Impulse Calculator

impulse = force × time = mass × change in velocity

F

The force, in newtons, with kilonewtons, pounds-force and kilogram-force in the same field. It is the force during the contact, which for an impact is not the weight of anything — it is the force the object feels while it is being stopped. Leave this pair empty if you are working from mass and velocity instead

Δt

How long the force acts, in seconds, minutes or hours. It is in the numerator here, which is what separates impulse from force: the same force for twice as long delivers twice the impulse. The field accepts zero, and zero is the right answer rather than an error — no duration means no impulse. It does not accept a negative time

m

The mass the force is acting on, in kilograms, with grams, tonnes and pounds in the same field. This is the second route to the same quantity and it needs the velocity change rather than the time, so fill in this pair only if you are leaving the force pair empty

Δv

How much the velocity changes, in metres per second, with km/h, mph, ft/s and knots in the same field. It is the difference between before and after, not either one of them: a car stopping from 54 km/h has a change of 54 km/h, and one that bounces back at the same speed has twice that. It is a size, so deceleration and acceleration are not distinguished here

J

The impulse, shown in N·s, in kg·m/s and in lbf·s. The first two rows are always the same number — the newton-second and the kilogram-metre per second are the same unit, so this is the theorem made visible rather than a repeated result. The third row is that same quantity for imperial readers, and it is about 4.45 times smaller because a pound-force is larger than a newton

Use this page when you know a force and how long it acts, or a mass and how much its speed changed, and you want the accumulated effect rather than the force itself: what a kick delivers to a ball, what a rocket's thrust adds up to over a burn, what a bat transfers to a pitch, or — the case where it matters most — how large the force must be to stop something in a given time. That last question is what the table below answers, and the answer is uncomfortable: the same 900 N·s that a crash barrier absorbs over four tenths of a second would be forty times the force if the stopping happened against a steering wheel in a hundredth. Two practical notes. The page wants a complete pair of numbers from one route; if you fill in a force and no time, it reports that rather than ignoring the force you typed, because a number in an input box that leaves no trace in the result reads as a bug. And the difference between this page and the momentum page is a difference of question, not of unit: momentum is the state of an object at an instant, p = m × v, while impulse is what happened over a stretch of time to change that state. The two are linked by exactly the equality this page displays.

1. The defaults: 1000 N for 0.1 seconds

   1. Force 1000 N, acting for 0.1 s. The other pair is left empty
   2. Impulse: J = F × Δt = 1000 × 0.1 = 100 N·s
   3. The change in momentum is the same 100, because 1 N·s is exactly 1 kg·m/s
   4. In imperial units: 100 ÷ 4.4482 = 22.48 lbf·s

   The first two rows being identical is the point of the page and not a rounding artefact: 1 N·s and 1 kg·m/s are the same unit written two ways, and the theorem that says so is Newton's second law integrated over the contact time. Read the row as a sentence — this kick delivered 100 N·s, so the object's momentum changed by 100 kg·m/s — and the two labels stop looking redundant. The third row is the same quantity divided by 4.4482, which is the number of newtons in a pound-force; an imperial reader would call this 22.5 pound-second.

2. The other route: a 145 g baseball, 40 m/s

   1. Mass 0.145 kg — a baseball — and a change in velocity of 40 m/s
   2. Impulse: J = m × Δv = 0.145 × 40 = 5.8 N·s
   3. Momentum change: the same 5.8 kg·m/s
   4. In imperial units: 5.8 ÷ 4.4482 = 1.30 lbf·s

   This is the route to use when the force is the thing you cannot measure. Nobody can put a sensor inside a bat-ball collision, but the ball's mass is on the specification and its speed before and after can be filmed or radared, so the impulse comes out of quantities that are easy to get. Notice how unimpressive 5.8 N·s looks next to the 100 of the first example, and then consider what it implies: if the bat is in contact for a thousandth of a second, the average force is 5.8 ÷ 0.001 = 5800 N, which is 590 kilograms of force delivered by a hand.

3. Both routes at once: a 60 kg person stopping from 15 m/s

   1. First route: 9000 N acting for 0.1 s gives 900 N·s
   2. Second route: 60 kg changing velocity by 15 m/s gives 900 kg·m/s
   3. The two agree, so the page accepts them and reports 900 N·s
   4. In imperial units: 900 ÷ 4.4482 = 202.33 lbf·s

   15 m/s is 54 km/h and this is the airbag row of the table below: the same collision is described once by the force and once by what happened to the person, and the calculation confirms they are two views of one event. That check is worth doing on purpose when you have both sets of numbers, because a mismatch means one of the four inputs is wrong — 60 kg and 15 m/s have to imply the same impulse as the force and the time, and if it reports that they disagree, it is telling you the truth rather than picking the pair it likes better.

The force in J = F × Δt is the average force over the contact, and the two routes agree only on that average. A real impact has a peak far above it — a bat on a ball, a hammer on a nail and a car on a barrier all rise and fall within the contact time — so the 5800 N in the second example is a mean, and the peak may be several times that. This matters because parts break at the peak. Nothing here accounts for the direction of anything: the page computes sizes, so a bounce back at the same speed and a stop from the same speed are not distinguished even though the second one has twice the change in velocity and twice the impulse. Both outputs are shown to two decimal places, so an impulse below 0.005 N·s prints as 0.00 — that is the floor of this display and not a rounding error, and a smaller impulse needs a different instrument rather than more digits. Finally, the page assumes a single collision with a single contact time. A person restrained by a seatbelt and then by an airbag is two impulses in sequence, and collapsing that into one average over the whole event understates the force at the beginning.

What is the impulse formula?

J = F × Δt, a force multiplied by how long it acts, which comes out in newton-seconds. The equivalent form is J = m × Δv, and the two are always equal — that equality is the impulse momentum theorem, and it is Newton's second law integrated over the time the force is applied. 1000 N for 0.1 s is 100 N·s, and so is a 100 kg object changing speed by 1 m/s. The page shows both answers side by side in N·s and kg·m/s, which is why those two rows always match.

What is the difference between impulse and momentum?

Impulse is a process and momentum is a state. Momentum is m × v — how much motion an object has right now, at one instant — and it is the other calculator's subject. Impulse is the accumulation of a force over a stretch of time, and the theorem says the impulse delivered to an object equals the change in its momentum. So if you are asked what a kicked ball's momentum is, that is a state; if you are asked what the kick put into it, that is an impulse. Their units coincide exactly, which is why the confusion is so common, but the questions are different.

Which two fields do I fill in?

One complete pair, from either route. Force and time give the impulse by the definition; mass and change in velocity give it by the theorem. Fill in both pairs and the page checks them against each other, which is a genuinely useful thing to do. Fill in only one field of a pair and it reports an error rather than using it — a force sitting in an input box with no time next to it cannot produce an answer, and silently discarding a number the user typed looks like a fault rather than a rule.

How do I work out the impact force in a crash?

Divide the impulse by the time the stop takes: the average force is J ÷ Δt. The table on this page takes one crash — a 60 kg person stopping from 15 m/s, which is a change in momentum of 900 kg·m/s — and varies only the stopping time. Against a steering wheel in 0.01 s the average force is 90000 N; into an airbag over 0.1 s it is 9000 N; against a crash barrier over 0.4 s it is 2250 N. The momentum change is fixed by the crash and the force is set entirely by how long the stop is stretched over, which is all a crumple zone does.

Is it impulse or change in momentum — are they the same thing?

The impulse delivered equals the change in momentum, and that is a theorem rather than a definition, but the two labels describe different halves of the event. The impulse is what the force did; the change in momentum is what happened to the object. They are numerically identical because 1 N·s is exactly 1 kg·m/s — not approximately, exactly, since a newton is a kilogram-metre per second squared — which is why the page can display one number under both labels without any conversion between them.

Can the impulse be negative or zero?

Zero, yes, and it is a real answer: a force of nothing, or a force acting for no time, delivers no impulse, and the page prints 0 rather than complaining. Negative impulse is a real thing too — it means the force pointed opposite to the direction you called positive — but this page computes sizes, so it does not accept a negative input. If the object bounces back, add the speed before and the speed after rather than subtracting: a ball arriving at 20 m/s and leaving at 10 m/s the other way has a change in velocity of 30 m/s.

• Linear Momentum and Force (College Physics 2e, §8.1) — momentum as m·v, the impulse-momentum theorem, and why the two routes give the same quantity — OpenStax
• Impulse (physics) — the newton-second, its exact equality with the kilogram-metre per second, and the distinction between impulse and momentum — Wikipedia
• NIST Guide to the SI, Chapter 4: The Two Classes of SI Units and the SI Prefixes — derived units with special names, such as the newton, and how they reduce to the base units that N·s and kg·m/s are both built from — National Institute of Standards and Technology