When can a force be an Impulsive force?

Question

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?

Answer 1

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: $\Delta \mathbf L=\displaystyle \int \mathbf F(t)~\mathrm dt.$

Answer 2

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.

Answer 3

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 $\tau\,.$

Answer 4

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:

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.