Asked by Timileyin
A ball of a mass 100g falls from a height of 3m unto a horizontal surface and bounce to the height of 12m calculate the change in the momentum of the ball when it strikes the surface.
Answers
Answered by
Anonymous
Do you mean bounce to 1.2 meter? I hope.
(1/2) m v^2 at floor = m g h at top
so
v = sqrt (2 g h) (but you knew that)
so
v down = sqrt (2*9.8*3)
v up = sqrt(2*9.8*1.2)
opposite signs so add
change in momentum = m * [ sqrt (2*9.8*3) + sqrt(2*9.8*1.2) ]
(1/2) m v^2 at floor = m g h at top
so
v = sqrt (2 g h) (but you knew that)
so
v down = sqrt (2*9.8*3)
v up = sqrt(2*9.8*1.2)
opposite signs so add
change in momentum = m * [ sqrt (2*9.8*3) + sqrt(2*9.8*1.2) ]
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