Answer:
Explanation:
A 200-g ball falls vertically downward, hitting the floor with a speed of 2.5 m/s and rebounding upward with a speed of 2.0 m/s. a. Determine the change in momentum of the ball.. b. If the ball is in contact with the floor for 0.02 ms (milliseconds), what is the average force applied to the ball
Given data
mass= 200g= 0.2kg
initial velocity= 2.5m/s
final velocity= 2m/s
time= 0.02ms
time= 0.00002 seconds
ΔP= mΔv
ΔP= 0.2*2.5-2
ΔP= 0.2*0.5
ΔP=0.1kgm/s
F= mv/t
F=0.1/0.00002
F=5000N
