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A large force which is applied for a short time is called an Impulsive force according to books. But is it the only condition for an Impulsive force? Are there some other conditions? Can we say that that a large force acting for a large time be impulsive?

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    $\begingroup$ what do u mean impulsive force or impulse, many people mix that up ?? $\endgroup$
    – Vishnu JK
    Oct 7, 2016 at 9:55

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See JohnRennie's answer here for an explanation.

An impulse is the integral of any force, large or small (even gravity), over time.

A force is an abstract entity, a potentiality, the impulsive force is that force actually acting for any length of time, and producing actual consequences : change of momentum.

Any force becomes an impulse whenever you take into account the length of time it acted.

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    $\begingroup$ Hi @user104572; so don't you think impulsive force is defined as the force which acts for an infinitesimal period of time? That's the point of impulsive force, isn't it? $\endgroup$
    – user36790
    Oct 7, 2016 at 8:23
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    $\begingroup$ @MAFIA36790, where did you get that impression? $\endgroup$
    – user104372
    Oct 7, 2016 at 8:24
  • $\begingroup$ It is written in my book; it's a pretty common application to model impulsive force with dirac delta function after introducing it. I wouldn't mind if you respond in the chat rather than here if you want to. $\endgroup$
    – user36790
    Oct 7, 2016 at 8:31
  • $\begingroup$ Now, it's looking good; +1. $\endgroup$
    – user36790
    Oct 7, 2016 at 8:50
  • $\begingroup$ There is no such thing as an impulsive force: there is a force who can act , and an impulse of that force, which acts for a definite time. infinitesimal is not definite, same as short $\endgroup$
    – user104372
    Oct 8, 2016 at 5:23
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An impulsive force is any force acting for a limited time. It is called impulsive because it transfers a definite linear momentum (impulse) to the body it is working on.

The transferred linear momentum $\Delta \mathbf L$ is given by the integral over the time during which the force $\mathbf F(t)$ is acting: $\Delta \mathbf L=\displaystyle \int \mathbf F(t)~\mathrm dt.$

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  • $\begingroup$ @ MAFIA36790 - I first used the term "impulse" for linear momentum and was not sure whether this was commonly understood in English. In German the technical term for linear momentum is "Impuls". $\endgroup$
    – freecharly
    Oct 7, 2016 at 13:01
  • $\begingroup$ Impulse is not momentum; but change in momentum. $\endgroup$
    – user36790
    Oct 7, 2016 at 13:05
  • $\begingroup$ @MAFIA36790 - Thank you, then as you said, it was correct anyway $\endgroup$
    – freecharly
    Oct 7, 2016 at 13:09
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As freecharly stated in his post, impulsive force is defined to be the force which acts for an infinitesimally short interval of time and yet is responsible for a finite change in momentum of the system on which the impulsive force is applied.

As implied from Newton's Second Law of Motion, impulse is defined to be $$\textrm{Impulse} \equiv \Delta p = \int_{t_1}^{t_2} \mathbf F(t)~\mathrm dt\tag I\,.$$

Now, consider a force, which acts in an infinitesimally short time-interval $|t - \tau| < \dfrac\epsilon2: \epsilon\gt 0$ where $\tau$ is the instant the impulse occurs.

From $\rm(I),$ $$\Delta p = \int_{\tau- \frac\varepsilon 2}^ {\tau+ \frac\varepsilon 2}~ \mathbf F(t)~\mathrm dt\tag{I.a} $$

Now, since, the force $\mathbf F$ is non-zero only near $t= \tau$ and is zero for every other instants, this provides us an opportunity to model the impulsive force using $\delta(t-\tau),$ famously known as Dirac Delta function, a general distribution as it is zero everywhere except at $\tau$ i.e. $$\delta(t-\tau )= 0, ~~~~t\neq \tau\,.$$

So, we define the impulsive force as $$\mathbf F(t) = \mathcal F~\delta(t-\tau)$$ where $\mathcal F$ has units of force-time.

From $\rm(I.a),$ $$\Delta p = \int_{\tau- \frac\varepsilon 2}^ {\tau+ \frac\varepsilon 2}~ \mathcal F~\delta(t-\tau)~\mathrm dt = \mathcal F\tag{I.a.i} $$

The whole point of defining the impulsive force by $\delta$ is that $\bf F$ is practically non-zero in $t= \tau$ and is zero at every other instants.

Can we say that that a large force acting for a large time be impulsive?

It should be kept in mind that the impulsive force acts for such a short interval of time that it is non-zero only at $\tau\,.$

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The idea of an impulse, $\int_t F(t) dt$ in mechanics comes from Newton's second law $F= \frac{dp}{dt} \Rightarrow \Delta p = F \Delta t$ ie it is something to do with change of momentum of a body.

Now often it is the case that one is interested in what happens before and after a collision but one is not interested in what happens during the collision particularly if the time during which the collision occurs is much less than the times scale before and after the collision.

Suppose a mass of $1$ kg, starting from rest, is subjected to a force of $10$ newtons for $1$ second.
The impulse is $10 \times 1 = 10$ Ns and as this is equal to the change in momentum of the body the bodies final speed is $10$ m/s.
However it does not reach that speed until one second after the froce was first applied.

Now apply a force of $20$ N on the body for $\frac 1 2 $ seconds.
The impulse is the same ($10$ Ns), the change in momentum is the same ($10$ Ns) and so is the final speed ($10$ m/s).
The only difference is that now the body reaches $10$ m/s half a second after the force was first applied.

So repeat the process:

enter image description here

As you can see from the graphs the impulse in the same and so is the final state, speed is $10$ m/s but as the time over which the force acts decreases so the time taken to reach the final state.

You will note that the impulse is the area under a force against time graph.
In collision problems it is convenient to say that the force acts over a very small period of time but what is important to me is not the actual magnitude of the force or the actual time over which it acts but the area under the force against time graph which is the impulse.
A mathematically convenient way of doing this is to write the impulse in terms of the delta function $10 \; \delta(0)$ Ns as the delta function has an area of one.

So in such an example a certain amount of momentum is being transferred to a body in next to no time.

So it is not a force which acts for no time, it is a force which acts fro a time much less that the time scale under consideration and as far as the change in momentum of the falling body is concerned it the force, which happens to be constant in this case) times the time over which the facts which is important.

There is no point in speculating about a force that acts for no time as that force does nothing. However Mathematics via the delta function allows you to apply an impulse to a body which changes the momentum of the body in a time period which is not of interest to you.

Now in the case of free fall the evaluation of the impulse is relatively easy in that the force, $mg$, does not change with time and one tends not to use the word impulse in such a situation.
It certainly does not make the problem easier to solve by saying that the mass is subjected to a whole series of delta functions as it falls although in a sense you do exactly that when an integration is performed.
In fact in solving such a problem the word momentum is probably never used?

So whether you call it a force acting over a period of time or an impulse is entirely up to you, the problem at hand and convention.

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  • $\begingroup$ Thanks for the detailed answer! So in physics problems involving collision and other forces like gravity, friction etc, we do consider collision duration $\Delta t \to 0$ ? So we can say impulse of impulsive forces has a definite value even as $\Delta t \to 0$ , while impulse of non-impulsive forces vanish as $\Delta t \to 0$ ? $\endgroup$
    – jonsno
    Oct 17, 2017 at 8:07
  • $\begingroup$ The term “force” is usually used instead of “non-impulsive force” as the opposite the term “impulsive force” which implies that the force is applied over a short interval of time. So for ball colliding in the air the period of them actually touching each other is much, much less than their time of flight. $\endgroup$
    – Farcher
    Oct 17, 2017 at 8:18
  • $\begingroup$ So this analysis can be used as a good approximation in actual cases. Thanks! $\endgroup$
    – jonsno
    Oct 17, 2017 at 8:45

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