I am a high school student and I am confused in one thing:"Why do we always assume in problems that if things are initially in contact with each other then they would be like that always?"

1)Suppose we have two blocks(A and B) initially in contact with each other and we apply force "F" on block "A".something like this At first, I used to think that the contact force between block A and B would rise to the point when they acquire the same acceleration and at this moment I thought they would move together so at this point the contact force would become constant as there is no "deformation-reformation" thing happening after this moment. But if that's the case then at this moment the velocity of block A would still be higher than block B because its initial acceleration was more. If I have to plot a curve of acceleration-time it would be something like this:acceleration-time graph . the shape of the curve is random but I think it doesn't matter to my point(unless curve A crosses B and go even below it), the area under curve A would be higher than B so its velocity should be higher even at the moment when both have acquired the same acceleration. So, the deformation-reformation should still be happening as A is still approaching B and they would "never" reach to a common acceleration. So, why do we "assume" in problem solving that if these block are initially in contact, they would maintain that contact? But in reality, we have seen them moving together so how does this happen?

  1. This question came to my mind because I was thinking of a similar question of "water in a bucket revolving in a vertical plane" . We know that the water doesn't spill out from the bucket because the internal forces changes in a way so as to make the exact centripetal force required for water to move along with bucket. But the real question is: "How the internal forces "knows" that they have to maintain thing in contact at any moment in time?"

Please answer in simple language if possible. Or if its a difficult concept for a high school student to learn ,please answer only in brief and let me know.

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    $\begingroup$ The curves do cross. This allows B to catch up to A's initially higher acceleration and speed. $\endgroup$
    – Rick
    Dec 7, 2022 at 18:25
  • $\begingroup$ As a counterexample, consider Newton's Cradle with three balls: a ball to the left of A, A, and B. When the first ball hits A, a force is applied to A, causing B to fly away from A. $\endgroup$ Dec 8, 2022 at 2:25

2 Answers 2


Why do we always assume in problems that if things are initially in contact with each other then they would be like that always?

Because it is simpler.

At this point in your studies you are not sufficiently advanced to realistically model the interaction of the two blocks. So we simplify it for you. The purpose of the simplification is not that it is realistic but that it makes the problem solvable for someone with your background.

The idea is that in order for you to eventually get to the point that you can handle more realistic physics you need "a very particular set of skills". Developing those skills requires working problems. Those problems don’t need to be realistic, they need to help you develop the skills. So we deliberately sacrifice realism in that pursuit.

A more realistic, but still simplified, approach would be to treat the interaction as a sort of spring. Even more realistic would be to treat it as a spring and dashpot. Even more realistic would be to use some experimentally measured force-deformation profile. But all of these can wait until you have acquired the basic skills necessary. And the simplified model is a good “first order” approximation to all of these more realistic models.

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    $\begingroup$ This sort of underplays the fact that, at low force levels, the model the OP is trying to solve generates answers that are extremely close to correct. And all of physics is full of these situations; higher order effects may exist, but we can neglect them in the model. Teaching someone to only use lower order physics is reasonable, unless you think we should start with relativistic effects instead of (lower order) Newtonian approximations. $\endgroup$
    – Yakk
    Dec 7, 2022 at 21:04
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    $\begingroup$ @Yakk that is a strange comment. You seem to be objecting to my answer by saying things that I already said in the answer. I explicitly said it is a good approximation and I explicitly said we need to teach the simple stuff first. I am not sure where in my answer you think I disagreed with either of the points in your comment. $\endgroup$
    – Dale
    Dec 7, 2022 at 22:40
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    $\begingroup$ I don't see where you stated clearly that the simple stuff is not only simple, but also quite accurate. The OP's solution isn't a toy math exercise, but actually solves (with reasonably high accuracy) what happens in a reasonable, physical situation. Maybe you are trying to state that in the very last sentence of your answer, but it isn't clear. Is the last word a typo? Or "easy assumption" your name for the OP's solution? If so, it is awkward; introduced in the conclusion, and worded strangely. And "they" in the last sentence; what that pronoun refers to is ambiguous. $\endgroup$
    – Yakk
    Dec 7, 2022 at 23:36
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    $\begingroup$ Your opening sentence also needs work. "At this point in your studies you are not sufficiently advanced to realistically model the interaction of the two blocks." -- I believe what they are modelling is realistic for a reasonable subset of situations. It isn't lack of realism -- it is a limited subset of realistic situations they can model. $\endgroup$
    – Yakk
    Dec 8, 2022 at 2:25
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    $\begingroup$ I disagree. The judgement about the realism of the model that I expressed there is simply echoing the concern of the OP. They are considering more realistic models of the collision. The simplified model is indeed unrealistic at the level that the OP is asking about. When an OP is wrong, then I will correct them. But when they express a valid opinion or judgement I prefer to accept it for the purpose of their question $\endgroup$
    – Dale
    Dec 8, 2022 at 3:53

I think by trying to analyse it in terms of "the contact force between block A and B would rise to the point when they acquire the same acceleration" (emphasis mine) you are making it more complex than it needs to be.

The sort of diagram you have drawn of the forces on A and B is more usually used when we're thinking about an instantaneous situation. Force and acceleration having single numerical magnitudes is also a momentary way of looking at things. At one moment in time there is a given force on the block A, which produces a given acceleration. If we're simplifying by considering the blocks are perfectly rigid then then we might say that the force on A "instantly" is transferred to B (via the contact forces), to work out the acceleration on B, and see that it is the same.

As you've realised there is something that doesn't quite make sense about this picture. No real physical object is "perfectly rigid", and forces can't be transmitted instantaneously. In reality there is a force on A which starts it moving1, and it pressing into block B causes a force on block B. But by talking about things like "block A pressing further into block B and causing an increasing contact force" we've moved away from the simple instantaneous viewpoint. Now we don't have forces and accelerations with single numerical magnitudes, we have a dynamic system with forces and accelerations that change over time. This means we need more information; it now matters what kind of force is being applied to A (is it a constant force like a magnetic field, or a quick impulsive force like a collision with something hard on the left side, or something else). The material properties of A and B matter. The situation of them moving with the same velocity (and so having experienced the same average acceleration) is not an instant result, it's a situation that is reached after we wait a while.

An intuitive way to think about it might be this: If A is put into motion to the right that will increase the contact force on B. The contact force will persist for a period of time (first increasing then decreasing) until until B has the same velocity as A (not acceleration). If there are no more forces being applied at that point, then A and B are both moving with the same velocity and so will remain in contact (if they still are; we're assuming that happens but it depends on the particulars). And at the point where they have attained identical velocity, then both have had the same average acceleration since the starting time. If you're thinking that A would have had a higher velocity initially then B must have later accelerated faster to catch up (otherwise they would still be "squashed together" and the contact force would still be accelerating B more); so the line on your graph of the acceleration of B must go above that of A at some time; they don't simply converge the way you have drawn them.

It should be clear that my simple description there is involving forces and accelerations that evolve over time. Modelling that is far beyond a single static force diagram. I think your confusion is more to do with trying to reconcile the dynamic view of the situation (involving block deformation and contact forces waxing and waning) with the instantaneous force diagram view of the situation (where a force is applied and assumed to transmit instantly). The "perfectly rigid blocks and instantly transmitting forces" view of your initial problem is a simplification; in that model the blocks don't squash together, there's no gradually increasing contact force, and we're not thinking about forces applied over time (which we'd need to know how much total change in velocity is applied; we're only thinking about the instantaneous acceleration resulting from the instantaneous force; this could change the blocks velocity by a lot if it's applied continuously for a long time, or barely at all if it's applied for a tiny fraction of a second).

  • $\begingroup$ I also think if A has more velocity then B it would continue to apply force on B it means acceleration of A will decrease further an accelertion of B will increase but whne they will reach the common velocity the acceleration of B would me higher so it will sperarate from A so the contact force will decrease gradually and so the acceleration of B will decrease and acceleration of A will increase, so these line of A and B would never meet they would just cross each other again and again and they will "never" reach to a common velocity and acceleration(but their average over time can be same) $\endgroup$ Dec 16, 2022 at 6:29
  • $\begingroup$ ....but that is purely a thought of reality...I wanna know do science courses discusses these thing in higher education? $\endgroup$ Dec 16, 2022 at 6:29
  • $\begingroup$ @ArunBhardwaj If they reach a common velocity, then they are travelling at exactly the same speed. A is only just keeping up with B, not pressing into it and accelerating it anymore. And if there's no more acceleration that means (by definition) that there's no more change in speed, so they will continue travelling at the same speed. $\endgroup$
    – Ben
    Dec 16, 2022 at 14:21

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