Re: Today's jaw dropper

["Daniel L. MacIsaac" <Dan.MacIsaac@nau.edu>]

> Here's a gem from the "How to mess with student's minds" file:
>
> Chapter 4, Problem 19 in Serway's, 4th Edition asks students to calculate
> the muzzle velocity and time of flight for a bullet which is fired
> horizontally by an astronaut standing on the Moon and which is then to
> "travel completely around the Moon and reach the astronaut at a point 10.0
> cm below its initial height."
>
> Making the assumptions that this problem seems to call for, I get
>
> time of flight = 0.35 s
> muzzle velocity = 0.10 c
>
>  A. John Mallinckrodt      http://www.intranet.csupomona.edu/~ajm

I got one for you from Wilson's COLLEGE PHYSICS 3rd Ed, Q15.64.  The 
problem states that an electron is moved tthrough a potention of 10^8 V
in a lightning discharge.  Ignoring collisions, what KE does it gain?  
The supposed e- gains 10^8 eV of KE.  

Now, my reasoning is that an e- masses about 5x10^5 eV.  Guess how fast an 
electron with a KE of 10^8 eV is going.