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GR0177 #4
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Alternate Solutions |
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Comments |
archard
2010-04-11 14:54:59 | Another quick way is to remember that kinetic energy can be written as  , and since momentum is conserved, the final:initial kinetic energy ratio is }{2(m1+m2)}) =  . Which means the fractional kinetic energy loss is  |  | wittensdog
2009-10-08 16:29:07 | Here's something that's surprisingly useful. When a moving body hits a second one at rest, and they stick, the fractional loss of kinetic energy is,
m2 / ( m1 + m2)
If you just memorize that, all sorts of problems involving sticky collisions become way faster, even ones where you have falling masses on pendulums and so forth. | | herrphysik
2006-09-19 16:14:15 | Small typo: You're missing a 2 in the left side of the initial KE equation. |  |
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