I'm sorry for this, as I know there are already many questions on here concerning inelastic collisions already, but I believe this one is actually slightly different. Although this has indeed the best response so far! Just wanted to say that.

I (hope at least) understand why kinetic energy is not conserved in inelastic collisions, that is not my issue here. I also believe I understand why conservation of momentum is true in both cases. I am confused about the following though:

Conservation of energy doesn't work, because some of the energy gets converted to deformation / binding energy. Conservation of momentum works, because momentum is, of course, conserved, while some of the energy goes to deformation and binding. Now, this energy which is "lost" to binding and deformation, doesn't that depend on the material and the sort of binding happening? For example, if a bullet lodges into a block of wood, woodn't (see what I did there?) that take more deformation energy than 2 sticky objects sticking together after the one hits the other or if the bullet came twice as far into the block because it was of a softer material?

So how how does the $m_a*v_a=m_b*v_b$ equation know, how much energy was lost to accomodate for the different binding or deformation energies required for different materials?

The only thing I could think of, is to assume that the entire original kinetic energy remains as kinetic energy in the result system, which it of course does not.

So why is the amount of energy "lost" to other stuff constant for each final energy value? If a block of glue hits another block of glue, the deformation and binding energy is not the same as a bullet hitting a block of wood at the same speed, is it?

(Yes, I know energy is not lost, thus the double quotes)

EDIT: Example: Say you have a bullet waying $m_a = 0.5kg$ hit a block of wood waying $m_b = 5kg$. After the collision, the bullet is stuck into the block and the total thing has a $E_{kin}=10J$. Now find the bullet's speed. The solution should be this:
$m_a * v_a = m_b * v_b | / m_a$
$v_a = \frac{m_b * v_b}{m_a}$
so now you do
$v_b = \sqrt{\frac{E_{kin}}{0.5m_b}}=\sqrt{\frac{10J}{0.5 * 5kg}}=\sqrt{4\frac{m^2}{s^2}}=2\frac{m}{s}$
which means that
This calculation makes sense to me: The momentum is conserved and some energy is lost. But if I now calculate the kinetic energy of the bullet: $E_{kinbullet}=0.5m_av_a^2=0.5*0.5kg*(20\frac{m}{s})^2=100J$
This energy loss I also understand, the energy goes towards reshaping, heating and so on the block of wood.
What confuses me is, that the amount of energy which goes towards reshaping and heating and so on is the same for every material. If the block were made of rubber for example (not elastic rubber, the more sticky kind), it would also loose 90J of energy to the rubber. How is that?

Yes, I know this is asked quite commonly, but I had to. I couldn't find any good explanation unfortunately :(


2 Answers 2


First, you need to be careful in how you apply the conservation of momentum. In an inelastic collision (where both objects stick together) you need to look at the total momentum before the collision and the total momentum after the collision. This is not $m_a * v_a = m_b * v_b$ as you suggested above.

Rather, the correct equation is:$$m_a * v_i + 0 = (m_a + m_b) * v_f$$,

where $i$ represents the initial condition (prior to the collision) and $f$ is the final condition (after the collision).

Initially there are two particles, one moving and one stationary, so the momentum before the collision is all due to the moving object, in your example, the bullet with a momentum of $m_a * v_a$. But after the collision the two objects are stuck together so their combined mass is $m_a + m_b$. For your example above this would result in a speed after the collision of $1.91 m/s$ and in initial speed of the bullet of $20.98 m/s$.

The reason this is close to your calculation of $2 m/s$ and $20 m/s$ is because the mass of your two objects are very different $0.5kg$ vs. $5.0kg$. If they had been more equal to each other your calculation would have shown greater error.

Next, to your core question. You make the assumption that every interaction between two objects with the velocity and mass that you have assigned to the bullet and the block of wood will end the same way. And, then you ask the question; How can that be? That is, per you example, that $90J$ (Note: this is actually closer to $100J$ when using the correct equation for momentum) of energy will be "lost" regardless of the composition of the colliding objects (a bullet vs. a rubbery object).

This is where you make a bit of a conceptual error, because not all inelastic collisions result in the two objects sticking to each other. For example, in a collision between two cars a significant amount of damage can be done to the cars (making it an inelastic collision) but each of the cars can still have their own unique momentum (and kinetic energy) after the collision. And, this could result in any number of different situations with different energy loses.

If, however, you are only concerned with collisions that result in the "binding together" of the two objects, then yes, the energy loss will be the same regardless of the composition of the objects involved.

How can this be? Because in your example of a bullet driven deeply into a block of wood vs. two objects with a rubber surface you need to take into account the surface area over which the force is applied. So, yes, the bullet and the sticky rubbery surfaces would both result in the same "loss" of energy if the two objects truly stick together after the collision.

But, as stated above, you cannot easily know in advance if the two rubbery objects will actually stick together (given the initial conditions of your experiment) or if they will collide, loose some energy in the collision, but then separate from each other.

In summary, you cannot be assured in advance, that the initial conditions of your collision will result in an inelastic collision resulting in two objects binding together into one. Therefore, the notion that somehow the momentum must know in advance how much kinetic energy will be lost in order to account for the proper momentum is an inaccurate assessment of how collisions occur. The only guiding principle is that the momentum before the collision will equal the momentum after the collision. If kinetic energy is also conserved, then it is an elastic collision. If kinetic energy is not conserved, then it is an inelastic collision. But, there are lots of ways in which a collision can be inelastic and these can all result in different energy loses.

  • $\begingroup$ Thanks for this answer! I think I understand it now. So just to reiterate (for a totally inelastic collision): The momentum is always conserved, in any collision. The energy that remains then, is used for deformation and such. This is always the same, independent of the material or the type of binding. Is that correct? It stills seems a bit fishy to me that the amount of energy "lost" is always the same - the area could be bigger or smaller... but it just doesn't really make sense to me... guess I'll have to live with it though :( $\endgroup$
    – Robbe
    Jul 7, 2021 at 13:48

Lets start from the beginning: Physics theories are mathematical models that fit existing measurements and observation and, very important predict correctly new measurements and observations.

In order to pick from the plethora of mathematical solutions those that can be used to consistently fit the data extra axiomatic statements are used, as strong as the mathematical axioms. These are called laws, postulates,principles and are always true.

The law of conservation of energy is one of these , and together with the conservation of momentum and angular momentum are mainstays of all branches of physics.

The progress in experimental physics and the need to keep energy conservation as a law has led to special relativity and the algebra of the four vectors, necessary to understand all atomic, nuclear and elementary particle phenomena. Our present day cosmology has defined dark energy and dark matter in order to keep the conservation of energy law true, which means that maybe for general relativity , large masses and energy a new look might be needed on the energy conservation law, but your question is not at this level.

For classical kinematics, conservation of energy has to take into account all forms of energy in the summation,in the system under observation. There is no separate conservation of kinetic energy, potential energy,heat energy...

If talking of matter, thermodynamic forms of energy have to be considered in the summation. If the scattering is inelastic, part of the kinetic energy becomes potential energy (sping type deformations), or even heat (permanent deformation) and radiation energy, in the case of extended objects. All these depend on the constants of the material , see for specific heat for example.

Edit after your edit.

Your first equation does not follow the momentum conservation law, because you divide by the input mass. The units are also wrong, left and right side. Conservation of momentum between two interacting masses is simply $m_a*v_a=m_b*v_b$ $v$s vectors, $p_a$ before the interaction $p_b$ after. The $p$ vectors.

  • $\begingroup$ Yes, I do understand that. But where does the amount of energy that goes to deformations or bindings appear in the solution equations? I don't understand how the binding or deformation energy could be the same for every different type of material. $\endgroup$
    – Robbe
    Jul 6, 2021 at 19:29
  • $\begingroup$ It is not the same, every material has its values , and there are tables to be used in engineering calculations. there will be particular equations for particular cases. In the first link see the fourth paragraph. $\endgroup$
    – anna v
    Jul 6, 2021 at 19:40
  • $\begingroup$ So why can one use the conservation of momentum solution for inelastic collision, when the amount of energy lost is always different? If you have the mass of both objects given, and are meant to calculate the speed of the first object, and with that its kinetic energy, how does that work? You don't know how much of the energy is lost when not looking at the material, yet the solution requires nothing in that direction. $\endgroup$
    – Robbe
    Jul 6, 2021 at 19:45
  • $\begingroup$ Only in special cases before and after momentum can be measured. Hit a wall with a pillow, what is the conservation of momentum? If it is a ball, elastic, one can say the mass of the wall is so large that the reflected ball carries the equal and opposite momentum and the wall takes up (not able to measure it) what is needed for conservation of momentum. $\endgroup$
    – anna v
    Jul 6, 2021 at 19:50
  • $\begingroup$ Now in deformation momentum is taken away by motion of atoms and molecules and eventually becomes heat radiated which radiation does carry momentum . $\endgroup$
    – anna v
    Jul 6, 2021 at 19:59

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