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

Since the collision is elastic, I thought total kinetic energy of the system is 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?


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).

enter image description here

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 "hat 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.

enter image description here

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.


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).

  • 1
    $\begingroup$ This sounds right, the key words being "throughout the duration". But I agree it's dangerous to speculate. $\endgroup$ – garyp Dec 7 '16 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.


The issue is the word throughout, no doubt.

When one says the usual statement: "kinetic energy is conserved in elastic collisions", then one 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.

The word "throughout" is therefore the reason. Kinetic energy is not conserved throughout the collision only afterwards.


protected by Qmechanic Dec 7 '16 at 7:50

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