*One of the most unaddressed questions in high school physics is regarding impulse, so much so that most people don't even think to ask it. Here's an answer.*

## Definition

*Before getting into the "Why?", we must have the "What?".*

Every physics student learns (Newton's second law!Newton's second.)(newtons-second) pretty early.

(start newtons-second)

For every object, there is a value called [*inertial mass* \(m\)][(as opposed to *gravitational mass,* which dictates how strongly it pulls other objects with gravity, and appears to coincidentally be the same thing.)] which dictates how much it resists [changes in motion.][(That is, changes from moving at a constant velocity, in a straight line, as dictated by Newton's first law.)] The exact effect is given by the equation:

\[ F = m a \]

The formula states that for [a particular force \(F,\)][(the "cause" of a change in motion)] you divide by the mass to get the resulting [acceleration \(a,\)][(the time derivative of velocity, which dictates the change in velocity over time)] so objects with more mass have smaller changes in motion.

(stop newtons-second)

The concept of impulse is given by (taking Newton's second and integrating!Integral definition.)(impulse-calculus) over time. (Or, if you don't have calculus yet...!Algebra definition.)(impulse-algebra)

(start impulse-calculus)

Remember that acceleration is the time derivative of velocity, so Newton's second law is

\[ F = m\frac{dv}{dt} \]

where the mass is constant, but the force and velocity are functions of time. Taking the integral over some time interval \(t_i\le t\le t_f,\) we get that

\[ \int_{t_i}^{t_f} F dt = \int_{t_i}^{t_f}m\frac{dv}{dt}dt = m(v_f - v_i) = m\Delta v \]

That is, if you find the total force acting on an object over a time interval, you get the mass times the change in velocity. The left-hand side is the quantity we call impulse, usually given in units of [\(\text{N}\cdot\text{s}.\)][(Newton-seconds, that is, force units times time units)]

(stop impulse-calculus)

(start impulse-algebra)

Without calculus, we assume that the force is constant over a time interval. For a particular object with fixed mass, Newton's second law \(F = ma\) then shows that the resulting acceleration is also constant over the time interval. When acceleration is constant, we get to use the standard kinematic equations so that [\(a = \frac{\Delta v}{\Delta t}.\)][(That is, the constant acceleration is equal to the change in velocity divided by the change in time, each calculated by subtracting the initial value from the final value.)] Then:

\[ F = ma = m\frac{\Delta v}{\Delta t} \\ F\Delta t = m\Delta v \]

That is, the force times the change in time gives the mass times the change in velocity. The left-hand side is the quantity we call impulse, usually given in units of [\(\text{N}\cdot\text{s}.\)][(Newton-seconds, that is, force units times time units)]

(stop impulse-algebra)

## So why name it?

*If impulse is just force times time, and its use comes directly from Newton's second law, why learn it as a separate concept, with its own name and everything?*

The main reason is that after integrating Newton's second, (the other side is the change in momentum.)(change-in-momentum)

(start change-in-momentum)

The right side comes out to \(m\Delta v,\) which is equal to \(\Delta(mv),\) since mass is constant. Since \(mv=p,\) the momentum, this is exactly the change in momentum. (When mass is not constant and we have calculus...)(mass-varies)

(start mass-varies)

There are a [few cases][(most notably, rocketry)] where the mass of the object being considered changes dramatically during the time interval in which a force is applied. When that is the case, Newton's second law actually changes so that

\[ F = \frac{dp}{dt} \]

That is, the force is the rate of change of momentum. The product rule of derivatives shows that this form reduces to the original when mass is constant. (How?)(newtons-second-product-rule)

(start newtons-second-product-rule)

First, apply the definition of momentum \(p=mv\) and product rule:

\[ F = \frac{dp}{dt} = \frac{d}{dt}(mv) = m\frac{dv}{dt} + \frac{dm}{dt}v \]

and if \(m\) is constant, then its derivative is zero:

\[ F = m\frac{dv}{dt} + (0)v = m\frac{dv}{dt} = ma \]

and you are back to the original form.

(stop newtons-second-product-rule)

Further, if you take the integral of this expression now, you directly get that impulse is equal to the change in momentum.

(stop mass-varies)

(stop change-in-momentum)

The second reason is that in this form, it's clear how (Newton's third law implies conservation of momentum.!Conservation of momentum.)(newtons-third)

(start newtons-third)

Newton's third law is that every action (force) has an equal but opposite reaction (force). These force pairs always act between the same pair of objects, but in the opposite order, simply stated as, "When I push you, you push me back."

If we want to study the effect of Newton's third on two objects that are [only interacting with each other,][(so they have an equal-but-opposite force pair between them and no other forces)] all we know is that their mass-times-accelerations are equal by Newton's second. Since the objects are likely to have different masses, that makes their accelerations almost certainly different.

However, there actually is another constant for both objects when they interact: the time interval we are studying. And that's what defining the impulse deals with, by moving that time interval in Newton's second over to the other side, incorporating it with the force. Since Newton's third applies to the force at every instant, the impulse pairs are equal but opposite over any time interval, as well. And thus, the change in momentum for the two objects together is zero, and momentum is conserved. (If there are more than two objects...)(closed-system)

(start closed-system)

In this case, the condition for conservation of momentum is basically the same: we need [the objects in our system to only be interacting with each other.][(that is, every force on an object in the system is caused by another object also in the system)] A system that meets this requirement is called a *closed system,* and it must be the case that momentum is conserved when the system is closed, since all impulses on the objects come in equal-but-opposite pairs.

Further, note that if the system is not closed, then there is a force acting on an object we're observing whose reaction force we're not taking into account. For any time interval, the total impulse delivered by these *external forces* is exactly how much momentum is added to the system.

(stop closed-system)

(stop newtons-third)