Redo the analysis of the Rutherford experiment, this time using the concept of the center-of-momentum reference frame.
Let m = the mass of the alpha particle and M = the mass of the gold nucleus. Consider the specific case of the alpha particle
rebounding straight back. The incoming alpha particle has a momentum p1, the outgoing alpha particle has a momentum p3,
and the gold nucleus picks up a momentum p4.
(a) Determine the velocity of the center of momentum of the system.
(b) Transform the initial momenta to that frame (by subtracting the center-of-momentum velocity from the original
velocities).
(c) Show that if the momenta in the center-of-momentum frame simply turn around (180°), with no change in their
magnitudes, both momentum and energy conservation are satisfied, whereas no other possibility satisfies both
conservation principles. (Try drawing some other momentum diagrams.)
(d) After the collision, transform back to the original reference frame (by adding the center-of-momentum velocity to
the velocities of the particles in the center-of-mass frame). Although using the center-of-momentum frame may be
conceptually more difficult, the algebra for solving for the final speeds is much simpler.