Question

A proton is fired from far away toward the nucleus of a mercuryatom. Mercury is element number 80, and the diameter of the nucleusis 14.0 fm. If the proton is fired at a speed of 3.4x10^7, what is its closest approach to the surface of the nucleus?

a) Assume the nucleus remains at rest.
The unit is (fm)

Answer 1

Actually, the equation that's required is the work-energy theorem,

W_ext + W_nc = ?K + ?U

where W_ext is the work done by external forces on a system, W_nc is the work done by non-conservative forces within a system, ?K is the change in kinetic energy of objects within a system, and ?U is the change in potential energy due to conservative forces within a system.

Eventually, of course, you'll end up with your formula,

1/2mv^2 = kq1q2/r

but if you start with your formula, you're going to find the physics consists of loads of disconnected formulas to memorize. If you start with the work-energy theorem, you'll find that physics consists of a few fundamental principles, that can be expressed mathematically by formulas you don't need to memorize because you'll find yourself using the same ones over and over until they become second nature.

With that out of the way, remember that "r" isn't the answer. It's the distance of the proton from the center of the mercury nucleus, but the question asks for the distance from the surface.

(Note that 1/2mv^2 = kq1q2/r assumes the mercury nucleus is fixed in place, which is a reasonable approximation, but which isn't generally true.)