Skip to main content
Admissions Open 2026-27Ravindra Higher Secondary School (Est. 1988) | Waidhan, Singrauli (MP)
+91 9826986106• Student Portal• Study Notes
Ravindra Higher Secondary School Logo
Ravindra Higher Secondary SchoolWaidhan, Singrauli (M.P.)
Home
Contact
Home
Study Portal
Class 9 Science
Force and Laws of Motion - Newton's three laws of motion, concept of inertia, momentum, and real-life applications
Back to All Study GuidesOpen in Interactive App
ScienceClass 9Force and Laws of Motion

Force and Laws of Motion - Newton's three laws of motion, concept of inertia, momentum, and real-life applications

2026-09-0711 min readRHS Academic Faculty
Overview & Key Summary:Chapter Guide: Force and Laws of Motion (Class 9 Science) Have you ever wondered why you tend to fall forward when a fastmoving bus suddenly slams its brakes? Or why a cricket fi...

Chapter Guide: Force and Laws of Motion (Class 9 Science)

Have you ever wondered why you tend to fall forward when a fast-moving bus suddenly slams its brakes? Or why a cricket fielder pulls his hands backward while taking a high catch? Sir Isaac Newton observed these exact everyday occurrences and gave us three magical rules—known as Newton’s Laws of Motion—that explain how everything around us moves.

By the end of this tutorial, you will master these laws, understand inertia and momentum, and easily solve numerical problems just like a physics pro! Let's dive in.


1. What is Force?

In simple terms, a force is a push or a pull exerted on an object.

When you push a heavy box, kick a football, or pull open a door, you are applying a force.

Effects of Force

A force cannot be seen, but its effects can be observed. A force can:

  1. Change the state of rest or motion of a body (make a stationary ball move, or stop a rolling ball).
  2. Change the speed of a moving body.
  3. Change the direction of motion.
  4. Change the shape and size of an object (like stretching a rubber band or squeezing a sponge).

Balanced vs. Unbalanced Forces

  • Balanced Forces: When two or more equal forces act on an object in opposite directions, the net force (FnetF_{\text{net}}Fnet​) is zero. Balanced forces do not change the state of rest or uniform motion of an object.
    • Example: A game of tug-of-war where both teams pull with equal strength—the rope doesn't move!
  • Unbalanced Forces: When the forces acting on an object are unequal, the net force is greater than zero. An unbalanced force causes a change in speed, direction, or state of rest.
    • Example: Pushing a stalled car—if your push overcomes friction, the car accelerates.

2. The Concept of Inertia and Newton's First Law

Before Sir Isaac Newton, the great Italian scientist Galileo Galilei suggested that objects continue to move with constant speed if no external force acts on them. Newton refined this concept into his First Law of Motion.

What is Inertia?

Inertia is the natural tendency of an object to resist any change in its state of rest or uniform motion. In simple words, objects are lazy—they want to keep doing whatever they are already doing!

Mass is the measure of Inertia: Heavy objects have more inertia than light objects. It is much harder to push a massive truck than a small bicycle because the truck has greater mass and, therefore, greater inertia.

Types of Inertia

  1. Inertia of Rest: The tendency of an object to remain at rest.
    • Example: When a bus suddenly starts, passengers jerk backward. Why? The lower part of the body in contact with the bus moves forward, but the upper body wants to remain at rest due to inertia of rest.
  2. Inertia of Motion: The tendency of an object to remain in uniform motion.
    • Example: When a running bus stops suddenly, passengers jerk forward. The feet come to rest with the bus, but the upper body keeps moving forward due to inertia of motion.
  3. Inertia of Direction: The tendency of an object to maintain its direction of motion.
    • Example: When a car takes a sharp turn, passengers lean outwards because their body tries to continue moving in a straight line.

Newton's First Law of Motion

Statement: An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.

Because this law defines the property of inertia, it is also called the Law of Inertia.


3. Momentum: The "Quantity of Motion"

Imagine a table tennis ball hitting your arm—it doesn't hurt. But if a cricket ball hits your arm at the same speed, it hurts a lot! Why? Because the cricket ball has more mass.

Now imagine a bullet thrown by hand versus a bullet fired from a gun. The gun bullet can penetrate deeply because of its extremely high velocity.

This tells us that the effect of a force depends on both mass (mmm) and velocity (vvv). Scientists combined these two into a single physical quantity called Momentum.

Definition & Formula

Momentum (ppp) of an object is defined as the product of its mass (mmm) and its velocity (vvv).

Momentum (p)=mass (m)×velocity (v)\text{Momentum } (p) = \text{mass } (m) \times \text{velocity } (v)Momentum (p)=mass (m)×velocity (v)

p=m⋅vp = m \cdot vp=m⋅v

  • SI Unit: kilogram-meter per second (kg⋅m/s\text{kg}\cdot\text{m/s}kg⋅m/s)
  • Nature: It is a vector quantity (it has both magnitude and direction, pointing in the same direction as the velocity).

4. Newton's Second Law of Motion

While the First Law tells us what happens when no force acts, the Second Law tells us how much force is needed to produce a change in motion.

Statement: The rate of change of momentum of an object is directly proportional to the applied unbalanced force in the direction of the force.

Mathematical Derivation of F=maF = maF=ma

Let an object of mass mmm have an initial velocity uuu. An unbalanced force FFF is applied on it for time ttt, changing its velocity to vvv.

  1. Initial momentum (p1p_1p1​) = m⋅um \cdot um⋅u
  2. Final momentum (p2p_2p2​) = m⋅vm \cdot vm⋅v
  3. Change in momentum (Δp\Delta pΔp) = p2−p1=m(v−u)p_2 - p_1 = m(v - u)p2​−p1​=m(v−u)
  4. Rate of change of momentum = m(v−u)t\frac{m(v - u)}{t}tm(v−u)​

According to Newton's Second Law: F∝m(v−u)tF \propto \frac{m(v - u)}{t}F∝tm(v−u)​

Since acceleration a=v−uta = \frac{v - u}{t}a=tv−u​, we can write: F∝m⋅aF \propto m \cdot aF∝m⋅a

To convert the proportionality into an equation, we insert a constant kkk: F=k⋅m⋅aF = k \cdot m \cdot aF=k⋅m⋅a

In the SI system, the unit of force is chosen such that k=1k = 1k=1. Therefore:

F=m⋅aF = m \cdot aF=m⋅a

Force=Mass×Acceleration\text{Force} = \text{Mass} \times \text{Acceleration}Force=Mass×Acceleration

Units of Force

  • SI Unit: kg⋅m/s2\text{kg}\cdot\text{m/s}^2kg⋅m/s2, which is named Newton (N\text{N}N) in honor of Sir Isaac Newton.
  • 1 Newton Definition: 1 N1\text{ N}1 N is the force that produces an acceleration of 1 m/s21\text{ m/s}^21 m/s2 in a body of mass 1 kg1\text{ kg}1 kg. 1 N=1 kg×1 m/s21\text{ N} = 1\text{ kg} \times 1\text{ m/s}^21 N=1 kg×1 m/s2

Real-Life Application of Newton's Second Law

  • Catching a Cricket Ball: A fielder pulls his hands backward while catching a fast ball. By increasing the time (ttt) taken to stop the ball, he reduces the rate of change of momentum (Δpt\frac{\Delta p}{t}tΔp​), which significantly decreases the force (FFF) exerted on his hands, preventing injury!

5. Newton's Third Law of Motion

Statement: To every action, there is always an equal and opposite reaction, and they act on two different bodies.

Force exerted by A on B (FAB)=−Force exerted by B on A (FBA)\text{Force exerted by A on B } (F_{AB}) = -\text{Force exerted by B on A } (F_{BA})Force exerted by A on B (FAB​)=−Force exerted by B on A (FBA​)

Crucial Point to Remember!

Action and reaction forces are equal in magnitude and opposite in direction, but they NEVER cancel each other out because they act on two different objects.

Real-World Examples of the Third Law:

  1. Walking on the floor: You push the ground backward with your foot (Action), and the ground pushes your foot forward with equal force (Reaction).
  2. Recoil of a Gun: When a bullet is fired from a gun, it exerts a forward force on the bullet (Action). The bullet exerts an equal backward force on the gun (Reaction), causing the gun to recoil.
  3. Rowing a Boat: The rower pushes the water backward with oars (Action), and the water pushes the boat forward (Reaction).
  4. Rocket Launch: Hot gases produced by burning fuel rush downwards out of the nozzle (Action), pushing the rocket upwards into space (Reaction).

Quick Summary Table

LawPopular NameCore IdeaKey FormulaEveryday Example
1st LawLaw of InertiaObjects keep doing what they are doing unless a force acts.Net F=0⇒a=0F = 0 \Rightarrow a = 0F=0⇒a=0Dust flying off a beaten carpet
2nd LawLaw of Force & AccelerationForce equals rate of change of momentum.F=maF = maF=maFielder catching a high cricket ball
3rd LawLaw of Action & ReactionForces always exist in equal & opposite pairs on different bodies.FAB=−FBAF_{AB} = -F_{BA}FAB​=−FBA​Recoil of a heavy rifle when fired


Common Student Mistakes to Avoid

  1. Confusing Key Terminology: Interchanging closely related scientific terms (e.g. mass vs. weight, reflection vs. refraction, or oxidation vs. reduction).
  2. Incomplete Chemical Equations or Formulas: Forgetting to balance chemical equations or omitting physical states (s, l, g, aq) in reaction steps.
  3. Diagram Labeling Errors: Drawing scientific diagrams without proper arrows showing light rays, electric current flow, or organ functions.
  4. Neglecting SI Units in Physics Problems: Calculating work, force, or energy without converting values into standard SI units first.

Practice Questions with Detailed Solutions

Let's test your understanding with these NCERT-standard questions!

Question 1 (Numerical - Second Law)

A constant force of 5 N5\text{ N}5 N acts on a body of mass m1m_1m1​, producing an acceleration of 10 m/s210\text{ m/s}^210 m/s2. The same force produces an acceleration of 20 m/s220\text{ m/s}^220 m/s2 when applied to another mass m2m_2m2​. What acceleration would this force produce if both masses were tied together?

Solution:

  • Step 1: Find mass m1m_1m1​ Using F=m1⋅a1F = m_1 \cdot a_1F=m1​⋅a1​: 5=m1×10  ⟹  m1=510=0.5 kg5 = m_1 \times 10 \implies m_1 = \frac{5}{10} = 0.5\text{ kg}5=m1​×10⟹m1​=105​=0.5 kg

  • Step 2: Find mass m2m_2m2​ Using F=m2⋅a2F = m_2 \cdot a_2F=m2​⋅a2​: 5=m2×20  ⟹  m2=520=0.25 kg5 = m_2 \times 20 \implies m_2 = \frac{5}{20} = 0.25\text{ kg}5=m2​×20⟹m2​=205​=0.25 kg

  • Step 3: Calculate total combined mass (MMM) M=m1+m2=0.5 kg+0.25 kg=0.75 kgM = m_1 + m_2 = 0.5\text{ kg} + 0.25\text{ kg} = 0.75\text{ kg}M=m1​+m2​=0.5 kg+0.25 kg=0.75 kg

  • Step 4: Find acceleration (aaa) for combined mass Using F=M⋅aF = M \cdot aF=M⋅a: 5=0.75×a  ⟹  a=50.75=50075=6.67 m/s25 = 0.75 \times a \implies a = \frac{5}{0.75} = \frac{500}{75} = 6.67\text{ m/s}^25=0.75×a⟹a=0.755​=75500​=6.67 m/s2

Answer: The combined mass will have an acceleration of 6.67 m/s26.67\text{ m/s}^26.67 m/s2.


Question 2 (Conceptual - First Law)

Why is it advised to tie luggage kept on the roof of a bus with a rope?

Solution: When the bus is at rest and suddenly starts moving forward, the luggage tends to remain at rest due to the inertia of rest. As a result, it may slip backward and fall off.

Similarly, when the moving bus suddenly applies brakes to stop, the luggage tends to maintain its forward state of motion due to the inertia of motion. Consequently, it can slide forward and fall off the roof.

To prevent the luggage from falling off during sudden starts, stops, or sharp turns, it is securely tied with a rope.


Question 3 (Numerical - Momentum & Force)

A motorcar of mass 1200 kg1200\text{ kg}1200 kg is moving along a straight line with a uniform velocity of 90 km/h90\text{ km/h}90 km/h. Its velocity is slowed down to 18 km/h18\text{ km/h}18 km/h in 4 seconds4\text{ seconds}4 seconds by an unbalanced external force. Calculate:

  1. Initial momentum
  2. Final momentum
  3. Magnitude of the force applied

Solution:

  • Step 1: Convert velocities to SI units (m/s\text{m/s}m/s)

    • Initial velocity (uuu) = 90 km/h=90×518=25 m/s90\text{ km/h} = 90 \times \frac{5}{18} = 25\text{ m/s}90 km/h=90×185​=25 m/s
    • Final velocity (vvv) = 18 km/h=18×518=5 m/s18\text{ km/h} = 18 \times \frac{5}{18} = 5\text{ m/s}18 km/h=18×185​=5 m/s
    • Mass (mmm) = 1200 kg1200\text{ kg}1200 kg
    • Time (ttt) = 4 s4\text{ s}4 s
  • Step 2: Calculate Initial Momentum (p1p_1p1​) p1=m×u=1200 kg×25 m/s=30,000 kg⋅m/sp_1 = m \times u = 1200\text{ kg} \times 25\text{ m/s} = 30,000\text{ kg}\cdot\text{m/s}p1​=m×u=1200 kg×25 m/s=30,000 kg⋅m/s

  • Step 3: Calculate Final Momentum (p2p_2p2​) p2=m×v=1200 kg×5 m/s=6,000 kg⋅m/sp_2 = m \times v = 1200\text{ kg} \times 5\text{ m/s} = 6,000\text{ kg}\cdot\text{m/s}p2​=m×v=1200 kg×5 m/s=6,000 kg⋅m/s

  • Step 4: Calculate Applied Force (FFF) F=p2−p1t=6,000−30,0004=−24,0004=−6000 NF = \frac{p_2 - p_1}{t} = \frac{6,000 - 30,000}{4} = \frac{-24,000}{4} = -6000\text{ N}F=tp2​−p1​​=46,000−30,000​=4−24,000​=−6000 N

(Note: The negative sign indicates that the force applied is a retarding force acting opposite to the direction of motion.)

Answer:

  1. Initial Momentum = 30,000 kg⋅m/s30,000\text{ kg}\cdot\text{m/s}30,000 kg⋅m/s
  2. Final Momentum = 6,000 kg⋅m/s6,000\text{ kg}\cdot\text{m/s}6,000 kg⋅m/s
  3. Magnitude of Force = 6000 N6000\text{ N}6000 N (in opposite direction to motion)

Keep practicing these concepts, observe the physical world around you, and remember: Physics isn't just in books—it's happening every time you step out, jump, or play sports! Happy learning!

Exam Preparation & Frequently Asked Questions (FAQ)

Q1. How should I revise Force and Laws of Motion for the Class 9 Science examination?

Focus on mastering core textbook definitions, practicing 3-4 numerical problems daily with pen and paper, and reviewing previous year CBSE/NCERT board exam questions.

Q2. What are the key concepts that carry maximum marks in this chapter?

Pay special attention to core definitions, step-by-step derivations, solved textbook examples, and practical real-world applications outlined in your NCERT curriculum.

Q3. How can I avoid losing marks in long answer questions?

Always structure your answers with clear subheadings, write step-by-step working for numerical problems, state given values clearly, and highlight your final answers with correct SI units.

Verified NCERT & Board Exam Aligned Material
Ravindra Higher Secondary School, Waidhan
Previous GuideGravitation - Universal law of gravitation, acceleration due to gravity, mass versus weight, and principles of buoyancyNext GuideArithmetic Progressions - Finding the nth term and calculating the sum of first n terms of an Arithmetic Progression

Related Study Notes

ScienceClass 9

Structure of the Atom

Structure of the Atom - Thomson and Rutherford atomic models, Bohr model of the atom, distribution of electrons in orbits, valency, atomic number, mass number, and isotopes

Read Article
ScienceClass 8

Friction

Friction - Types of friction including static, sliding, and rolling friction, factors affecting friction, and fluid friction

Read Article
ScienceClass 9

Atoms and Molecules

Atoms and Molecules - Laws of chemical combination, atomic and molecular mass, writing chemical formulae, and the mole concept

Read Article

NCERT Study Guide Directory

Textbook solutions, chapter notes & practice worksheets by grade

Interlinked Syllabus
Class 10 NCERT Guides14 chapters
  • Triangles
  • Circles
  • The Human Eye and the Colourful World
  • Carbon and its Compounds
  • Magnetic Effects of Electric Current
  • Arithmetic Progressions
  • Electricity
  • Light - Reflection and Refraction
  • Life Processes
  • Acids, Bases and Salts
  • Chemical Reactions and Equations
  • Introduction to Trigonometry
  • Quadratic Equations
  • Real Numbers
Class 9 NCERT Guides11 chapters
  • Structure of the Atom
  • Atoms and Molecules
  • Work and Energy
  • Gravitation
  • → Force and Laws of Motion (Science)
  • Motion
  • The Fundamental Unit of Life
  • Matter in Our Surroundings
  • Coordinate Geometry
  • Number Systems
  • Polynomials
Class 8 NCERT Guides11 chapters
  • Algebraic Expressions and Identities
  • Friction
  • Squares and Square Roots
  • Practical Geometry
  • Sound
  • Combustion and Flame
  • Coal and Petroleum
  • Microorganisms: Friend and Foe
  • Linear Equations in One Variable
  • Understanding Quadrilaterals
  • Rational Numbers
Class 7 NCERT Guides8 chapters
  • Acids, Bases and Salts
  • Heat
  • Nutrition in Animals
  • Nutrition in Plants
  • Perimeter and Area
  • Integers
  • Rational Numbers
  • Simple Equations
Class 6 NCERT Guides8 chapters
  • Algebra
  • Decimals
  • Fractions
  • Knowing Our Numbers
  • Electricity and Circuits
  • Components of Food
  • Getting to Know Plants
  • Separation of Substances
Ravindra Higher Secondary School Logo

Ravindra Higher Secondary School

Waidhan, Singrauli (M.P.)

We Serve Society By Serving People

Established in 1988, Ravindra Higher Secondary School (RHS Waidhan) is dedicated to delivering excellence in education, character building, and holistic growth for students in Waidhan, Singrauli (MP).

Quick Links

  • Home Page
  • About RHS & Leadership
  • Academic Programs & Curriculum
  • Admissions Process 2026-27
  • Campus & Facilities
  • Faculty & Staff Members
  • Photo & Video Gallery
  • Notice Board & Announcements
  • Contact & Location

Shift & Office Hours

KG to Class 5th (Morning Shift)

07:30 AM – 11:30 AM

Class 6th to 12th (Afternoon Shift)

12:00 PM – 05:00 PM

Administrative Office Hours

Mon – Sat: 09:00 AM – 04:00 PM

Address & Location

  • Ravindra Higher Secondary School, Main Campus, Waidhan, Singrauli, Madhya Pradesh – 486886
  • +91 9826986106
  • rhswaidhan@gmail.com

© 2026 Ravindra Higher Secondary School, Waidhan, Singrauli. All rights reserved.

Privacy Policy•Contact Us•Student Portal