Asked by Angela
A superball of mass 0.15 kg drops 2.3 m to the floor and rebounds perfectly. It is in contact with the floor for 0.012 s. What is the average force during the impact?
Answers
Answered by
Damon
average force = change in momentum / time
change in momentum = 2 m v because it goes from v down to v up
To get v:
(1/2) m v^2 = m g h
so
v^2 = 2 g h
change in momentum = 2 m v because it goes from v down to v up
To get v:
(1/2) m v^2 = m g h
so
v^2 = 2 g h
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