After falling from rest to a height of 30m, a 0.50kg ball rebounds upward, reaching a height of 20m. If the contact between ball and ground lasted 2.0m/s. What average force was exerted on the ball?
Mass of the ball, m = 0.5 kg
Initial height = 30 m
Final height = 20 m
Contact time, t = 2 ms = 2 x 10^-3 s
Velocity during descent, V1 = `sqrt(2gh) = sqrt(2xx9.81xx30) = 24.3 m/s`
Velocity during ascent, V2 = `sqrt(2gh) = sqrt(2xx9.81xx20) = 19.8 m/s`
Momentum change during the impact = m(V1-V2) = 0.5 (24.3-(-19.8)) kg m/s
= 0.5 x 44.1 kg m/s = 22.05 kg m/s
(note that downward velocity is assumed positive and upward as negative and hence the negative sign)
Now, force is rate of change of momentum.
Thus, force exerted by the ball = momentum change/impact duration
= 22.05/(2x10^-3) N = 11025 N.
Hope this helps