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
$\begin{align}
\mathbf{J} &= \int_{t_1}^{t_2} \frac{d\mathbf{p}}{dt}\, dt
=\Delta \mathbf{p} \end{align}$
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
Best Answer
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:
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.
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
$$\bar a = \frac{10 \frac{m}{s}}{0.1s} = 100 \frac{m}{s^2} $$