A 14.6-g marble is dropped from rest onto the floor 1.23 m below. If the marble bounces straight upward to a height of 0.665 m, what is the magnitude of the impulse delivered to the marble by the floor.

Question

If the marble had bounced to a height of 0.884 m, what would the magnitude of the impulse delivered to the marble by the floor be?

Answer 1

m1v1 = (0.0146)(v1) (vertically downward is neg direction)
v1 = √2gh (where h = 1.23)
v1 = 4.91 m/s
m1v1 = (0.0146)(4.91) = - 0.0717 kg-m/s (Momentum Before Floor)

m2v2 = (0.0146)v2 (vertically upward)
v2 = √2gh (where h=.665)
v2 = 3.61 m/s
m2v2 = (0.0146)(3.61) = 0.0527 kg-m/s (Momentum After Floor)

Change in Momentum is:
m2v2 - m1v1 = 0.0527 - (- 0.0717) = 0.124 kg-m/s (upward direction)
Impulse = 0.124 N s (upward direction)

Do the same thing with .884 instead of .665 for the second question.