Quick Momentum Question

[man0005]

A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m s–1 and rebounds at 8.0 m s–1.
7 The time of contact between the ball and the floor during the bounce was 0.050 s.
a Calculate the average net force acting on the ball during its contact with the floor.
b Calculate the average force that the floor exerts on the ball.
c Calculate the average force that the ball exerts on the floor.

[Whatlol]

7 (a)

F = ma
F =( m x change in velocity) / t
F = (0.08 x -18) / 0.05              (taking negative to be up)
F = 28.8 N Up
To one sig figure, 29 N up



b).
     I = Ft
     F = I/t
     F = change in momentum / t
     F = (-0.08x8 -10x0.08) / 0.05
     F = 29 N Up

This tells you that the force exerted on the ball is 29N up

c). The force that the ball exerts on the floor will be equal in magnitude but opposite in direction to that in part b).
So 29N down

[man0005]

ah k, i got the same
except the answers said 29 for a then 30 for b and c so i wasnt sure
==

[Whatlol]

ah k, i got the same
except the answers said 29 for a then 30 for b and c so i wasnt sure
==

really? hmm i dont think it should be

[man0005]

yeah maybe they just rounded idk

[Shark 774]

It's 30 because:
Fg = mg = 0.78 N
F (total) = Fg + Fn
This gives Fn = 29 - (-0.78) = 29.78up