Newton’s Third Law for JEE & CBSE Class 11 PCM: From Theory to Top‑Score Tricks

In your 4th Physics chapter, Laws of Motion, the Third Law of Motion is seen as
“To every action, there is always an equal and opposite reaction.”

Simple explanation:

If object A pushes on object B, then object B pushes back on object A with the exact same strength or force of the same magnitude. Only the direction is opposite.

Core concept to remember about Newton’s 3rd Law of Motion: Forces act on different bodies, so they never cancel on the same object. 

This is the most crucial point that sometimes confuses students. 

Remember that action and reaction forces always act on different objects. So they can never cancel each other out. 

So when you are studying the motion of one object, you only consider the forces acting on that object. The reaction force that the object exerts on something else is irrelevant to its own motion.

The law is the same, but the questions asked in your Class XI annual exams will be more conceptual than numerical and application-based in JEE or similar tests. 

In CBSE School Tests and Exams, Class XI

Your school examiners may test your grasp of this Newtonian law of motion with such themes of questions.

• Whether action and reaction forces act on the same or different bodies, or whether they cancel each other.
• Stating and explaining Newton's Third Law with an example, or identifying the action-reaction pair in a simple scenario, like a book resting on a table.

You would also have to be clear with practising the NCERT solutions for Chapter 4 to solve the related numerical problems on Newton’s 3rd Law. 

JEE Main Angle

JEE usually clubs different concepts from later chapters in your Physics textbook to test your knowledge on the applications of the Third Law. You can expect multistep problems from here. 

Consider reviewing the below examples, as this chapter as a whole carries ~6.6% weightage in the JEE. 

• You may have to find the tension in strings and reactions at pulleys, for instance.  For that, you’ll need to recognise that action–reaction pairs (tension forces) act on different bodies, including the block and the pulley. 
• Some questions may need you to apply Newton’s third law to the ejection of gases and the resulting thrust on the rocket. 

Check out the previous JEE Main question papers. 

Some critical concepts you must be able to visualise when learning about the 3rd law of Motion by Newton. 

Action–Reaction Pairs 

An action-reaction pair is a set of two forces that are equal in magnitude, opposite in direction, and act on two different interacting objects.

• Example 1: Skateboard Push: When you push a wall while on a skateboard (Action: your hand pushes the wall), the wall pushes you back with an equal force, causing you to accelerate away (Reaction: wall pushes your hand). 
• Example 2: Swimming: A swimmer pushes water backwards with their hands and feet (Action). The water, in turn, pushes the swimmer forward with an equal force (Reaction), propelling them through the pool.

Non‑Cancellation Explained

So, why don't these equal and opposite forces just cancel out, resulting in zero movement? Because they act on different objects.

Mini-quiz

Why don't the teams stay put if the force from Team A on the rope is equal and opposite to the force from Team B on the rope?

Answer: You have to consider the forces on each team. Team A wins if the force of the ground on their feet (a reaction force) is greater than the pull from Team B. The action-reaction pair is between the hand and the rope. It’s not between the two teams directly.

Force Types Must Match 

The action and its reaction are always the same type of force.

Action Force	Reaction Force	Explanation
Gravitational Pull from Earth on the Moon	Gravitational Pull from the Moon on Earth	Both are gravitational forces.
Normal Force from Table on the Book	Normal Force from Book on Table	Both are contact/electromagnetic forces.
Tension Force from Rope on Block	Tension Force from Block on Rope	Both are tension forces.

While preparing for competitive tests, you have to deal with application-based questions. Here are some tips and tricks to avoid mistakes. 

When it’s about Recoil and Propulsion

A rocket pushes hot gas down (action). The gas pushes the rocket up (reaction). 

This reaction force is the thrust. 

So, where is the direction?

The Verbal Trick

To find the direction of motion, always "think backwards!" Figure out what is being pushed and in which direction. The object will move in the opposite direction.

When You’re Solving Problems Based on Pulley & Tension Systems

In a pulley system, the rope exerts a downward tension force on the block it's holding. 

By the Third Law, the block exerts an equal upward pull on the rope.

The FBD Trick

"Don't forget the rope's action-reaction on each pulley wheel." The rope pulls down on the pulley, and the pulley support pulls up on the rope.

When You’ve to Find the Collision & Momentum Link

During a collision between ball A and ball B, the force on A by B is equal and opposite to the force on B by A.

Quick Derivation Box Trick

Since F_AB = -F_BA

(Δp_A / Δt) = -(Δp_B / Δt)

This directly leads to the Law of Conservation of Momentum: Δp_A + Δp_B = 0, or the total momentum of the isolated system remains constant.

Don’t miss out on these mistakes, though…

Here is a worked example that shows how you can go about solving complex problems.  

Problem Statement: Two blocks, each with a mass of 2 kg and 3 kg, are in contact on a frictionless surface. A force of 10 N is applied to the 2 kg block. Find the force between the blocks.

Solution:

• System acceleration: $a=\frac{10}{2+3}=2\mathrm{m}/{\mathrm{s}}^2$ .
• For 3 kg block: Force between blocks, $F_{\text{contact}}=m_3\cdot a=3\cdot 2=6\mathrm{N}$ .
• By the Third Law of Motion, the 2 kg block experiences 6 N in the opposite direction. 

Step-by-Step Explanation

Step 1: Draw Separate FBDs (Free-Body Diagrams)

Block 1 (m₁ = 2 kg):

Applied Force (F = 10 N) →

Contact Force from Block 2 (F_contact) ←

Weight (2g) ↓

Normal Force from ground (N₁) ↑

Block 2 (m₂ = 3 kg):

Contact Force from Block 1 (F_contact) →

Weight (3g) ↓

Normal Force from ground (N₂) ↑

Step 2: Write Equations (Horizontal Motion)

Using Newton's Second Law (F_net = ma) for each block:

For Block 1 (2 kg): The net force is the applied force minus the resisting contact force from Block 2.

F - F_contact = m₁a

10 - F_contact = 2a

For Block 2 (3 kg): The only horizontal force causing it to accelerate is the contact force from Block 1.

F_contact = m₂a

F_contact = 3a

Step 3: Solve for 'a' then 'F_contact'

First, find the acceleration (a) of the combined system.

a = F_total / m_total = 10 N / (2 kg + 3 kg)

a = 10 N / 5 kg

a = 2 m/s²

Now, use the equation for Block 2 (which is simpler) to find the contact force (F_contact).

F_contact = 3a

F_contact = 3 kg × 2 m/s²

F_contact = 6 N

Step 4: Interpretation

The force exerted by Block 1 on Block 2 is 6 N to the right. By Newton's Third Law, the force exerted by Block 2 on Block 1 is also 6 N, but in the opposite direction (to the left). Our step-by-step equations have correctly calculated this internal contact force.

Exam Tip: Circle your FBDs to avoid sign errors, and clearly label which forces act on which objects.

For quick reference on the Third Law of Motion before your exams, focus on these. 

The Third Law of Newton states, "For every action, there is an equal and opposite reaction."

Key Rules of Newton's Third Law 

• Forces come in pairs. 
• The pair is equal in magnitude, but opposite in direction. 
• The forces act on DIFFERENT objects. 

Common Misconception on Applying Newton's 3rd Law

• Trying to cancel action-reaction on one FBD.
• Confusing gravity-normal force on one object as a Third Law pair.

Newton's Third Law of Motion Formula

$${\stackrel{\to }{F}}_{AB}=-{\stackrel{\to }{F}}_{BA}$$

Before your exam question that comes on Newton's Third Law, also ask yourself if you can confidently affirm this checklist.

• Can I draw the FBD for two blocks in contact?
• Can I identify the action-reaction forces in a pulley system?
• Do I remember that thrust is the reaction to expelled mass?
• Am I confident that action-reaction forces never cancel each other out?
  
  
  

Here are some detailed notes on Physics Class 11 chapters. You will find topic-wise guides for each as well. They are updated and aligned with the NCERT syllabus. 

Find some overviews on what your current syllabus holds.