Final answer:
In the head-on elastic collision between two identical iron nuclei, 100% of the kinetic energy is transferred from the moving nucleus to the stationary one.
Step-by-step explanation:
The question concerns a head-on elastic collision between two iron nuclei (56Fe), where we are asked to determine the percent of kinetic energy transferred from a moving nucleus to a stationary one. In an elastic collision, both momentum and kinetic energy are conserved. For identical masses, the moving object transfers all of its kinetic energy to the stationary object in a head-on collision. Thus, 100% of the kinetic energy is transferred to the previously stationary iron nucleus.
However, the question provides additional context and examples that are not directly related to elastic collisions between identical masses, like a billiard ball hitting the bumper of a pool table or skaters catching each other. In those cases, the collisions can be elastic or inelastic and involve objects with different masses, which affects the kinetic energy transfer differently.
For the iron nucleus collision, since the question stipulates that the iron nuclei are identical and that the collision is elastic and one-dimensional, we use the conservation of kinetic energy principle. We can represent the initial kinetic energy of the moving nucleus as KEi and acknowledge that this energy will entirely be the kinetic energy KEf of the previously stationary nucleus after the collision. When equal masses collide head-on elastically, they simply exchange velocities, leading to an energy transfer of 100%.