Subtopics - Newton's Laws of Motion (NEET)
Four major blocks: the language of force and inertia with Newton's three laws and their derivations, linear momentum and impulse with the conservation principle for isolated systems, FBD-based analysis of connected bodies and the Atwood machine, and apparent weight in an accelerating lift with rocket propulsion.
1) Newton's Three Laws, Inertia and Force
Defines inertia (resistance to change of state, proportional to mass, no units or dimensions) and states Newton's First Law (law of inertia): a body remains at rest or in uniform motion along a straight line unless acted upon by an external net force, mathematically: a = 0 if F = 0. Newton's Second Law provides the quantitative definition: F = dp/dt = ma for constant mass. Newton's Third Law: for every action there is an equal and opposite reaction; action and reaction always act on DIFFERENT bodies, so they cannot cancel each other. Also covers inertia of rest, motion, and direction with daily-life examples (bus-passenger, gun recoil, swimming).
2) Linear Momentum, Impulse and Conservation
Defines linear momentum p = mv (vector, [MLT⁻¹], SI unit kg m/s), derives impulse J = ∫F dt = F_avg × ∆t = ∆p (impulse-momentum theorem), and proves the law of conservation of linear momentum: for an isolated system (no net external force) total momentum remains constant. Applications: gun recoil (m_G v_G = −m_B v_B), rocket propulsion (thrust = −u dm/dt; instantaneous velocity v = u ln(m₀/m) − gt), and all collision problems. Machine-gun force formula: F = mnv for n bullets of mass m fired per second at speed v.
3) Free Body Diagrams and Connected Bodies
Systematic FBD technique for solving Newton's second law for multiple-body systems. Key results: blocks in contact F pushed into n bodies gives a = F/(m₁+m₂+…); massless-string connected blocks: T = m₁m₂F / (m₁+m₂)(m₁+m₂) × factors; Atwood machine: a = (m₂−m₁)g / (m₁+m₂), T = 2m₁m₂g / (m₁+m₂). Inclined plane: a = g sinθ (smooth); block on smooth horizontal surface with applied force at angle θ: a = F cosθ / m, R = mg − F sinθ (upward pull) or R = mg + F sinθ (downward push). Massive string introduces position-dependent tension.
4) Apparent Weight in Lift and Spring/Physical Balances
Defines apparent weight as the normal reaction N from the contact surface (what the weighing machine reads). For a body of mass m in a lift: (1) Rest or uniform motion: N = mg. (2) Accelerating upward at a: N = m(g + a) — feels heavier. (3) Accelerating downward at a: N = m(g − a) — feels lighter. (4) Free fall (a = g downward): N = 0, weightlessness. (5) Deceleration upward: N = m(g − a). Spring balance in non-inertial frame: reads apparent weight, not true weight. Physical balance (beam balance): compares mass directly — unaffected by frame of reference. True weight = √(W₁W₂) for false balance with unequal arms.
Newton's Laws of Motion Download Notes & Weightage Plan
For each topic in the Newton's Laws of Motion chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.
Newton's Three Laws, Inertia and Force
The conceptual and definitional foundation: inertia types, qualitative first law, quantitative second law F=ma, and action-reaction pairs on different bodies.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Action-reaction pairs identification is the single highest-yield conceptual point. Second: inertia type illustrated by given scenario (e.g., bus starts → inertia of rest). Third: F=ma units and dimensions.
- High-risk Area: Claiming N (normal) and mg (weight) on a book are an action-reaction pair. They are NOT — they act on the same body (the book). The action-reaction pair of mg is the gravitational pull on Earth by the book (upward).
- Best Practice Style: For every action-reaction question draw two separate FBDs — one for each body. Mark the action force on one body and find its reaction on the other. This mechanically prevents the common error.
Linear Momentum, Impulse and Conservation
Quantitative impulse-momentum theorem J = F∆t = ∆p and the conservation law for isolated systems, with gun-recoil and rocket-propulsion applications.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Gun recoil formula v_G = −(m_B/m_G)v_B. Machine-gun force F = mnv. Impulse = area under F-t graph. Conservation only when F_ext = 0 — identify isolation condition first.
- High-risk Area: Applying conservation of momentum when friction or an inclined-plane gravity component acts on the system. These external forces violate isolation. NEET deliberately presents borderline scenarios where students forget to check.
- Best Practice Style: For every conservation problem write 'Step 1: CHECK ISOLATION. External forces present? List all of them.' Only if all external forces are zero (or cancel by symmetry) proceed with conservation. This step-1 discipline catches 80% of application errors.
Free Body Diagrams and Connected Bodies
The computational engine of the chapter: FBD technique for blocks in contact, string-connected blocks, and the Atwood machine — all requiring F=ma per body and constraint equations.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Atwood: a=(m₂−m₁)g/(m₁+m₂) and T=2m₁m₂g/(m₁+m₂). These two results give instant answers to 70% of Atwood MCQs. For other pulley variants: always derive from FBD rather than memorising formulae.
- High-risk Area: Assuming tension is same throughout a string when pulley is massive. For massive pulley T₁ ≠ T₂; the difference (T₁−T₂) = Ma/2 is the torque equation. NEET occasionally gives massive-pulley data to test whether student distinguishes the two cases.
- Best Practice Style: For every pulley problem: (1) draw FBD for each mass and the pulley, (2) write F=ma for each mass, (3) write torque equation for pulley if massive, (4) solve simultaneously. Never look up a formula table — derive every time and the answer follows from the process.
Apparent Weight in Lift and Spring/Physical Balances
Application of F=ma to an accelerating reference frame (lift): apparent weight table, spring balance reading in inertial vs non-inertial frames, and physical balance frame-independence.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: The two lift formulas: N = m(g + a) for upward acceleration; N = m(g − a) for downward acceleration. Weightlessness condition: a = g (free fall). These three results cover every NEET lift MCQ.
- High-risk Area: Confusing N = m(g + a) with N = m(g − a) when problem states 'decelerating' rather than 'accelerating'. A lift decelerating while going upward has net downward acceleration, so N = m(g − a). This direction-of-deceleration subtlety is the most common error in this topic.
- Best Practice Style: Always determine actual direction of acceleration vector from the described motion BEFORE applying the formula. Draw small diagram: lift moving up/down + decelerating/accelerating → mark a vector direction. Then N = m(g ± a) follows from whether a is upward (+) or downward (−).
Newton's Laws of Motion Chapter NEET Traps & Common Mistakes (Topic-Wise)
Each subtopic below is of the Newton's Laws of Motion chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.
Mistake Snapshot (What Students Do Wrong)
- Claiming N and mg on a book are an action-reaction pair:: N (normal force exerted by table on book) and mg (weight of book due to Earth) both act on the SAME body (the book). Action-reaction pairs always act on DIFFERENT bodies. N's reaction pair is the force the book exerts on the table (downward); mg's reaction pair is the gravitational pull the book exerts on Earth (upward). NEET tests this distinction regularly in first-principles conceptual MCQs.
- Saying action and reaction cancel and therefore net force on system is zero:: Action and reaction act on different bodies, so they appear in different free body diagrams. Within a single-body FBD, there is only one of the pair. They can never cancel for the purposes of that body's motion. Students who add them up (F + (−F) = 0) and conclude the object cannot accelerate are applying the forces to the wrong object.
Book rests on table: Weight W = mg acts on book (downward, Earth on book). N = normal force acts on book (upward, table on book). Since book is in equilibrium N = mg — but these are NOT action-reaction pair, they are equilibrium of forces ON THE SAME BODY. The true reaction to W is book pulling Earth upward with same magnitude; the true reaction to N is book pushing table downward with same magnitude.
How NEET Frames The Trap
NEET asks: 'Which of the following is the action-reaction pair for weight of a book on a table?' Provides options mixing up N and mg on the same body vs forces on different bodies. This is a guaranteed 1-mark conceptual MCQ in a majority of recent papers.
Q. A book rests on a table. The weight of the book acts on the table. According to Newton's third law, the reaction to this force is:
A. Normal force exerted by table on book B. Weight of table acting downward C. Gravitational pull of book on Earth D. Friction force between book and table
Trick: Weight = gravitational pull of Earth on book. By Newton's third law the reaction is the gravitational pull of book on Earth (equal magnitude, opposite direction, acting on Earth). Option A is N — the equilibrium partner of weight on the same body (book), NOT the third-law reaction pair. Option C is correct.
Mistake Snapshot (What Students Do Wrong)
- Applying N = m(g + a) when lift is decelerating upward (should be N = m(g − a)):: A lift moving upward but decelerating has actual acceleration directed DOWNWARD. Therefore the correct equation is N = m(g − a), meaning apparent weight decreases. Students who see 'decelerating' and 'moving upward' incorrectly apply N = m(g + a) because the motion is upward, ignoring that acceleration direction is downward.
- Concluding apparent weight is zero whenever an object is in a moving lift:: Apparent weight N = 0 occurs ONLY in free fall (a = g downward). For any other acceleration value, N ≠ 0. Constant-velocity lift: N = mg (no change). Upward acceleration: N > mg. Downward acceleration a < g: N is still positive but less than mg. Students confuse 'lighter feeling' with 'weightlessness'.
Lift moving upward decelerating at 3 m/s². Mass m = 60 kg. Actual acceleration = 3 m/s² downward. Correct: N = m(g − a) = 60(10 − 3) = 420 N. Wrong answer with N = m(g + a) = 60(10 + 3) = 780 N. NEET invariably provides both 420 N and 780 N as options. The word 'decelerating while moving up' is the clue — net acceleration is downward.
How NEET Frames The Trap
NEET states: 'A lift is moving upward with deceleration a. What is the apparent weight of a person of mass m?' Correct answer is m(g−a). Distractor is m(g+a) — wrong because motion direction ≠ acceleration direction.
Q. A man of mass 60 kg is standing in a lift. The lift is moving downward with an acceleration of 4 m/s². What is his apparent weight? (Take g = 10 m/s²)
A. 840 N B. 600 N C. 360 N D. 240 N
Trick: Lift accelerating downward: N = m(g − a) = 60(10 − 4) = 60 × 6 = 360 N. Option C is correct. Option A (840 N) uses N = m(g + a) = 60 × 14 — wrong, that applies to upward acceleration. Option B (600 N) uses N = mg — wrong, that applies to rest or uniform velocity.
Mistake Snapshot (What Students Do Wrong)
- Applying conservation of momentum to a system where friction or gravity component acts as external force:: Conservation of linear momentum holds ONLY when the net external force on the system is zero. If a block slides on a rough horizontal surface, friction is an external force — momentum is NOT conserved. Similarly for two blocks on an inclined plane where gravity component acts along the slope.
- Treating the gun-bullet system as non-isolated because the explosion happens:: The explosion force between gun and bullet is INTERNAL to the gun-bullet system. Before and after firing, the net external force on the system is considered zero (horizontal direction), so horizontal momentum is conserved. Students confuse 'explosion' with 'external force' and refuse to apply conservation.
Two blocks (m₁ = 2 kg, m₂ = 3 kg) collide on a rough surface. Students apply m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This is WRONG — friction is external. Correct statement: momentum change of system = impulse of external forces (friction × time). Only for collision time → 0 (impulsive) can friction's impulse be neglected and conservation approximately applied.
How NEET Frames The Trap
NEET gives two cars colliding on a road with friction and asks final speed using momentum conservation. Correct approach: check isolation. If the collision time is given as 'very short', friction impulse is negligible and approximate conservation is valid; if not specified, flag as non-isolated.
Q. A bullet of mass 20 g is fired horizontally from a gun of mass 2 kg. The bullet leaves the gun with a velocity of 300 m/s. What is the recoil velocity of the gun?
A. 1.5 m/s B. 3 m/s C. 0.3 m/s D. 6 m/s
Trick: Gun-bullet system is isolated horizontally (no external horizontal force). Initial momentum = 0. Final: 2 × v_G + 0.02 × 300 = 0. Therefore v_G = −0.02 × 300 / 2 = −3 m/s. Magnitude = 3 m/s. Option B correct. Option A (1.5 m/s) results from using 150 m/s (half bullet speed) — arithmetic error. The negative sign indicates gun recoils opposite to bullet direction.
Mistake Snapshot (What Students Do Wrong)
- Applying the standard Atwood formula T = 2m₁m₂g/(m₁+m₂) when pulley is massive or string has mass:: The standard Atwood formula assumes massless frictionless pulley. When pulley has mass M, the rotational inertia must be included: a = (m₁−m₂)g / (m₁+m₂+M/2). The standard formula overestimates acceleration and gives wrong tension values.
- Assuming both sides of string have same tension when pulley is massive:: For a massive pulley T₁ ≠ T₂. The net torque (T₁−T₂)R = Iα = (MR²/2)(a/R) gives T₁−T₂ = Ma/2. Using T₁ = T₂ for massive pulley ignores the angular acceleration of the pulley and gives wrong answers for both T₁ and T₂.
m₁ = 3 kg, m₂ = 5 kg, pulley M = 4 kg. Standard (wrong): a = (5−3)×10/(3+5) = 20/8 = 2.5 m/s². Correct (massive pulley): a = 2×10/(3+5+2) = 20/10 = 2 m/s². NEET provides 2.5 m/s² as distractor; correct answer 2 m/s² requires the M/2 correction.
How NEET Frames The Trap
NEET specifies pulley mass M explicitly in the problem. If student uses the standard formula (ignore M), they get a distractor answer. The key: pulley mass present → add M/2 to denominator.
Q. In an Atwood machine, masses m₁ = 2 kg and m₂ = 4 kg are connected over a massless frictionless pulley. What is the acceleration of the system? (g = 10 m/s²)
A. 3.33 m/s² B. 5 m/s² C. 6.67 m/s² D. 10 m/s²
Trick: Standard Atwood (massless pulley): a = (m₂−m₁)g/(m₁+m₂) = (4−2)×10/(2+4) = 20/6 = 3.33 m/s². Option A correct. Option C (6.67) uses (m₁+m₂)g/(m₁+m₂) — wrong, divides by total mass without the difference. Option B (5 m/s²) uses only m₂g/(m₁+m₂) = 40/8 = 5 — the single-block-hanging formula, wrong for Atwood.