A common example question we get in exams involves something like the following: Calculate the velocity of a block with mass X$kg$, when it is struck by a bullet of mass Y$kg$ travelling at Z$m/s$. The block is initially at rest and the bullet embeds into the block during the collision.

To solve this we usually find the momentum of the bullet $(Pkg/m/s = Ykg * Zm/s)$ and then find the velocity of the block/bullet combination by reversing it: $(Am/s = Pkg/m/s / (X+Y)kg)$.

We will then be asked to calculate if the collision is elastic or inelastic, which we will do by calculating the mechanical energy of the bullet before, and the mechanical energy of the bullet/block combination after the collision.

My question is how is it possible to have a different energy before and after? We frequently end up with different energies before and after, however nowhere in our equations do we take into account loss of energy in the form of friction, sound, heat etc.

TL;DR: How do we get different before/after energies in a collision without taking into account external energy loss? Is the equation wrong?

  • $\begingroup$ I've noticed that many texts discuss conservation of mechanical energy, and then mumble some words about energy conversion. Some texts explicitly formulate a more global conservation law. The difficulty with the former is that it causes confusion like yours. The difficulty with the latter is that it's practically impossible to enumerate all of the energy conversion channels. Still, I lean toward favoring the latter. $\endgroup$ – garyp Aug 4 '16 at 12:57

First of all, total energy (of the block + bullet + surroundings) is conserved (where by 'surroundings', I mean anything that isn't the block or bullet).

However, as you've rightly said

We frequently end up with different energies before and after

These 'before' and 'after' energies are the kinetic energies of the block + bullet.

Having different 'before' and 'after' energies defines an inelastic collision. Having the 'before' energy = the 'after' energy defines an elastic collision.

In the case of an inelastic collision, the difference between the before and after energies goes into the surroundings in forms such as heat and sound (friction creates heat) or into the block/bullet in forms other than kinetic energy (again e.g. heat).

So, while total energy of the entire system (block + bullet + surroundings) is conserved, the difference between the before and after (kinetic) energy of the block + bullet goes into things like heat and sound. In other words, in an elastic collision, no energy is 'lost' as heat or sound, while in an inelastic collision, that is exactly where the difference in energy of the block + bullet goes.

  • $\begingroup$ However we never take any of those into account in our equation, nowhere do we calculate frictional energy loss, or heat or any such thing, essentially in our collision according to the equation, our system is entirely isolated, so how can we use it to validate elasticity or lack thereof in our collision? $\endgroup$ – SchoolJava101 Aug 5 '16 at 9:03
  • $\begingroup$ @SchoolJava101 No, we don't - total energy is conserved. This means that total energy before = total energy after. So initial KE (of block + bullet) - final KE (of block + bullet) gives the energy that has been output as everything else. This is what's known in this context as 'energy loss', although the energy has just gone into other forms (e.g. heat) - it hasn't actually disappeared. If this 'energy loss' is 0, then the collision is an elastic one by definition, otherwise it is an inelastic collision. We don't explicitly calculate all other energy because we've got the total energy $\endgroup$ – Mithrandir24601 Aug 5 '16 at 20:31

The energy loss is accounted for by the assumption that the bullet stays stuck in the block. This means that the bullet loses most of its kinetic energy. In most cases, this kinetic energy loss is not fully converted into kinetic energy loss - as the kinetic energy of the bullet before the collision is usually larger than the kinetic energy of bullet and block afterwards (which, as you have said, can be calculated by using momentum conservation). So we have something like $$E_{kin; after}(bullet + block) < E_{kin; before}(bullet) $$ To regain energy conservation (which of course still holds), we need to account for the energy "losses" (i.e., energy that is converted into forms of energy that are not mechanical), as you have said yourself: $$E_{kin; after}(bullet + block) + E_{dissipated} = E_{kin; before}(bullet) $$ The problem is that it is quite a lot harder to accurately account for the processes by which energy is lost - so calculating that ab initio would be quite a challenge. As we know that momentum conservation holds, we can just calculate the kinetic energy difference before and afterwards and as we further know that energy conservation holds, we know that any discrepancy must have been converted into other forms of energy. So we don't calculate this dissipated energy (usually), but it is implicitly there.


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