If the net external force on a system is zero, the total linear momentum of the system remains constant. p_initial = p_final. This is a direct consequence of Newton's Second and Third laws applied to a closed system; basis of collision and recoil analysis.
-- NCERT Class 11 Physics, Ch. 4, p. 9Conservation Linear Momentum
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
The trap that costs marks here: treating momentum as a number instead of a vector. A body at rest explodes into three fragments — you add their momenta as magnitudes and get a nonzero total. But momentum conservation is a vector equation: the x-components sum to zero and the y-components sum to zero, independently. Every NEET explosion or recoil problem tests whether you decompose or just add magnitudes.
The law. Newton's second law in the form F = dp/dt tells you that if the net external force on a system is zero, the total linear momentum does not change (NCERT Class 11 Physics Chapter 4, page 9). Internal forces — collision impacts, spring forces, explosion gases — always cancel by Newton's third law. They redistribute momentum among parts of the system but cannot change the total.
The condition is strict. "Net external force = 0" means zero over the time interval you care about. During a collision lasting milliseconds, gravity's impulse is negligible compared to the collision force, so momentum is conserved across the collision instant. But if you track a projectile for several seconds of free fall, gravity's impulse is not negligible — momentum is not conserved over that interval.
Component-wise application. Write Σ mᵢvᵢ (before) = Σ mᵢvᵢ (after) along each axis separately. In explosion problems, the initial momentum is zero (body at rest), so the vector sum of all fragment momenta must also be zero. Two fragments flying perpendicular do not cancel each other's momenta by simple subtraction — you need Pythagoras on the resultant.
Watch out: impulse (J = Δp) has the same units as momentum (kg·m/s), not the same units as force (N). Confusing impulse with force inflates or deflates your answer by a factor of seconds.
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 1Direct ApplicationPractice
A 10 kg body at rest explodes into two fragments of mass 4 kg and 6 kg. If the 4 kg fragment moves at 6 m/s, what is the speed of the 6 kg fragment?
MCQ 2Easy RecallPractice
Linear momentum of a system is conserved when:
MCQ 3Easy RecallPractice
The SI unit of impulse is:
MCQ 4CalculationPractice
A ball of mass 0.2 kg is dropped from 5 m and rebounds to 1.25 m. Taking g = 10 m/s², the impulse exerted by the floor on the ball is:
MCQ 5CalculationPractice
A 12 kg body at rest explodes into three equal fragments. Two fragments fly off perpendicular to each other, each at 8 m/s. The speed of the third fragment is:
MCQ 6Direct ApplicationPractice
A bullet of mass 20 g moving at 500 m/s embeds itself in a 4.98 kg wooden block at rest. The velocity of the block-bullet system immediately after impact is:
MCQ 7Direct ApplicationPractice
A cannon of mass 1000 kg fires a shell of mass 1 kg at 500 m/s. The recoil speed of the cannon is:
MCQ 8Concept TrapPractice
A student applies conservation of momentum to a ball thrown vertically upward, claiming its momentum is the same at launch and at the highest point. This is incorrect because:
Quick recall before you leave
Worked Example
- 1
Given
A body of mass 5m, initially at rest, explodes into three fragments of mass m, 2m, and 2m. The two fragments of mass 2m each fly off at speed v, perpendicular to each other.
- 2
Required
Find the speed of the fragment of mass m.
- 3
Concept
Conservation of linear momentum. The body is initially at rest, so total initial momentum = 0. After explosion, the vector sum of all fragment momenta must equal zero.
- 4
Formula
Σ mᵢ**v**ᵢ = 0 (since **p**_initial = 0)
- 5
Substitution
Let fragment 1 (mass 2m) move along +x at speed v, and fragment 2 (mass 2m) move along +y at speed v. Fragment 3 (mass m) moves at speed u in some direction. x-component: 2m·v + 0 + m·u_x = 0 → u_x = −2v y-component: 0 + 2m·v + m·u_y = 0 → u_y = −2v
- 6
Calculation
Speed of fragment 3: u = √(u_x² + u_y²) = √((2v)² + (2v)²) = √(4v² + 4v²) = √(8v²) = 2v√2 Note: The mass ratio 5m = m + 2m + 2m is an exact counting relationship. The factor 2 and √2 are mathematical constants. Neither contributes to significant-figure considerations.
- 7
Final answer
The fragment of mass m flies off at speed **2√2 v** in the direction diagonally opposite to the resultant momentum of the other two fragments (at 225° from the +x axis, or equivalently at 45° below the −x axis).
- 8
Common trap
Adding momenta as scalars: 2mv + 2mv = 4mv, then setting m·u = 4mv → u = 4v. This is wrong because the two 2m fragments move perpendicular to each other, not in the same direction. Their combined momentum magnitude is √((2mv)² + (2mv)²) = 2mv√2, not 4mv. The correct answer is 2√2 v, not 4v.
- 9
Similar NEET-style question
A body of mass 3m at rest breaks into three equal fragments. Two fragments move with speed v at 60° to each other. Find the speed of the third fragment. *Approach:* Resolve the two momenta along and perpendicular to their bisector. The resultant of the two equal fragments along the bisector = 2 × mv × cos 30° = mv√3. The perpendicular components cancel. The third fragment moves opposite to the bisector at speed √3 v. ---
Before solving, remember these
Formulas
9 formulas — click to collapse
Centripetal acceleration in uniform circular motion
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
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
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
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) | N*s = kg*m/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
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
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
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
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.
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: 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).
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.
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).
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.
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.
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.
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.
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).
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.
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
9 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
10.75
20.25
30.40
40.5
NTA Answer: Option 1(final)
NEET 2025
184 NS
221 NS
37 NS
40
NTA Answer: Option 2(final)
NEET 2025
1200 3N
2100 N
3100 3N
4200 N
NTA Answer: Option 3(final)
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 2023
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
1- 5 - 3 2 v
2v
32 v
42 2v
NTA Answer: Option 4(final)
NEET 2021
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
1g/10
2g
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.9
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