A glass marble is dropped down an elevator shaft and hits a thick glass plate on...
A glass marble is dropped down an elevator shaft and hits a thick glass plate on top of an elevator that is descending at a speed of 3.00 m/s. The marble hits the glass plate 8.0 m below the point from which it was dropped. If the collision is elastic, how high will the marble rise, relative to the point from which it was dropped?
Homework Answers
Accordin to law of conservation of energy
KE _1 = m g h1 + ( 1/ 2) m v ^2
m g h2 = m gh1 + ( 1/ 2) m v ^2
h2 = ( 9.8 m/s^2) (8.0 m ) + ( ( 3.0 m/s )^2 / 2 ) / ( 9.8 m/s^2)
=8.45 m
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