terms of mass and velocity:

$ m_A \vec{v}_{A, \text{initial}} + m_B \vec{v}_{B, \text{initial}} = m_A \vec{v}_{A, \text{final}} + m_B \vec{v}_{B, \text{final}} $

Applications and Examples

The conservation of momentum is a powerful principle used to analyse various situations, especially collisions and explosions, where internal forces are dominant and external forces can often be neglected or considered negligible over the short duration of the event.

Example 1. A 5 kg object moving at 10 m/s collides head-on with a 2 kg object moving at 3 m/s in the opposite direction. If they stick together after the collision, what is their common velocity?

Answer:

This is a collision, and assuming no significant external forces (like friction) during the brief moment of collision, the system of the two objects is isolated. Momentum is conserved.

Let object 1 be the 5 kg object and object 2 be the 2 kg object.

Initial momentum of object 1: $ \vec{p}_{1i} = m_1 \vec{v}_{1i} $. Let's take the direction of the 5 kg object's motion as positive.

$ \vec{p}_{1i} = (5 \, \text{kg}) \times (10 \, \text{m/s}) = 50 \, \text{kg m/s} $ (in the positive direction)

Initial momentum of object 2: $ \vec{p}_{2i} = m_2 \vec{v}_{2i} $. The 2 kg object moves in the opposite direction, so its velocity is -3 m/s.

$ \vec{p}_{2i} = (2 \, \text{kg}) \times (-3 \, \text{m/s}) = -6 \, \text{kg m/s} $ (in the negative direction)

Total initial momentum of the system: $ \vec{P}_{total, i} = \vec{p}_{1i} + \vec{p}_{2i} = 50 \, \text{kg m/s} + (-6 \, \text{kg m/s}) = 44 \, \text{kg m/s} $.

After the collision, the two objects stick together, forming a single combined object with a total mass $ m_{total} = m_1 + m_2 = 5 \, \text{kg} + 2 \, \text{kg} = 7 \, \text{kg} $. Let their common final velocity be $ \vec{v}_f $.

Total final momentum of the system: $ \vec{P}_{total, f} = m_{total} \vec{v}_f = (7 \, \text{kg}) \vec{v}_f $.

By the Law of Conservation of Momentum:

$ \vec{P}_{total, i} = \vec{P}_{total, f} $

$ 44 \, \text{kg m/s} = (7 \, \text{kg}) \vec{v}_f $

$ \vec{v}_f = \frac{44 \, \text{kg m/s}}{7 \, \text{kg}} \approx 6.29 \, \text{m/s} $

The common velocity of the combined object after the collision is approximately $ 6.29 \, \text{m/s} $ in the positive direction (the direction of the initial 5 kg object).

Example 2. A bomb of mass 10 kg at rest explodes into two fragments. A 4 kg fragment moves with a velocity of 6 m/s. What is the velocity of the other fragment?

Answer:

Initially, the bomb is at rest, so its total momentum is zero ($ \vec{P}_{initial} = m_{bomb} \times \vec{v}_{initial} = 10 \, \text{kg} \times 0 \, \text{m/s} = 0 $).

After the explosion, the bomb breaks into two fragments. Let fragment 1 have mass $ m_1 = 4 $ kg and velocity $ \vec{v}_1 = 6 $ m/s. The mass of the other fragment (fragment 2) is $ m_2 = m_{bomb} - m_1 = 10 \, \text{kg} - 4 \, \text{kg} = 6 $ kg. Let its velocity be $ \vec{v}_2 $.

The explosion is an internal process, and we can consider the bomb as an isolated system just before and after the explosion (assuming gravity is negligible during the explosion event). Momentum is conserved.

Total final momentum: $ \vec{P}_{final} = \vec{p}_1 + \vec{p}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2 $.

By Conservation of Momentum:

$ \vec{P}_{initial} = \vec{P}_{final} $

$ 0 = m_1 \vec{v}_1 + m_2 \vec{v}_2 $

Let's assume the 4 kg fragment moves in the positive direction ($ v_1 = +6 $ m/s). Then:

$ 0 = (4 \, \text{kg})(+6 \, \text{m/s}) + (6 \, \text{kg})\vec{v}_2 $

$ 0 = 24 \, \text{kg m/s} + (6 \, \text{kg})\vec{v}_2 $

$ (6 \, \text{kg})\vec{v}_2 = -24 \, \text{kg m/s} $

$ \vec{v}_2 = \frac{-24 \, \text{kg m/s}}{6 \, \text{kg}} = -4 \, \text{m/s} $

The velocity of the other fragment (6 kg) is $ 4 \, \text{m/s} $ in the opposite direction to the 4 kg fragment.

Conservation of momentum is a consequence of the fact that forces between interacting particles are equal and opposite (Newton's Third Law). It holds true regardless of the type of interaction (collisional or non-collisional) as long as the system is isolated.