Hammer strike
F = 100 N, Δt = 0.5 s → J = 50 N·s.
Impulse
A force acting over time changes momentum. Enter F and Δt, or m, v₀ and vₖ, and you get impulse J (= Δp).
The same idea lives in the momentum tool. Stopping distance at known deceleration: decelerated motion.
Result
Enter values — the result shows up here.
How it works
Impulse is J = F·Δt. In SI the unit is the newton-second (N·s), equal to the momentum unit kg·m/s. Newton's second law says momentum changes when a constant force acts over Δt: the change is exactly F·Δt.
Hence the second mode: J = Δp = m(vₖ − v₀). A 1000 kg car braking from 20 m/s to rest sheds 20000 kg·m/s of momentum, so the brakes (and tyres) must deliver that much impulse.
This is not the same as plain F = m·a: here you care about force×time (or Δp), not instantaneous a alone.
The change-in-momentum page describes the same idea; this is the main calculator. Keep time in seconds.
The chart shows what average |F| would produce the same |J| for different Δt: shorter time → larger force.
How to use
- Choose mode: F·Δt or momentum change (m, v₀, vₖ).
- Enter values in the units shown on the labels.
- Read impulse J: the primary result (= |Δp|).
- In F·Δt mode you also see average |F|.
- Compare the examples: ball, car, lift, hammer.
Formula
J = F·Δt
J = Δp = m(vk − v0)
Unit: N·s = kg·m/s.
Real-life examples
Car braking (Δp)
m = 1000 kg, v₀ = 20 m/s → 0.
Tennis serve
F ≈ 400 N for 0.02 s on the ball.
Lift start
m = 800 kg, v₀ = 0 → 2 m/s.
Throwing a ball
F = 50 N, Δt = 0.15 s.
Tram brakes
m = 20000 kg, 10 m/s → 2 m/s.
Pushing a cabinet
F = 200 N for 1.2 s.
Cyclist slowing
m = 80 kg, 8 m/s → 3 m/s.
Short impulse
F = 1000 N, Δt = 0.05 s.
Ways to use this calculator
- School problems on impulse and momentum.
- Estimate average force from Δp and contact time.
- Braking: how much impulse must be scrubbed off.
- Rough collision models as F·Δt.
- Lift / elevator starts.
- Sport: racket–ball contact.
- Soft vs hard landings (different Δt).
- Check that F·Δt matches m·Δv.
- Warm-up for collision dynamics.
Frequently asked questions
How is impulse different from force?
Force is “how hard” at an instant. Impulse accumulates force over time: the same F for a longer Δt changes momentum more.
Why is the unit N·s?
Because J = F·Δt. 1 N·s = 1 kg·m/s, the momentum unit, so J = Δp.
When should I use the Δp mode?
When you know mass and speeds but not force or contact time. Typical braking “from v₀ to vₖ”.
Does the sign of impulse matter?
Yes: it follows Δv. We show |J| and note the opposite direction when Δp < 0.
How does this relate to F = m·a?
For constant F: a = F/m and Δv = a·Δt, so m·Δv = F·Δt. Both modes agree.
Is this the same as “change in momentum”?
Same idea. This is the main calculator; the other page points here.
Is driver reaction time included?
No. Braking impulse is the force while tyres scrub speed. Reaction time adds distance before braking starts.
What about a time-varying force?
Strictly ∫F dt. Here we use average F or pure Δp, which is enough for school estimates.
Which speed units?
Whatever the field label shows (m/s or ft/s in US mode). Don’t mix km/h with seconds without converting.
Can mass be zero?
No: Δp would be meaningless. F·Δt mode does not need mass.
What next after J?
See momentum p = mv and braking distance in decelerated motion if you know a.