Definition
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. 55Impulse
8 MCQs5 revision cards9-step worked example
Source: NCERT Laws of MotionPYQ coverage: NEET 2021, 2025Official key: NTA-verifiedLast updated: 24 Sep 2026
Impulse, explained for NEET
Impulse: the quantity NEET tests when force acts briefly
When a cricket bat strikes a ball, the contact lasts a few milliseconds. The force during that instant is enormous but unmeasurable in real time. What you can measure — and what NEET asks about — is the product of that average force and the contact time. That product is impulse.
Newton's Second Law in its momentum form gives the definition directly. Force equals the rate of change of momentum: F = dp/dt. Multiply both sides by the time interval Δt and you get J = F · Δt = Δp (NCERT Class 11 Physics Chapter 4, page 55). Impulse equals the change in linear momentum. The SI unit of impulse is N·s, which is dimensionally identical to kg·m/s — the unit of momentum.
The high-frequency trap on impulse: direction reversal at a bounce. When a ball hits a floor and rebounds, velocity reverses sign. If you define downward as positive, the ball arrives at +v₁ and leaves at −v₂. The change in momentum is m(−v₂ − v₁) = −m(v₁ + v₂). The magnitude of impulse is m(v₁ + v₂), not m(v₁ − v₂). Students who forget the sign reversal subtract instead of adding the speeds, losing marks to negative marking.
A second common confusion: impulse is not force. Impulse has dimensions of momentum (kg·m/s); force has dimensions kg·m/s². Dropping the Δt factor turns a correct impulse answer into the wrong unit — an error that NEET distractors exploit.
When force varies with time, impulse equals the area under the F–t curve: J = ∫F dt. For a constant or average force, this reduces to J = F_avg · Δt.
Watch out: impulse is a vector. Its direction is the direction of the average force, which is the direction of Δp — not the direction of motion.
Can you answer these Impulse 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 impulse is:
Show answer and why every option is right or wrong
Answer: C. Impulse J = F·Δt has unit N·s. Since 1 N = 1 kg·m/s², we get N·s = kg·m/s, which is the unit of momentum. This is stated in NCERT Class 11 Physics Chapter 4, page 55.
Why A is wrong: A gives the unit of force (newton = kg·m/s²), not impulse. Confusing force with impulse is a common unit error.
Why B is wrong: B gives the unit of energy (joule = kg·m²/s²). Impulse and energy have different dimensions.
Why D is wrong: D gives watt·second = joule, which is energy. Impulse has dimensions of momentum, not energy.
MCQ 2Easy RecallPractice
Impulse delivered to a body equals:
Show answer and why every option is right or wrong
Answer: B. By Newton's Second Law in momentum form, J = F·Δt = Δp. Impulse equals the change in momentum, not the momentum itself (NCERT Class 11 Physics Chapter 4, page 55).
Why A is wrong: A confuses impulse with the work-energy theorem. Impulse relates to momentum change, not energy change.
Why C is wrong: C gives the total momentum p = mv, not the change in momentum Δp. Impulse equals Δp, not p.
Why D is wrong: D describes work (force × displacement), not impulse (force × time). Different physical quantities.
MCQ 3Easy RecallPractice
Which of the following is dimensionally equivalent to impulse?
Show answer and why every option is right or wrong
Answer: D. Impulse has dimensions [MLT⁻¹], identical to momentum (mass × velocity). Both are measured in kg·m/s (equivalently N·s).
Why A is wrong: A has dimensions [MLT⁻²]. Force differs from impulse by a factor of time.
Why B is wrong: B has dimensions [ML²T⁻³]. Power is energy per unit time, unrelated to impulse.
Why C is wrong: C has dimensions [ML²T⁻²]. Energy and impulse have different dimensional formulas.
MCQ 4Direct ApplicationPractice
A 0.50 kg ball moving at 6.0 m/s is brought to rest by a fielder in 0.10 s. The average force exerted by the fielder on the ball is:
Show answer and why every option is right or wrong
Answer: A. Impulse J = Δp = m(v_f − v_i) = 0.50 × (0 − 6.0) = −3.0 kg·m/s. Average force F = J/Δt = 3.0/0.10 = 30 N (magnitude). The force opposes the motion.
Why B is wrong: B results from computing m × v = 0.50 × 6.0 = 3.0 and forgetting to divide by Δt. That gives impulse magnitude, not force.
Why C is wrong: C results from dividing by 10 s instead of 0.10 s, or from an arithmetic slip in the decimal.
Why D is wrong: D results from using Δt = 0.01 s instead of the given 0.10 s. Always read the contact time carefully.
MCQ 5Direct ApplicationPractice
A force of 10 N acts on a body of mass 2.0 kg for 3.0 s. If the body was initially at rest, its final momentum is:
Show answer and why every option is right or wrong
Answer: C. J = F·Δt = 10 × 3.0 = 30 N·s. Since initial momentum is zero, final momentum = Δp = 30 kg·m/s.
Why A is wrong: A results from computing F/m = 10/2 = 5.0, which is the acceleration in m/s², not the momentum.
Why B is wrong: B results from computing F × Δt × m = 10 × 3.0 × 2.0 = 60, incorrectly multiplying by mass a second time. Impulse already equals Δp.
Why D is wrong: D results from computing m × Δt = 2.0 × 3.0 = 6.0, which has no physical meaning in this context.
MCQ 6Direct ApplicationPractice
A 0.20 kg ball hits a wall at 5.0 m/s perpendicular to the wall and rebounds at 3.0 m/s. The magnitude of impulse on the ball is:
Show answer and why every option is right or wrong
Answer: B. Take toward-wall as positive. v_i = +5.0 m/s, v_f = −3.0 m/s (rebound reverses direction). Δp = m(v_f − v_i) = 0.20 × (−3.0 − 5.0) = 0.20 × (−8.0) = −1.6 kg·m/s. Magnitude = 1.6 kg·m/s. The speeds ADD because velocity reverses — this is the rebound sign-reversal trap.
Why A is wrong: A results from computing m × (v₁ − v₂) = 0.20 × (5.0 − 3.0) = 0.40, subtracting speeds instead of adding them. The velocity direction reverses on rebound, so the change is v₁ + v₂, not v₁ − v₂ (trap: sign reversal at bounce).
Why C is wrong: C results from computing m × v₁ = 0.20 × 5.0 = 1.0, using only the initial momentum and ignoring the rebound entirely.
Why D is wrong: D results from computing m × (v₁ − v₂) × 2 or a similar arithmetic error that still misses the correct sign treatment.
MCQ 7CalculationPYQ Pattern
A ball of mass 0.10 kg is dropped from a height of 5.0 m and rebounds to a height of 3.2 m. Taking g = 10 m/s², the magnitude of impulse imparted by the ground to the ball is:
Show answer and why every option is right or wrong
Answer: D. Speed at impact: v₁ = √(2 × 10 × 5.0) = 10 m/s. Speed 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(v₁ + v₂) = 0.10 × (10 + 8.0) = 1.8 kg·m/s.
Why A is wrong: A results from computing m × (v₁ − v₂) = 0.10 × (10 − 8.0) = 0.20, subtracting speeds. The velocity reverses direction on bounce, so the magnitudes add: Δp = m(v₁ + v₂) (trap: rebound sign reversal).
Why B is wrong: B results from computing m × v₁ = 0.10 × 10 = 1.0, using only the downward momentum and ignoring the rebound contribution.
Why C is wrong: C gives a value in newtons (force unit), not in kg·m/s (impulse unit). This error conflates impulse with force — impulse has dimensions of momentum (trap: impulse vs force unit confusion).
MCQ 8Concept TrapPYQ Pattern
A body of mass m moving due north at speed v turns due east at the same speed v after a brief impulsive force. The direction of the impulse is:
Show answer and why every option is right or wrong
Answer: A. Impulse direction = direction of Δp = p_f − p_i. p_i = mv (north), p_f = mv (east). Δp = mv (east) − mv (north) = mv (east) + mv (south). The resultant points south-east at 45°. The force is NOT in the direction of final motion — it is in the direction of the change in momentum.
Why B is wrong: B assumes force acts in the direction of initial velocity, which would maintain the original motion, not change it.
Why C is wrong: C results from averaging the initial and final velocity directions. The impulse direction is found by vector subtraction p_f − p_i, not by bisecting the two velocity vectors.
Why D is wrong: D assumes force acts in the direction of final velocity. Force direction equals the direction of Δp = p_f − p_i, not the direction of p_f alone (trap: confusing Δp direction with velocity direction).
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Impulse: quick recall before you leave
How do you solve a Impulse question? A worked example
Pattern: Drop-and-rebound impulse (NEET pattern: impulse drop rebound, observed in NEET 2021 and 2025)
- 1
Given
A rubber ball of mass m = 0.15 kg is dropped from height h₁ = 4.05 m onto a hard floor. It rebounds to height h₂ = 1.80 m. Take g = 10 m/s² (exact, problem-defined).
- 2
Required
Find the magnitude of impulse imparted by the floor to the ball.
- 3
Concept
Impulse equals the change in momentum: J = Δp = m(v_after − v_before). Velocity reverses direction at the bounce, so the speeds add in magnitude.
- 4
Formula
J = m(v₁ + v₂), where v₁ = √(2gh₁) and v₂ = √(2gh₂).
- 5
Substitution
v₁ = √(2 × 10 × 4.05) = √81.0 = 9.0 m/s
v₂ = √(2 × 10 × 1.80) = √36.0 = 6.0 m/s - 6
Calculation
Taking downward as positive: v_before = +9.0 m/s (downward), v_after = −6.0 m/s (upward).
Δp = m(v_after − v_before) = 0.15 × (−6.0 − 9.0) = 0.15 × (−15.0) = −2.25 kg·m/s
Magnitude of impulse = 2.25 kg·m/s.
Note: g = 10 m/s² is an exact problem-defined constant and does not limit the significant-figure count. The given data (0.15 kg, 4.05 m, 1.80 m) each have 3 significant figures, so the answer is reported to 3 significant figures. - 7
Final answer
|J| = 2.25 kg·m/s (equivalently 2.25 N·s), directed upward (away from the floor).
- 8
Common trap
The high-frequency error is computing m(v₁ − v₂) = 0.15 × (9.0 − 6.0) = 0.45 kg·m/s — subtracting speeds instead of adding them. The velocity reverses direction on rebound, so Δv = v₁ + v₂, not v₁ − v₂. This incorrect value (0.45) typically appears as a distractor in NEET options.
- 9
Similar NEET-style question
A 0.20 kg ball is dropped from 1.25 m and bounces back to 0.80 m. Find the impulse on the ball from the ground (g = 10 m/s²). [Answer: v₁ = 5.0 m/s, v₂ = 4.0 m/s, J = 0.20 × (5.0 + 4.0) = 1.8 kg·m/s upward.]
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What to remember before solving Impulse questions
Which Impulse 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 Impulse?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
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: 14 exam traps and mistakes · 8 formulas · 9 question patterns from its other lessons.
Impulse questions from past NEET papers
2 questions from NEET 2021, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
184 NS
221 NS
37 NS
40
NTA Answer: Option 2(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)
How does NEET ask about Impulse?
1 recurring pattern from past papers — click to collapse
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.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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