An object is attached to a vertical spring and slowly lowered to its equilibrium position. this stretches the spring by an amount d. if the same object is attached to the same vertical spring but permitted to fall instead, through what maximum distance does it stretch the spring?

An alpha particle with kinetic energy 12.5 mev makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude l = p0 b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10−12 m. (assume that the lead nucleus remains stationary and that it may be treated as a point charge. the atomic number of lead is 82. the alpha particle is a helium nucleus, with atomic number 2.) what is the distance of closest approach?