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November 22[edit]

Static forces versus forces in motion[edit]

I have a simple thought experiment here. A one gram elastic ball strikes a solid wall with 1 kg of force and unsurprisingly rebounds symmetrically with respect to the angle of trajectory. But now suppose instead that the ball is simply pressed against the wall with 1 kg of force at the same angle as before then quickly released. Does the ball more or less rebound in an identical fashion in both such cases? Earl of Arundel (talk) 20:09, 22 November 2021 (UTC)[reply]

As thought experiments go this is a bit over the place. kg is a unit of mass not force. A force of 1kgf will produce an acceleration of 1000 m/s/s on the ball, so you are talking about a very dynamic event. The force will vary with time, and collisions or impacts are notoriously hard to analyse in the real world. However, in a perfectly elastic solution, if the peak force was 1 kgf, then there is no intrinsic difference between squishing the ball slowly, and then releasing it instantaneously, compared with bouncing the ball with the same peak force. In the real world of course there are many reasons why the subsequent rebound speed would differ, but you knew that. Another point of view, which may confuse more than it helps, suppose we took a snapshot at the point of maximum force on the wall, after the pressing force is released. It has a location, a velocity of zero, and an acceleration of -1000 m/s/s in both cases, there is no measurement that could be made that would be able to determine what the previous path of the ball had been. Greglocock (talk) 21:28, 22 November 2021 (UTC)[reply]

Yes I was most definitely being just a bit too loose with the terminology there. That is indeed a very important distinction to make, to wit, mass != force! Earl of Arundel (talk) 16:41, 23 November 2021 (UTC)[reply]

Perfectly elastic macroscopic bodies exist only in thought experiments as the limit case of their elasticity tending to infinity. The kinetic energy of the striking ball is returned in the kinetic energy of the ball rebounding, but at the moment of symmetry it is stored as potential energy, which is only possible if the ball (or wall) are somewhat deformable. Is the deformation caused by pressing the ball against the wall the same as that at the moment of symmetry? If the ball is at rest while being "simply pressed", the reactive force is in equilibrium with the pressing force, which means it points in the opposite direction, and the ball will initially, upon release, accelerate in that direction. So if the wall is vertical and the ball was pressed down, say at an angle of −45° with the horizontal, it will now go up at an angle of +45°. That is not the angle of the rebounding ball; if moving down coming in, it keeps going down on the rebound. --Lambiam 07:45, 23 November 2021 (UTC)[reply]

I see. So the depressed ball is in equalibrium, the compressive forces contained therein being basically equal and opposite to eachother, whereas in the case of an elastic ball in motion an additional non-zero vector component of velocity causes the ball to reflect symmetrically with respect to the surface normal upon impact. Well that makes sense. Thanks! Earl of Arundel (talk) 16:41, 23 November 2021 (UTC)[reply]