Newton 2

8 MCQs2 revision cards9-step worked example

Source: NCERT Laws of MotionPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 6). 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.)
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.
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.
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.

Practice 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:

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:

MCQ 3Easy RecallPractice

Newton's Second Law in its most general form is:

MCQ 4Easy RecallPractice

The SI unit of force is the newton. 1 N is equivalent to:

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):

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):

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:

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:

Quick recall before you leave

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: 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.]

Before solving, remember these

Law

Newton's Second Law of Motion

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. 6

Formula

Newton's Second Law (vector form)

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. 6

Key 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. 16

Formulas

9 formulas — click to collapse

Centripetal acceleration in uniform circular motion

a_{c} = \frac{v ^{2}}{r} = ω^{2} r

An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.

Symbol	Quantity	SI Unit
a_c	Centripetal acceleration	m/s^2
v	Tangential speed	m/s
r	Radius of circle	m
omega	Angular speed	rad/s

Valid when

Speed v is constant (uniform circular motion)
r and the centre are well-defined (instantaneous radius of curvature for general curved motion)

Do NOT use when

Non-uniform circular motion (then there is also a tangential acceleration component)

Maximum safe speed on a banked road (with friction)

v_\max = \sqrt{\tfrac{gr(\mu_s + \tan\theta)}{1 - \mu_s \tan\theta}}

On a road banked at angle theta from horizontal with tyre-road friction coefficient mu_s, this is the maximum speed for safe negotiation of a curve of radius r.

Symbol	Quantity	SI Unit
v_max	Maximum safe speed	m/s
g	Gravitational acceleration	m/s^2
r	Radius of the curve	m
mu_s	Coefficient of static friction	(dimensionless)
theta	Banking angle	rad/deg

Valid when

Banked turn at angle theta (theta = 0 reduces to level-road formula)
1 - mu_s*tan_theta > 0 (formula breaks down for very steep banks at high friction)
Optimum/no-friction speed v_o = sqrt(g*r*tan_theta) is a SPECIAL CASE

Do NOT use when

Banked angle so steep that 1 - mu_s*tan_theta <= 0 (use centripetal limit form)
Friction direction reversed (very low speed on a steep bank — vehicle slides inward)

Centripetal force in uniform circular motion

F_{c} = \frac{m v ^{2}}{r} = m ω^{2} r

The net inward force required to keep a body of mass m moving in a circle of radius r at speed v is m*v^2/r. This 'centripetal' force is NOT a new fundamental force — it is whichever real force (tension, friction, gravity, etc.) provides the inward acceleration.

Symbol	Quantity	SI Unit
F_c	Centripetal force	N
m	Mass of the body	kg
v	Tangential speed	m/s
r	Radius of the circle	m
omega	Angular speed	rad/s

Valid when

Speed is uniform (a_t = 0, only radial acceleration matters)
Identify the real force that provides F_c (tension, friction, normal component, etc.)

Conservation of linear momentum

\sum F_{ext} = 0 \Rightarrow i \sum m_{i} v_{i} = const.

If the net external force on a system of particles is zero, the total linear momentum of the system is conserved (vector equality of total p before and after any internal interaction). Basis of all collision and recoil analysis.

Symbol	Quantity	SI Unit
F_ext	Net external force on system	N
p_total	Sum of m_i*v_i over all particles	kg*m/s

Valid when

Net EXTERNAL force is zero (internal forces always cancel by Newton's 3rd law)
Conservation is vector — apply componentwise (x and y separately)
Holds independent of whether collisions are elastic or inelastic

Do NOT use when

External impulses present (gravity over a long time, friction)

Impulse of a force

J = F Δ t = Δ p

Impulse equals the product of force and the time interval over which it acts. By Newton's Second Law, impulse equals the change in linear momentum during that interval.

Symbol	Quantity	SI Unit
J	Impulse (vector)	Ns = kgm/s
F	Force (treated as average)	N
Delta_t	Time interval	s
Delta_p	Change in momentum	kg*m/s

Valid when

Useful when forces are large but act briefly (collisions, bat-on-ball, kicks)
Direction of impulse is the direction of average force

Kinetic friction

f_{k} = μ_{k} N

When two surfaces slide relative to each other, kinetic friction opposes the motion with magnitude proportional to the normal force.

Symbol	Quantity	SI Unit
f_k	Kinetic friction force	N
mu_k	Coefficient of kinetic friction	(dimensionless)
N	Normal force	N

Valid when

Surfaces ARE sliding
Direction: opposite to instantaneous relative velocity of one surface vs the other

Maximum safe speed on a level circular road

v_\max = \sqrt{\mu_s g r}

Static friction is the only force available to provide centripetal acceleration on a level road. Setting mu_s*N = m*v^2/r and N = m*g gives this maximum-safe speed bound.

Symbol	Quantity	SI Unit
v_max	Maximum safe speed (no skid)	m/s
mu_s	Coefficient of static friction (tyre vs road)	(dimensionless)
g	Gravitational acceleration	m/s^2
r	Radius of the circular path	m

Valid when

Road is level (no banking)
Tyres do not slide (static friction regime)
Driver maintains uniform speed on the curve

Do NOT use when

Banked road (use the banked-road formula)
Slippery / wet road where mu_s is reduced

Newton's Second Law of Motion

F = m a (or) F = \frac{d p}{d t}

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)

Maximum static friction

f_{s, m a x} = μ_{s} N

The maximum value of static friction between two surfaces in contact equals the coefficient of static friction times the normal force. Below f_s_max, static friction self-adjusts to whatever value is needed to prevent relative motion.

Symbol	Quantity	SI Unit
f_s_max	Maximum static friction	N
mu_s	Coefficient of static friction	(dimensionless)
N	Normal force	N

Valid when

Surfaces in contact, no relative motion (impending motion limit)
Below f_s_max, actual static friction = applied tangential load (self-adjusting)
f_s_max is independent of the apparent area of contact (Coulomb-Amontons assumption)

Do NOT use when

Surfaces are sliding (use kinetic friction f_k = mu_k * N instead)
Lubricated / fluid-friction conditions

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

16 items — click to collapse

Trap

Uniform circular motion: velocity vs speed constancy

Category: Similar Terms

Student claims velocity is constant in uniform circular motion (it's not — direction changes).

When it triggers

Question asks 'in uniform circular motion at constant speed, which is also constant?'

How to avoid

In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.

Trap

Force direction: Δp vs v

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.

Trap

Friction force vs friction-limited acceleration

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.

Trap

Impulse on bounce: direction reversal in velocity

Category: Sign Convention

Student forgets that velocity DIRECTION reverses on bounce; computes |v₁ - v₂| instead of |v₁ + v₂|.

When it triggers

Question describes ball dropped from height h₁, rebounding to height h₂; asks for impulse on ball.

How to avoid

Impulse J = Δp = m(v_after - v_before). Take down as positive: v_before = +v₁, v_after = -v₂. So J = m(-v₂ - v₁) = -m(v₁ + v₂); magnitude = m(v₁ + v₂).

Trap

Incline-with-friction: missing μ cos θ term

Category: Sign Convention

Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.

When it triggers

Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.

How to avoid

On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.

Trap

Ladder problem: pivot choice for torque equation

Category: Overthinking

Student takes torque about the centre of mass (introducing all 4 forces with non-zero moment arms) instead of about the floor contact (where 2 forces have zero moment arm and the equation simplifies).

When it triggers

Question gives a uniform rod or ladder leaning against a wall; asks for friction coefficient or limiting condition.

How to avoid

Pick the pivot to ELIMINATE unknown forces from the torque equation. Floor-contact pivot: normal force and friction at floor contribute zero torque; only weight (mid-length) and wall normal (top) appear. Result: μ_min = 1 / (2 tan θ).

Trap

Momentum vector vs scalar in explosions

Category: Sign Convention

Student adds magnitudes of momenta of fragments instead of vectors; ignores cancellation when fragments fly perpendicular or in opposite directions.

When it triggers

Question describes a body at rest exploding into multiple fragments with given mass ratios and partial velocity info.

How to avoid

Total momentum is a VECTOR. Initial p = 0; therefore Σ m_i v_i = 0 as a vector equation. Decompose along chosen axes (often natural symmetry axes); sum = 0 in each.

Trap

Treating Atwood pulley as single mass instead of two F=ma equations

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).

Trap

Atwood pulley: T and a sub-formulas

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.

Trap

Two-block contact-force confusion

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.

Trap

Lift-power: friction term overlooked

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.

Mistake

Student identifies the weight of a book and the normal force from the table on the book as a Newton's-third-law action-reaction pair (they are not — both act on the SAME body, the book).

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.

Mistake

Student adds 'centripetal force' as a separate, additional force on the free-body diagram of a body in circular motion (instead of recognising it as the NET inward radial force already supplied by tension/friction/gravity etc.).

Root cause: concept gap

Correction

Centripetal force is NOT a new fundamental force. It is the NET inward radial component of the real forces (tension, friction, normal, gravity, etc.). On a free-body diagram, draw only the real forces; their net inward component must equal m*v^2/r.

Wrong option pattern

Distractor sums tension + 'centripetal force' as separate inward forces.

Mistake

Student equates impulse with force (units N = kgm/s^2 versus impulse Ns = kg*m/s).

Root cause: unit error

Correction

Impulse J has dimensions of momentum (kg*m/s) and equals F*Delta_t. Force has dimensions kg*m/s^2. Confusing them inflates or deflates an answer by a factor of seconds. Always check units before declaring an answer.

Wrong option pattern

Distractor reports an answer in newtons where the correct answer is in N*s (or vice versa).

Mistake

Student applies 'conservation of linear momentum' to a system over a time interval during which a significant external force (gravity, normal reaction with non-zero net) acts.

Root cause: concept gap

Correction

Linear momentum is conserved only when the net EXTERNAL force is zero. Gravity over a finite time changes momentum. For collision problems we use conservation because the collision happens over a brief time interval where external impulses are negligible compared to internal collision forces.

Wrong option pattern

Distractor sets initial momentum = final momentum for a free-fall problem where gravity has acted for several seconds.

Mistake

Student uses f_s = mu_s * N for static friction whenever surfaces are in contact, even when applied tangential force is small. (mu_s * N is the MAXIMUM static friction, not the actual value.)

Root cause: formula misuse

Correction

Static friction is SELF-ADJUSTING: f_s exactly cancels the applied tangential force up to a ceiling f_s_max = mu_s * N. Below the ceiling, f_s = applied force. At the ceiling, motion is impending. Substituting mu_s * N too early over-estimates the friction.

Wrong option pattern

Distractor uses mu_s * N for the static friction force in a no-slip scenario where the applied force is well below threshold.

Past Year Questions

10 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (µ k) between the object and the rough surface is close to

10.75

20.25

30.40

40.5

NTA Answer: Option 1(final)

NEET 2025

A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s2)

184 NS

221 NS

37 NS

NTA Answer: Option 2(final)

NEET 2025

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take g = 10 m/s2)

1200 3N

2100 N

3100 3N

4200 N

NTA Answer: Option 3(final)

NEET 2024Revised key

A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is :

1Zero

24 N

36 N

410 N

NTA Answer: Option 3(revised_final)

NEET 2023

A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is

1Along northward

2Along north-east

3Along south-west

4Along eastward

NTA Answer: Option 2(final)

NEET 2023

Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 (g = 10 m s–2).

1150 m s–2

21.5 m s–2

350 m s–2

41.2 m s–2

NTA Answer: Option 2(final)

NEET 2022

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms–1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 m s–2)

123500

223000

320000

434500

NTA Answer: Option 4(final)

NEET 2022

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is

1- 5 - 3 2 v

32 v

42 2v

NTA Answer: Option 4(final)

NEET 2021

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the m/s2) ball is (g = 10 nearly

11.4 kg m/s

20 kg m/s

34.2 kg m/s

42.1 kg m/s

NTA Answer: Option 3(final)

NEET 2020

4 kg U 6 kg U U 39. Two bodies of mass 4 kg and 6 kg are tied to the U U U – U U U ends of a massless string. The string passes over U U ( U )– L fl fl U (g) a pulley which is frictionless (see figure). The fl U — acceleration of the system in terms of acceleration due to gravity (g) is :

1g/10

3g/2

4g/5

NTA Answer: Option 4(final)

How NEET usually asks this

10 recurring patterns from past papers — click to collapse

Particle in uniform circular motion at constant speed; identify which of speed / velocity / acceleration / kinetic energy is constant. Speed and KE constant; velocity (vector) and acceleration (centripetal) are NOT constant — direction changes.

InterpretationEasy

Common distractors

treats velocity as scalar

Conflates speed (scalar) with velocity (vector)

A body at rest explodes into 3 fragments; mass ratios given; some fragments' velocities given; find velocity of remaining fragment. Apply vector momentum conservation: Σ m_i v_i = 0 (since initial p = 0). Common shape: 2:2:1 mass ratio, two equal-mass fragments fly perpendicular with given speed; third fragment recoils.

Multi StepMedium

Common distractors

scalar sum of momenta

Treating momentum as scalar; ignores vector cancellation

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

Ball of mass m dropped from height h₁; rebounds to height h₂ (≤ h₁); find impulse on ball from ground. Use v² = 2gh to get speeds at impact and rebound; impulse J = m(v₂ - (-v₁)) = m(v₁ + v₂) where v₁ = √(2g h₁) and v₂ = √(2g h₂). Sign of velocity flips on rebound.

Multi StepEasy

Common distractors

subtracts v2 from v1 instead of adding

Forgetting velocity DIRECTION reversal at bounce

Two inclines of equal length L and same angle θ (e.g. 45°); one rough (with μ), one smooth. Compare time-of-descent or final velocity at bottom. Smooth: a = g sin θ. Rough: a = g(sin θ - μ cos θ). Use L = ½ a t² or v² = 2aL.

Multi StepMedium

Common distractors

ignores mu cos theta term

Forgetting the friction-along-incline component

Uniform rod / ladder of mass M and length L leans against a smooth vertical wall at angle θ; floor is rough with friction coefficient μ. Find limiting condition for static equilibrium. Apply ΣF = 0 (3 equations: horizontal, vertical, torque about base). Wall provides only horizontal reaction (smooth); floor provides normal + friction. μ_min = 1 / (2 tan θ).

Multi StepHard

Common distractors

torque about wrong pivot

Default pivot at center of mass adds complexity

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.6

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Newton 2

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Newton's Second Law

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