posted by Yasmina on .
A rubber ball, with a mass of 40.0 grams is dropped from rest from a height of 1.60 m above the floor. It hits the floor, and then reaches a maximum height of 90.0 cm when it comes back up again. In this problem, use g = 10.0 m/s2.
a) The collision with the floor causes some mechanical energy to be lost (this energy generally ends up as thermal energy). How much mechanical energy is lost in this case?
b) What is the ball's speed, just as it leaves the floor on its way up?
c) What is the magnitude of the impulse experienced by the ball for the entire time it is in contact with the floor?
a. V^2 = Vo^2 + 2g*h
V^2 = 0 + 20*1.6 = 32
V = 5.66 m/s. = Final velocity before striking the gnd.
Vo^2 = V^2 - 2g*h
Vo^2 = 0 - 20*0.9 = 18
Vo = 4.24 m/s = Velocity at which it leaves the gnd.
Energy Lost=0.5m*V^2-0.5m*Vo^2 =
0.02*5.66^2 - 0.02*4.24^2=0.281 Joules
b. Vo = 4.24 m/s.
c. Impulse = m*V = 0.04 * 5.66 = 0,2264