I'm new to physics and trying to understand Newton's second law $F = ma$ but I don't think I'm grasping the concept of force very well. I've read other questions and answers on this law and it is my understanding, for now, that $F = ma$ is a "definition" of force based on the empirical "law" that the product $m \times a$ remains constant when the same amount of "force" (here the term used colloquially before formally defining it and assigning SI unit) is applied while varying the mass. But still, I don't seem to understand the full dynamics the law implies.

Suppose at time $t=0$, an external (?) force of $1$N is applied to a resting mass ($x'(0) = 0$) on a frictionless single-dimensional space with initial displacement $x(0) = 0$. The force applied is instantaneous in the sense that the bullet or whatever object that applied the force ricochets or disappears right away after contact with the resting mass. Here's my short train of thought that I needs guidance on:

  1. Since the force was instantaneous, $F(0) = 1$ and $F(t) = 0$ for $t > 0$. Then the second law will imply that the mass doesn't move at all since $p(t) = \int_0^t F(v) dv = 0$, meaning zero velocity.
  2. No no, that doesn't make sense. The initial condition $F(0) = 1$ applied to the resting mass will give instantaneous acceleration $a(0) = 1/m$. But how does acceleration or velocity evolve? It seems odd if acceleration is constant over time, but if it's not constant, then I don't see how to proceed to derive the trajectory.

I concluded I'm either misunderstanding what force is or I lack the kinematics toolbox or I am just completely loony and misguided overall. I'm just beginning to self-study basic physics and I feel like I'm starting off on the wrong foot. So here's me asking for some guidance. Please correct me and fill me in with what I am lacking/misunderstanding here.

  • $\begingroup$ Instead of an (unphysical) "instantaneous" force like the one you're dealing with, why don't you try to solve the the problem with a constant force of (say) 1 N, acting for some finite time (say, 1 second)? $\endgroup$ – Philip Jul 3 at 15:42
  • $\begingroup$ @Philip I guess the last point in BioPhysicist's answer relates to your comment. I'm curious what you mean by instantaneous force (or impulse, I guess) being unphysical. Does it mean that all interactions in physics, or at least Newtonian mechanics, are continuous? I'd like to grasp some general view so I know where I'm headed as I learn. $\endgroup$ – xqst Jul 3 at 16:47
  • $\begingroup$ A 1N force lasting 0 seconds imparts 0 momentum to the object. $\endgroup$ – John Alexiou Jul 4 at 0:50

Warning: Non-rigorous math up ahead.

Usually we handle instantaneous things in physics with the Dirac Delta function, which is qualitatively an infinite spike at some given point. We can use this here to express the instantaneous force in terms of an impulse per time: $$F(t)=J_0\delta(t)$$ so that the total change in momentum of the object is given by $$\Delta p=\int F(t)\,\text dt=\int J_0\delta(t)\,\text dt=J_0$$

The difference between using the Dirac Delta function and doing what you propose is that $F(0)\neq1\,\rm N$ in the above analysis. Instead, $F(0)\to\infty$. This is the only way you can make an instantaneous force do something. This is because there is no area under a finite spike as you are proposing. In other words, if we use your proposed force

$$F(t) = \begin{cases} 1\,\rm N, & t=0 \\ 0\,\rm N, & \text{otherwise} \end{cases}$$

then nothing happens because $$\int F(t)\,\text dt=0$$

Therefore, your number 1 is actually correct for what you propose. The reason this probably doesn't sit right with you is because no force is truly instantaneous, and Dirac Delta functions are actually just idealizations to make the math work out nicely. In reality we will have something like $F(t)=F_0\cdot g(t)$, where $g(t)$ is some sort of finite function that takes the form of a thin spike.

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  • $\begingroup$ Thanks, this helps a lot! I'd like a few more clarification please. Does this mean that the notion of force has a measure-theoretic extension? Will this be the way forces are dealt with if I continue learning about the subject? Also, which view is typically taken when considering a physical model/theory of collision: impulse using a Dirac Delta or the more "realistic" smoothing? Is it a matter of taste? $\endgroup$ – xqst Jul 3 at 16:53
  • $\begingroup$ You can disregard the questions above (unless you'd like to add or share your thoughts), @John Alexiou helped to clarify. I'll still leave the comment undeleted to reflect on my ignorance in the future. $\endgroup$ – xqst Jul 4 at 6:06

I think you are essentially talking about impulse. You are right that $\int_0^t F(u)du$ would be zero if the force is applied for an instant, but any real force is applied for at least some time and so the integral cannot be zero. However, it may be very small as in your case where the force is not of a very large magnitude.

The same goes with acceleration, so the acceleration corresponding to that force will have a very small but finite integral with respect to time i.e $\Delta v = \int_0^t a(u)du \neq 0$. After the force has been applied, the acceleration will be zero so the velocity will be constant (and not the acceleration). The exact velocity of the object will depend on the integral $\int_0^t a(u)du$. Usually, the more useful formula in these cases is not the integral calculated explicitly but through the averages, $$\int_0^t f(u)du = f_{avg}\Delta t$$.

For more information you can check out the Wikipedia article on impulse.

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  • $\begingroup$ Thanks! Just one clarifying question: are you saying the average is a more useful quantity in this case because the force is applied for too short a time period so the actual global trajectory (? distribution? what would you call it?) of the force $F(u), 0 \le u \le t$ is not so informative? $\endgroup$ – xqst Jul 3 at 17:22

Velocity is the integral of acceleration with time, and if your acceleration curve has a finite value at $t=0$ and zero otherwise, the area under the curve is zero. This means, your first interpretation is correct, and the body will not move.

The rule to remember is:

$$ \Delta \text{(momentum)} = \text{(impulse)} $$

The change in momentum equals the impulse. And impulse is defined as the area under the force curve $$J = \int F\,{\rm d}t$$

In short, if a finite force is applied on an infinitesimal time then the result is nil

$$\Delta \text{(momentum)} = \lim_{\epsilon \rightarrow 0} \int_0^\epsilon F \, {\rm d}t = \lim_{\epsilon \rightarrow 0} F (\epsilon - 0) = 0$$

Only if the force is finite over a finite time there is an observable effect. Only when $J = F \Delta t$, we have

$$\Delta (m v) = J = F \Delta t \Rightarrow m \Delta v = F \Delta t \Rightarrow \Delta v = \tfrac{F}{m} \Delta = a\, \Delta t$$

But if you want to assume this event happens over infinitesimally small time $\Delta t \rightarrow 0$, you have to make the force unknowingly large $F \rightarrow \infty$ such that through mathematical trickery you get $$ J = \int F \,{\rm d}t > 0$$

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    $\begingroup$ Your answer along with @BioPhysicist's answer seem to suggest that while mathematical idealization do allow defining force on a Dirac measure ($F_0$N on a point mass at $t=0$), force is in its nature (?) defined on the Lebesgue measure. So in order for a realistic continuous approximation of the "discrete force" to have the same impact in terms of impulse, the force in its natural definition should be extremely large. Here I'm thinking a continuous approximation $F(t) = F_0 g_{\epsilon}(t)$ where $g_{\epsilon}(t) > 0$ on $0 < t < \epsilon$ and (continued) $\endgroup$ – xqst Jul 4 at 3:49
  • $\begingroup$ (continued) $\int_0^{\epsilon} g_{\epsilon}(t) dt = 1$ for all $\epsilon > 0$. This would make $F(t)$ arbitrarily huge at some $t$ for an arbitrarily small $\epsilon$ but the impulse is the same in both cases: $\int F_0 d\delta(t) = \int F(t) dt = F_0$. Am I understanding the point correctly? $\endgroup$ – xqst Jul 4 at 3:49
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    $\begingroup$ @xqst - exactly you got it. $\endgroup$ – John Alexiou Jul 4 at 4:26
  • $\begingroup$ That was helpful, thanks! On hindsight I think that was essentially @BioPhysicist's point. Guess my brain just needed additional input to fully comprehend. Yours was totally a valuable addition, but because it's materially equivalent, I'll accept the other one as the answer for the sake of the order of time. $\endgroup$ – xqst Jul 4 at 6:13

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