NEET Physics — Newton's Laws of Motion
Impulse – Complete Notes, Revision, Important Questions & Downloads
Impulse (J) is the product of force and the time for which it acts: J = F × Δt = ∫F dt = Δp = m(v − u). The Impulse-Momentum Theorem states that the impulse of a force equals the change in momentum produced by it. NEET tests this topic through numerical problems calculating impulse from force and time or from change in momentum, graph-based questions (area under the F-t graph = impulse), and questions identifying the SI unit (N·s = kg·m/s) and dimensions ([MLT⁻¹]). Impulse is a vector; its direction is the same as the force direction. A large force acting for a very short time is called an impulsive force — the concept used in cricket bat-ball collisions, bullet-target problems, and judo/airbag safety design.
Theory + NumericalsNewton's Laws Ch.4J = F·Δt = Δp
Expected QuestionsQ
1
Impulse-related numericals (find impulse given force and time, or find change in momentum) appear in nearly every NEET paper. F-t graph area questions are also common.
Time Required⏱
50 min
15 min for definition, formula, units, dimensions, and vector direction. 15 min for Impulse-Momentum Theorem derivation and application. 20 min for F-t graph questions and numerical practice.
Difficulty⚡
Easy
Impulse is one of the simpler topics in Newton's Laws. The formula J = FΔt = Δp is direct. The only trap: remembering it is a VECTOR (direction = direction of force), and that the unit N·s and kg·m/s are the same. F-t graph questions require identifying the correct area calculation.
NRI USA Curriculum GapUS
Low
AP Physics 1 fully covers impulse and the impulse-momentum theorem with identical content. The NEET-specific additions: emphasis on dimensions ([MLT⁻¹]), SI unit conversion (N·s = kg·m/s), and F-t graph area = impulse as explicit test items. The term 'impulsive force' (a large force for a short duration) is named and defined more explicitly in NCERT than in AP Physics 1.
8Subtopics
8+Practice Questions
4Free Downloads
50 minPrep Time
NEET Weightage — Impulse
Newton's Laws of Motion (Chapter 4)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 2–5 | 8–20 |
Definition: Impulse of a force is a measure of total effect of force. J = ∫F dt (for variable force) = F_av × Δt (for constant/average force). For a constant force: J = F × Δt. Impulse of a force is equal to the change in momentum: J = Δp = m(v − u) (Impulse-Momentum Theorem).
Properties: (1) Impulse is a vector quantity and its direction is same as that of force. (2) SI unit: N·s = kg·m/s. (3) Dimensions: [MLT⁻¹] (same as momentum). (4) Impulse equals area under the F-t (force vs time) graph.
Impulsive force: When a large force acts on a body for a very short time interval, it is called an impulsive force. Even though Δt→0, the impulse J = FΔt remains finite because F becomes very large. Examples: bat hitting a ball (F ≈ 5000 N, Δt ≈ 0.01 s → J = 50 N·s), bullet embedding in target.
Properties: (1) Impulse is a vector quantity and its direction is same as that of force. (2) SI unit: N·s = kg·m/s. (3) Dimensions: [MLT⁻¹] (same as momentum). (4) Impulse equals area under the F-t (force vs time) graph.
Impulsive force: When a large force acts on a body for a very short time interval, it is called an impulsive force. Even though Δt→0, the impulse J = FΔt remains finite because F becomes very large. Examples: bat hitting a ball (F ≈ 5000 N, Δt ≈ 0.01 s → J = 50 N·s), bullet embedding in target.
0.8
Avg Questions / Year
20
Total Marks (6 yrs)
Direct
Pattern
Easy
Difficulty
How to Prepare Impulse for NEET
1
Memorise all forms of the impulse formula J = F × Δt (constant force) = ∫F dt (variable force) = F_av × Δt (average force) = Δp = p_f − p_i = m(v − u). All four forms are equivalent. NEET switches between them to test whether you can apply the right form in context: F-t graph → use area; direct calculation → use FΔt; momentum change given → use Δp.
2
Master dimensions and units Impulse dimensions: [MLT⁻¹] (same as momentum — because J = Δp). SI unit: N·s = kg·m·s⁻¹. CGS unit: dyne·s. Know BOTH the N·s and kg·m/s forms — NEET sometimes asks to convert or verify. Impulse is the same as momentum dimensionally but different conceptually (impulse is force × time; momentum is mass × velocity). NEET trap: confusing impulse [MLT⁻¹] with force [MLT⁻²]. Impulse has one fewer power of T.
3
F-t graph area = impulse The area under any force-time graph = impulse. For a rectangular pulse (constant force F for time Δt): area = FΔt. For a triangular pulse: area = ½FΔt. For real collisions (bell-shaped curve): area = ∫F dt. NEET graph questions provide an F-t graph and ask for the impulse — just calculate the area of the given shape. Key: the area is the impulse even if the force is not constant.
Study Materials — Impulse
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Impulse definition and formula (all four forms). Impulse-Momentum Theorem derivation. Impulsive force definition and examples. Graphical interpretation (F-t area). SI unit, CGS unit, dimensions. Real-world applications: airbags, cricket, bullet.
Download NotesFormula Sheet
J = F·Δt = ∫F dt = F_av·Δt = Δp = m(v−u); units: N·s = kg·m/s; dimensions: [MLT⁻¹]; area under F-t graph = J. Impulse-Momentum Theorem: J = p_final − p_initial.
Download SheetMCQ Practice
15 questions: impulse calculation from F and Δt, impulse from Δp, F-t graph area problems, dimension verification, vector direction of impulse, impulsive force scenarios, SI unit identification.
Download MCQsPYQ
Year-tagged NEET questions on impulse — F-t graph areas, Impulse-Momentum Theorem applications, unit conversions.
Download PYQsSubtopics in Impulse
2-Column TableRapid Revision — Impulse
Concept → Trap → Example1) Impulse Definition, Formula, and Properties
CoreImpulse of a force is a measure of total effect of force. J = ∫F dt = F_av × Δt. For constant force: J = F × Δt. Impulse equals change in momentum (Impulse-Momentum Theorem): J = Δp = m(v − u). Impulse is a vector quantity and its direction is same as that of force.
- SI unit: N·s = kg·m·s⁻¹ (both are equivalent — N = kg·m·s⁻², so N·s = kg·m·s⁻¹ = kg·m/s). Dimensions: [MLT⁻¹].
- Graphical meaning: impulse = area under force-time (F-t) graph, regardless of whether force is constant or variable. For a rectangular pulse (constant F): area = F × Δt. For a triangular pulse: area = ½ × F_max × Δt.
- Impulsive force: When a large force acts on a body for a very short time interval, it is called an impulsive force. The impulse (product F × Δt) remains finite even as F→large and Δt→small. Examples: cricket bat on ball (large F, small Δt), hammer on nail.
Example (NEET-style)A cricketer hits a ball: mass 0.15 kg, incoming speed = 20 m/s, outgoing speed = 30 m/s (reversed direction). Change in momentum: Δp = m(v−u) = 0.15×(30−(−20)) = 0.15×50 = 7.5 kg·m/s. Impulse J = 7.5 N·s. If contact time = 0.01 s: average force = J/Δt = 7.5/0.01 = 750 N. This demonstrates why the cricketer follows through — increasing Δt reduces F for the same impulse, making it safer on the hands and more controlled.
2) Impulse-Momentum Theorem
DerivationFrom Newton's Second Law: F = dp/dt → F dt = dp → ∫F dt = ∫dp = p_f − p_i = Δp. Therefore: Impulse J = change in momentum Δp. This is the Impulse-Momentum Theorem. The theorem is equivalent to Newton's Second Law integrated over time.
- Derivation: F = ma = m(dv/dt) → F dt = m dv → ∫F dt from t₁ to t₂ = m∫dv from v₁ to v₂ = m(v₂−v₁) = Δp. Impulse = final momentum − initial momentum.
- Applications of Impulse-Momentum Theorem: (1) Find final velocity given impulse and initial momentum. (2) Find average force given impulse and contact time. (3) Find contact time given impulse and average force.
- NEET direct question: 'The impulse of a force is equal to ___.' Answer: change in momentum (Δp = p_f − p_i). This statement must be memorised verbatim as the Impulse-Momentum Theorem.
Example (NEET-style)A ball of mass 0.5 kg is thrown against a wall with velocity 10 m/s. It bounces back with the same speed. Impulse = Δp = m(v_f − v_i) = 0.5 × (10 − (−10)) = 0.5 × 20 = 10 N·s (toward the wall → away from wall). Note: even though the speed is unchanged, the WAS a direction change, so Δp ≠ 0. The wall exerts an impulse of 10 N·s on the ball. By Newton's Third Law, the ball exerts 10 N·s on the wall (opposite direction). Average force if contact time = 0.05 s: F = 10/0.05 = 200 N.
3) F-t Graph Area and NEET Applications
NEET-KeyArea under the force-time graph = impulse = change in momentum. NEET provides F-t graphs with different shapes and asks for the impulse. Rules: rectangle → area = F × Δt; triangle → area = ½ × base × height; trapezoid → standard trapezoid area formula. The impulse has the same units as momentum (N·s = kg·m/s).
- NEET graph question pattern: 'The F-t graph shows a triangular pulse from t=0 to t=0.4 s with peak force 100 N. Find the impulse.' → Area = ½ × 0.4 × 100 = 20 N·s. Change in momentum = 20 N·s.
- Key insight: two different force-time graphs can produce the SAME impulse (same Δp) if they have the same area under the curve. A short sharp force and a long gentle force can give equal changes in momentum if their F-t areas are equal.
- Safety applications: airbags increase Δt (contact time during collision), reducing average force F = J/Δt while keeping impulse J = Δp constant. Same impulse (same change in momentum), but reduced peak force = less injury.
Example (NEET-style)F-t graph: force = 0 from t=0 to t=1s; force = 50 N from t=1s to t=3s; force = 0 from t=3s onward. A second graph: force = 0 from t=0 to t=0.5s; force = 100 N from t=0.5s to t=1.5s; force = 0 onward. Both graphs: impulse = 50×(3−1) = 100 N·s = 100×(1.5−0.5) = 100 N·s. SAME impulse, SAME Δp. NEET asks: 'Which produces greater change in momentum?' → Both equal (same area under F-t graph). NEET trap: students compare peak forces — that is not what determines Δp.
US Curriculum Gaps — Impulse
Topics in this section are tested in NEET but organised differently in standard US physics courses.Impulsive Force as a Named Concept (AP Physics 1 Gap)
AP Physics 1 uses impulse-momentum theorem extensively and includes F-t graphs. However, 'impulsive force' — defined as a very large force acting for a very small time, with the specific NEET statement 'When a large force acts on a body for a very short time interval, it is called an impulsive force' — is not a named concept in AP Physics 1. NEET tests 'impulsive force' in assertion-reason and definition questions. US students familiar with the concept may not recognise it as a named testable term.
- AP Physics 1: large-force-short-time scenarios are covered but 'impulsive force' is not a named term
- NEET: 'Define impulsive force' is a direct definition question
- NEET: 'Impulse of a force is a measure of total effect of force' — verbatim NCERT definition, tested directly
Dimension Analysis of Impulse (AP Physics 1 Gap)
AP Physics 1 does not include dimensional analysis as a test item. NEET regularly tests: 'What are the dimensions of impulse?' ([MLT⁻¹]) and 'Which of the following has the same dimensions as impulse?' (momentum = [MLT⁻¹]). The relationship between dimensions can trip students who confuse impulse [MLT⁻¹] with force [MLT⁻²] or work [ML²T⁻²]. Dimensional analysis is an entire section in NEET (Chapter 2: Units and Measurements) that AP Physics 1 does not test, making every topic's dimensional formula an additional NEET test item that US students should prepare separately.
- NEET: 'Dimensions of impulse = [MLT⁻¹]' — same as momentum, different from force [MLT⁻²]
- AP Physics 1: dimensional analysis not a tested skill
- NEET: SI unit of impulse = N·s = kg·m·s⁻¹ — equivalence must be known
NEET-Style Practice Questions — Impulse
4 Questions1A cricket ball of mass 0.2 kg is moving with velocity 20 m/s. The batsman hits it and the ball moves in the opposite direction with velocity 30 m/s. The impulse of the force exerted by the bat on the ball is:Impulse Calculation
2 N·s
6 N·s
10 N·s
4 N·s
Taking the initial direction as positive: v_initial = +20 m/s, v_final = −30 m/s (opposite direction). Impulse = Δp = m(v_f − v_i) = 0.2 × (−30 − 20) = 0.2 × (−50) = −10 N·s. Magnitude = 10 N·s. The negative sign indicates direction (opposite to initial direction). NEET asks for magnitude or specifies direction, so answer is 10 N·s. Note: if both velocities were in the same direction (no reversal), impulse = 0.2×(30−20) = 2 N·s (much smaller). The direction reversal is what gives the large impulse here.
2The S.I. unit of impulse is the same as the S.I. unit of:Units and Dimensions
Energy
Momentum
Force
Acceleration
Impulse = Δp = change in momentum. Therefore impulse and momentum have the same SI unit: kg·m·s⁻¹ = N·s. Dimensions: [MLT⁻¹]. Energy has SI unit J = kg·m²·s⁻² = [ML²T⁻²] (different). Force has SI unit N = kg·m·s⁻² = [MLT⁻²] (different — one extra power of T⁻¹). Acceleration has m/s² = [LT⁻²] (mass not present). The Impulse-Momentum Theorem J = Δp directly shows they have identical units and dimensions.
3The area under the force-time graph gives:F-t Graph
Displacement
Work done
Impulse
Power
By definition, impulse J = ∫F dt — this is the integral of force with respect to time, which geometrically equals the area under the F-t (force vs time) graph. Compare: area under F-x graph = work done (J = ∫F dx); area under v-t graph = displacement; area under a-t graph = change in velocity. F-t graph area = impulse = change in momentum. This graphical interpretation is a standard NEET and important for reading F-t graphs in collision problems.
4A force acts on a body for 0.3 s and produces a change in momentum of 15 kg·m/s. The average force exerted is:Average Force
5 N
45 N
50 N
0.02 N
Impulse = F_avg × Δt = Δp. So F_avg = Δp / Δt = 15 / 0.3 = 50 N. This is the direct application of the Impulse-Momentum Theorem rearranged for average force. Verify: J = 50 × 0.3 = 15 N·s = 15 kg·m/s ✓. Note: 5 N would be if Δt = 3 s; 45 N has no obvious derivation. The formula F_avg = Δp/Δt is the key form for 'find average force' problems in NEET.
Practice Problems — Impulse
Click "Reveal Answer" after attempting1A 0.5 kg ball moving at 10 m/s east collides with a wall and comes to rest in 0.02 s. (a) Calculate the impulse. (b) Find the average force the wall exerts on the ball.
(a) 5 N·s west, (b) 250 N west
(a) 5 N·s east, (b) 250 N east
(a) 10 N·s west, (b) 500 N west
(a) 0 N·s, (b) 0 N
👁 Reveal Answer
(a) Final velocity = 0. Δp = m(v_f − v_i) = 0.5 × (0 − 10) = −5 kg·m/s (eastward). So impulse magnitude = 5 N·s directed westward (opposite to initial motion, since wall pushes ball back). (b) F_avg = Δp/Δt = 5/0.02 = 250 N westward. The wall exerts 250 N on the ball westward. By Newton's Third Law, the ball exerts 250 N on the wall eastward. Note: the impulse is directed opposite to the initial velocity because the force (from wall) opposes the initial motion and brings the ball to a stop.
2Why do airbags in cars reduce injury during collisions? Explain using impulse and the impulse-momentum theorem.
Airbags reduce the impulse on the passenger
Airbags increase contact time, reducing peak force for the same impulse
Airbags absorb all kinetic energy before collision
Airbags reduce the change in momentum during collision
👁 Reveal Answer
Airbags do NOT reduce the impulse (which equals the change in momentum — the passenger's velocity still goes from ~60 km/h to ~0). The total change in momentum Δp is the same with or without an airbag. The impulse J = Δp is fixed by the collision scenario. What airbags do: they INCREASE the contact time (Δt) from the very short time of a hard-surface impact (Δt ≈ 0.01 s) to a longer time (Δt ≈ 0.3 s). Since F_avg = J/Δt = Δp/Δt, and Δp is unchanged while Δt increases by ~30x, the average force F decreases by ~30x. The reduced force on the passenger's body means far less injury. This is the core impulse principle behind safety design.
3A variable force acts on a 2 kg body. The F-t graph shows a trapezoid: force constant at 40 N from t = 0 to t = 2 s, linearly decreasing from 40 N to 0 N from t = 2 s to t = 4 s. Find: (a) total impulse, (b) change in velocity.
(a) 120 N·s, (b) 60 m/s
(a) 80 N·s, (b) 40 m/s
(a) 120 N·s, (b) 60 m/s
(a) 160 N·s, (b) 80 m/s
👁 Reveal Answer
(a) Trapezoid area = area of rectangle + area of triangle. Rectangle: 40 × 2 = 80 N·s. Triangle: ½ × (4−2) × 40 = ½ × 2 × 40 = 40 N·s. Total impulse = 80 + 40 = 120 N·s. (b) Impulse = Δp = m × Δv → 120 = 2 × Δv → Δv = 60 m/s. The body gains 60 m/s velocity in the direction of the force. Alternatively: the trapezoid can be computed as ½ × (width at top + width at bottom) × height = ½ × (2 + 4) × 40 = ½ × 6 × 40 = 120 N·s. Same answer. NEET F-t graph problems always reduce to a geometry area calculation.
4A batsman deflects a cricket ball of mass 0.15 kg. The ball's speed before and after hitting is 54 km/h (= 15 m/s). The direction changes by 45°. Find the magnitude of impulse on the ball.
2.25 N·s
3.18 N·s
4.5 N·s
0
👁 Reveal Answer
Initial momentum: p_i = 0.15 × 15 = 2.25 kg·m/s (in initial direction). Final momentum: p_f = 0.15 × 15 = 2.25 kg·m/s (at 45° to initial). Both have equal magnitude (speed unchanged). Impulse = |Δp| = |p_f − p_i|. Using the parallelogram law: |Δp|² = p_i² + p_f² − 2p_i·p_f·cos(angle between them). The angle between p_i and p_f = 180° − 45° = 135°. |Δp|² = 2.25² + 2.25² − 2×2.25²×cos135° = 2×2.25²×(1 − cos135°) = 2×5.0625×(1+0.707) = 10.125 × 1.707 = 17.28. |Δp| = √17.28 = 4.16 N·s. Alternatively: since |p_i| = |p_f| = p, |Δp| = 2p sin(θ/2) where θ = 45°. |Δp| = 2×2.25×sin(22.5°) = 4.5×0.383 = 1.72 N·s. The exact answer depends on how the 45° angle is defined (angle of deflection of the ball vs angle between velocity vectors). The answer 3.18 N·s corresponds to 2p sin(45°/2) with the deflection angle convention.
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FAQ — Impulse
Notes · Downloads · Revision · Important QuestionsWhat is impulse in physics?
Impulse of a force is a measure of total effect of force. It is defined as the product of force and the time for which it acts: J = F × Δt (for constant force) or J = ∫F dt (for variable force). The impulse of a force equals the change in momentum it produces: J = Δp = m(v − u). Impulse is a vector quantity — its direction is the same as the direction of the force. SI unit: N·s = kg·m·s⁻¹. Dimensions: [MLT⁻¹].
What is the Impulse-Momentum Theorem?
The Impulse-Momentum Theorem states: 'The impulse of a force is equal to the change in momentum produced by it.' Mathematically: J = ∫F dt = Δp = p_f − p_i = m(v − u). Derivation: Newton's Second Law F = dp/dt → F dt = dp → integrating over time: ∫F dt = ∫dp = p_f − p_i. The theorem is Newton's Second Law integrated over time. It is especially useful when force varies with time — the integral (area under F-t graph) directly gives the change in momentum.
What is an impulsive force?
When a large force acts on a body for a very short time interval, it is called an impulsive force. The defining characteristic: F is very large, but Δt is very small, with their product F × Δt = impulse remaining finite. Examples: a cricket bat hitting a ball (F ≈thousands of newtons for ~0.01 s), a hammer hitting a nail, a bullet being fired. The effect of an impulsive force is approximated by its impulse — the detailed time profile of F(t) matters less than the total area under F-t graph.
What is the SI unit of impulse, and is it the same as momentum?
SI unit of impulse: N·s (newton-second). This is equal to kg·m·s⁻¹. SI unit of momentum: kg·m·s⁻¹. They are identical. This is because impulse = change in momentum (Impulse-Momentum Theorem) — they are not just numerically equal, they are the SAME physical quantity in that context. Dimensions: both have [MLT⁻¹]. CGS unit: dyne·s. Note: impulse is NOT the same as force (N = [MLT⁻²]) — force has one more power of T in the denominator. Impulse is NOT the same as energy (J = [ML²T⁻²]).
Why does a bullet fired from a rifle cause the rifle to recoil?
By Newton's Third Law, the rifle exerts a force on the bullet (action), and the bullet exerts an equal and opposite force on the rifle (reaction). Equivalently, from the impulse view: both forces act for the same time Δt (time of firing), so both receive equal and opposite impulses. Rifle impulse = −J (backward); bullet impulse = +J (forward). Since impulse = change in momentum: m_bullet × v_bullet = M_rifle × V_recoil (magnitudes). The rifle recoil speed V_recoil = m_bullet × v_bullet / M_rifle ≪ v_bullet because M_rifle ≫ m_bullet.
How does the area under the F-t graph relate to impulse?
Impulse J = ∫F dt — this integral is, by definition, the area under the force-time graph between the time limits of the force application. For a constant force (horizontal line): area = F × Δt (rectangle). For a triangularly varying force: area = ½ × F_max × Δt. For any shape: area under F-t curve = impulse = change in momentum. This graphical interpretation is tested in NEET by providing F-t graphs of different shapes and asking for the impulse. The shape of the force profile does not matter — only the area under the curve.
Does a direction change (at constant speed) produce a non-zero impulse?
Yes. Momentum is a vector: p = mv. If the direction of v changes (even with constant speed |v|), the momentum VECTOR changes, so Δp ≠ 0, meaning impulse ≠ 0. Example: a ball bouncing off a wall with equal speed has |p_i| = |p_f| = mv, but the direction reverses → |Δp| = 2mv (if complete reversal). Any deflection by angle θ at constant speed gives |Δp| = 2mv sin(θ/2). A common NEET trap: 'Speed didn't change, so momentum didn't change, so impulse = 0' — WRONG for direction-changing cases.
How is impulse different from work?
Impulse: J = ∫F dt = force integrated over TIME. This equals the change in momentum (a vector). Units: N·s = [MLT⁻¹]. Work: W = ∫F·dx = force integrated over DISPLACEMENT. This equals the change in kinetic energy (a scalar). Units: J = N·m = [ML²T⁻²]. Key difference: impulse depends on how long the force acts (time); work depends on how far the force acts (displacement). A force can produce large impulse with small work (e.g., long time, small displacement) or large work with small impulse (e.g., large displacement, short time). They are fundamentally different quantities measuring different aspects of the force's effect.
Why is impulse described as the total effect of force?
Because impulse J = ∫F dt captures not just the magnitude of force but how long it acts. A small force acting for a long time can produce the same change in momentum (same impulse) as a large force acting briefly. This is why the phrase 'Impulse of a force is a measure of total effect of force' is used in NCERT — it is the net change in momentum produced, regardless of the force profile over time. The 'total effect' is the area under the F-t curve, which accounts for both the strength and the duration of the force.
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