Physics · Ch 4 — Laws of Motion
Necessity of Defining Impulse
Necessity of Defining Impulse
Impulse earns its own name and definition because, in many everyday situations, an appreciably large force acts for an extremely short interval of time -- too short to measure the force and the time independently and reliably -- while the resulting CHANGE IN MOMENTUM is perfectly measurable. Real-life illustrations include hitting a ball with a bat, kicking a football, hammering a nail, and a ball bouncing off a hard surface: in every case the contact time is negligibly small and hard to record directly, but the resulting change of momentum of the object is a well-defined, measurable quantity -- so it becomes convenient to treat that change in momentum, the impulse, as the physical quantity of interest, rather than trying to separately pin down the force and the (very short) time.
On a force-versus-time (F-t) graph with the force axis starting at zero, the force during a typical collision (e.g. bat on ball) rises from zero to a peak and falls back to zero over the short contact duration; the SHADED AREA under this curve equals the impulse. A softer object gives a longer collision time and a correspondingly smaller peak force, keeping the total area (impulse) the same -- exactly why a wicket-keeper eases their hands back (increasing the effective collision time) while catching a fast-moving ball, reducing the peak force felt. This mirrors two other area-under-a-curve results already met: displacement as the area under a velocity-time graph, and work done as the area under a force-displacement graph (both, similarly, requiring the relevant axis to start at zero). …
Worked out. An oxygen molecule (mass 5.35x10^-26 kg) travelling at 400 m/s collides head-on and elastically with a nitrogen molecule (mass 4.65x10^-26 kg) travelling at 500 m/s in the opposite direction; using the elastic-collision final-velocity formulas the example finds v1 (oxygen) = -437 m/s and v2 (nitrogen) = 463 m/s (i.e., both molecules effectively rebound), computes the impulse received by each molecule from its own change in momentum (magnitude about 4.478x10^-23 N s, equal and opposite for the two, confirming zero net impulse for the isolated pair), and then divides by the given 1 ms collision duration to get the average force experi …