The rate of change of momentum of a body is directly proportional to the applied external force, and the change occurs in the direction of the applied force. F = dp/dt. For constant mass, F = m a — the applied force equals mass times acceleration.
-- NCERT Class 11 Physics, Ch. 4, p. 54Newton 2
8 MCQs2 revision cards9-step worked example
Source: NCERT Laws of MotionPYQ coverage: NEET 2020, 2023, 2024Official key: NTA-verifiedLast updated: 25 Sep 2026
Newton 2, explained for NEET
Here is the trap that costs marks on Newton's Second Law questions: you read "force," you see a mass, and you write F = ma for one object — but the question asked for the contact force between two objects in a system, or for the force direction when momentum changes. The formula is simple. The application requires you to decide what system you're writing F = ma for and what F means in that system.
Newton's Second Law states that the net external force on a body equals the rate of change of its linear momentum (NCERT Class 11 Physics Chapter 4, page 54). For constant mass: F_net = ma. Both F and a are vectors — the acceleration points in the direction of the net force, not necessarily in the direction of motion.
The momentum form, F = dp/dt, is the more general statement. It handles variable-mass problems and, critically, tells you that force direction equals the direction of Δp, not the direction of v. A body moving south that turns east at the same speed experiences a net force along the Δp vector (which points north-east), not along the final velocity.
Three high-frequency NEET traps on this topic:
-
Force direction ≠ velocity direction. Force is along Δp = m(v_f − v_i). Draw both velocity vectors and subtract. (Trap: picking the direction of motion instead.)
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Friction force vs friction-limited acceleration. When a body sits on an accelerating vehicle, maximum acceleration before sliding = μg (no mass in the answer). The friction force is μmg. Confusing the two gives an answer off by a factor of m.
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Two-block contact force. When force F pushes block A against block B, the system acceleration is F/(m_A + m_B). The contact force on B is m_B × a, not F. Applying the full external force to each block separately is wrong — that ignores Newton's Third Law at the contact surface.
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Atwood pulley: write two equations, not one. For two unequal masses on a pulley, write F = ma separately for each mass with tension T as the unknown. The shortcut a = (m₁ − m₂)g/(m₁ + m₂) works, but only if you keep track of which mass is heavier. Losing the sign gives wrong tension.
Can you answer these Newton 2 MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Concept TrapPYQ Pattern
A 0.15 kg ball moving east at 20 m/s is struck by a bat. After the strike, the ball moves north at 20 m/s. The direction of the net impulse on the ball is:
Show answer and why every option is right or wrong
Answer: D. Impulse J = Δp = m(v_f − v_i). Take east as +x, north as +y. v_i = (20, 0), v_f = (0, 20). Δv = (−20, +20), which points midway between west and north — i.e., north-west. The net impulse (and hence net force during the strike) is directed north-west (NCERT Class 11 Physics Chapter 4, page 54: F = dp/dt, so force direction = direction of momentum change).
Why A is wrong: A is wrong because east is the direction of the initial velocity, not the direction of momentum change. Force is along Δp = m(v_f − v_i), not along v_i (trap: confusing force direction with velocity direction).
Why B is wrong: B is wrong because north is the direction of the final velocity v_f, not the direction of Δv. The change in velocity has components in both x and y (trap: confusing force direction with final velocity direction).
Why C is wrong: C is wrong because north-east would require both x and y components of Δv to be positive. But Δv_x = 0 − 20 = −20 (westward), so the vector points north-west, not north-east (trap: sign error in vector subtraction).
MCQ 2Direct ApplicationPYQ Pattern
A horizontal force of 30 N acts on a system of two blocks A (mass 2 kg) and B (mass 3 kg) in contact on a frictionless surface. The force is applied on A, which pushes B. The contact force between A and B is:
Show answer and why every option is right or wrong
Answer: A. System acceleration a = F/(m_A + m_B) = 30/(2 + 3) = 6 m/s². Contact force on B = m_B × a = 3 × 6 = 18 N (NCERT Class 11 Physics Chapter 4, page 54: apply F = ma to B alone with the contact force as the only horizontal force on B).
Why B is wrong: B is wrong because 30 N is the external force on the entire system, not the contact force between A and B. Applying the full external force to block B alone ignores that A also accelerates (trap: treating each block as receiving the full external force).
Why C is wrong: C is wrong because 12 N equals m_A × a = 2 × 6 = 12 N, which is the net force on block A, not the contact force on B. This reversal of which block's mass to use is a common error in two-block problems.
Why D is wrong: D is wrong because 6 N would result from using just the acceleration value (6 m/s²) without multiplying by the correct mass, or from an incorrect mass partition.
MCQ 3Easy RecallPractice
Newton's Second Law in its most general form is:
Show answer and why every option is right or wrong
Answer: D. The general form is F = dp/dt (NCERT Class 11 Physics Chapter 4, page 54). F = ma is the special case when mass is constant. The momentum form handles variable-mass systems (e.g., rockets).
Why A is wrong: A is wrong as the most general form because F = ma assumes constant mass. For variable-mass systems (rockets, conveyor belts), F = ma does not hold — F = dp/dt is needed.
Why B is wrong: B is wrong because m dv/dx is not a standard formulation of Newton's Second Law. While F = m(dv/dt) can be rewritten as F = mv(dv/dx) using chain rule, this is a derived mathematical manipulation, not the law's general statement.
Why C is wrong: C is wrong because mv is momentum p, not force. Force is the rate of CHANGE of momentum, not momentum itself. Confusing p with F gives wrong dimensions: kg·m/s vs kg·m/s².
MCQ 4Easy RecallPractice
The SI unit of force is the newton. 1 N is equivalent to:
Show answer and why every option is right or wrong
Answer: B. From F = ma, the unit of force = (unit of mass)(unit of acceleration) = kg × m/s² = kg·m/s² = 1 N (NCERT Class 11 Physics Chapter 4, page 54).
Why A is wrong: A is wrong because kg·m/s is the unit of momentum (p = mv), not force. Momentum and force differ by a factor of time: F = dp/dt.
Why C is wrong: C is wrong because kg·m²/s² is the unit of energy (joule), not force. Energy = force × distance, so J = N·m = kg·m²/s². Confusing force with energy is a dimensional error.
Why D is wrong: D is wrong because kg·m²/s is the unit of angular momentum (L = Iω or r × p), not force. This has dimensions of [M L² T⁻¹], while force has [M L T⁻²].
MCQ 5Direct ApplicationPYQ Pattern
A box of mass 5 kg sits on the floor of a truck. The coefficient of static friction between the box and floor is 0.4. The maximum acceleration of the truck for which the box does not slide is (take g = 10 m/s²):
Show answer and why every option is right or wrong
Answer: B. For the box to stay stationary relative to the truck, friction provides the horizontal force: f_s = ma_box. Maximum friction = μ_s mg. At the limit, μ_s mg = ma_max, so a_max = μ_s g = 0.4 × 10 = 4 m/s². The mass cancels — the answer does not depend on the box's mass (NCERT Class 11 Physics Chapter 4, page 54: F = ma applied to the box in the truck's reference).
Why A is wrong: A is wrong because 20 m/s² equals μ_s × m × g = 0.4 × 5 × 10 = 20. This is the maximum friction FORCE in newtons, not the maximum acceleration in m/s². Confusing force (N) with acceleration (m/s²) by forgetting to divide by mass is a high-frequency trap on this pattern.
Why C is wrong: C is wrong because 2 = μ_s × m = 0.4 × 5 uses the mass where g belongs; the mass cancels, leaving a_max = μ_s g.
Why D is wrong: D is wrong because 50 = mg = 5 × 10 is the box's weight in newtons, not an acceleration; friction is only the fraction μ_s of that weight, and dividing by the mass then gives μ_s g.
MCQ 6Direct ApplicationPYQ Pattern
In an Atwood machine, two masses m₁ = 5 kg and m₂ = 3 kg are connected by a massless inextensible string over a frictionless pulley. The acceleration of the system is (take g = 10 m/s²):
Show answer and why every option is right or wrong
Answer: C. Writing F = ma for each mass separately: m₁g − T = m₁a and T − m₂g = m₂a. Adding: (m₁ − m₂)g = (m₁ + m₂)a, so a = (m₁ − m₂)g/(m₁ + m₂) = (5 − 3) × 10/(5 + 3) = 20/8 = 2.5 m/s² (NCERT Class 11 Physics Chapter 4, page 54: F = ma applied to each body in the system).
Why A is wrong: A is wrong because 10 m/s² = g. This would be the acceleration only if one mass were zero (free fall). With two masses connected, the net driving force is (m₁ − m₂)g and the total inertia is (m₁ + m₂), giving a < g (trap: ignoring the lighter mass entirely).
Why B is wrong: B is wrong because 6.25 m/s² = m₁g/(m₁ + m₂) = 50/8 uses only the heavier mass's weight as the driving force. The net force is the difference of the two weights, (m₁ − m₂)g.
Why D is wrong: D is wrong because 3.75 m/s² = m₂g/(m₁ + m₂) = 30/8 uses only the lighter mass's weight as the driving force. The net force is (m₁ − m₂)g = 20 N.
MCQ 7Easy RecallPractice
A force F = 10 N acts on a body of mass 2 kg for 3 s. The impulse delivered to the body is:
Show answer and why every option is right or wrong
Answer: A. Impulse J = F × Δt = 10 × 3 = 30 N·s. Equivalently, J = Δp (change in momentum). The unit is N·s = kg·m/s, which is a momentum unit (NCERT Class 11 Physics Chapter 4, page 54: F = dp/dt, so FΔt = Δp).
Why B is wrong: B is wrong because 10 N is the force, not the impulse. Impulse = force × time, not force alone. The unit of impulse is N·s, not N.
Why C is wrong: C is wrong because 5 m/s² = F/m = 10/2 is the acceleration, not the impulse. Acceleration and impulse are different quantities with different dimensions (m/s² vs N·s).
Why D is wrong: D is wrong because the unit 'kg' is a unit of mass, not impulse. The numerical value 30 is correct (F × Δt = 30), but the unit must be N·s or kg·m/s, not kg.
MCQ 8CalculationPYQ Pattern
Two blocks A (3 kg) and B (2 kg) are in contact on a frictionless surface. A force of 25 N is applied on block B, pushing both blocks. The contact force between A and B is:
Show answer and why every option is right or wrong
Answer: C. System acceleration a = F/(m_A + m_B) = 25/(3 + 2) = 5 m/s². Since the force is applied on B and pushes A, the contact force on A = m_A × a = 3 × 5 = 15 N. Alternatively, from B's perspective: F − contact force = m_B × a → 25 − contact force = 2 × 5 → contact force = 15 N. Both approaches give the same answer (NCERT Class 11 Physics Chapter 4, page 54).
Why A is wrong: A is wrong because 25 N is the total external force on the system, not the internal contact force. Applying the full force to each block ignores that both blocks share the acceleration (trap: treating full external force as the contact force).
Why B is wrong: B is wrong because 10 N = m_B × a = 2 × 5. This is the net force on block B from the external force minus the contact reaction, not the contact force itself. Using the wrong block's mass in the contact-force calculation is a common two-block error.
Why D is wrong: D is wrong because 5 N equals the numerical value of the acceleration (5 m/s²), not a force. This error arises from confusing the acceleration with the contact force or from dividing 25 N by 5 kg and stopping there without multiplying by the correct mass.
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Newton 2: quick recall before you leave
How do you solve a Newton 2 question? A worked example
Pattern: Two-block horizontal push (NEET pattern: two blocks horizontal push, observed 2024 Q3 Q7).
- 1
Given
A horizontal force F = 20 N is applied on block A (mass m_A = 4 kg), which pushes block B (mass m_B = 6 kg) on a frictionless horizontal surface. Both blocks move together.
- 2
Required
Find (a) the acceleration of the system and (b) the contact force between A and B.
- 3
Concept
Newton's Second Law applied twice: once to the system as a whole (to find acceleration), then to block B alone (to find the contact force). The contact force is an internal force — it does not appear in the system equation but is the only horizontal force on B.
- 4
Formula
System: F = (m_A + m_B) × a → a = F/(m_A + m_B)
Block B alone: contact force N_AB = m_B × a - 5
Substitution
a = 20/(4 + 6) = 20/10
N_AB = 6 × a - 6
Calculation
a = 2.0 m/s²
N_AB = 6 × 2.0 = 12 N
Note on exact values: F = 20 N and the masses 4 kg, 6 kg are problem-defined exact values. They do not limit significant figures. The answers 2.0 m/s² and 12 N are exact within the problem's framework. - 7
Final answer
(a) Acceleration = 2.0 m/s²
(b) Contact force between A and B = 12 N - 8
Common trap
Applying the full 20 N force to block B alone would give a_B = 20/6 ≈ 3.3 m/s², which is wrong — that ignores that block A is also accelerating and absorbs part of the applied force. The external force acts on the system; only the contact force acts on B (trap: two block internal vs external force).
- 9
Similar NEET-style question
A 50 N horizontal force pushes block P (mass 3 kg) against block Q (mass 7 kg) on a smooth surface. Find the contact force between P and Q. [Answer: a = 50/10 = 5 m/s²; contact force = 7 × 5 = 35 N.]
What to remember before solving Newton 2 questions
F = m a, where F is the net (resultant) external force on the body, m is its (constant) mass, and a is the resulting acceleration. Holds in any inertial reference frame. SI unit of force: newton (N) = kg·m·s⁻².
-- NCERT Class 11 Physics, Ch. 4, p. 54Key Fact
Free-body diagrams
Approach to mechanics problems: isolate each body, draw all external forces on it as vectors (free-body diagram), apply Newton's Second Law along chosen axes. The choice of axes is free; aligning one axis with the direction of acceleration usually simplifies the algebra.
-- NCERT Class 11 Physics, Ch. 4, p. 65Which Newton 2 formulas do you need for NEET?
1 formula — click to collapse
Newton's Second Law of Motion
The net external force on a body equals the rate of change of its linear momentum. For a body of constant mass, this reduces to F = m*a — net force equals mass times acceleration. Both F and a are vectors; the acceleration is in the direction of the net force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Net external (vector) force | N |
| m | Mass of the body | kg |
| a | Acceleration (vector) | m/s^2 |
| p | Linear momentum (= m*v) | kg*m/s |
| t | Time | s |
Valid when
- F is the resultant (net) of all external forces, not any single force
- Mass is constant for the form F = m*a (use F = dp/dt for variable mass)
- Inertial reference frame (no pseudo-forces); add inertial corrections in non-inertial frames
Do NOT use when
- Frame is non-inertial (need pseudo-forces)
- Mass is varying significantly (use F = dp/dt)
- Quantum / relativistic regimes (Newtonian mechanics breaks down)
Where do students lose marks on Newton 2?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
6 items — click to collapse
Category: Sign Convention
Student picks the direction of MOTION as the direction of the net force, instead of the direction of the change in momentum (Δp). When velocity changes direction at constant speed, the force is perpendicular to BOTH the initial and final velocity vectors (in the limit) — it's the direction of Δv.
When it triggers
Question describes a body changing direction (e.g. turning) and asks for the force direction or the direction of Δp.
How to avoid
F ∝ Δp = m Δv = m (v_f − v_i). Always draw v_i and v_f as vectors and subtract; the result (v_f − v_i) is the direction of net force.
Category: Unit Conversion
Student computes μmg (friction FORCE in Newtons) when asked for the friction-limited ACCELERATION. The two differ by a factor of m: a = μg, F = μmg.
When it triggers
Question gives μ, g, and a body's mass and asks for the maximum acceleration of the supporting surface OR the friction force on the body.
How to avoid
Read carefully: 'maximum acceleration of the vehicle so the body stays still' = μg (no mass). 'Friction force on the body' = μmg (mass present). Units expose the error: N for force, m/s² for acceleration.
Category: Overthinking
Student tries to apply F=ma to the system as a whole (using net force = (m1-m2)g and total mass m1+m2) but loses track of the tension. The correct approach writes Newton's Second Law SEPARATELY for each mass and treats T as an unknown in two simultaneous equations.
When it triggers
Question contains a frictionless pulley with two unequal masses tied to a string. Asks for tension T or acceleration a (or both).
How to avoid
Draw a free-body diagram for EACH mass. Write F=ma per body, treating T as the same magnitude on both sides of the string. Solve simultaneously: a = (m1 - m2) g / (m1 + m2); T = 2 m1 m2 g / (m1 + m2).
Category: Negative Marking
Multi-mass pulley problem requires computing acceleration first, then tension. T = 2 m1 m2 g/(m1+m2). Sign errors in m1−m2 propagate.
When it triggers
Atwood machine or pulley system with multiple masses.
How to avoid
Compute a = (m1-m2)g/(m1+m2) first with m1 the heavier mass and downward as positive. Then T from F=ma on either mass. Always re-check by plugging back into both Newton's 2nd Law equations.
Category: Overthinking
Student applies the full external force F to a single block instead of recognising the system needs to be analysed for the contact (internal) force.
When it triggers
Question gives horizontal force F on block A which pushes block B; asks for contact force between A and B or acceleration.
How to avoid
System acceleration: a = F / (m_A + m_B). Contact force on B from A = m_B × a. The full F acts on the system, not on each block independently.
Category: Overthinking
Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.
When it triggers
Question describes a lift moving at constant speed with explicit friction force on cable or guides.
How to avoid
At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.
More in Laws of Motion: 10 exam traps and mistakes · 8 formulas · 5 question patterns from its other lessons.
Newton 2 questions from past NEET papers
3 questions from NEET 2020, 2023, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
1Zero
24 N
36 N
410 N
NTA Answer: Option 3(revised_final)
NEET 2023
1Along northward
2Along north-east
3Along south-west
4Along eastward
NTA Answer: Option 2(final)
NEET 2020
1g/10
2g
3g/2
4g/5
NTA Answer: Option 4(final)
How does NEET ask about Newton 2?
5 recurring patterns from past papers — click to collapse
A body moving in one direction suddenly changes velocity direction (same or different speed); find the direction of the net force. Force direction = direction of momentum CHANGE (Δp = p_f − p_i), NOT direction of motion. Common shape: 'moving south, suddenly turning east at same speed' → Δp vector points north-east.
InterpretationEasy
Common distractors
force along final velocity
Default to thinking force points in direction of motion
force along initial velocity
Newton-1 misread: object 'wants' to keep moving in original direction
Atwood-style pulley with two unequal masses connected by an inextensible massless string over a frictionless pulley. Apply F = ma to each mass separately along the string direction. The tension is the same throughout the string; the magnitudes of acceleration are equal but oriented oppositely. Solve simultaneous equations for tension T and acceleration a. Common shape: given two masses m1, m2 and asked for a or T, with options testing common confusions (g vs a in equations, treating the system as one body).
Multi StepMedium
Common distractors
uses g where a belongs
Forgetting that the system accelerates, so weight is balanced by net force minus T
confuses tension with weight
Treating T = m·g for one of the masses (which is true only when a=0)
A body rests on the floor of an accelerating vehicle; find the maximum vehicle acceleration before the body slides. Static friction provides the horizontal force on the body; max accel = μ_s g. Above this, body slides backward relative to the vehicle. Common shape: μ_s, g given; find a_max.
Direct ApplicationEasy
Common distractors
uses mu times g times mass
Confusing force (μ·N = μmg) with acceleration
Horizontal force F applied to block A (mass m_A); A pushes B (mass m_B) in front. Find acceleration of system AND contact force between A and B. System: a = F/(m_A + m_B); contact force on B from A = m_B × a = m_B F / (m_A + m_B).
Multi StepEasy
Common distractors
treats each block with full F
Forgetting Newton's 3rd law internal force decomposition
Lift moving up at constant speed v with total mass M; friction force f opposes motion. Power required from cable = (Mg + f) × v. Common shape: M = 2000 kg, v = 1.5 m/s, f given; find motor power.
Multi StepEasy
Common distractors
forgets friction term
Ignoring opposing force
Sources
NCERT refs: Class 11 Physics Chapter 4, p.54
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.
Test yourself on this topic with real past-paper questions:
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