Example 1
A constant force of 100 N acts for 0.5 s. This is the reference example on this page: 100 · 0.5 = 50 N·s.
|J| = 50 N·s
Impulse Calculator
Enter a force and a time, or a mass and two speeds. The calculator finds the impulse J = F·Δt or J = m(vₖ − v₀) and shows the absolute value, the signed impulse, and the same number in kg·m/s. For 100 N and 0.5 s the result is 50 N·s. The model treats force as an average over Δt and does not include stopping distance or the integral of a varying force ∫F dt.
Result
Enter a force and a time, or a mass and two speeds. The impulse, the sign of the change, and the average-force chart appear here.
Momentum p = m·v alone is on the momentum calculator. Force from a known Δp and Δt is on the F = Δp/Δt calculator. Distance at constant deceleration is on the decelerated-motion calculator.
How it works
F100 N
Δt0.5 s
J50 N·s
The impulse of a force equals the change of momentum. For a constant force the textbook line is J = F·Δt = Δp. That is the impulse–momentum theorem: the time integral of force equals the change of momentum. In the second law of the Principia (1687) Newton ties the change of the quantity of motion to the impressed force. For a constant force that integral reduces to the product F·Δt. For 100 N and 0.5 s the steps are short. Step 1: 100 · 0.5 = 50. The result is 50 N·s, which is also 50 kg·m/s, because 1 N = 1 kg·m/s².
The calculator uses the numbers you enter. In F·Δt mode those are force in newtons and time in seconds. In momentum-change mode those are mass in kilograms and the initial and final speeds in meters per second. Enter 50 km/h as 13.9, not as 50. 72 km/h is 20 m/s. A comma and a period are the same separator. Zero and a negative time are rejected. Zero and a negative mass are rejected. A minus sign on the force or on a speed means the opposite direction along the chosen axis.
What the result panel shows. The panel reports the absolute value |J|, average |F| in F·Δt mode, the signed impulse, the time or the speed change, and the same |J| in kg·m/s. For 100 N and 0.5 s the average |F| is 100 N, because dividing |J| by Δt returns the absolute value of the force you entered. For 1,000 kg and speeds of 20 m/s and 0, the impulse is J = 1,000 · (0 − 20) = −20,000 N·s. The main panel shows 20,000 N·s and a note about the opposite direction. Momentum-change mode has no average force, because that form has no time Δt.
Limits of the model. The formula assumes a constant force or a known speed change of one body. The calculator does not integrate a varying F(t), does not compute a stopping distance, and does not report momentum at one instant, p = m·v. Momentum of 70 kg at 1.4 m/s, which is 98 kg·m/s, is on the momentum page. Force from a known Δp and Δt is on the F = Δp/Δt page. Distance at a known deceleration is on the decelerated-motion page. The chart draws bars of average |F| = |J|/t at 0.1 s, 0.25 s, 0.5 s, 1 s, 1.5 s, and 2 s. That comparison shows how large the average force would be if the same |J| were spread over a shorter or a longer time.
The unit switch in the header converts force to pounds-force, mass to pounds, speed to ft/s, and impulse to lbf·s, and it converts numbers already entered in the fields of the active mode. Time stays in seconds. The switch does not by itself convert kilometers per hour to meters per second. The values in this text stay in SI as the reference. In SI the number of newton-seconds and the number of kg·m/s are equal. After a switch to US units, the lbf·s figure and the lb·ft/s figure differ, because a pound-force and a pound-mass are not the same unit.
The examples on this page use the same formulas. 100 N for 0.5 s gives 50 N·s. 400 N for 0.02 s gives 8 N·s. That is an order of magnitude for a short contact, not a measurement of one kick. 800 kg from 0 to 2 m/s gives 1,600 N·s. 50 N for 0.15 s gives 7.5 N·s. 20,000 kg from 10 m/s to 2 m/s gives −160,000 N·s. 200 N for 1.2 s gives 240 N·s. 80 kg from 8 m/s to 3 m/s gives −400 N·s. 1,000 N for 0.05 s gives 50 N·s, the same value as 100 N for 0.5 s. The “Fill this example” button enters those numbers in the fields and refreshes the panel and the chart. The fields start empty.
Formula
J = F · Δt (constant force)
J = Δp = m (vk − v0)
[J] = N·s = kg·m/s
Model: the impulse–momentum theorem for a constant force or for a known change of speed. The full statement is J = ∫ F dt = Δp. This calculator does not evaluate the integral of a varying F(t). In F·Δt mode the step is the product F·Δt. In momentum-change mode the step is the product m·(vₖ − v₀). Then the absolute value, and the same number in kg·m/s.
Theory, the unit, and the domain
The impulse–momentum theorem
The impulse of a force is the time integral of that force: J = ∫ F dt. When the force is constant over Δt, the integral equals the product F·Δt. Newton’s second law in the form F = dp/dt, integrated over that interval, gives J = Δp. In the second law of Philosophiæ Naturalis Principia Mathematica (1687) Newton writes that the alteration of motion is proportional to the motive force impressed. He does not print the equation J = F·Δt there. The school product is a later shorthand of that relation for a constant force.
Two modes, one result
F·Δt mode multiplies the force you enter by the time you enter. For 100 N and 0.5 s the product is 50 N·s. Momentum-change mode multiplies mass by the difference of the speeds. For 1,000 kg, 20 m/s, and 0 the product is −20,000 N·s. Both modes return an impulse. The calculator does not convert one mode into the other when the other mode’s data are missing. A field the active mode does not use does not enter the formula.
The newton-second and the kilogram meter per second
1 N = 1 kg·m/s², so 1 N·s = 1 kg·m/s. The definitions of the units are in the SI Brochure (BIPM, 9th edition). NIST SP 330 is the U.S. guide to the same system. The panel reports the absolute value in N·s and the same number in kg·m/s. After the header switches to US units, impulse is in lbf·s and the momentum equivalent is in lb·ft/s. Those two U.S. figures are not the same number, because a pound-force and a pound-mass differ by the standard acceleration of gravity. In SI the numbers match.
The sign, and one body
A negative force, or a final speed smaller than the initial speed, gives a negative J. The main panel shows the magnitude and a note about the opposite direction. A separate field keeps the sign. The form describes one body on one line. It does not assemble a two-body collision, and it does not compute a coefficient of restitution. You enter the speed difference yourself, or the force and the time.
Average force and the chart
In F·Δt mode the average-|F| field is the absolute value of the force you entered, because |F·Δt| / Δt = |F|. It is not a separate measurement. Momentum-change mode has no time, so an average force does not appear. Average force F = Δp/Δt, when the momentum change and the time are known, is a separate calculator. The chart takes the |J| already computed and divides it by 0.1 s, 0.25 s, 0.5 s, 1 s, 1.5 s, and 2 s. A shorter time for the same impulse means a larger average force. For |J| = 50 N·s and 0.1 s the average is 500 N. For 2 s it is 25 N.
Outside the formula
The calculator does not compute distance, a stopping time at a known deceleration, kinetic energy, or momentum p = m·v at one instant. At a constant deceleration of 7 m/s², a body slowing from 14 m/s stops in 2 s over 14 m. This page does not return those two numbers. It does not integrate a force that varies with time, and it does not describe deformation in a collision. The energy of the motion is on the kinetic-energy page. Force at constant mass and a known acceleration is on the F = m·a page.
Sources
The definition, the unit, and the link to the second law come from mechanics textbooks and from the historical source: H. D. Young and R. A. Freedman, University Physics, the chapter on momentum and impulse; D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, the chapter on momentum and impulse; R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Vol. I, Chapter 9; I. Newton, Philosophiæ Naturalis Principia Mathematica, 1687, the second law; BIPM, SI Brochure, 9th edition. This page is educational.
How to use
- Choose a mode. Use F·Δt when you know the force and how long it acts. Use the momentum change when you know the mass and the initial and final speeds.
- In F·Δt mode, enter the force and the time, for example 100 and 0.5. Zero and a negative time are rejected.
- In momentum-change mode, enter the mass, v₀, and vₖ, for example 1000, 20, and 0. Zero and a negative mass are rejected.
- Click Calculate, or wait for the live result. For 100 N and 0.5 s the calculator shows 50 N·s.
- Read the panel: the absolute value, the signed impulse, the time or the speed change, and the same number in kg·m/s. Clearing the form removes the numbers you entered and hides the panel.
Symbol glossary
J is impulse, F is force, Δt is time, m is mass, and v₀ and vₖ are the speeds before and after the change. For 100 N and 0.5 s the result is 50 N·s.
- J
- Impulse [N·s], equal to F·Δt or to m(vₖ − v₀). For 100 N and 0.5 s it is 50 N·s.
- F
- Force [N] in F·Δt mode. A minus sign means the opposite direction. Average |F| in this mode is the absolute value of that force.
- Δt
- Duration of the force [s]. It must be greater than zero. Momentum-change mode has no time field.
- Δp
- Change of momentum, equal to J. For 1,000 kg from 20 m/s to 0 it is −20,000 kg·m/s.
- v₀
- Initial speed [m/s] in momentum-change mode. Enter 72 km/h as 20, not as 72.
- vₖ
- Final speed [m/s]. Together with v₀ and the mass it gives J without a force.
- |J|
- Absolute value of the impulse. The main panel shows it as a non-negative number, with a note when the sign is negative.
Worked examples
Example 2
A body of mass 1,000 kg slows from 20 m/s to 0. The momentum change is 1,000 · (0 − 20) = −20,000 N·s. The main panel shows the absolute value and a note about the opposite direction.
J = −20,000 N·s
Example 3
A short contact on the order of 400 N for 0.02 s, for example a kick of a ball. The product is 8 N·s. That is an order of magnitude, not a measurement of one kick.
|J| = 8 N·s
Example 4
A mass of 800 kg goes from 0 to 2 m/s. The momentum change is 800 · 2 = 1,600 N·s.
J = 1,600 N·s
Example 5
A force of 50 N acts for 0.15 s. The impulse is 50 · 0.15 = 7.5 N·s.
|J| = 7.5 N·s
Example 6
A mass of 20,000 kg slows from 10 m/s to 2 m/s. The momentum change is 20,000 · (2 − 10) = −160,000 N·s.
J = −160,000 N·s
Example 7
A force of 200 N acts for 1.2 s. The impulse is 200 · 1.2 = 240 N·s.
|J| = 240 N·s
Example 8
A mass of 80 kg slows from 8 m/s to 3 m/s. The momentum change is 80 · (3 − 8) = −400 N·s.
J = −400 N·s
Example 9
A force of 1,000 N acts for 0.05 s. The impulse is 50 N·s, the same value as 100 N for 0.5 s, with a different force and a different time.
|J| = 50 N·s
Ways to use this calculator
- You estimate the impulse of a constant force when you know newtons and seconds, for example 100 N for 0.5 s.
- You check the momentum change of one body when you know the mass and two speeds, for example 1,000 kg from 20 m/s to rest.
Common questions
What does this calculator compute, and from which formula?
The calculator returns the impulse of a force in one of two modes. For a constant force it computes J = F·Δt. When the mass and two speeds are known, it computes J = Δp = m(vₖ − v₀). For 100 N and 0.5 s the result is 50 N·s. For 1,000 kg, 20 m/s, and 0 the result is −20,000 N·s, and the main panel shows the absolute value 20,000 N·s with a note about the opposite direction.
Which mode should I choose?
Use F·Δt mode when the force and the time it acts are known. Use momentum-change mode when the mass and the initial and final speeds are known. The calculator does not choose the mode for you. A field the active mode does not use does not enter the formula.
Which units should I enter?
Enter force in newtons, time in seconds, mass in kilograms, and speed in meters per second. 50 km/h is 13.9 m/s, not the number 50 in the speed field. 72 km/h is 20 m/s. After you switch the header to US units, force is in pounds-force, mass is in pounds, speed is in ft/s, and impulse is in lbf·s. Time stays in seconds. A comma and a period are the same decimal separator.
Is N·s the same unit as kg·m/s?
Yes, in SI. Because 1 N = 1 kg·m/s², 1 N·s equals 1 kg·m/s. That is why impulse and a change of momentum share a measure. The panel reports |J| in N·s and the same value in kg·m/s. After a switch to US units, the lbf·s number and the lb·ft/s number differ, because a pound-force and a pound-mass are not the same unit.
How is impulse different from momentum?
Momentum at one instant is p = m·v. Impulse is the change of that momentum, or the product of force and time. 70 kg at 1.4 m/s has a momentum of 98 kg·m/s. The momentum calculator computes that product. This page subtracts the final speed from the initial speed, or multiplies F by Δt.
What does a minus sign mean, and what does the main panel show?
The sign of J follows the sign of the force, or the sign of vₖ − v₀. A negative result means momentum falls along the chosen line. The main panel shows the absolute value and a note about the opposite direction. A separate field reports the signed impulse. For 1,000 kg from 20 m/s to 0 the impulse is −20,000 N·s.
What happens at zero time, zero mass, or empty fields?
Zero and a negative time are rejected in F·Δt mode. Zero and a negative mass are rejected in momentum-change mode. Empty fields do not open the result panel. A force of zero with a positive time gives J = 0. The fields start empty.
What do the panel and the chart show?
Besides |J|, the panel reports average |F| in F·Δt mode, the signed impulse, the time or the speed change, and |J| in kg·m/s. In F·Δt mode the average |F| equals the absolute value of the force you entered. Momentum-change mode has no average force, because Δt is missing. The chart draws bars of average |F| = |J|/t at 0.1 s, 0.25 s, 0.5 s, 1 s, 1.5 s, and 2 s. It is not a plot of a varying force F(t). The “Fill this example” button enters the example numbers and refreshes the panel and the chart.
Does the result include stopping distance or a varying force?
No. The calculator does not integrate F(t) and does not compute a distance. At a constant deceleration of 7 m/s² and an initial speed of 14 m/s, the stopping distance is 14 m and the time is 2 s. Those quantities are on the decelerated-motion page. Force from a known momentum change and a known time is on the separate F = Δp/Δt page.
Does the calculator convert kilometers per hour?
No. Enter speed in meters per second. 72 km/h is 20 m/s, and about 50 km/h is 13.9. The header switch converts m/s to ft/s and converts the number already entered. It does not by itself convert kilometers per hour to meters per second.
How do I calculate impulse with J = F·Δt?
Multiply the average force by the time it acts. For 100 N and 0.5 s: 100·0.5 = 50 N·s.
What is the impulse for 100 N acting for 0.5 s?
50 N·s, which is also 50 kg·m/s. That is the momentum change such an average force delivers in that time.
How do I find impulse from mass and a speed change?
In mΔv mode: J = m(vₖ − v₀). Velocity signs matter on a rebound; the panel shows signed impulse and |J|.
Does a longer contact time lower average force for the same Δp?
Yes when impulse is fixed: |F| = |J|/Δt. Doubling Δt halves average |F|. The calculator does not find stopping distance s = v²/(2a).
Does this tool solve a two-body collision?
No. It finds the impulse on one body from F·Δt or mΔv. Conservation of momentum for two masses and collision energy are not on this page.
Sources
The definition of impulse, the unit, and the link to a change of momentum come from physics and metrology sources. The page does not invent publication identifiers.
- Wikipedia — Impulse (physics)
- Feynman Lectures on Physics, Vol. I, Ch. 9 — Newton’s Laws of Dynamics
- BIPM — The International System of Units (SI), 9th edition
- NIST SP 330 — The International System of Units
- Newton, I., Philosophiæ Naturalis Principia Mathematica, 1687, the second law.
- Young, H. D., and Freedman, R. A., University Physics, chapter on momentum and impulse.
Page updated November 3, 2026.