How this instrument works
Momentum is mass in motion: p = mv. Impulse is what changes it — the net effect of a force acting over some stretch of time, defined as J = m(v₂ − v₁). The formula looks like a rearranged version of momentum because that is exactly what it is: subtract the momentum you started with from the momentum you ended with, and whatever pushed or pulled the object is credited with that difference. A catcher's mitt, a car's crumple zone, and a rocket's exhaust are all, physically, machines for delivering or absorbing impulse.
The identity comes straight from Newton's second law. Written as force equals the rate of change of momentum, F = dp/dt, integrating both sides across an interval gives ∫F dt = Δp. The left side is impulse by definition — force accumulated over time — and the right side is exactly m(v₂ − v₁) when mass stays constant. That is why the unit works out to newton-seconds as readily as kilogram-metres per second: N·s and kg·m/s are the same physical unit wearing two different outfits.
The formula is silent about the shape of the push — a sharp one-millisecond jolt and a smooth quarter-second shove deliver the same impulse if they produce the same velocity change, because the calculator only sees the endpoints. That silence is also the limit: it tells you nothing about the peak force involved, only the total. A padded stop and a rigid one carrying the same mass through the same speed change need the same impulse but very different force, since force is impulse divided by however long the contact lasts.
- Enter the Mass of the object — kilograms or grams — the m in J = m(v₂ − v₁).
- Enter Initial velocity, the speed just before the interval you're studying; leave it at 0 m/s for an object starting from rest.
- Enter Final velocity, the speed just after. If the object reverses direction — a ball bouncing back — enter that velocity as negative.
- Switch either velocity field to ft/s if that matches your data; the two fields do not need to share a unit.
- Read Impulse (change in momentum) in kilogram-metres per second, numerically identical to newton-seconds.
Worked example — a 2 kg sled accelerated from rest to 10 m/s
Crash-test labs certify airbag timing with a guided sled mass launched from rest. Take a 2 kg sled brought from 0 m/s to 10 m/s: with m = 2, v1 = 0, and v2 = 10, the formula gives J = 2 × (10 − 0) = 20 kg·m/s. That figure is fixed the moment the mass and the two speeds are fixed — it does not care whether the sled reaches 10 m/s in 20 milliseconds or 2 seconds.
What the airbag changes is how that fixed 20 kg·m/s gets delivered, not its size. A rigid dashboard would deliver the same 20 kg·m/s in a few milliseconds, producing a force spike in the thousands of newtons; a properly timed bag spreads it over roughly 60 to 80 milliseconds instead, cutting the peak force by an order of magnitude. J stays at 20 kg·m/s throughout — only F = J ⁄ t falls as the contact time t grows.
Questions
Is impulse the same thing as force?
No. Force is an instantaneous quantity measured in newtons; impulse is force accumulated over an interval, measured in newton-seconds — the same unit as kilogram-metres per second. A small force sustained for a long interval can deliver the same impulse as a large force applied briefly, which is why an airbag or a padded landing, by stretching the contact time, lowers the peak force for an unchanged impulse.
Why is impulse m(v₂ − v₁) instead of just mv?
Because impulse measures a change, not a single state. Momentum p = mv describes the object at one instant; impulse J = Δp = m(v₂ − v₁) describes what happened to that momentum across an interval. Integrating Newton's second law, F = dp/dt, across the interval gives ∫F dt = Δp, and since mass is constant here, that reduces to mass times the velocity difference.
Can the result come out negative?
Yes, and the sign is meaningful, not an error. It shows the net impulse acted opposite to whichever direction you called positive. Enter v1 = 10 m/s and v2 = −10 m/s — a ball bouncing straight back — with a 2 kg mass, and J comes out to −40 kg·m/s: twice the size of accelerating from rest, because the velocity swings across a full 20 m/s span instead of 10.
Does a bigger collision force always mean a bigger impulse?
No, and this is the most common mix-up. Impulse depends only on the endpoints, m(v₂ − v₁), not on how hard the push felt. A slow, cushioned stop and a sudden, rigid one carrying the same mass through the same velocity change deliver exactly the same impulse; only the peak force and the duration differ, because force and time trade off while their product — the impulse — stays fixed.
Does this formula work for a rocket burning fuel?
Not on its own. J = m(v₂ − v₁) assumes the mass m stays constant across the interval. A rocket sheds propellant continuously, so part of the momentum change belongs to the exhaust, not the rocket body — that case needs the rocket equation, which handles a varying mass. Use this calculator when the object's mass genuinely does not change during the interval: a crash-test sled, a thrown ball, a struck cue ball all qualify.
What units does the impulse come out in?
Kilogram-metres per second (kg·m/s), built from mass in kilograms and velocity in metres per second. That is dimensionally identical to the newton-second (N·s) used in most mechanics textbooks, since one newton is one kg·m/s². Enter mass in grams or velocity in feet per second if that is how your data arrived; the instrument converts internally before applying the formula.