Impulse and Change of Momentum

Q: Is an impulse of 1 N s the same as 1 kg m/s?

Yes, exactly, with no conversion. A newton is kg m s−2 by the SI Brochure's Table 4, so a newton second is kg m/s. Papers use N s when they are thinking of a force acting for a time and kg m/s when they are thinking of a body in motion, but the numbers are identical and either is acceptable in an answer.

Impulse is the change in momentum, and the marks are in the signs. A ball that arrives at 25 m/s and leaves at 15 m/s the other way has changed its momentum by 40 m/s worth, not by 10 — this takes signed velocities, gives you the impulse, the average force and how many times the body’s own weight that force is.

Last updated 11 Nov 2026

Signed velocities in, impulse and average force out

J = Δp = m(v − u), and F = J/Δt

Mass of the bodyIn kilograms. Grams are the standard trap on this topic because so many of the bodies are small: a cricket ball is about 0.16 kg, a tennis ball 0.058, a golf ball 0.046. Typing 160 instead of 0.16 multiplies the force by a thousand.

Velocity before, signedIn m/s, and SIGNED along whichever direction you choose as positive. The choice is yours; the consistency is not. Pick a direction, call it positive, and stick to it for both velocities — the sign of the answer then tells you which way the impulse pointed.

Velocity after, signedIn m/s, with the same positive direction as the field above. If the body reverses, this is NEGATIVE where the first was positive, and that is the whole point of the page: a reversal makes the change of momentum larger than either momentum on its own, not smaller. Zero means the body stops.

Time the force acts forIn seconds, and this is where the impressive numbers come from. Contact times in impacts are short — a few milliseconds for a bat on a ball, tens of milliseconds for a car’s crumple zone — and the average force is the impulse divided by this. Halve the contact time and you double the force for the same change in momentum, which is the engineering idea behind every airbag, crash helmet and crumple zone there is. Zero is accepted and leaves the force rows blank, because an impulse delivered in no time at all implies an infinite force.

-6.4000N sExample

A 0.16 kg cricket ball arriving at 25 m/s, leaving at 15 m/s the other way, over 15 ms

One quantity under two names, and the signs that decide the answer

J = F Δt = Δp = mv − mu · F_avg = Δp / Δt · 1 N s = 1 kg m/s

J = Δp: the impulse-momentum principle, and it is an identity rather than a formula to apply: the impulse of the resultant force IS the change in momentum. Edexcel’s Further Mechanics 1 calls it “the impulse-momentum principle” and lists it as content; Cambridge 9702 gets at the same idea by asking you to “define and use force as rate of change of momentum”
mv − mu: the order matters: after minus before. Writing it the other way round reverses the sign of every answer. And both terms are SIGNED — the single most common error on this topic is to take magnitudes and subtract, which for a rebounding body gives an answer that is both too small and pointing the wrong way
N s and kg m/s: the same unit twice. A newton second is kg m s⁻² × s = kg m/s, so an impulse and a momentum are measured in identical units — which is the clearest possible hint that they are the same kind of quantity. Use N s for the impulse and kg m/s for the momentum if you like, but never treat them as needing a conversion
F_avg: the average force, which is all a single division can give you. The force during a real impact rises and falls, and the peak is commonly two to three times the average. For the same impulse, force and time trade off exactly: double the time and halve the force
F = dp/dt: Newton’s second law in its general form, and the one the impulse-momentum principle comes straight out of. It survives a changing mass, which F = ma does not, so it is the version that works for rockets, for a conveyor being loaded and for a hosepipe pushing back on your hands

Worked example

A 0.16 kg cricket ball arriving at 25 m/s, leaving at 15 m/s the other way, over 15 ms
CHOOSE A POSITIVE DIRECTION and keep it. Take the ball's incoming direction as positive. Then u = +25 m/s and v = −15 m/s, because it goes back the way it came. Everything below follows from getting that minus sign in.
THE TWO MOMENTA. Before: mu = 0.16 × 25 = +4.0000 kg m/s. After: mv = 0.16 × (−15) = −2.4000 kg m/s. The momentum has not merely shrunk, it has changed sign.
THE IMPULSE. J = mv − mu = −2.4000 − (+4.0000) = −6.4000 N s. Note the size: 6.4 is LARGER than either momentum on its own, because reversing takes more than stopping. Drop the signs and subtract magnitudes and you get 0.16(15 − 25) = −1.6 N s, which is four times too small; the page prints that wrong answer as a row so the two can be seen together.
THE AVERAGE FORCE. F = J/Δt = −6.4000 / 0.015 = −426.667 N, directed back along the ball's original path. The ball's own weight is 0.16 × 9.80665 = 1.569 N, so the bat applies about 272 times the ball's weight — which is why gravity is ignored during an impact calculation and why the peak force, two or three times this average, is what breaks bats.
AND THE ENERGY, WHICH IS NOT CONSERVED. Kinetic energy before: ½ × 0.16 × 25² = 50.0000 J. After: ½ × 0.16 × 15² = 18.0000 J. So 32 J has gone into deformation, sound and heat. Momentum changed by a definite impulse and energy simply disappeared; the two conservation statements are not the same statement.

Signed velocities, the unit that is the same unit twice, and why Δt is the only thing you can change

The impulse of a force is the change in momentum it produces, and almost every lost mark on this topic is a sign. J = mv − mu, with both velocities signed along one direction you have chosen. For a body that keeps going the same way, the signs look after themselves and the arithmetic is unremarkable. For a body that reverses — a ball off a bat or a wall, a particle rebounding, a vehicle thrown back in a collision — the two momenta have opposite signs and the change is the SUM of their magnitudes. The classic wrong answer takes the two speeds, subtracts them, and reports a quarter of the truth. This page prints that wrong answer as its own row, labelled, because seeing the two side by side is what fixes the habit.

A newton second and a kilogram metre per second are the same unit. N s = kg m s⁻² × s = kg m/s, so impulse and momentum are not merely related, they are measured identically. If you ever find yourself looking for a conversion factor between them, the thing to change is the expectation that there is one. This also explains why F Δt and Δp can appear on opposite sides of the same equation without anything being converted.

For a fixed change in momentum, force and time trade off exactly — and this is the most useful thing on the page. If you are travelling at 30 m/s and you are going to stop, the impulse is settled: it is your mass times 30, and nothing can reduce it. What is not settled is how long the stopping takes. Stop in 0.1 s against a wall and the average force is ten times what it is stopping in 1 s against a crumple zone. Airbags, crash helmets, crash mats, the bend in a cricketer’s knees as they land and the give in a boxer’s neck all do the same single thing: they extend Δt. None of them removes any impulse at all.

Impulse is also the right tool when the force is not constant, which is most of the time. During a real impact the force rises from nothing to a peak and falls back, and nothing you could sensibly call “the force” exists. The impulse does: it is the area under the force-time graph, and it is exactly the change in momentum however jagged the graph is. That is why F = Δp/Δt gives an average and never a peak, and why quoting the average as though it were the maximum understates a real load by a factor of two or three.

Momentum is conserved; kinetic energy is not. In the worked example above, the ball loses 32 of its 50 J, to deformation, sound and heat. The momentum, by contrast, changed by precisely the impulse the bat delivered — and if you drew the boundary round the bat, the batter and the earth as well, the total momentum would not have changed at all. Keeping the two conservation statements apart is what makes collision problems tractable: momentum always, energy only when the collision is elastic. The CBSE Class XI list puts impulse and the “law of conservation of linear momentum and its applications” together in its Laws of Motion unit for that reason.

Frequently asked questions

What is the difference between impulse and momentum?

Momentum is a property of a moving body, mv. Impulse is what a force does to it over an interval, F Δt, and it equals the CHANGE in momentum. They share a unit — a newton second is a kilogram metre per second — because they are the same kind of quantity, one as a state and one as a change. A body has momentum; a collision delivers an impulse.

Why is the impulse bigger when the ball bounces back?

Because reversing is more than stopping. Stopping a ball of momentum +4 kg m/s takes an impulse of −4. Sending it back out at 2.4 takes that 4 to bring it to rest and another 2.4 to drive it the other way: 6.4 in total. In the signed arithmetic it happens automatically, J = (−2.4) − (+4) = −6.4. This is also why a bouncy ball exerts more force on a wall than a lump of putty of the same mass and speed, which is a favourite exam question.

Is an impulse of 1 N s the same as 1 kg m/s?

Yes, exactly, with no conversion. A newton is kg m s⁻² by the SI Brochure’s Table 4, so a newton second is kg m/s. Papers use N s when they are thinking of a force acting for a time and kg m/s when they are thinking of a body in motion, but the numbers are identical and either is acceptable in an answer.

How do airbags and crumple zones work, in these terms?

They increase Δt and nothing else. Your change in momentum in a crash is fixed by your mass and your speed, and no device can shrink it. Since F = Δp/Δt, stretching the stopping time from a few hundredths of a second to a few tenths cuts the average force by the same factor. That is the entire mechanism, and it is why a longer crumple zone is a safer one.

Is the average force the same as the maximum force?

No, and the difference matters. The force in a real impact starts at zero, rises to a peak and falls back; the average is the impulse divided by the whole duration, and the peak is commonly two to three times that. The impulse is the AREA under the force-time graph, which is well defined no matter what shape the graph has. Use the average for a school answer and treat any design or safety judgement as needing the peak.

Do I need to include the weight of the body during the impact?

Almost never. In the worked example the bat applies 427 N on average while the ball weighs 1.57 N, so the weight is 0.4 per cent of the impact force and disappears into the rounding. The page prints the ratio so you can check it rather than assume it. Over a LONG interaction — a ball in free flight, a rocket burning for a minute — gravity’s impulse is not negligible at all, and then it belongs in the sum.

Related calculators

References

PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26 (read 11 November 2026). Unit III, Laws of Motion, reads in part: “Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse”, “Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication” and “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Quoted to establish what is in scope, not to claim any endorsement by the board.
Cambridge International AS & A Level Physics 9702, syllabus for examination in 2025, 2026 and 2027, Cambridge Assessment International Education (read 11 November 2026). Two exclusions on this syllabus are worth more than most inclusions, because they explain why a question from one board looks unanswerable to a student taught on another. Topic 3.2 asks only for “a qualitative understanding of frictional forces and viscous/drag forces including air resistance” and states that “no treatment of the coefficients of friction and viscosity is required”. Topic 3.3 states that “knowledge of the concept of coefficient of restitution is not required”, while still requiring that for an elastic collision “the relative speed of approach is equal to the relative speed of separation”. Its data sheet gives the acceleration of free fall as 9.81 m s⁻².
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0), specification Issue 4, June 2023 (read 11 November 2026). Further Mechanics 1 carries what the physics syllabuses leave out: topic 1.1 is “Momentum and impulse. The impulse-momentum principle”, topic 4.1 is “Direct impact of elastic spheres. Newton’s law of restitution”, with the note that “Students will be expected to know and use the inequalities 0 ≤ e ≤ 1”, and topics 4.2 and 5 extend to successive and oblique impacts. So the coefficient of restitution is examinable — in mechanics, under mathematics, on a different paper from the physics.
The International System of Units (SI), 9th edition, 2019, Bureau International des Poids et Mesures (read 11 November 2026). Table 4 gives the newton as kg m s⁻², which is the only definition of force this site uses: a newton is not a unit of mass and a kilogram is not a unit of force. Appendix 1 lists the 3rd CGPM of 1901, whose declaration both separated mass from weight and fixed the conventional value of g_n — the two things a force question most often conflates. Cited, not reproduced.
Fundamental Physical Constants — Complete Listing, NIST Reference on Constants, Units and Uncertainty, 2022 CODATA adjustment (read 11 November 2026). The line this batch depends on is “standard acceleration of gravity 9.806 65 (exact) m s⁻²”. The word exact is the point: g_n is a conventional value, not a measured one, so it carries no uncertainty and it is not the g where you are standing. The same listing gives the Newtonian constant of gravitation as 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² with an uncertainty of 0.000 15 × 10⁻¹¹, which is 22 parts per million — the worst-known of the constants a school paper quotes, and a useful reminder that “constant” and “exact” are different claims.