Definition
Linear momentum
The linear momentum p of a body is the product of its mass m and velocity v: p = m v. Momentum is a vector with the direction of v. SI unit: kg·m·s⁻¹.
-- NCERT Class 11 Physics, Ch. 4, p. 54Momentum
8 MCQs6 revision cards9-step worked example
Source: NCERT Laws of MotionPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 25 Sep 2026
Momentum, explained for NEET
Momentum is a vector quantity: p = mv, where m is mass and v is velocity. The direction of momentum is the direction of velocity — not speed, not force, not acceleration. This vector nature is where NEET traps live.
Newton's Second Law in momentum form (NCERT Class 11 Physics Chapter 4, page 54): the net external force on a body equals the rate of change of its momentum, F = dp/dt. For constant mass, this reduces to F = ma. But the momentum form is the more general statement and is essential for variable-mass problems and impulse calculations.
The direction trap. When a body changes direction — say, moving south and then turning east at the same speed — the net force is NOT along the final velocity. Force is along Δp = p_final − p_initial. You must draw both momentum vectors and subtract. This is a high-frequency NEET trap: students default to "force points where the body is going."
Impulse (NCERT Class 11 Physics Chapter 4, page 55): impulse J = FΔt = Δp. It has the same units as momentum (kg·m/s or N·s). When a ball bounces off a floor, velocity reverses direction. Taking downward as positive: v_before = +v₁, v_after = −v₂. So Δp = m(−v₂ − v₁) = −m(v₁ + v₂). The magnitude is m(v₁ + v₂), not m(v₁ − v₂). Forgetting the sign reversal on rebound is a common mark-losing error.
When momentum is NOT conserved. Momentum conservation requires zero net external force. Gravity acting over a long time interval is an external force — momentum is NOT conserved during free fall. In collisions, we invoke conservation because the collision time is so brief that external impulses are negligible compared to the large internal collision forces.
Watch out: impulse has units of momentum (N·s = kg·m/s), not force (N). Confusing the two inflates or deflates your answer by a factor of seconds.
Can you answer these Momentum MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The SI unit of linear momentum is:
Show answer and why every option is right or wrong
Answer: A. Linear momentum p = mv has dimensions [M][L][T⁻¹], giving SI unit kg·m/s (NCERT Class 11 Physics Chapter 4, page 53).
Why B is wrong: B (kg·m/s²) is the unit of force (Newton), not momentum. Momentum has one fewer power of time in the denominator.
Why C is wrong: C (N·s²) has dimensions [M][L][T⁻²]·[T²] = [M][L], which is not momentum.
Why D is wrong: D (J·s) has dimensions [M][L²][T⁻¹], which is angular momentum (or Planck's constant), not linear momentum.
MCQ 2Easy RecallPractice
Newton's Second Law in its most general form states that the net external force on a body equals:
Show answer and why every option is right or wrong
Answer: A. F = dp/dt — the net external force equals the time rate of change of linear momentum (NCERT Class 11 Physics Chapter 4, page 54). F = ma is the constant-mass special case.
Why B is wrong: B is the power (rate of change of KE = dW/dt = P), not force.
Why C is wrong: C describes momentum itself (p = mv), not the force law. Force is the rate of CHANGE of momentum, not momentum.
Why D is wrong: D would have dimensions (N·s)/(kg) = m/s, which is velocity — not force.
MCQ 3Easy RecallPractice
Impulse has the same dimensions as:
Show answer and why every option is right or wrong
Answer: C. Impulse J = FΔt = Δp. Its dimensions are [M][L][T⁻¹], identical to momentum (NCERT Class 11 Physics Chapter 4, page 55).
Why A is wrong: A is wrong. Force has dimensions [M][L][T⁻²]; impulse has dimensions [M][L][T⁻¹]. They differ by a factor of [T]. (Trap: mistake: impulse vs force — confusing impulse units N·s with force units N.)
Why B is wrong: B is wrong. Energy has dimensions [M][L²][T⁻²], which differs from impulse [M][L][T⁻¹].
Why D is wrong: D is wrong. Power has dimensions [M][L²][T⁻³], not [M][L][T⁻¹].
MCQ 4Direct ApplicationPractice
A 0.50 kg ball moving at 6.0 m/s due east is struck so that it moves at 6.0 m/s due north immediately afterward. The direction of the impulse delivered to the ball is:
Show answer and why every option is right or wrong
Answer: C. C is correct. Impulse is the CHANGE in momentum, Δp = p_final − p_initial, so it has to be found by vector subtraction. Taking east as +x and north as +y: p_initial = (3.0, 0) kg·m/s and p_final = (0, 3.0) kg·m/s, so Δp = (−3.0, +3.0) kg·m/s. Equal parts west and north puts it 45° north of west. The westward part is what stops the original eastward motion. (trap: force direction deltap vs velocity)
Why A is wrong: A is wrong because due north is the direction of the FINAL velocity. Impulse points along Δp, not along p_final; choosing north ignores the eastward momentum that had to be cancelled.
Why B is wrong: B is wrong because due east is the direction of the INITIAL velocity. The impulse must oppose that motion, not follow it.
Why D is wrong: D is wrong because 45° north of east is the direction of p_final + p_initial, from adding the two momentum vectors instead of subtracting. The sign of the eastward component is the whole difference between this answer and the right one.
MCQ 5Direct ApplicationPractice
A 2.0 kg object at rest is acted upon by a net force of 4.0 N for 3.0 s. The magnitude of the impulse delivered to the object is:
Show answer and why every option is right or wrong
Answer: B. Impulse J = FΔt = 4.0 × 3.0 = 12.0 N·s (NCERT Class 11 Physics Chapter 4, page 55). Equivalently, J = Δp = mv_final − 0 = 2.0 × 6.0 = 12.0 N·s.
Why A is wrong: A (6.0) results from computing m × Δt = 2.0 × 3.0, multiplying the mass by the time instead of the force by the time. Impulse is F × Δt.
Why C is wrong: C (8.0) results from computing F × m = 4.0 × 2.0, confusing impulse (F·Δt) with F × mass.
Why D is wrong: D (24.0) results from computing F × Δt × m = 4.0 × 3.0 × 2.0, incorrectly multiplying by mass an extra time.
MCQ 6Direct ApplicationPractice
A body of mass 0.20 kg is dropped from a height of 5.0 m onto a floor and rebounds to a height of 3.2 m. Taking g = 10 m/s², the magnitude of the impulse exerted by the floor on the body is:
Show answer and why every option is right or wrong
Answer: B. Speed just before hitting floor: v₁ = √(2 × 10 × 5.0) = 10 m/s. Speed just after rebound: v₂ = √(2 × 10 × 3.2) = 8.0 m/s. Taking downward as positive: v_before = +10, v_after = −8.0. Impulse magnitude = m|v₂ − v_before| = 0.20 × (10 + 8.0) = 0.20 × 18 = 3.6 kg·m/s. (Trap: trap: impulse rebound sign reversal — velocity reverses direction on bounce, so momenta ADD in magnitude.)
Why A is wrong: A (0.40) results from computing m(v₁ − v₂) = 0.20 × (10 − 8) = 0.40, forgetting the direction reversal at bounce. The velocities do not subtract — they add because the bounce reverses direction. (Trap: sign reversal on rebound.)
Why C is wrong: C (2.0) results from computing m × v₁ = 0.20 × 10 = 2.0, using only the pre-impact momentum and ignoring the rebound momentum entirely.
Why D is wrong: D (1.2) results from a calculation error, possibly using incorrect heights or forgetting to take the square root in v = √(2gh).
MCQ 7CalculationPractice
A body moving due south at speed v is acted upon by a force that changes its velocity to the same speed v due east. If the mass of the body is m, the magnitude of the change in momentum is:
Show answer and why every option is right or wrong
Answer: D. Take east as +x, north as +y. p_initial = (0, −mv) (due south). p_final = (+mv, 0) (due east). Δp = (mv, mv). |Δp| = mv√2. Speed is unchanged, but momentum is a vector — direction change means Δp ≠ 0 (trap: force direction deltap vs velocity).
Why A is wrong: A (zero) assumes that because speed is unchanged, momentum is unchanged. Momentum is a vector; direction change means Δp ≠ 0.
Why B is wrong: B (mv) results from subtracting magnitudes: |p_final| − |p_initial| = mv − mv = 0, then incorrectly guessing mv. Or from forgetting to apply vector subtraction.
Why C is wrong: C (2mv) results from adding magnitudes: mv + mv = 2mv. This would be correct only if the velocity completely reversed (180° turn), not a 90° turn.
MCQ 8Concept TrapPractice
A ball is thrown vertically upward. A student claims that during the ball's flight (after leaving the hand, before landing), the linear momentum of the ball is conserved. This claim is:
Show answer and why every option is right or wrong
Answer: C. Momentum conservation requires zero net external force. Gravity acts on the ball throughout its flight, continuously changing the magnitude and direction of its momentum. The ball is NOT an isolated system — Earth exerts a gravitational force on it (mistake: mistake: momentum not conserved during gravity).
Why A is wrong: A is wrong. A single body is not automatically isolated. 'Isolated' means no net external force. Gravity is an external force acting on the ball.
Why B is wrong: B is wrong. Although no contact forces act during flight (ignoring air resistance), gravity is a non-contact external force that acts throughout. Conservation of momentum requires ALL external forces to be zero, not just contact forces.
Why D is wrong: D is wrong in emphasis. Even in a vacuum with zero air resistance, gravity alone prevents momentum conservation. Air resistance is not the primary reason.
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Momentum: quick recall before you leave
How do you solve a Momentum question? A worked example
Pattern: Ball dropped from height, rebounds — find impulse (NEET pattern: impulse drop rebound, observed in NEET 2021 and 2025).
- 1
Given
• Mass of ball: m = 0.10 kg• Drop height: h₁ = 4.0 m• Rebound height: h₂ = 1.0 m• g = 10 m/s² (exact, problem-defined)
- 2
Required
Magnitude of impulse exerted by the floor on the ball.
- 3
Concept
Impulse = change in momentum. The ball's velocity reverses direction at the floor. We must account for this sign change. Using v² = 2gh to find speeds just before impact and just after rebound.
- 4
Formula
v = √(2gh) for speed at ground level.
J = Δp = m(v_after − v_before), with a consistent sign convention. - 5
Substitution
Speed before impact: v₁ = √(2 × 10 × 4.0) = √80
Speed after rebound: v₂ = √(2 × 10 × 1.0) = √20
Taking downward as positive:• v_before = +v₁ (moving down)• v_after = −v₂ (moving up)
J = m(v_after − v_before) = 0.10 × (−v₂ − v₁) = −0.10(v₁ + v₂) - 6
Calculation
v₁ = √80 = 4√5 ≈ 8.944 m/s
v₂ = √20 = 2√5 ≈ 4.472 m/s
|J| = 0.10 × (8.944 + 4.472) = 0.10 × 13.416 ≈ 1.34 N·s
Note on exact values: g = 10 m/s² is an exact problem-defined constant and does not limit significant figures. The heights 4.0 m and 1.0 m have 2 significant figures, so the final answer is reported to 2 significant figures. - 7
Final answer
|J| ≈ 1.3 N·s (to 2 significant figures)
The impulse is directed upward (away from the floor). - 8
Common trap
The high-frequency error is computing |J| = m(v₁ − v₂) = 0.10 × (8.944 − 4.472) = 0.45 N·s. This SUBTRACTS the speeds instead of ADDING them. The subtraction ignores the direction reversal at the bounce — velocity flips from downward to upward, so the change in velocity is v₁ + v₂, not v₁ − v₂.
- 9
Similar NEET-style question
A 0.25 kg ball falls from 10 m and bounces back to 2.5 m. Taking g = 10 m/s², find the impulse on the ball from the ground. (Answer: the speeds are √200 ≈ 14.14 m/s and √50 ≈ 7.07 m/s; impulse magnitude = 0.25 × (14.14 + 7.07) ≈ 5.3 N·s.)
---
What to remember before solving Momentum questions
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. 54Definition
Impulse
Impulse J is the change in momentum produced by a force acting for a finite time: J = F Δt = Δp. Useful when forces are large but act briefly (collisions, kicks). SI unit: N·s = kg·m·s⁻¹.
-- NCERT Class 11 Physics, Ch. 4, p. 55Which Momentum formulas do you need for NEET?
1 formula — click to collapse
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
Where do students lose marks on Momentum?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
3 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: 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).
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).
More in Laws of Motion: 13 exam traps and mistakes · 8 formulas · 7 question patterns from its other lessons.
Momentum questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
1
2
3
4
NTA Answer: Option 3(final)
How does NEET ask about Momentum?
3 recurring patterns from past papers — click to collapse
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
Sources
NCERT refs: Class 11 Physics Chapter 4, p.54 | Class 11 Physics Chapter 4, p.55
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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