The impulse-momentum theorem links a force and the time it acts to the velocity change it causes: FΔt = m(v − u). Know any four of its five quantities and you get the fifth, which suits impacts: speeds before and after are easy to measure with radar or video, the force is not. This calculator solves for average force, contact time, mass, initial velocity or final velocity, and keeps strict track of direction.
How to use the impulse-momentum calculator
- Under Solve for, choose Average force, Contact time, Mass, Initial velocity or Final velocity.
- Decide which direction counts as positive before typing anything, then enter Initial velocity (u) and Final velocity (v) with signs relative to that choice.
- Fill in the other known values: average force in N, kN, lbf, kgf, dyn or ozf; contact time in ms, s, µs or min; mass in kg, g, mg, t, lb, oz, US tons or stone.
- Choose the answer’s unit under Show the result in. Next to the answer you get the impulse, the momentum change, the initial and final momentum, the average acceleration in m/s² and in g, and, when force is one of your inputs, its magnitude in pounds-force.
Impulse-momentum formula
The calculator uses one rearrangement per unknown:
Δt = m(v − u) ÷ F
m = FΔt ÷ (v − u)
v = u + FΔt ÷ m
u = v − FΔt ÷ m
F is the average net force over Δt; u and v are the velocities before and after.
Sign convention for rebounds and reversals
Velocity and force are signed, and every value must use the same positive direction:
- Anything moving in the positive direction has a positive velocity; anything moving the other way, a negative one.
- A force pushing toward the positive direction is positive. A braking force on an object moving positively is negative.
- In a rebound, u and v have opposite signs, so v − u adds the two speeds. That is why bounces and hits generate much larger forces than simple stops.
Contact time and mass must be positive, so whenever you solve for one of them, the force and the velocity change (v − u) must share a sign. The calculator rejects inputs that break this rule, and it refuses equal initial and final velocities when solving for time or mass.
Worked example: bat meets baseball
Take “toward the outfield” as positive.
A 0.145 kg baseball arrives at 40 m/s (about 89 mph) heading toward the batter, so u = −40 m/s. It leaves the bat at 50 m/s (about 112 mph) toward the outfield, so v = +50 m/s. Bat and ball touch for 0.7 ms.
Velocity change: 50 − (−40) = 90 m/s.
Momentum change: 0.145 × 90 = 13.05 kg·m/s, from −5.8 to +7.25 kg·m/s.
Average force: 13.05 ÷ 0.0007 = 18,643 N, about 18.6 kN or 4,191 lbf.
Average acceleration: 90 ÷ 0.0007 ≈ 128,571 m/s², roughly 13,111 g.
Ignoring the signs and entering u = +40 m/s would make v − u only 10 m/s and the force about 2,071 N, roughly nine times too small. The ball does not just stop; it is stopped and then thrown back, and the bat pays for both.
Second example: braking time and speed
A 1,400 kg car travels at 90 km/h (25 m/s) in the positive direction. Its brakes and tires supply an average net retarding force of 7,000 N, entered as −7,000 N because it opposes the motion.
- Solve for contact time with v = 0: Δt = 1,400 × (0 − 25) ÷ (−7,000) = 5 s. The average deceleration is 5 m/s², about 0.51 g.
- Solve for final velocity with Δt = 2 s: v = 25 + (−7,000 × 2) ÷ 1,400 = 15 m/s, or 54 km/h.
Enter +7,000 N when solving for the stopping time and the calculator rejects it: a force pushing the car forward could never bring it to rest.
When the theorem applies
- Net force only. If gravity, friction or drag act alongside the force you care about, the result is their combined push, which matters for slow events more than for a bat and ball.
- Constant mass. Objects that shed mass while moving, such as rockets burning fuel, need the rocket equation instead.
- One axis at a time. For an angled hit, split the velocities into components and solve each axis separately.
- Averages, not peaks. The force is a time average; the instantaneous peak is higher.
Related calculators
If you already know the impulse instead of the velocities, the impulse calculator is quicker. The momentum calculator handles p = mv for a single state, and the force calculator solves F = ma when the acceleration is known directly.
Frequently asked questions
What is the impulse-momentum theorem?
It states that the net impulse on an object equals its change in momentum: FΔt = m(v − u). It follows directly from Newton's second law, F = ma, once the acceleration is written as (v − u) ÷ Δt.
Why does a rebound need a negative velocity?
When an object reverses direction, its initial and final velocities have opposite signs, so v − u becomes the sum of the two speeds. Typing both as positive numbers gives only their difference. In the baseball example that mistake makes the force about nine times too small.
Which force does the calculator find?
The average net force on the object during the contact time. For a bat and ball, gravity and air drag are tiny next to the contact force, so the result is effectively the bat's push on the ball. By Newton's third law, the ball pushes back on the bat just as hard.
Why are my inputs rejected when I solve for time or mass?
Time and mass must come out positive, so the force has to point the same way as the velocity change (v − u). A braking force on a car moving in the positive direction must therefore be entered as negative. The calculator also stops when v equals u, because with no momentum change there is no time or mass to find.
What does the 'g felt by the object' figure mean?
It is the average acceleration, |v − u| ÷ Δt, divided by standard gravity (9.80665 m/s²). It shows how violent the event is for the object itself: more than 13,000 g for a batted baseball, about half a g for a car braking firmly.