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The question I asked was disputed amongst XVIIe century physicists (at least before the invention of calculus).

Reference: Spinoza, Principles of Descartes' philosophy ( Part II: Descartes' Physics, Proposition XIX). Here, Spinoza, following Descartes, denies that a body, the direction of which is changing, is at rest for some instant.

https://archive.org/details/principlesdescar00spin/page/86

How is it solved by modern physics?

If the object is at rest at some instant, one cannot understand how the movement starts again ( due to the inertia principle).

If the object is not at rest at some instant, it seems necessary that there is some instant at which it goes in both directions ( for example, some moment at which a ball bouncing on the ground is both falling and going back up).

In which false assumptions does this dilemma originate according to modern physics?

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    $\begingroup$ "If the object is at rest at some instant , one cannot understand how the movement starts again ( due to the inertia principle)." this is wrong, see the second law of newton! $\endgroup$ – Wolphram jonny Nov 9 at 20:03
  • $\begingroup$ Wolphram jonny.- Could you please explain with some details the way in which the false assumption produces a fallacious dilemma here? Or could you produce a counter-reasoning? $\endgroup$ – Eleonore Saint James Nov 9 at 20:06
  • $\begingroup$ If you had to find the distance a ball would travel if thrown vertically upwards a velocity of $15ms^-1$ (On earth and ignoring air resistance) how would you calculate it? $\endgroup$ – Ubaid Hassan Nov 9 at 21:02
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    $\begingroup$ The problem hinges on the assertion "If the object is at rest at some instant , one cannot understand how the movement starts again ( due to the inertia principle)". This is empirically wrong (items caught without effort at the apex of their free fall, or children touched lightly on the cheek at the extreme of an otherwise wild swing on a swing set are clearly at rest). The assertion is also fundamentally impossible: If resting objects could not be put into directed motion by applying a force, nothing would ever move. So can you give us the citation for this claim? $\endgroup$ – Peter - Reinstate Monica Nov 10 at 13:17
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After the invention of modern calculus and notions like continuity and differentiability, the answer is quite trivial in Newton's formulation of mechanics assuming the body is moving along a line. The second derivative of the position should be always defined as it equals the total force acting on the body. Therefore the first derivative must be continuous. This derivative is the velocity. You are assuming that it changes its sign passing from time $t$ to time $t'$. A continuous function defined on an interval which changes its sign at the endpoints of the interval must vanish somewhere in the interval. That is an elementary result of Calculus. In summary, the body must be at rest at some time, if accepting the modern version of Newton's formulation of classical mechanics.

The objection that if the body stops at a certain instant, then the direction of the motion immediately after that instant cannot be decided is untenable. The direction is actually decided by the acceleration, that is by the total force acting on the body at the said instant.

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    $\begingroup$ Is this not also the mean value theorem...? $\endgroup$ – Stian Yttervik Nov 11 at 10:25
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    $\begingroup$ @StianYttervik Seems more like the intermediate-value theorem. $\endgroup$ – chrylis -on strike- Nov 11 at 10:37
  • $\begingroup$ Yes, in Italy we call that theorem the "intermediate-value theorem"... $\endgroup$ – Valter Moretti Nov 11 at 10:41
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I assume you are talking about an object that reverses direction through a collision. Take the case of a ball bouncing off the ground. As it hits the ground, the ball deforms and the ground compresses a tiny amount. The resistance of the ball to deformation and the resistance of the ground to compression decelerate the ball to the point at which its centre of mass is stationary, and thereafter accelerate the ball upwards.

The 'dilemma' is the mistaken result of assuming that no force acts on the object at the point at which its motion is stopped.

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  • $\begingroup$ I think the dilemma is when considering a perfectly elastic collision in which neither the ball nor ground compresses $\endgroup$ – Mars Nov 11 at 6:24
  • $\begingroup$ @Mars You have to be very careful with side conditions that need to be absolute. There is no such thing as an incompressible material, just materials that are very hard to deform. But even that tiny deformity will push back, and accelerate your bouncing ball back up. Or if it absorbs the force of deformation, it's no longer an elastic collision. $\endgroup$ – Emil Bode Nov 12 at 17:20
  • $\begingroup$ @EmilBode That's why in reality, there's no dilemma. In the concept of a perfectly elastic collision, one would think that there would be a frame where the velocity is zero and there is no compression. If so, how does it then start to move again?? Sam's answer explains it-- in a perfectly elastic collision, there is actually no frame where you have zero velocity. No matter how small of a dt, before you will be moving downward, after you will be moving upwards $\endgroup$ – Mars Nov 13 at 0:30
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To explain this, I shall use the same example of a ball bouncing on the ground. In a perfectly ideal world, the ball will never be at rest throughout the bounce. There will be a time $t$ when the ball is going downwards. At time $t+dt$, the ball will be going upwards. This assumes that the coefficient of elasticity of the ball is exactly $1$ and the ball and ground are extremely rigid.

However this cannot happen in real life as there is no body is perfectly elastic. Also, the above case would also imply that the force applied by the ground on the ball would be ${\infty}$.

Practically speaking, the ball hits the ground and gets deformed. The velocity slowly decreases to $0$ as the kinetic energy gets used up in changing the shape. The ball will be at rest at a particular moment before bouncing up again.

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    $\begingroup$ This answer would be more accurate if you said, "...center of mass..." You can't really talk about the velocity of the whole ball while it is in contact with the ground because different parts of it will have different velocities at any given instant during that interval. $\endgroup$ – Solomon Slow Nov 9 at 20:49
  • $\begingroup$ The ball will be at rest at a particular moment... before decompressing and bouncing up again $\endgroup$ – Mars Nov 11 at 6:28
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It boils down to the direction of the force applied.

if the force works perfectly against the movement -><- then indeed the object comes to rest - but it would only remain at said rest, if the forces now are in equilibrium. Usually when throwing a ball upward - the gravity doesn't stop working magically ... so the net-force is still pointing downward thus overcome the short rest where apparent velocity is 0.

in all other cases the force you apply and the movement together distinguish the new movement path and speed along that path.

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