Prerequisites
Before studying this topic, make sure you understand:
- Newton’s Second Law - Force and acceleration
- Vectors - Force vectors and addition
- Newton’s First Law - Inertia and equilibrium
The Hook: Why Doesn’t a Gun Recoil Backward When Fired?
In Pathaan or Jawan, when Shah Rukh Khan fires a gun, the gun recoils backward. But why doesn’t the gun fly backward as fast as the bullet flies forward? They’re both experiencing equal and opposite forces according to Newton’s Third Law, right?
The answer reveals the beautiful connection between Third Law and Second Law!
The Big Questions:
- Why do forces always come in pairs?
- If action and reaction are equal, how does anything ever move?
- Why don’t action-reaction cancel each other out?
The Core Concept: Forces Come in Pairs
Newton’s Third Law Statement
$$\boxed{\vec{F}_{AB} = -\vec{F}_{BA}}$$In words: For every action, there is an equal and opposite reaction.
More precisely:
When object A exerts a force on object B, object B simultaneously exerts an equal and opposite force on object A.
Key characteristics of action-reaction pairs:
- Equal in magnitude
- Opposite in direction
- Same type of force (both gravitational, or both normal, etc.)
- Act on DIFFERENT bodies
- Act SIMULTANEOUSLY
Interactive Demo: Visualize Action-Reaction Pairs
Explore how action and reaction forces work on different objects.
Action and reaction act on DIFFERENT bodies, so they can NEVER cancel each other!
This is why objects can still accelerate even though forces are equal and opposite.
Understanding Action-Reaction Pairs
Example 1: Book on Table
Action-Reaction Analysis:
Pair 1:
- Action: Book pulls Earth downward (gravitational force by book on Earth)
- Reaction: Earth pulls book downward (weight of book, $mg$)
Pair 2:
- Action: Book pushes table downward (normal force by book on table)
- Reaction: Table pushes book upward (normal reaction, $N$)
Question: The weight ($mg$) and normal reaction ($N$) are equal and opposite. Are they action-reaction?
Answer: NO! They both act on the SAME body (the book). Action-reaction pairs MUST act on different bodies.
$mg$ and $N$ are equal because the book is in equilibrium (not accelerating vertically), not because they’re action-reaction pairs.
Example 2: Walking
When you walk forward:
- Action: Your foot pushes ground backward
- Reaction: Ground pushes your foot forward
The forward push from the ground accelerates you forward!
Why is it hard to walk on ice?
On ice, friction is very low, so when you push backward, the ice can’t push you forward with much force. Action-reaction pairs still exist — but they’re both very small!
Why Action-Reaction Don’t Cancel
They Act on Different Bodies!
Consider a horse pulling a cart:
- Action: Horse pulls cart forward with force $F$
- Reaction: Cart pulls horse backward with force $F$
For the cart:
$$F_{\text{horse on cart}} - f_{\text{cart}} = m_{\text{cart}} \cdot a_{\text{cart}}$$For the horse:
$$F_{\text{ground on horse}} - F_{\text{cart on horse}} - f_{\text{horse}} = m_{\text{horse}} \cdot a_{\text{horse}}$$The system moves because:
- Ground pushes horse forward (friction)
- This forward push is greater than the backward pull from cart
- Different net forces on different bodies → different accelerations
Classic paradox: If horse pulls cart with force F, and cart pulls horse backward with force F, how do they move?
Answer: The horse’s motion depends on ALL forces on the horse:
- Ground pushes horse forward (friction): $F_{\text{ground}}$
- Cart pulls horse backward: $F$ (reaction to horse pulling cart)
Net force on horse = $F_{\text{ground}} - F$ (if this is positive, horse accelerates forward)
The cart’s motion depends only on forces on cart:
- Horse pulls cart forward: $F$
- Friction opposes: $f_{\text{cart}}$
Net force on cart = $F - f_{\text{cart}}$ (if positive, cart accelerates forward)
Gun-Bullet Recoil Revisited
When a gun fires:
- Action: Gun exerts force $F$ on bullet (forward)
- Reaction: Bullet exerts force $F$ on gun (backward)
Both experience the same force for the same time.
$$\text{Impulse on bullet} = F \cdot \Delta t = m_b v_b$$ $$\text{Impulse on gun} = F \cdot \Delta t = m_g v_g$$Since impulses are equal:
$$m_b v_b = m_g v_g$$ $$\boxed{v_g = \frac{m_b}{m_g} v_b}$$Example:
- Bullet: $m_b = 0.02$ kg, $v_b = 500$ m/s
- Gun: $m_g = 2$ kg
The gun recoils much slower because it has much greater mass!
Notice: This is actually conservation of momentum!
Initially: Total momentum = 0 (both at rest) Finally: $m_b v_b - m_g v_g = 0$ (opposite directions)
Newton’s Third Law leads to momentum conservation for isolated systems.
Internal vs External Forces
Internal Forces
Forces between parts of a system (action-reaction pairs within the system).
Key property: Internal forces always come in action-reaction pairs, so they cancel out when considering the system as a whole.
Cannot change: System’s total momentum
External Forces
Forces exerted by objects outside the system.
Can change: System’s total momentum and motion
Scenario: You’re inside a stationary train. Can you make it move by pushing on the walls?
Answer: NO!
When you push the wall, the wall pushes you back with equal force (Third Law). These are internal forces. Your momentum increases in one direction, wall’s momentum increases in opposite direction — net momentum of system (you + train) remains zero.
To move the train: You need an EXTERNAL force (friction between wheels and track when engine starts).
Common Action-Reaction Pairs
| Action | Reaction |
|---|---|
| Earth pulls you down (weight) | You pull Earth up |
| You push wall to the right | Wall pushes you to the left |
| Rocket expels gas downward | Gas pushes rocket upward |
| Hammer hits nail (force on nail) | Nail hits hammer (force on hammer) |
| Bat strikes ball | Ball strikes bat |
| Sun attracts Earth | Earth attracts Sun |
| Tire pushes road backward | Road pushes tire forward |
Applications of Third Law
1. Rocket Propulsion
Rocket expels hot gases downward with high velocity.
- Action: Rocket pushes gas downward
- Reaction: Gas pushes rocket upward
This works in space (no air needed) because of the Third Law!
Thrust force:
$$F_{\text{thrust}} = v_{\text{exhaust}} \frac{dm}{dt}$$2. Swimming
Swimmer pushes water backward.
- Action: Hands push water backward
- Reaction: Water pushes swimmer forward
3. Recoil of Gun
Covered earlier — gun and bullet receive equal impulses in opposite directions.
4. Jumping
When you jump:
- Action: You push ground downward
- Reaction: Ground pushes you upward
The upward push accelerates you into the air!
Memory Tricks & Patterns
Mnemonic for Third Law
“EASY - Equal And Simultaneous, Yes!”
- Equal in magnitude
- And opposite in direction
- Simultaneous (at the same time)
- Yet on different bodies
Identifying Action-Reaction Pairs
Formula to check: If force 1 is “A on B”, then its reaction is “B on A”
| Force | Its Reaction |
|---|---|
| Earth on book (weight) | Book on Earth |
| Hand on wall | Wall on hand |
| Bat on ball | Ball on bat |
| Rocket on gas | Gas on rocket |
Golden rule: Swap the two objects!
Common Mistakes to Avoid
Wrong: Weight ($mg$) and normal reaction ($N$) on a book on a table are action-reaction pairs.
Correct: They’re NOT action-reaction because:
- Both act on the SAME body (the book)
- They’re equal because the book is in equilibrium, not because of Third Law
Actual action-reaction pairs:
- Weight (Earth on book) ↔ Book’s pull on Earth
- Normal by table on book ↔ Normal by book on table
Wrong: Since action and reaction are equal and opposite, net force is zero, so nothing moves.
Correct: Action and reaction act on DIFFERENT bodies, so they never cancel. Each body experiences only ONE of these forces.
Confusing question: Which force is the action and which is the reaction?
Truth: It doesn’t matter! The terms are interchangeable. If you call F₁ the action, then F₂ is the reaction. Or vice versa — both are correct.
The important point: They’re a pair of equal and opposite forces on different bodies.
Wrong: The action is gravitational, so the reaction must be normal force.
Correct: Action and reaction must be the SAME TYPE of force.
- Gravitational action ↔ Gravitational reaction
- Normal action ↔ Normal reaction
- Tension action ↔ Tension reaction
Practice Problems
Level 1: Foundation (NCERT)
A book of mass 2 kg rests on a table. Identify all action-reaction pairs.
Solution:
Pair 1 (Gravitational):
- Action: Earth pulls book downward with force $mg = 2 \times 10 = 20$ N
- Reaction: Book pulls Earth upward with force 20 N
Pair 2 (Normal/Contact):
- Action: Book pushes table downward with force 20 N
- Reaction: Table pushes book upward with force 20 N (normal reaction)
Note: The weight (20 N down) and normal reaction (20 N up) on the book are NOT action-reaction. They’re equal because book is in equilibrium.
Answer: Two action-reaction pairs identified above
Explain using Newton’s Third Law how you walk forward.
Solution:
When you walk:
- Your foot pushes the ground backward (action)
- Ground pushes your foot forward (reaction)
The forward push from the ground accelerates you forward!
Why it works:
- The friction between your foot and ground allows you to push backward
- The ground’s reaction force (forward) acts on you, causing forward motion
- Without friction (like on ice), you can’t push backward effectively, so no forward force
Answer: Foot pushes ground backward; ground pushes foot forward (reaction)
A gun of mass 5 kg fires a bullet of mass 50 g at 400 m/s. Find the recoil velocity of the gun.
Solution:
Using momentum conservation (consequence of Third Law):
Initial momentum = 0 (both at rest) Final momentum = 0 (momentum conserved)
$$m_b v_b + m_g v_g = 0$$ $$m_b v_b = -m_g v_g$$ $$v_g = -\frac{m_b}{m_g} v_b$$Converting: $m_b = 50$ g $= 0.05$ kg
$$v_g = -\frac{0.05}{5} \times 400 = -\frac{1}{100} \times 400 = -4 \text{ m/s}$$Negative sign indicates opposite direction to bullet.
Answer: 4 m/s (backward)
Level 2: JEE Main
A man of mass 60 kg stands on a weighing scale in an elevator. The elevator is accelerating upward at 2 m/s². What does the scale read? Identify all action-reaction pairs. (g = 10 m/s²)
Solution:
Scale reading = Normal force exerted by scale on man
For the man (taking upward as positive):
$$N - mg = ma$$ $$N = m(g + a) = 60(10 + 2) = 60 \times 12 = 720 \text{ N}$$Action-reaction pairs:
Pair 1: Earth-Man gravitational
- Action: Earth pulls man down with $mg = 600$ N
- Reaction: Man pulls Earth up with 600 N
Pair 2: Man-Scale contact
- Action: Man pushes scale down with 720 N
- Reaction: Scale pushes man up with 720 N (this is $N$)
Pair 3: Scale-Elevator contact
- Action: Scale pushes elevator down with 720 N
- Reaction: Elevator pushes scale up with 720 N
Answer: Scale reads 720 N; three action-reaction pairs identified
A bullet of mass 20 g traveling at 500 m/s embeds itself in a wooden block of mass 980 g resting on a frictionless surface. Find: (a) Final velocity of the block-bullet system (b) Was Newton’s Third Law applicable during collision?
Solution:
(a) Using momentum conservation:
$$m_b u_b + m_{block} \times 0 = (m_b + m_{block}) v$$Converting: $m_b = 0.02$ kg, $m_{block} = 0.98$ kg
$$0.02 \times 500 = (0.02 + 0.98) v$$ $$10 = 1 \times v$$ $$v = 10 \text{ m/s}$$(b) Yes, Newton’s Third Law was applicable!
During collision:
- Bullet exerted force $F$ on block (forward)
- Block exerted force $F$ on bullet (backward)
These forces were equal and opposite (Third Law).
The bullet decelerated drastically (from 500 to 10 m/s) due to the backward force from block. The block accelerated (from 0 to 10 m/s) due to the forward force from bullet.
Answers: (a) 10 m/s (b) Yes, Third Law applied during collision
Two blocks A (3 kg) and B (5 kg) are connected by a string. A horizontal force of 40 N is applied on block B. Find the tension in the string and accelerations. (Frictionless surface)
Solution:
For the system (both blocks together):
$$F = (m_A + m_B) a$$ $$40 = (3 + 5) a$$ $$a = 5 \text{ m/s}^2$$For block A (only tension acts):
$$T = m_A \times a = 3 \times 5 = 15 \text{ N}$$Verification for block B:
$$F - T = m_B \times a$$ $$40 - 15 = 5 \times 5$$ $$25 = 25$$✓
Action-reaction pair:
- Tension on A (by string): 15 N to the right
- Tension on B (by string): 15 N to the left
These are equal and opposite (Third Law), but act on different bodies!
Answers:
- Tension = 15 N
- Acceleration = 5 m/s²
Level 3: JEE Advanced
A man of mass 60 kg stands at one end of a boat of mass 140 kg and length 4 m, which is at rest in still water. He walks to the other end of the boat. By what distance does the boat move? (Neglect water resistance)
Solution:
Key concept: No external horizontal force, so horizontal position of center of mass doesn’t change.
Initially: Taking left end of boat as origin.
- Man at $x_m = 0$
- Boat’s CM at $x_b = 2$ m (center of boat)
Finally: Let boat move distance $d$ to the left.
- Man at position $4 - d$ (relative to ground)
- Boat’s CM at position $2 - d$
Since CM doesn’t move:
$$x_{CM,f} = x_{CM,i}$$ $$\frac{520 - 200d}{200} = 1.4$$ $$520 - 200d = 280$$ $$200d = 240$$ $$d = 1.2 \text{ m}$$Newton’s Third Law application: When man walks right on boat, he pushes boat left. Boat pushes man right with equal force. These internal forces don’t change the system’s CM position.
Answer: Boat moves 1.2 m (in opposite direction to man’s walk)
A uniform chain of mass $m$ and length $L$ is placed on a smooth table with length $b$ hanging over the edge. The chain is released. Find the normal reaction from table when a length $x$ has slipped off.
Solution:
At any instant:
- Length on table: $(L - x)$
- Length hanging: $x$
Mass on table: $m_1 = \frac{m(L-x)}{L}$ Mass hanging: $m_2 = \frac{mx}{L}$
For the hanging part:
$$\frac{mx}{L} \cdot g - T = \frac{mx}{L} \cdot a$$where $T$ is tension at the edge.
For the part on table:
$$T = \frac{m(L-x)}{L} \cdot a$$Adding both equations:
$$\frac{mgx}{L} = ma$$ $$a = \frac{gx}{L}$$For the part on table (vertical forces):
$$N = \frac{m(L-x)}{L} \cdot g - T\sin 0°$$But wait, the chain is accelerating horizontally! For the part on table:
Vertical equilibrium:
$$N = \frac{m(L-x)g}{L}$$Actually, we need to account for the kink at the edge. The tension has a vertical component.
More careful analysis:
Consider the entire chain:
$$N + T_{\text{vertical}} = mg - ma_{\text{vertical}}$$For the chain on table, there’s no vertical acceleration:
$$N = \frac{m(L-x)}{L} g$$But this doesn’t account for the impulse at the edge!
Correct approach using momentum change:
As chain slips, the part leaving the table loses normal support. The vertical impulse creates:
$$\boxed{N = \frac{m(L-x)g}{L} - \frac{mv^2}{L}}$$where $v$ is the velocity at that instant.
From energy considerations: $v^2 = \frac{g(x^2 - b^2)}{L}$
$$N = \frac{mg(L-x)}{L} - \frac{mg(x^2-b^2)}{L^2}$$Answer: $N = \frac{mg}{L^2}[L(L-x) - (x^2-b^2)]$
A rocket of initial mass $m_0$ ejects gas at constant rate $\alpha$ (kg/s) with velocity $u$ relative to rocket. Find the velocity of rocket at time $t$ in gravity-free space.
Solution:
At time $t$:
- Mass of rocket: $m = m_0 - \alpha t$
- Velocity: $v$
In next time interval $dt$:
- Gas ejected: $dm = \alpha \, dt$
- Rocket’s mass becomes: $m - dm$
- Rocket’s velocity becomes: $v + dv$
Conservation of momentum:
Initial: $mv$
Final: $(m - dm)(v + dv) + dm(v - u)$
where $(v - u)$ is velocity of gas in ground frame (rocket moving at $v$, gas ejected backward at $u$ relative to rocket).
$$mv = (m-dm)(v+dv) + dm(v-u)$$ $$mv = mv + m\,dv - dm\cdot v - dm\cdot dv + dm\cdot v - dm\cdot u$$ $$0 = m\,dv - dm\cdot dv - dm\cdot u$$Neglecting second-order term $dm \cdot dv$:
$$m\,dv = u\,dm$$ $$dv = u \frac{dm}{m}$$Integrating from $t=0$ to $t=t$:
At $t=0$: $m = m_0$, $v = 0$ At $t=t$: $m = m_0 - \alpha t$, $v = v(t)$
$$\int_0^v dv = u \int_{m_0}^{m_0-\alpha t} \frac{dm}{m}$$ $$v = u \ln m \Big|_{m_0}^{m_0-\alpha t}$$ $$v = u \left[\ln(m_0 - \alpha t) - \ln m_0\right]$$ $$\boxed{v = u \ln\left(\frac{m_0}{m_0 - \alpha t}\right)}$$Third Law in action: Rocket pushes gas backward with force → Gas pushes rocket forward with equal force!
Answer: $v = u \ln\left(\frac{m_0}{m_0 - \alpha t}\right)$
Quick Revision Box
| Concept | Key Point | Formula/Fact |
|---|---|---|
| Third Law | Equal & opposite forces | $\vec{F}_{AB} = -\vec{F}_{BA}$ |
| Key feature | Act on DIFFERENT bodies | Never cancel each other |
| Always | Same type of force | Both gravitational or both normal, etc. |
| Timing | Simultaneous | Occur at same instant |
| Internal forces | Come in pairs | Don’t change system’s momentum |
| Recoil | Gun-bullet example | $m_1v_1 = m_2v_2$ |
| Rocket | Third Law application | Gas pushes rocket up |
When to Use This Law
Use Newton’s Third Law when:
- Identifying force pairs in a system
- Analyzing recoil problems (gun, rocket)
- Internal forces in connected systems
- Momentum conservation problems (Third Law → conservation)
- Conceptual questions about forces
Use with Second Law for:
- Finding tensions in strings connecting bodies
- Multi-body problems (apply both laws to each body)
- Constraint motion problems
Avoid confusion with:
- Equilibrium forces (those are on SAME body, not action-reaction)
JEE Exam Strategy
Weightage
- JEE Main: 1-2 conceptual questions per year
- JEE Advanced: Often combined with momentum conservation (2-4 marks)
Common Question Types
- Identify action-reaction pairs (easy, 1 mark)
- Recoil problems (gun, rocket) (moderate, 2-3 marks)
- Man on boat/plank (classic, 3-4 marks)
- Conceptual: why forces don’t cancel (easy-moderate, 2 marks)
- Combined with momentum (advanced, 4+ marks)
Time-Saving Tricks
Trick 1: For recoil: $m_1v_1 = m_2v_2$ (momentum conservation from Third Law)
Trick 2: To identify action-reaction: Swap the two objects!
- “A on B” → Reaction is “B on A”
Trick 3: If question asks about equilibrium forces being action-reaction, answer is always NO (equilibrium forces are on same body)
Trick 4: Internal forces always come in action-reaction pairs → don’t change system’s total momentum
Teacher’s Summary
- Forces always come in pairs — you can’t have a single isolated force in the universe
- Equal, opposite, simultaneous — same magnitude, opposite direction, at same instant
- Act on DIFFERENT bodies — this is why they don’t cancel; each body feels only one force
- Same type of force — gravitational ↔ gravitational, normal ↔ normal, etc.
- Leads to momentum conservation — internal forces (action-reaction pairs) don’t change system’s total momentum
- Recoil, rockets, walking — all are applications of Third Law
“Action and reaction are like two sides of a coin — you can’t have one without the other!”
Related Topics
Within Laws of Motion
- Newton’s Second Law — Often used together with Third Law
- Impulse and Momentum — Third Law leads to momentum conservation
- Newton’s First Law — The three laws form complete framework
Connected Chapters
- Conservation of Momentum — Direct consequence of Third Law
- Collisions — Third Law during impact
- Rocket Propulsion — Classic application
Math Connections
- Vectors — Force pairs as equal and opposite vectors
- Logarithms — Rocket equation derivation