To every action there is an equal and opposite reaction. Equivalently: the forces between two bodies are mutual, equal in magnitude, opposite in direction, and act on different bodies. Action and reaction always act simultaneously on different bodies.
-- NCERT Class 11 Physics, Ch. 4, p. 56Newton 3
8 MCQs3 revision cards9-step worked example
Source: NCERT Laws of MotionOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Newton 3, explained for NEET
Here is the trap that costs marks on Newton's Third Law: confusing equilibrium with action-reaction. A book sits on a table. Its weight acts downward; the normal force acts upward. Students label these an "action-reaction pair." They are not. Both forces act on the same body — the book. That is equilibrium, not the third law.
Newton's Third Law (NCERT Class 11 Physics Chapter 4, page 56) states: when body A exerts a force on body B, body B simultaneously exerts an equal-and-opposite force on body A. The two forces always act on different bodies. They are always of the same type (both gravitational, both contact-normal, both tension, etc.) and they exist simultaneously — you cannot have one without the other.
For the book on a table:
- The third-law partner of the book's weight (Earth pulls book down) is the book pulling the Earth up — a gravitational pair acting on two different bodies.
- The third-law partner of the table's normal force on the book is the book pressing down on the table — a contact pair acting on two different bodies.
This distinction matters in NEET because distractors routinely label two forces on the same body as a third-law pair. The test: do the two forces act on different bodies? If both act on the same object, it is equilibrium under the first or second law, not the third law.
Newton's Third Law also underpins why internal forces cancel in a system. When you push block A against block B, A pushes B forward and B pushes A backward — equal and opposite, on different bodies. For the system as a whole, these internal forces sum to zero. Only external forces accelerate the system. This is the bridge between the third law and the contact-force problems (two-block push) that appear in NEET.
Watch-out: the third law has no exceptions in classical mechanics. It holds whether bodies are in contact or not, whether they are accelerating or at rest, and whether the forces are gravitational, electromagnetic, or tension-based.
Can you answer these Newton 3 MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A book of mass 2 kg rests on a table. Which of the following is the Newton's third law reaction to the gravitational force exerted by the Earth on the book?
Show answer and why every option is right or wrong
Answer: C. C is correct. Newton's Third Law pairs always act on different bodies and are of the same type. The gravitational pull of Earth on the book (weight) is paired with the gravitational pull of the book on the Earth — both gravitational, acting on different bodies (NCERT Class 11 Physics Chapter 4, page 56).
Why A is wrong: A is wrong because the normal force and the weight both act on the book (same body). They form an equilibrium pair under Newton's First/Second Law, not a third-law pair. (trap: mistake: action reaction same body)
Why B is wrong: B is wrong because friction acts between the book and the table surface and is a contact force — it is not the same type of force as the gravitational force, and in this scenario the book is at rest with no applied horizontal force so static friction is zero.
Why D is wrong: D is wrong because the weight of the table on the floor involves the table and the floor — neither of which is the partner body in the book-Earth gravitational interaction.
MCQ 2Easy RecallPractice
A person pushes a wall with a force of 50 N. The wall does not move. What is the force exerted by the wall on the person?
Show answer and why every option is right or wrong
Answer: B. B is correct. By Newton's Third Law, the wall exerts an equal (50 N) and opposite force on the person. This holds regardless of whether either body moves (NCERT Class 11 Physics Chapter 4, page 56).
Why A is wrong: A is wrong because Newton's Third Law guarantees a reaction force whenever there is an action force — a zero reaction would violate the law.
Why C is wrong: C is wrong because the reaction force is always opposite in direction to the action force, not in the same direction.
Why D is wrong: D is wrong because the reaction force is exactly equal in magnitude to the action force — not less than it. Newton's Third Law does not allow partial reactions.
MCQ 3Concept TrapPractice
A horse pulls a cart forward. According to Newton's Third Law, the cart pulls the horse backward with an equal force. The horse-cart system still accelerates forward because:
Show answer and why every option is right or wrong
Answer: D. D is correct. The horse pushes the ground backward; the ground pushes the horse forward (third-law pair). This forward friction on the horse is the external force that accelerates the horse-cart system. The horse-cart internal forces cancel. The system accelerates because the net external force (ground friction on horse minus any resistance on cart) is nonzero (NCERT Class 11 Physics Chapter 4, page 56).
Why A is wrong: A is wrong because Newton's Third Law states action and reaction are always exactly equal in magnitude — neither is greater. (trap: mistake: action reaction same body — misidentifying which forces to compare)
Why B is wrong: B is wrong because the comparison conflates two different third-law pairs. The horse-ground pair and the horse-cart pair are separate interactions. Acceleration comes from the net external force on the system, not from one pair being 'larger' than another.
Why C is wrong: C is wrong because Newton's Third Law applies in all scenarios — accelerating, decelerating, or at rest. There are no exceptions in classical mechanics.
MCQ 4Direct ApplicationPYQ Pattern
Two blocks A (mass 3 kg) and B (mass 2 kg) are placed in contact on a smooth horizontal surface. A horizontal force of 10 N is applied on block A, pushing it toward B. What is the contact force exerted by block A on block B?
Show answer and why every option is right or wrong
Answer: A. A is correct. System acceleration a = F / (m_A + m_B) = 10 / (3 + 2) = 2 m/s². Contact force on B from A = m_B × a = 2 × 2 = 4 N. By Newton's Third Law, B pushes back on A with 4 N (NCERT Class 11 Physics Chapter 4, page 56; pattern: two-block horizontal push).
Why B is wrong: B is wrong. This value (6 N) could arise from incorrectly computing m_A × a = 3 × 2 = 6 N and labeling it as the contact force on B, but m_A × a gives the net force on A, not the force A exerts on B.
Why C is wrong: C is wrong because 10 N is the external force on the entire system, not the contact force between the blocks. Using F = 10 N as the contact force means treating the full external force as if it acts only between A and B. (trap: two block internal vs external force)
Why D is wrong: D is wrong. This value (2 N) equals the acceleration (2 m/s²) numerically but in the wrong unit/context — it confuses the acceleration value with the force on B.
MCQ 5Direct ApplicationPractice
A block of mass 5 kg is placed on a smooth horizontal floor. A horizontal force F = 15 N pushes the block against a wall, and the block remains stationary. What is the magnitude of the normal force exerted by the wall on the block?
Show answer and why every option is right or wrong
Answer: B. B is correct. The block is in equilibrium horizontally: the applied force (15 N, toward the wall) is balanced by the normal force from the wall (15 N, away from the wall). By Newton's Third Law, the block pushes the wall with 15 N and the wall pushes the block with 15 N. The block's weight (5 × 10 = 50 N) acts vertically and is balanced by the floor's normal force — it does not affect the horizontal equilibrium (NCERT Class 11 Physics Chapter 4, page 56).
Why A is wrong: A is wrong. 5 N is just the mass value in kg, not a force. The horizontal equilibrium requires the wall's reaction to equal the applied 15 N.
Why C is wrong: C is wrong. 10 N does not correspond to any correct calculation in this problem. The horizontal balance is between 15 N applied and 15 N reaction.
Why D is wrong: D is wrong. 50 N is the weight of the block (mg = 5 × 10 = 50 N), which acts vertically. The horizontal normal force from the wall equals the applied horizontal force, 15 N, not the weight. (trap: mixing vertical weight with horizontal contact force)
MCQ 6Easy RecallPractice
In a tug-of-war, team A pulls team B with a force of 500 N through the rope. Which statement about the force exerted by the rope on team A is correct?
Show answer and why every option is right or wrong
Answer: D. D is correct. When team A pulls the rope toward itself with 500 N, by Newton's Third Law the rope pulls team A toward team B with 500 N. The reaction is equal in magnitude, opposite in direction, and acts on the other body (team A vs. the rope). Whether the system accelerates toward one side depends on the ground friction under each team's feet — the third-law forces in the rope are always equal (NCERT Class 11 Physics Chapter 4, page 56).
Why A is wrong: A is wrong because the reaction force always acts in the direction opposite to the action force, not the same direction.
Why B is wrong: B is wrong because the magnitude of the third-law reaction force is always exactly equal to the action force. It does not depend on who is winning; it depends only on the force being applied at that instant.
Why C is wrong: C is wrong because Newton's Third Law guarantees a reaction on team A the instant team A exerts a force on the rope — 'doing the pulling' does not exempt you from the reaction.
MCQ 7CalculationPYQ Pattern
A horizontal force F is applied to block A (mass m₁ = 4 kg) which pushes block B (mass m₂ = 6 kg) on a frictionless surface. If the contact force between A and B is 12 N, what is the applied force F?
Show answer and why every option is right or wrong
Answer: A. A is correct. The contact force on B = m₂ × a, so a = 12 / 6 = 2 m/s². Then F = (m₁ + m₂) × a = (4 + 6) × 2 = 20 N. By Newton's Third Law, B pushes back on A with 12 N, and the net force on A is F − 12 = m₁ × a = 4 × 2 = 8 N, which is consistent.
Why B is wrong: B is wrong. 12 N is the contact force between the blocks, not the external applied force. The applied force must accelerate both blocks, so it is larger than the contact force. (trap: two block internal vs external force)
Why C is wrong: C is wrong. This value (24 N) could arise from incorrectly doubling the contact force (2 × 12 N), which has no physical basis in this problem.
Why D is wrong: D is wrong. 30 N might come from an incorrect formula such as F = contact force × (m₁ + m₂)/m₁ with an arithmetic error, or from confusing the total mass ratio.
MCQ 8Direct ApplicationPractice
A swimmer pushes the wall of a pool with her feet exerting a force of 80 N. She accelerates away from the wall. Her mass is 40 kg. Assuming negligible water resistance during the push, what is her acceleration during the push?
Show answer and why every option is right or wrong
Answer: C. C is correct. By Newton's Third Law, the wall pushes the swimmer with 80 N opposite to her push. This is the net external force on the swimmer. By Newton's Second Law, a = F/m = 80/40 = 2 m/s² (NCERT Class 11 Physics Chapter 4, pages 54 and 8).
Why A is wrong: A is wrong. 0.5 m/s² would result from computing m/F = 40/80 instead of F/m = 80/40 — an inversion error.
Why B is wrong: B is wrong. 4 m/s² might arise from halving the mass erroneously (using 20 kg) or doubling the force — neither is justified by the problem.
Why D is wrong: D is wrong. 80 m/s² would result from using F × m = 80 × 40... which is dimensionally incorrect, or from confusing force with acceleration. (trap: confusing F and a when mass is a round number)
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Newton 3: quick recall before you leave
How do you solve a Newton 3 question? A worked example
- 1
Given
• Block A: mass m_A = 5 kg• Block B: mass m_B = 3 kg• Applied horizontal force on A: F = 24 N• Surface: smooth (frictionless)• Blocks are in contact; A pushes B.
- 2
Required
(a) Acceleration of the system
(b) Contact force between A and B - 3
Concept
Newton's Second Law applied to the system gives the common acceleration. Newton's Third Law tells us the contact force between A and B is an internal action-reaction pair — equal magnitude, opposite directions, acting on different bodies.
- 4
Formula
• System: F = (m_A + m_B) × a → a = F / (m_A + m_B)• Contact force on B from A: F_contact = m_B × a• By Newton's Third Law, B pushes back on A with the same magnitude F_contact.
- 5
Substitution
• a = 24 / (5 + 3) = 24 / 8• F_contact = 3 × a
- 6
Calculation
• a = 3 m/s²• F_contact = 3 × 3 = 9 N
Note on exact values: all masses (5 kg, 3 kg) and the applied force (24 N) are problem-defined exact values. They do not limit significant figures. - 7
Final answer
(a) Acceleration = 3 m/s²
(b) Contact force between A and B = 9 N (A pushes B forward with 9 N; B pushes A backward with 9 N) - 8
Common trap
A common confusion is applying the full 24 N as the contact force between A and B. The 24 N is the external force on the system — the contact force is only the portion needed to accelerate block B alone: m_B × a = 9 N, not 24 N.
Verification: Net force on A = F − F_contact = 24 − 9 = 15 N. Check: m_A × a = 5 × 3 = 15 N. Consistent. - 9
Similar NEET-style question
A force of 20 N is applied horizontally to block P (mass 6 kg) which pushes block Q (mass 4 kg) on a smooth floor. Find the contact force between P and Q.
Answer: a = 20/10 = 2 m/s²; contact force = 4 × 2 = 8 N.
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What to remember before solving Newton 3 questions
Figure 4.2 illustrates a book at rest on a horizontal table: weight W = mg acts downward; the table exerts an upward normal reaction R. With the book at rest (in equilibrium), R = W. This is NOT Newton's third law — both forces act on the SAME body (the book). Action-reaction pairs always act on DIFFERENT bodies.
-- NCERT Class 11 Physics, Ch. 4, p. 52Where do students lose marks on Newton 3?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Root cause: concept gap
Correction
Action-reaction pairs ALWAYS act on DIFFERENT bodies. The pair to the book's weight is the gravitational pull the book exerts on the Earth. The pair to the normal force from table on book is the force the book exerts on the table. Equal-and-opposite forces on the SAME body are an equilibrium statement, not third-law statement.
Wrong option pattern
Distractor labels two forces on the same body as a Newton's-third-law pair.
More in Laws of Motion: 15 exam traps and mistakes · 9 formulas · 10 question patterns from its other lessons.
Newton 3 questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 4, p.56
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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