Calculate change in momentum from mass and velocities, force × time, or p2 − p1, with impulse and direction shown in kg·m/s and N·s.
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Change in Momentum Formula
The first three solve modes find momentum change. Four further modes rearrange p₁ = m(v₂ − Δv) to find initial momentum, final velocity, velocity change or mass. Use signed components on one axis; the mass-and-velocity relations assume constant positive mass and classical, nonrelativistic motion.
Δp = m(v₂ − v₁)
Δp = F·Δt
Δp = p₂ − p₁
- Δp — change in momentum, in kg·m/s (equivalent to N·s)
- m — mass of the object, in kg
- v₁, v₂ — initial and final velocities, in m/s
- F — average net force during the interval, in N
- Δt — duration the force is applied, in s
- p₁, p₂ — initial and final momentum, in kg·m/s
The three momentum-change methods map to these formulas:
- Mass + velocities uses Δp = m(v₂ − v₁). Enter mass and both signed velocities; the sign of each velocity sets direction.
- Force × time uses Δp = F·Δt (the impulse-momentum theorem). Add an optional mass and you also get Δv = Δp/m.
- p₂ − p₁ uses Δp = p₂ − p₁ when you already know the momentum values directly.
Reference Values and Unit Conversions
Use these for unit checks. Scenario magnitudes below use the stated illustrative inputs, not empirical typical values. Signed change may be negative; magnitude is nonnegative.
| Unit | Equivalent in kg·m/s |
|---|---|
| 1 N·s | 1 kg·m/s |
| 1 g·m/s | 0.001 kg·m/s |
| 1 lb·ft/s | 0.138254954376 kg·m/s |
| 1 slug·ft/s | 4.4482 kg·m/s |
| 1 lbf·s | 4.4482 kg·m/s |
| Scenario | Illustrative |Δp| (kg·m/s) |
|---|---|
| Tennis ball returned at the net (0.058 kg, 25 m/s reversed) | ~2.9 |
| Baseball hit back to pitcher (0.145 kg, 40 m/s to −45 m/s) | ~12.3 |
| Soccer ball kicked from rest (0.43 kg, 0 to 25 m/s) | ~10.8 |
| Car braking to stop (1500 kg, 20 m/s to 0) | 30,000 |
| Rifle bullet leaving barrel (0.01 kg, 0 to 800 m/s) | 8 |
The four initial-momentum relationships are p₁ = m(v₂ − Δv), v₂ = p₁/m + Δv, Δv = v₂ − p₁/m, and m = p₁/(v₂ − Δv). Mass must be positive. Zero initial momentum with zero initial velocity does not identify a unique mass; nonzero initial momentum with zero initial velocity has no finite mass solution. Blank optional mass omits velocity change in force-and-time mode.
Worked Examples and FAQ
Example 1. A 0.16 kg hockey puck moving at 2.5 m/s is stopped by a goalie's glove. Using mass and velocities: Δp = 0.16 × (0 − 2.5) = −0.40 kg·m/s. The negative sign shows the impulse acted opposite the puck's motion.
Example 2. A bumper applies an average force of 800 N for 0.12 s on a cart. Using force and time: Δp = 800 × 0.12 = 96 N·s = 96 kg·m/s. If the cart's mass is 40 kg, then Δv = 96 / 40 = 2.4 m/s.
Is impulse the same as change in momentum? Yes. The impulse-momentum theorem states J = F_avg Δt = Δp along the chosen axis, so 1 N·s equals 1 kg·m/s. Use average net force over the interval.
Why does the answer come out negative? Velocity and force have direction. If the final velocity is smaller than the initial (or points the other way), Δp is negative. The sign tells you which direction the net impulse acted.
Do I need SI units? No. Pick any unit from each dropdown. The calculator converts to SI internally and reports Δp in kg·m/s, N·s, lb·ft/s, and lbf·s.
What if the force is not constant? Use the average force over the interval. Δp = F_avg × Δt still gives the correct total change in momentum, even when the actual force varies.
