Impulse Momentum Calculator
Calculate impulse and change in momentum with J = F dt = m(v2 - v1), from mass, initial and final velocity, and contact time.
Impulse Momentum Calculator
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Formula: J = FΔt = Δp = m(v₂-v₁)
Worked example — 5800 N average force
Formula
J = FΔt = Δp = m(v₂-v₁)
Impulse equals the change in momentum. Longer contact time reduces peak force.
Worked Examples
Example 1: Baseball hit
Problem:0.145 kg ball, 0→40 m/s in 0.001s
Solution:J=0.145×40=5.8 N·s, F=5800 N
Result:5800 N average force
Frequently Asked Questions
What is impulse in physics?
Impulse (J) is the product of the net force and the time interval over which it acts: J = FΔt. It equals the change in momentum (Δp = mΔv), measured in Newton-seconds (N·s) or kg·m/s.
How do you calculate impulse and momentum change?
Use J = Δp = m(v₂ - v₁). Multiply the object's mass by the change in velocity. For a 0.5 kg ball going from 0 to 20 m/s: J = 0.5 × 20 = 10 N·s.
What units are used for impulse and momentum?
Both impulse and momentum share the same SI units: kg·m/s or equivalently N·s. Force is in Newtons (N), time in seconds (s), mass in kilograms (kg), and velocity in m/s.
Why do airbags and padded surfaces reduce injury?
Airbags extend the collision time (Δt). Since J = FΔt = constant, a longer Δt means a smaller average force F on the body. A 10 N·s impulse in 0.01 s gives 1000 N; in 0.1 s it gives only 100 N.
What is the difference between impulse and average force?
Impulse is the whole area under the force-versus-time curve, measured in N s, while average force is that area divided by the contact time. The calculator reports the average because collision forces are rarely constant. A 0.145 kg baseball taken from rest to 40 m/s carries an impulse of 5.8 N s; delivered in 0.001 s that averages 5800 N, but the instantaneous peak force partway through the contact is considerably higher.
How does the impulse-momentum theorem follow from Newton's second law?
Newton's second law in its general form is F = dp/dt. Integrating both sides over the interval of contact gives the integral of F dt on the left, which is the impulse J, and the change in momentum delta-p on the right. For constant mass this reduces to the familiar F = m x a, but the impulse form is more general because it handles forces that vary during the interaction without needing their detailed shape.
How does impulse explain conservation of momentum in a collision?
By Newton's third law the two colliding bodies push on each other with equal and opposite forces for exactly the same duration, so they receive equal and opposite impulses. That means delta-p for one body is the negative of delta-p for the other, and the total momentum of the pair is unchanged when no outside force acts. This holds for elastic and inelastic collisions alike, even though kinetic energy is only conserved in the elastic case.
Why does the sign of the velocities matter so much?
Momentum is a vector, so both velocities must be measured along one chosen positive direction. A 0.2 kg ball arriving at 5 m/s and rebounding at 5 m/s has a final velocity of -5 m/s, giving delta-p = 0.2 x (-5 - 5) = -2 kg m/s, an impulse of magnitude 2 N s. Had it simply stopped dead, the impulse would be only 1 N s - a bounce demands twice as much from the surface.
How is impulse used to design a safer landing or catch?
The momentum change is fixed by the mass and speed, so the only lever is the stopping time: F_avg = delta-p/delta-t. A 70 kg person landing at 5 m/s must absorb 350 N s. Stopping in 0.05 s with locked knees means 7000 N; bending to stretch it to 0.5 s means only 700 N. The same reasoning drives crumple zones, crash mats, and moving your hands back when catching a ball.
When is an impulse approach better than an energy approach?
Use impulse when you know how long a force acted and want the resulting velocity change, especially for collisions where the force profile is unknown but the contact duration is measurable. Use work and energy when the force is known over a distance instead. In an inelastic collision much of the kinetic energy is lost to deformation and heat, so energy bookkeeping fails while momentum bookkeeping stays exact.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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