The elusive difference between impulse and momentum

1) In classical mechanics, impulse is the product of a force, F, and the time, t, for which it acts. The impulse of a force acting for a given time interval is equal to the change in linear momentum produced over that interval...($t_1-t_2$).The SI unit of impulse is the newton >second (N·s) or, in base units, the kilogram meter per second (kg**·m/s)**.

2) A resultant force causes acceleration and a change in the velocity of the body for as long as it acts. A resultant force applied over a longer time therefore produces a bigger change in linear momentum than the same force applied briefly: the change in momentum is equal to the product of the average force and duration. Conversely, a small force applied for a long time produces the same change in momentum—the same impulse—as a larger force applied briefly.Therefore

$\begin{align} \mathbf{J} &= \int_{t_1}^{t_2} \frac{d\mathbf{p}}{dt}\, dt =\Delta \mathbf{p} \end{align}$

source

I read a couple of previous answers, but that hasn't helped me understand: one repeats the above definition, the second says: "In the Newtonian point of view, impulse and change of momentum are different concepts...."

wiki's definition is as confusing: $J = \Delta p$

In order to simplify most, let's consider the unitary mass (m = 1) since velocity is what we call the momentum of unitary mass. Impulse is equal to the change of momentum/velocity/ : $J =[1*] \Delta v$ ( * sec)?

Summing up, considering m=1, we have:

• Velocity = $v =[1*] v$ (m) $v$


• Momentum = $p =[1*] v$ (m) $v$


• Acceleration = $a = [1] \Delta v$ change of velocity (m) $v/s$


• Force = $F =[1*] \Delta v$ change of momentum (m) $v/s$


• Impulse of a force = $J =[1*] \Delta v$ change of momentum, (m) $ = v$

Unless I made some mistakes, impulse is equal to momentum and not to change of momentum. Where did I go wrong, or, what is the final word?

2) as to the second period, I thought that the proportion between longer time and bigger change is valid only if the force gives constant acceleration, like gravity. How can that definition apply to collisions, where a ball gives a fixed amount of momentum which cannot be increased by duration? A cue ball hitting another ball gives constant acceleration? A bowler throwing a bowl on a lane gives constant acceleration? Does it matter if his arm swings for 1 or two seconds?.

I am confused and making confusion. Can you clarify my doubts?

update:

Your problem is that acceleration isn't the change in velocity

what is change of velocity then? if a football is at rest and I kick it and it aquires v=10m/s, haven't I accelerated it over a period of time? isn't that difference of $\Delta v= +10 m/s$ acceleration?,

but (taking there is a mistake there) my question was not about acceleration but:

. Unless I made some mistakes, impulse is equal to momentum and not to change of momentum. what is the final word?

• is change of momentum/velocity the same as momentum/velocity? how can they have the same units?

that statement applies to forces that can be sustained over some time.

I said: consider the hand of a bowler, it pushes a bowl with a force. If he pushes it for half a second or a second the change of momentum is the same, what changes is only that he can aim at the target with more precision

Answer

Op has read the previous answers and there is confusion, the answers either just repeat the definition or say:

The impulse takes into consideration both the effect of the force on the system, and the duration of time for which the force acts. --

or

In the Newtonian point of view, impulse and change of momentum are different concepts...

(and then an identical alternative is offered)....

Alternatively, if you take the point of view where "force" is defined as ma, then impulse and change of momentum of the body have the same values as a consequence of definitions only. But I wouldn't say this means that impulse and change of momentum are the same concepts,. --

or

so, I *don'*t think one can successfully sustain "Force really is the derivative of momentum."

and here:

Only now we can state a physical law saying that the impulse is always equal to the change in momentum, ... J and Δp are still different concepts; it's only through Newton's second law that they become equal

In order to give a comprehensible answer to the question, some historical background is indespensable: you can see here that there is no Newton second law and that force equal to $ma$ was a 'law' defined some two century later, when force was also modelled after gravity and work after the lifting of buckets from mines, against gravity. Keeping this in mind we can examine what happens in these formulas and why it happens. $$F = ma; W = [F . d] = E; J = [F . t] = \Delta p$$

As you may have read in the link, Newton made already a distinction between 'motion' and 'measure of motion' = momentum, and this is a first difference, but there is another more fundamental difference, which we might call of 'category', between the left and the right side of the equations: the 'concept' on the left indicates a 'cause' and the one on the right its 'effect'.

What is it all about? Energy, the transfer of energy from A to B, that is the bottom line, and, when you give (kinetic) energy to a body you are giving it, at the same time, velocity/momentum, as we are talking, here, about impulse and collisions. In the case of work the effect, the 'concept' involved is directly,energy as its measure is in joules, which is useful for all kinds of energy and, therefore, also when work is done against a force and gives no momentum.

In the case of impulse, momentum has been chosen, preferred for various reasons but nothing forbids you to choose 'energy' also in this case.

Let's try to make it even clearer. Consider the definitions (from wiki, the arrow shows the direction of transfer):

• $F$ : a force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate, hence $F = \rightarrow E_k$


• $W$ : work is the energy associated with the action of a force, a force is said to do work when acting on a body there is a displacement, hence $[F.d] = E . d \rightarrow = \leftarrow W = \leftarrow E$


• $J$ : impulse is the product of a force, F, and the time, t, for which it acts. The impulse of a force acting for a given time interval is equal to the change in momentum. $[F.t] = J = \rightarrow p$, since $p * v/2 = E_k$ , then $$[E_k .t] \rightarrow = \frac{E_k}{v/2} J \rightarrow = \leftarrow \frac{E_k}{v/2} \rightarrow p$$

Keep also in mind that in collisions the formula F.t is a pro-forma as time is usually considered istantaneous. One of the reasons of the choice of this formula is that what we know in collisions is v, velocity of cue ball A (= 3m/s) and m, its mass (= 2Kg), therefore it is easier to get momentum (= 6) and assume that the same amount will be the change of momentum of ball B.

In conclusion, what is the answer? Impulse is the cause and (change of) momentum is the effect (of the second order) of a transfer of energy, impulse is a 'concept' = force', momentum is the measure both of 'motion' and of impulse' : energy (J) , momentum (2E/v kgm/s) transferred from A to B. A possesses energy/momentum which, the moment it is 'donated' becomes force/impulse of force, when it is 'received' by B, is called work done/energy transferred/change of momentum. The moment ball B meets a ball C its (change of) momentum becomes itself an 'impulse'.