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
 A: Your problem is that acceleration isn't the change of velocity, it's the change of velocity divided by the change in time. I still agree with one of the answers you link in saying that impulse and change in momentum are different concepts.
Suppose we didn't know anything about Newton's second law. Then we could (maybe) imagine the concept of force and define impulse as $J = \int F\ dt$. Then impulse would be the average force applied over a period of time multiplied by the time. It wouldn't be a very useful concept, because we wouldn't know what to do with forces, yet it would exist. We could also define momentum as $p = mv$, and define the change of momentum $\Delta p$ of a body that undergoes some physical process. We have two concepts, $J$ and $\Delta p$, apparently unrelated. Then Newton comes along and tells us that after doing lots of experiments, he found out that it always happened that $J = \Delta p$. Only now we can state a physical law that says that the numerical values of impulse and change in momentum are always equal, or, as it is more commonly said, force is mass times acceleration. $J$ and $\Delta p$ are still different concepts; it's only through Newton's second law that we say that in any physical situation their values will be the same; when we don't want to say all that we simply say that they are equal, even though they're still different concepts.
As for the second question, yes, that statement applies to forces that can be sustained over some time. Obviously you can't increase the duration of a collision, so the amount of time is fixed. You could still increase the impulse by making the balls collide with a larger force.
A: 
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?

I'm not sure I completely understand your notation in your reasoning preceding the statements quoted above but 3rd line certainly doesn't look correct:

Acceleration = a=[1]Δv change of velocity (m) v/s

The correct equation is
$$\bar a = \frac{\Delta v}{\Delta t} $$
where $\bar a$ is the average acceleration.  With this correction, the final equation becomes
$$J = (m)\bar a \Delta t = (m) \Delta v = \Delta p$$
This seems so straightforward that I suspect I don't understand what you're actually trying to show.


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 Δv=+10m/s acceleration?

Yes, you accelerated it over a period of time and no, the difference in velocity is not acceleration.
(Average) acceleration is, as I wrote above, the ratio of the change in velocity to the elapsed time over which the change occurred.
So, if the change in velocity is 
$$\Delta v = 10 \frac{m}{s}$$
one does not know the average acceleration unless one also knows the elapsed time $\Delta t$ since the average acceleration is given by
$$\bar a = \frac{\Delta v}{\Delta t}$$
Clearly, the average acceleration is inversely proportional to the elapsed time so, the smaller the elapsed time, the larger the average acceleration.
To be concrete, let us say that the foot was in contact with the football for $100 ms$.  Then, the average acceleration of the football is
$$\bar a = \frac{10 \frac{m}{s}}{0.1s} = 100 \frac{m}{s^2} $$ 
A: 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'.
A: I came across your question trying to figure out how to derive force (accelerating force) a photon exerts on a sun sail. Different from spheres of iron and zagging bowler   hats ("ziggidizagg-why...", remember the "space" girls:-)) photons (not: phonons:-))  do not have mass, however, they have got impulse (up-beat, yeah;:-)).
From there I wondered, as Force equals mass times accleration, and by definition there is no mass with photons, how to "get there": How can I derive an equation of force (of acceleration) if  there is only given impulse i.e. momentum, p equals mass times velocity?
My understanding is your question reflects this "acceleration in no time". It appears that there must be some period of time given, "period".
I propose some simple solution: it's in the units.
If you got just one single impulse/momentum, you have included in the unit one second of time.
It must be given as a seconde premise that the bowler hat you throw had been at rest before, or, you must been given its velocity (and mass) before receiving the given impulse/momentum
Obviously your question focuses on the difference, the delta. In case there is just one impulse, momentum, no period of time of acceleration given (how long have you been doing this, how long has this been going on), the one second in the unit "is" the time period of acceleration.
Thus, in my opinion m times velocity becomes m times acceleration. Some impulse of 5 m/s becomes force of 5 m/s square.
