Question

Two balls each of mass, m = 0.2kg, are dropped from the same height, h = 2 m, above the ground. Ball A contacts the floor and stops. Ball B bounces up to a height of 1.5 m above the ground. What is the impulse exerted by the floor on ball B?

Answer 1

here is how i would answer this. let me know if you think I'm right.

Answer 2

yes that is correct

Answer 3

impulse=change in momentum
velocity of ball just bfore hitting the ground=v1=-\[\sqrt{2*a*s}\]=-\[\sqrt{2*9.8*2}\]=-\[\sqrt{39.2}\]
velocity just bfore ball starts moving up=v2=\[\sqrt{2*a*s}\]=\[\sqrt{2*9.8*1.5}\]=\[\sqrt{29.4}\]
so change in momentum=m(v2-v1)=0.2*(\[\sqrt{29.4}\]+\[\sqrt{39.2}\])

Answer 4

Awesome I'm glad you found my answer helpful.