In my textbook it is mentioned that “impulse is used to get an idea of about the change in dynamical state of a moving particle”,but what does impulse actually mean?
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I think of impulse as "the stuff that it takes to drive a nail into a piece of wood". If I try to push a nail into a piece of wood by hand, I can't. Even if I try to push with a hammer, I can't do it. I can apply a medium-sized force over a relatively long period. But if I swing the hammer, applying that medium-sized force over a relatively long period, I can build up momentum in the hammer head which, when it relatively instantaneously hits the nail, converts that momentum back into a much larger force, applied over a much shorter time, which is just what's needed to drive the nail. – Steve Summit Sep 19 '22 at 12:47
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Have any of the many answers helped? If not, why not? – Bob D Sep 20 '22 at 14:49
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An impulse is a change in momentum, usually with the connotation that it is very brief or sudden.
Dale
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If this were the definition of impulse, the impulse-momentum theorem would have no content. It is like saying that the net work acting on a point particle is its change in kinetic energy ... – user26872 Sep 19 '22 at 15:49
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If impulse is change in momentum, the impulse-momentum theorem states that the change in momentum is the change in momentum. Impulse is the time integral of force. It is only when we know this that the theorem has a nontrivial meaning. Can we give the OP an intuition about what the time integral of force is? – user26872 Sep 19 '22 at 17:43
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@user26872 the theorem is: $\int F \ dt = \Delta p$. Now if we call the left side “impulse”, then the theorem literally says “impulse is a change in momentum”, which is exactly what I said. So I am not sure why you think that stating a theorem somehow negates its content. It is a pretty nonsensical objection – Dale Sep 20 '22 at 00:09
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It is nonsensical is to confuse a theorem with a definition. Can we give the OP an intuition about what the time integral of force is? – user26872 Sep 20 '22 at 12:46
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And I didn't confuse the theorem with the definition. Read what I wrote. Did I say "impulse is defined as a change in momentum"? No, I said "impulse is a change in momentum", which is simply stating the theorem $\int F \ dt = \Delta p$ in English. The only mistake here is yours where you read a statement of a theorem as though it were a definition, and then proceeded to make a nonsense objection – Dale Sep 20 '22 at 12:59
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In the context of Mechanics it is $J = \int F(t)\,dt$, ie the area under a force, $F$, against time, $t$, graph and has the unit $\rm N\,s$.
Often it is taken to mean a force which acts over a very short period of time compared with other time scales relating to the system.
For example, when two billiard balls collide and the time of collision (when the balls are in contact) the forces acting on the billiard balls can be termed impulsive because they act over a much shorter period of time than the time taken for the balls moving from their original positions..
Thus in terms of the complete motion of the billiard balls the collision can be though of as occurring instantaneously.
The from Newton's second law, force is equal to the rate of change of linear momentum, $F=\frac{dp}{dt} \Rightarrow \int F\,dt = \int dp$, the impulse is equal to the change in momentum.
Farcher
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1The impulse definition is not restricted to short periods, as rocket engines have impulse ratings and burn over finite time. The short time restriction has to do with the nature of contact forces, and the assumption that nothing else changes during a collision. But that is not the only application of impulses. – John Alexiou Sep 18 '22 at 16:49
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1@JohnAlexiou Please read my second paragraph which indicates that impulses do not always occupy short intervals of time. – Farcher Sep 18 '22 at 17:41
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Both @Farcher and @Dale are correct as the full answer is the combination of their answers.
You can think of impulse as the currency of momentum if you want to make a money analogy. Momentum is how much money you have in your bank account, and impulse is the amount of money you deposit or withdraw.
Impulse has two descriptions
Physically an impulse is a way to describe a change in momentum, like with the analogy above.
$$ \Delta p = p_f - p_i = j \tag{1}$$
where $\Delta p = p_f - p_i$ is a change in momentum $p$, and $j$ is an impulse.
Mathematically it is the area under a force vs. time curve $$j=\int F\,{\rm d}t \tag{2}$$ if force is an applied force, and not restricted to short time periods, just as rocket engines have impulse values, and they burn over several seconds. Hence why the units of momentum are often stated as Newton-second $\text{[N s]}$.
But this begs the question, of what is momentum. I know mathematically you know momentum is $p=m v$, what does this mean physically?
The answer comes from (1), as I describe momentum as the quantity of impulse needed to completely stop a moving object. So I understand momentum, in terms of impulses and not the other way around.
Suppose you have a body with momentum $p_i \neq 0 $ and you want to find the impulse $j$ needed to completely stop the body and make $p_f = 0$
$$\Delta p = p_f -p_i = -p_i = j$$ $$ |p_i| = |j|$$ and this makes momentum measurable, because impulse is measurable.
Conservation of momentum is automatically enforced in this system, just as as with banks the total amount of money is conserved when you pay for something (what leaves your account goes into another account). An impulse leaving a body (and reducing its momentum) goes into another body (and increases its momentum) in a way that total momentum is conserved, and all as a consequence of Newton's 3rd law (equal and opposite impulse as a result of equal and opposite forces).
$$\require{cancel} \begin{array}{r|cc|c} \text{momentum} & \text{body 1} & \text{body 2} & \text{total} \\ \hline \text{before} & p_{1} & p_{2} & p_{1} + p_{2} \\ \text{after} & p_{1}-j & p_{2}+j & p_{1} \cancel{ -j }+ p_{2} \cancel{+j} = p_{1} + p_{2} \end{array}$$
Furthermore, the above can be extended into vector notation, where not only the magnitude and direction of momentum is defined, but also the point in space where momentum is defined. This leads to the law of conservation of angular momentum.
John Alexiou
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1I found the rocket burst example very nice. The (usual) billiard balls are too abstract for me - that example "feels" like the impulse would be instantaneous, which it can't be, is totally unobservable to the naked eye, and all that. A rocket engine running for 3 seconds is very observable, and very obvious. – AnoE Sep 19 '22 at 11:56
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1Also I like that you grab the initial formula (delta p = ... = j) and cheekily turn everything around to argue from impulse to momentum instead of the other way round. Very well done. :) – AnoE Sep 19 '22 at 11:57
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@AnoE - thank you. It came to me a while back when thinking about how to stop a spinning 3D object with a single impact. You can only do it for certain types of motion (where $v \cdot \omega =0$), and the magnitude and direction of the impulse is defined by the momentum vector, but the location of the impulse by the angular momentum vector. I am glad I was able to articulate my thoughts well enough were at least one person understood where I was going with this post. – John Alexiou Sep 19 '22 at 13:26
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Impulse is the momentum gained or lost by applying a given force for a given amount of time.
jelly ears
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In classical physics, the formal definition of the impulse is the integral of the force, 'F' over a certain time period t. Mathematically, it looks like this: ${\displaystyle \mathbf {J} =\int _{t_{1}}^{t_{2}}\mathbf {F} \,\mathrm {d} t}$.
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Newtons seconds law for a body is a law of motion which applies for all time instances. If we are to accumulate the information of all the forces over a period of time, we would have to integrate this law.
We have $ma= F$. The integral of LHS with time is just change in momentum and right we name as impulse.
tryst with freedom
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