## Introduction

Over the last 300 years, Newton’s laws of motion became one of the most famous laws of physics. The laws were stated by a famous English physicist and a mathematician, and his name is Sir Isaac Newton.

Here is the three (3) Newton’s laws of motion:

**A body will continue to be at rest or in a motion, until it’s acted upon by an external force. This is also known as the law of inertia (“Inertia of a body is its reluctance to start moving when at rest or stationary and its reluctance to stop moving once it is moving”).****The rate of change of momentum is proportional to the external force and takes place in the direction of the applied force.****When an external force is applied on a body, there is always an equal and opposite reaction. The law can also be defined as to every action, there is an equal and opposite reaction.**

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### Brief History of the Laws of Motion

Newton was born in England on 4th of January, 1643 in Woolsthorpe Manor House. During his time, children were trained to rear sheeps to earn a living. However, Newton was uninterested in making a living as a farmer. Instead, he was more attracted to books.

At young age, he found love in the books that were arranged in the shelves of his parent’s library. The famous scientist dedicated his time to observing the possible occurrences that happen between sunrise to sunset, spring to summer, autumn to winter. Newton tried to understand what triggers these changes, and it led him to understanding the gravity.

He observed the effect of sun, wind and water on materials like water clocks, windmills, and kites. Isaac finally realizes that the earth and the sky work on the same principle which is contrary to Aristotle’s understanding.

The influential English scientist devoted himself to reading the works of ancient Greeks, Islamic and medieval views on gravity. Newton asked himself as to why an apple fruit falls on the ground instead of going up to the sky. He provided mathematical evidence to Galilei Galileo and Johannes Kepler’s theory of gravity.

Isaac Newton started his journey of research at the age of 23. He came up with a book on classical mechanics, and the name of the book is “Principia Mathematica Philosophiae Naturalis.”. The meaning of the book in English is “Mathematical principles of natural philosophy”.

This book perfectly explained the 3 laws of motion. The three (3) laws of motion shaped our understanding of gravity and we call them Newton’s laws of motion.

### Newton’s Laws of Motion in our Everyday Life

We apply Newton’s laws of motion in our everyday life. For example, while playing soccer, we must apply force for the ball to move. Once we apply force on the ball, it will head towards our target due to momentum. The wide range of applications of newton’s laws are as follows:

- Basketball
- Rocket science
- Javelin throw
- Table tennis
- Squash sport
- Riding Bicycle
- Driving a Bus on the road
- Eating with a Spoon
- An airplane taking off
- A bag resting on a desk, and many more.

In this article, I will explain Newton’s laws of motion in detail.

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## Newton’s First (1st) Law of Motion: Law of Inertia

The first law of motion states that **A body will continue to be at rest or in motion, until it’s acted upon by an external force.** According to this law, when you hide an object in your room. The object will continue to remain in that place for millions of years, until someone or something applies a force to move the object from its original hiding position.

Therefore, when you keep something in a safe place. We can easily say that the object will remain in that place until you decide to change its position or someone moves it from its original position.

Additionally, the same law state that when you set your car in motion. The car will continue to move in a straight line and will never stop. It will wait for you to apply brakes before it can stop. Additionally, the car may also stop if the engine is bad or the fuel is empty. Hence, there must be an application of external force for the car to stop.

Also, if you don’t control the steering, it will continue to move in a straight line until you decide to go left or right. This scenario also explains Newton’s first law of motion or law of inertia.

### Understanding Inertia

Newton’s first law of motion is also known as the law of inertia. As defined in his book, **the law of inertia is the power by which an object, if it’s at rest, stays at rest, or if in motion, will travel in a straight line.**

To understand this law, try rolling a marble on smooth tiles, the marble will continue to move until an object blocks its movement. Alternatively, the friction of the tiles will make the speed of the marble keep decreasing until it comes to rest.

This scenario of a rolling marble against the tiles perfectly describes Newton’s first law of motion.

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## Newton’s Second (2nd) Law of Motion

The second law of motion states that when you apply an external force on an object **the rate of change of momentum is proportional to the applied force and it takes place in the direction of the force.**

Here is the best way to explain this law, when you push a shopping cart in the supermarket. The shopping cart will move because of the external force you apply on it, and the mass of the cart.

Therefore, the mass of the , mass of the body and the external force acting upon it plays a crucial role in setting a body into motion.

Here is another example, before buying anything, push the cart for a while. You will notice that it moves faster. Now, add some of the items you bought from the store into the cart. What did you notice? It’s heavier than it was, right? You may need to add extra force to push it this time around.

Ask yourself again, what brought this sudden change? Here is the answer to your question: “The more mass you add to the cart, the heavier it becomes, and the more force you will need to apply to move the cart”.

Hence, there exists a relationship between the mass of the body and the applied external force.

The mathematical expression (formula) for newton’s second law of motion is

Force (F) = Mass (m) x Acceleration (a)

Which can be written as

F = ma

The S.I unit of force is called Newton (Honors Isaac Newton) or kilogram meter per second square. The symbol of Newton is N, while the symbol for kilogram meters per second square is kgm/s^{2}

The unit of mass is in kilograms (kg)

The unit of acceleration is in meter per second square (m/s^{2})

### How to Derive Newton’s Second Law

According to Newton’s second law, the rate of change of momentum is proportional to the applied force. Therefore, this law can be translated into

Force (F) = change in momentum / time (t)

Change in momentum is the product of mass and velocity.

Thus,

Change in momentum = mass x velocity

Which can be further broken down into

Change in momentum = mv – mu [where v = final velocity and u = initial velocity]

Therefore, we can insert (Change in momentum = mv – mu) into [Force = change in momentum / time] to obtain

F = (mv – mu) / t

Hence,

F = m (v – u) / t

And acceleration is the rate of change of displacement

Acceleration, a = (v – u) / t

And we have F = ma which is Newton’s second law of motion.

### Momentum

This is the product mass and velocity of an object. The S.I unit of momentum is kilogram meter per second (kgm/s). The formula for calculating momentum is

Momentum (M) = mass (m) x velocity (v)

The rate of change of momentum = (change in momentum) / time

### Impulse of a Force

The impulse of a force is the instantaneous response of an object to the force that acts on it. We can also define impulse (I) as the product of the force and the time it takes momentum to change. The S.I unit of impulse is Newton-second (Ns).

Impulse (I) = Force (F) x Time (t)

### Calculations Involving Second Law of Motion

Here are some questions to help you understand the applications of Newton’s second law of motion to solve a problem.

#### Problem 1

An object of mass 7.5 kilograms has a force applied to it that pulls eastward with a magnitude of 50 Newtons and another applied to it that pulls westward with a magnitude of 150 Newtons. What is the westward acceleration of the object? Give your answer to one decimal place.

**Data**

Mass of the object is m = 7.5 N

Initial Force, F1 = 50 Newtons

Final force, F2 = 150 Newtons

**Unknown**

Acceleration of the object, a = ?

**Formula**

The formula we need to apply to solve this problem is

F_{Net} = ma

But F_{Net} = F2 – F1

Therefore, w can further break it down into

F_{2} – F_{1} = ma

##### Solution

We can now insert our data into F_{2} – F_{1} = ma

150 – 50 = 7.5 x a

Make a subject of the formula

a = (150 – 50) / 7.5

Hence,

a = 100/7.5 = 13 m/s^{2}

#### Problem 2

An object of mass 3 kilograms has a force of 9 Newtons applied to it. At what rate does the force accelerate the object?

**Data**

Mass of the object, m = 3 kg

Applied force, F = 9 N

**Unknown**

Acceleration, a = ?

**Formula**

The formula that can help us to solve this problem is

a = F/m

##### Solution

We can substitute our formula with our data

Acceleration, a = 9/3 = 3 m/s2

**Therefore, the acceleration (a) is 3 meters per second square.**

#### Problem 3

A boy pushes his toy car with a cat in inside. The mass of the toy car and the cat put together is 70 kilograms. The toy car accelerates at 0.9 meters per second square. Find the force that the boy applies to pull the toy car and the cat together.

**Data**

Mass of the toy car with a cat is m = 70 kg

Acceleration of the toy car, a = 0.9 m/s2

**Unknown**

Force, F = ?

**Formula**

We will use the formula F = ma to solve the problem

##### Solution

Insert the data into the above formula

F = 70 x 0.9 = 63 N

**Therefore, the applied force is 63 Newtons**

## Third Newton’s Law of Motion

The third Newton’s law of motion states that action and reaction are equal and opposite. This law explains that when a force is applied to a body, there is a chain of equal and opposite reactions that will follow.

For example, when you blow a balloon, and later release the air entrance. The balloon will run away from you.

What if we decide to insert the balloon inside a toy canoe? We will blow the balloon and position its air entrance to face the opposite direction of the canoe’s motion. After releasing the air entrance of the balloon, it will set the toy canoe in motion on water.

Therefore, we can see that when an object X applies a force on another object Y. Object Y will equally apply the same type of force on X. Both applied forces are equal in magnitude but opposite in direction.

We can demonstrate this mathematically as

Force X = – Force Y [ Force X is the action force, and Force Y is the reaction Force]

Another example of Newton’s third law of motion is how the rocket operates. When a rocket is launched into space, it will accelerate away from the earth surface. This is thanks to the burning fuel of the rocket that kept igniting fire at the back of the rocket to push it in the upward direction.

We have many examples to demonstrate the effect of this law in our everyday life. Among these examples are:

- The lift of an airplane
- Student’s school bag on the desk
- A hammer hitting a nail
- A bird flying
- A boy pushing against the wall
- Applying brakes on your car
- A duck swimming in the water

### Principle of Conservation of Momentum

We can state the principle of conservation of momentum in three ways:

- The principle states that when two bodies that are traveling in a straight line collide, the total force before collision is equal to the total force after collision provided no external force acts on the body.
- In any system of colliding bodies, the total momentum is always conserved provided that there is no net external force acting on the system.
- The total momentum of an isolated or closed system of colliding bodies remains constant.

Therefore, the mathematical definition of this principle is

m_{1}v_{1} = m_{2}v_{2} [Where m_{1}v_{1} is the initial momentum, and m_{2}v_{2} is the final momentum

Now, if we have two balls M and N exerting equal forces on each other. We can say that

F_{M} = – F_{N}

And since F = ma

Then, it’s easy to say that

m_{M}a_{M} = – m_{N}a_{N}

Additionally, a = (v – u) / t

Which implies

m_{M} [ (v_{m} – u_{m}) / t ] = – m_{N} [ (v_{N} – u_{N}) / t ]

And we will end up with

m_{M}v_{m} – m_{M}u_{m} = – ( m_{N}v_{N} – m_{N}u_{N})

The above equation will give us

m_{M}u_{m} + m_{N}u_{N} = m_{M}v_{m} + m_{N}u_{N}

**And the above expression shows that momentum before collision is equal to the momentum after collision.**

### Applications of Newton’s Third Law to Solve a Problem

Here are a few problems to help you understand how to solve problems that are related to newtons third law of motion, momentum, and impulse.

#### Problem 1

A body (P) has mass of 5 kilograms moving with a velocity of 30 m/s collides with another body (Q), moving in opposite direction with a velocity of 20 m/s. If both bodies now move in the direction of P at a velocity of 10 m/s. Calculate the mass of Q.

**Solution**

To solve this problem, we need to remember that

m_{1}v_{1} = m_{2}v_{2}

Which implies that

(m_{1} + m_{2})v = m_{1}u_{1} – m_{2}u_{2}

and m_{1}u_{1} – m_{2}u_{2} = Before collision

(m_{1} + m_{2})v = After collision

Therefore, since m_{1} = 5kg, u_{1} = 30 m/s, m_{2} = ? , u_{2} = 20 m/s, v = 10 m/s

We can now insert the above information into (m_{1} + m_{2})v = m_{1}u_{1} – m_{2}u_{2} to obtain

(5 + m_{2}) x 10 = (5 x 30) – (m_{2} x 20)

We will now have

50 + 10m_{2} = 150 – 20m_{2}

We can now collect like terms to get

150 – 50 = 20m_{2} + 10m_{2}

And it will become

100 = 30m_{2}

By making m_{2} subject of the formula, we will have

m_{2} = 100 / 30 = 3 kg

**Therefore, the mass of Q is 3 kilograms.**

#### Problem 2

A bullet of mass 0.05 kg is fired horizontally into a 10 kg block which is fired free to move. If both bullet block move with velocity 0.5 m/s after the impact, find the velocity with which the bullet hit the body.

**Data**

Mass of the bullet = 0.05 kg, mass of the block = 10 kg, and the mass of both bullet and the block = 10 + 0.05 = 10.05

Initial mass of the bullet = u

The momentum before collision = Momentum after collision

0.05 x u = 10.05 x 0.5

Hence,

u = 5.025 / 0.05 = 100.5 = 101 m/s

#### Problem 3

A rifle of mass 15 kilograms fires a bullet of mass 60 grams with a velocity of 200 m/s. Calculate the recoil velocity of the rifle.

##### Solution

m_{1} = 15 kg, and m_{2} = 60 g = (60/1000) kg = 0.06 kg,

v_{1} = ? and v_{2} = 200 m/s

We need to remember that Momentum before impact = Momentum after impact

Momentum before impact = 0

And the Momentum after impact = m_{1} v_{1} + m_{2} v_{2}

Hence,

m_{1} v_{1} + m_{2} v_{2} = 0

By inserting our data, we will have

15 x v_{1} + 0.06 x 200 = 0

Which will become

15 v_{1} + 12 = 0

After making v_{1} subject of the formula

v_{1} = – (12/8) = – 0.8 m/s

**Therefore, the recoil velocity is 0.8 meters per second **

## Isaac Newton Defined the Following Quantities in His Book

**Matter**: Anything that takes up space

**Mass**: The measure of a quantity or amount of matter

**Momentum**: The quantity of motion, which is the product of velocity (speed) and mass

**Inertia**: The power by which an object, if it is at rest, stays at rest, or if in motion, will travel in a straight line

**Force**: An action applied upon a body

**Centripetal Force**: An attraction toward the center of something (as in gravity)

We apply Newton’s laws of motion in our everyday activities.

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