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Suppose there is a perfectly elastic collision between 2 objects and kinetic energy of the system is completely conserved. Disregard the specifics of masses and velocities for the time being because all we need to really consider is the fact that for kinetic energy to be conserved in the collision there must be a transfer of energy from one object to the other, so all that is occurring is a redistribution of existing kinetic energy in the system under consideration.

This seems to imply that the work done by one object is equal to the negative work done by the second object. We know that the force generated by the collision is the same on both objects due to Newton's 3rd Law. We also know that the force interaction occurs for a singular duration and so the time of contact is the same for both objects. This assures conservation of momentum.

So, if the force is the same on both objects and the energy lost by one object is the energy gained by another object, does this mean that for perfectly elastic collisions the displacement occurring during contact is always the same for both objects? W = F*d = change in kinetic energy?

This seems somewhat confusing to me because if you have 2 masses where one is very small and the other is very large and you apply the same force to both for the same amount of time, the smaller mass will go through a much larger acceleration, covering a much larger distance in the same amount of time and therefore would have more work imparted to it, causing a greater change in kinetic energy. This is not the case however in an elastic collision.

Assume now that a very massive object is moving at speed toward a stationary object of small mass. They collide perfectly elastically and kinetic energy is conserved. It seems to me that because the moving object already has a starting velocity when it collides with the stationary one that as the stationary object of much lesser mass undergoes a very large acceleration it will end up covering the same displacement that the massive object already moving will cover as it slows down via much slower deceleration. Would this be what is actually happening? At some point they must be moving at equal velocities while still in contact and so the contact interface would be moving as well over some displacement.

I've seen a few authors or online contributors often refer to thinking of collisions in terms of colliding springs or colliding masses which have springs attached. On Physics LibreText, a popular open-source platform, a collision is discussed where 2 unequal masses are attached to a compressed spring and when the spring is allowed to decompress the smaller mass covers a larger distance due to larger acceleration in equal time with equal force. And so it is argued that the "collision" distributes the energy in different ways where the displacements are not the same depending on mass.

Is this a mistake? Or is this just a case that is different from an elastic collision?