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I have the following question, I have no relation to the study of Physics in any way, but the question has been teasing me for some time. Please accept my poor physical terminology. Here we go...

When a solid ball is thrown in a given angle with sufficient force, that ball will move in a some path. The ball will move until the energy passed to it is no longer effective against other forces affecting its movement.

I have the following questions about this scenario:

  1. Where does the 'solid' ball stores the force that was exerted on it?
  2. In what form is that energy stored?
  3. We pass force to the ball when it is hit; a quantity with a direction; to the ball, so where is the initial direction stored when the ball is hit?

Edit: Any web reference readable by an armature is highly appreciated.

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    $\begingroup$ You have to realize that the 'solid ball' is not a computer program. It does not need to store the direction of the force or the energy supplied to it, it just uses them. In other words, they are stored in its properties e.g. movement. Storage is not a helpful concept here perhaps because the process is instantaneous. $\endgroup$
    – Hasan
    Commented Jan 11, 2015 at 1:39
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    $\begingroup$ @Hasan Just FYI there are no instantaneous processes in nature. $\endgroup$
    – Asphir Dom
    Commented Jan 11, 2015 at 1:40
  • $\begingroup$ Yeah, a process by definition is not instantaneous. I was trying to go for kind of an ideal system to help build intuition. $\endgroup$
    – Hasan
    Commented Jan 11, 2015 at 7:28

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  1. The ball does not 'store' the force that was exerted on it. It stores kinetic energy and momentum associated with its motion, but it could have arrived at the same state of motion through many different force/impulse profiles.

  2. The ball's energy is stored in the kinetic energy of its molecules, which on average have the same velocity as the ball. In terms of where the kinetic energy is stored in each molecule, that's a hard one to answer. It's not really stored anywhere, i.e. it's not like each molecule has a container where it stores the KE. Velocity and mass (and hence KE) are intrinsic properties of the particles/molecules that make up all matter. Beyond that, you'd be getting more into Philosophy than Physics and I don't think I can help you with that.

    Regarding how the kinetic energy is equally transferred between the molecules that make up the ball: all of the molecules in a solid are inter-connected by chemical bonds which essentially rely on electrostatic forces. If any molecules were moving faster than their neighboring ones, then the chemical bonds would impart forces on those neighboring molecules and they would accelerate (i.e. KE would be transferred) until the molecules were all moving at the same velocity. When the ball is struck initially, not all of the molecules are struck, only a subset on the surface. The kinetic energy is then transferred throughout the rest of the ball by the interaction of the chemical bonds.

  3. The direction of motion is stored in the velocity and momentum of the ball. The direction of motion will change as the ball moves, but only if a force/acceleration acts upon it. Velocity is a vector quantity, which means it has a value and a direction. The velocity is an intrinsic property of the ball (and all of its molecules), so the fact that it has a velocity property implies that it has a direction. Again, the question of where it is stored is probably more a matter of Philosophy.

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  • $\begingroup$ +1, Thx for your answer. It adds value. As per (#2), so all the molecules of the ball store the energy...I find this hard to imagine, because I don't know the mechanism by which the energy would be equally transferred and where in the molecule would it be stored! As per (#3) I find your phrase "the direction of the motion is stored in the velocity" to be strange to me as well. I guess I need to read more in the light of your answer. $\endgroup$
    – NoChance
    Commented Jan 11, 2015 at 19:20
  • $\begingroup$ I edited my answer to try to help clarify those points. $\endgroup$
    – Time4Tea
    Commented Jan 11, 2015 at 19:47
  • $\begingroup$ Marked as answer - Great explanation. Thanks again. $\endgroup$
    – NoChance
    Commented Jan 11, 2015 at 20:56

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