# Why isn't kinetic energy conserved during an elastic collision between two particles?

Particle $$X$$ has mass $$m$$ and particle $$Y$$ has mass $$3m$$. They travel at equal speeds $$v$$ but in the opposite direction along a smooth horizontal surface, and then collide head-on elastically.

Since the collision is elastic, I thought the total kinetic energy of the system would be conserved. However, the following statement is correct:

The total kinetic energy of the system consisting of the two particles is not conserved throughout the duration of the collision.

Why is this the case?

• The total energy is conserved=kinetic energy + potential energy + rest masses + heat + .... Dec 7, 2016 at 1:51
• But in this case I don't care about total energy. Instead, I am only interested in kinetic energy. Dec 7, 2016 at 2:01
• Are you sure it's elastic? Dec 7, 2016 at 2:18
• Dec 7, 2016 at 2:23
• The key here is the word "throughout" Dec 7, 2016 at 7:32

I think that what happens during the collision is perfectly illustrated by some frames of a golf ball hitting a rigidly fixed steel structure (wall). Frame $$1$$ shows the ball with kinetic energy about to hit the wall.
Frame $$2$$ shows some of the kinetic energy of the ball being converted into elastic potential energy in the ball (and the wall).
Frame $$3$$ shows the ball at rest with the ball storing lots of elastic potential energy and having no kinetic energy.
Frames $$4$$ and $$5$$ show the ball moving away with the elastic potential energy being converted back to the same amount of kinetic energy as the ball had before the collision.
So during the collision kinetic energy was not conserved.

You may ask "what has happened to the momentum of the ball?
If the ball is the system then you can answer "the wall has exerted an impulse on the ball".
If the ball and the wall is the system you can answer that the centre of mass of the ball and wall (and Earth) system still has the same momentum as the ball had initially but because the wall is so much more massive then the ball you do not observe the wall moving.

Well in fact I was economical with the truth to illustrate what would happen if it was an elastic collision.
This is what really happened. The ball rebounded but the collision was not elastic because the translational kinetic energy of the centre of mass of the ball before the collision was not the same as after the collision.
As a result of the collision the ball started to oscillate about its centre of mass.
So there was some kinetic energy due to the oscillation of the ball but there was also elastic potential energy associated with the oscillations.
One would guess that these were damped oscillations and so the energy associated with the oscillations which eventually be dissipated as heat.
The end result would be a hooter ball which was not oscillating and had less translational kinetic energy before the collision.
An inelastic collision.
If you have ever played or watched squash you will have seen the players warming the squash ball up by making the ball rebound from the walls before actually starting a game?

There are many copies of this video shot at 70,000 frames per second with the golf ball incident at $$150$$ mph ($$240$$ km/hr.).
Here is the one I used.

The issue is the word throughout.

The usual statement: "kinetic energy is conserved in elastic collisions", means before and after the collision, not during.

During elastic collision the kinetic energies convert into potential energies. It springs back to same amount of kinetic energy afterwards, but not during.

It is difficult to comment on an unsourced statement, but maybe the statement means that during the collision, part of the total energy is potential energy (say, Coulomb energy of charged particles).

• This sounds right, the key words being "throughout the duration". But I agree it's dangerous to speculate. Dec 7, 2016 at 2:24

Consider the simpler example of two equal masses, moving with equal speeds in opposite directions, colliding elastically. We know that the masses, after the collision will be moving with the same speed as they began with in directions opposite their initial velocity. So if one of the masses $m_1$ began with velocity $v_o$, it will end with velocity $-v_o$. But since the impulses to the masses are not delivered instantly, at some point in time while the masses are in contact, they both have velocity $0$ and thus zero kinetic energy. I believe this is what they're talking about is the short time while the particles are in contact that some (at one instant all) of their kinetic energy is converted to elastic or some other type of potential energy.

In the head on collision you describe there has to be a point of closest approach at which point the relative velocity of the two particles is zero - that is to say they have the same velocity. If you solve for this velocity using conservation of momentum and then work out the total kinetic energy of the particles you will find it to be less than before the collision. To understand why, think about the forces between the particles and they work they do on each other.

In the first half of the collision, when the particles are approaching each other, they are applying forces on each other that are in the opposite direction to their motion. That means each particle does negative work on the other and thus their kinetic energies are reduced. In the second half of the collision, when the particles are moving away from each other, the forces are acting in the same direction of the particles motion thus doing positive work on each other and increasing each other's kinetic energy.

One of the characteristics of an elastic collision is that the magnitude of the force acting on the particles is determined only by the distance between them i.e. not by their velocities. If this condition is met then you know that all the work done by the particles going into the collision will be returned as they come out of the collision. In an inelastic collision the force repelling the particles as the go into the collision with be greater than than it is when they are separating. The magnitude of the positive work done on them as they move apart is less than the negative work done as the moved together with the result being a net loss in kinetic energy.

• I also tried this and got a similar result: if you solve for this velocity using conservation of momentum and then work out the total kinetic energy of the particles you will find it to be less than before the collision. Why can we assume momentum is conserved but not also the energy? I feel like if there is a loss in the system, it should have an effect on the momentum too. Sep 28, 2021 at 6:14
• Total energy is conserved all all points throughout the collision. It's the kinetic energy that is reduced during the collision. While conservation of momentum is a physical law conservation of kinetic energy is not. Sep 28, 2021 at 18:36
• Ok, but don't you feel like a change in total kinetic energy after collision is weird. Sep 29, 2021 at 19:49

Consider a 1kg mass moving at 1 m/s that collides elastically with a very large stationary mass with a 0.5 kg mass behind it.

During the collision, all of the momentum of the 1kg mass is transferred to the 0.5 kg mass via the very large one, like a lop-sided Newton's cradle.

According to the law of conservation of momentum, the smaller mass will leave at 2m/s and will gain 0.5 × 0.5 × (2 × 2) = 1 Nm of KE but the 1kg mass only lost 0.5 × 1 × (1 × 1 )= 0.5 Nm of KE. So apparently through this elastic collision we have managed to double our energy.

Money spinning ideas are racing through my head.

Unfortunately I'm not going to be getting rich out of this, because while the swing in momentum is a physical property of the masses, Kinetic Energy is just a derivation of this that doesn't represent a physical change in the energy of the masses.

There appears to be some strange fundamental phenomenon here: Consider 2 perfectly elastic bodies masses Ma and Mb and initial velocitie Va0 and Vb0. If they collide with no energy losses to heat, sound etc, and if total Kinetic energy is conserved, then Ma.Va0^2+Mb.Vb0^2 = MaVa1^2+Mb.Vb1^2. Va and Vb are final velocities. For a simple case where Vb0 = 0 and Ma = Mb. If final velocites are equal, Va1 = Vb1 = Va0/sqrt(2) = 0.707Va0. For the same case and applying conservation of total momentum final velocity Va1 = Vb1 = 0.5Va0. Any explanations?

• I think my error was to assume that the final velocities are equal. This is in general not the case. There exist final velocities after imact for the 2 elastic bodies which correspond to both conservation of momentum and of kinetic energy. Jun 17 at 17:23