Impulse-Momentum Theorem

Because the impulse — the product of force and time — is fixed by the situation. In a collision, the change in momentum Δp is determined by the initial and final velocities, both of which are known. Since J = F·t = Δp, if you increase t while keeping Δp the same, F must decrease proportionally. This is the principle behind every impact safety technology: an airbag extends the time over which the passenger's momentum changes from 50 km/h to zero from a few milliseconds (hitting the steering wheel) to about 100 milliseconds, reducing the average force by a factor of 20 or more. A crumple zone does the same for the car, a helmet for the head, and bending your knees does it for your joints when you jump and land.

Because the velocity change is larger when the direction reverses. If a ball hits a wall at speed v and sticks, its final velocity is zero, so Δv = −v and the magnitude is v. If the ball bounces back at the same speed, its final velocity is −v, so Δv = −v − v = −2v, and the magnitude is 2v — twice as large. The impulse is therefore twice as large, and the average force is twice as large for the same interaction time. This is why a bouncy ball hits harder than a lump of clay of the same mass and speed, and why materials that absorb energy without rebounding (like crumple zones) are better for safety than materials that bounce.

In the equation J = F·t, the quantity F is the average force over the interaction time t. In real impacts, the force is not constant — it rises, peaks, and falls — and the peak force can be several times larger than the average. The impulse is the area under the force-time curve, and the average force is that area divided by the total duration. The peak force is what determines whether a structure breaks or an occupant is injured, and it is the quantity that safety engineers work to reduce. Both the average force and the peak force matter: the average for the impulse calculation, and the peak for the strength and injury criteria.

For a bouncing ball with a coefficient of restitution e (where e = 1 is perfectly elastic and e = 0 is perfectly inelastic), the rebound speed is v_f = −e·v_i (assuming v_i is the approach speed in the negative direction). The velocity change is Δv = v_f − v_i = −e·v_i − v_i = −(1 + e)·v_i, so the impulse magnitude is J = m·(1 + e)·|v_i|. For a perfectly elastic bounce (e = 1), J = 2m|v_i|; for a perfectly inelastic impact (e = 0), J = m|v_i|. You can use this corrected velocity change as Δv in the calculator to handle energy loss during the bounce, and the resulting impulse will be less than the ideal elastic value.

The impulse-momentum theorem is a direct consequence of Newton's second law written in its most general form: F = dp/dt. If you multiply both sides by dt and integrate over the interaction time, you get ∫F dt = ∫dp = Δp, which is the impulse-momentum theorem. So the theorem is not a separate law of physics — it is the integral form of Newton's second law. This is why impulse-momentum analysis is so powerful: it allows you to analyze collisions and impacts without needing to solve the full differential equation of motion, because the integral of the force over time is exactly the change in momentum, regardless of the detailed shape of the force curve.

Impulse and work are parallel concepts: impulse is the integral of force over time, and work is the integral of force over displacement. Impulse changes momentum (a vector), while work changes kinetic energy (a scalar). The two are related through the motion: if a constant force F acts over a displacement d and produces a velocity change Δv, then the impulse is J = F·t and the work is W = F·d. The ratio J/W = t/d = 1/v_avg, where v_avg is the average velocity during the interaction. In problems where you know the force and the distance, work-energy is usually more convenient; where you know the force and the time (as in impacts), impulse-momentum is usually easier. Many problems can be solved either way, and the two approaches should give the same final velocities.