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When a ball hits a wall (ground), some quantity of its momentum must be transferred to the wall at rest. If else the ball would bounce back with the exact same speed as it did when it collided with the wall. This would result in the ball traveling up to the exact height from which it was dropped to the ground, which is not seen. Does this mean that the ground receives momentum but it gets converted into sound energy by the vibration of ground which had collided with the ball? Whose net result being that: the ball does not deform and the ground vibrates; the conservation momentum principle holds as momentum is transferred to the ground particles which vibrate to produce sound.

But, can someone explain the coefficient of restitution!

Edit: From what I understand now, to find the velocity of separation from the velocity of approach we must use an equation involving the coefficient of restitution by plugging in the values of coefficient of restitution and the relative velocity of approach!!

Am I right? Please let me know! This is the best possible explanation I could think of! If, I'm wrong may I know how you would say the kinetic energy of the ball gets converted into sound energy after the collision?

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    $\begingroup$ If you are trying to convert momentum into energy then something has gone wrong. $\endgroup$ – Aaron Stevens Jun 13 at 4:10
  • $\begingroup$ I was trying to say that the region of the ground that underwent the collision is the only place or location where the effect of the third law and hence conservation of momentum occurs. It cannot be true that the earth itself, and its center of mass make a translatory motion since earth is not a perfectly rigid body and plastic and elastic deformation are what usually predominantly occurs in such collision. The layers of mass on the region below the region where the collision occurred will resist the motion in various ways such as internal friction. Hope that clears up what I was trying to do. $\endgroup$ – mu-sic123 Jun 13 at 4:24
  • $\begingroup$ hence dissipation of the kinetic energy of the ground occurs which gets converted into other forms of energy as stated by the conservation of energy law. $\endgroup$ – mu-sic123 Jun 13 at 4:27
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When a ball hits a wall (ground), some quantity of its momentum must be transferred to the wall at rest.

That is correct.

If else the ball would bounce back with the exact same speed as it did when it collided with the wall. This would result in the ball traveling up to the exact height from which it was dropped to the ground, which is not seen.

That is not necessarily correct. Theoretically the ball could bounce back with exactly the same speed as it did when it collided with the wall or bounce back to the same height after impacting the ground. It would happen if the collision were perfectly elastic. The reason it is not seen is because no collisions are perfectly elastic. All collisions of macroscopic bodies are inelastic, some more inelastic than others. Some kinetic energy is always lost (converted to heat, sound, etc.).

Does this mean that the ground receives momentum but it gets converted into sound energy by the vibration of ground which had collided with the ball?

No.

First of all, it isn’t momentum that is converted into sound, heat or other energy forms. It is the kinetic energy lost in an inelastic collision. Momentum is always conserved.

I think you are assuming the velocity of the ground or the wall are zero after the collision. They are not. It’s just that the mass of the ground or wall, which is essentially that of the earth is so much greater than the ball that their velocities after the collision are too small to be observed.

Whose net result being that: the ball does not deform and the ground vibrates; the conservation momentum principle holds as momentum is transferred to the ground!

Conservation of momentum holds regardless of whether or not the ball deforms or the ground vibrates. Regarding deformation of the ball (or anything else) there basically two types, elastic and plastic. See answer to your question about coefficient of restitution below.

But, can someone explain the coefficient of restitution!

Basically, the coefficient of restitution is the ratio of the total initial kinetic energy of two colliding bodies after a collision to the total kinetic energy of the two colliding bodies before a collision. It is a number between zero for a completely inelastic collision (for example, something hits a wall and sticks to it) to 1 for a perfectly elastic collision. As I indicated above, all collisions at the macroscopic level are inelastic so the coefficient of restitution is always less than 1..

Taking the example of the ball hitting a wall and bouncing back. We know the kinetic energy of the wall is zero before the collision. Technically it is not zero after the collision because of conservation of momentum but as a practical matter it is considered zero when considering the coefficient of restitution of the ball.

The following responds to these follow up questions:

I was actually relating the conservation of momentum principle and the equations of kinematics! Is it better not to do that?

Kinematics is basically the analysis of motion in terms of displacement, time, velocity, and acceleration without regard to the forces involved that cause motion. It's OK to relate it to conservation of momentum since certainly velocity plays the primary role (in addition to mass) relating to conservation of momentum.

I'm also assuming the momenta of the wall before or after the collision are both zero... Where did I go wrong?

As I already indicated, assuming the momentum of the wall after the collision is zero is incorrect. Consider the following example.

A pitched baseball impacts a wall. It has a velocity of 40 m/s prior to impact (90 mph). The mass of the baseball is about 0.145 kg. Consider the wall to be an extension of the earth. The mass of the earth about 6 x $10^{24}$ kg. Let the collision of the ball be perfectly elastic, so its rebound velocity is the same. Since the ball reverses itself its change in momentum is -(0.145)(40)= -5.8 kg-m/s.

For conservation of momentum the change is momentum of the earth must be equal and opposite of the ball, or +5.8 kg-m/s. So the velocity of the earth following the collision is +5.8 kg-m/s divided by the mass of the earth, or V= 9.75 x $10^{-25}$ m/s.

Hope this helps.

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  • $\begingroup$ I was actually relating the conservation of momentum principle and the equations of kinematics! Is it better not to do that? I'm also assuming the momenta of the wall before or after the collision are both zero... Where did I go wrong? $\endgroup$ – mu-sic123 Jun 2 at 13:00
  • $\begingroup$ When you bounce a ball off a wall, the wall’s momentum changes. It’s mass is so big that you don’t see a velocity change. $\endgroup$ – Bob Jacobsen Jun 2 at 15:58
  • $\begingroup$ @mu-sic123 I have extensively revised my answer to respond to your comment, as well as to specifically address each question you had. Hope this helps clarify things for you. $\endgroup$ – Bob D Jun 2 at 21:25
  • $\begingroup$ So; to further clarify my doubts; Is it necessary that all particles of the massive body have to actually move, to result in a translatory motion of the body! "Let the collision ... be perfectly elastic" It's not so!, as far as I know; how else can the mechanism of conversion of kinetic into sound energy be explained!! Also, does plastic deformation have to take place for sound to be produced? I beg to differ! Drop a stone in a bucket of water to get the point I'm trying to argue! Anyway, let's part away with our differing opinions... Thanks for your contributions! $\endgroup$ – mu-sic123 Jun 4 at 16:43
  • $\begingroup$ @mu-sic123 When talking about macroscopic translational motion we are generally talking about the motion of center of mass of the system. When I said "let the collision be perfectly elastic" I did so to simplify the example. I already said there is no such thing as a perfectly elastic collision on a macroscopic scale and right at the beginning I included sound as a loss of kinetic energy. Where did I say "plastic deformation has to take place for sound to be produced"? I don't think we have different opinions, there's just a failure to communicate. But I agree let's leave it at that. $\endgroup$ – Bob D Jun 4 at 18:21
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An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction. _source: wilkipedia

In collisions of macroscopic bodies, some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect, and the bodies are deformed. The premises and observations:

Newton's third law holds without exception; Momentum is conserved without exception(Here the force on the ground is confined to the region where the collision occurred). But, here force (by the ball) causes a change in shape or even maybe a deformation in the body (as in a spring which gets compressed or stretched when a longitudinal force is applied on it) and so the kinetic energy will not be conserved.

We cannot observe any movement of the entire wall nor will the ball reach the original position; what is seen is the ball reaches a certain height lower (or distance less) than its original height due to a loss of kinetic energy from the ball (but, note that, the energy lost is not transferred to the wall as kinetic energy as it would have been if it were treated as if it were a single particle) and sound, a form of energy, is produced, arguably (with more than ample evidence), due to collision.

I am also just assuming, merely assuming that the earth is not a single particle... Woe on me if the earth is actually a single particle... Also I'm assuming the earth is not a perfectly rigid object, in which, if you would apply a force, it is necessary that the force changes its motion.

Therefore even though this collision causes deformation, the deformation is elastic in nature, similar to what occurs in a block-spring that is used to demonstrate oscillations (which is fundamental to the understanding of sound- production).

Please note:-
The explanation:

Some of the kinetic energy lost from the ball (obviously indicated by it reaching a lower height than from where it was originally dropped!) will change its form to other energy forms and thus get dissipated from the ball-wall system while the rest of the energy will be shared between the ball-wall system (mostly by the ball as we will see in moment). Meanwhile, most of the velocity will be imparted to the ball as it's much less massive as compared to the wall. Also, the wall (may) actually convert all the energy it receives during the collision into forms other than kinetic energy hence resulting in a lot of sound production due to vibrations of the part of the wall that is involved in the collision! I say so because it is very unlikely that all of the parts (section of layers) of mass contained in a comparatively much massive body such as a wall is affected due to a minute collision such as is the mentioned case. Hence the collision is only partially elastic and the momentum transferred to the wall does not cause in any observable movement as the momentum is transferred to a layer (section) of the mass of the wall undergoing collision and the momentum of its particles (atoms, say) is converted into vibratory motions of the layers of particles possessing mass and hence sound is produced and there is actually no movement of the wall and the collision is partially elastic... And to save the day momentum is conserved without exception at the particulate level through vibrations though no net translatory motion of the earth or wall occurs :D!!

Any error, small or large, in this very novel and original explanation by the author, is begged to be pardoned... :D!!

What a relief! But, please do bring any disputes into the comments!

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  • $\begingroup$ The following statement: "But, in this case, force (action by the ball) causes a change in shape or deformation in the body and so the kinetic energy will not be conserved". Is not necessarily true. Elastic deformation is fully recoverable and there is not loss of kinetic energy. Plastic deformation is not. $\endgroup$ – Bob D Jun 2 at 20:35
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    $\begingroup$ Sure, but the amount of kinetic energy converted into sound energy under ordinary conditions (explosions an exception) is generally so small compared to other forms of energy loss it can generally be ignored. $\endgroup$ – Bob D Jun 4 at 16:53
  • $\begingroup$ Where did I say that the deformation has to be plastic? I had only stated that the collision was partially inelastic and that the ground or earth does not necessarily have kinetic energy or momentum transferred to its center of mass causing its translatory motion!!!... @Bob D $\endgroup$ – mu-sic123 Jun 13 at 4:44

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