Hi,
Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:
A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s
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.
thanks
I = Δp = F(average) Δt = m Δv
Always assign a positive direction! Let Up be positive.
a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s
F(average) x 0.050 = 0.080 x 18.0
Therefore F (average) = 1.44 / 0.050 = 28.8 N
Therefore average net force is 28.8 N in the upwards direction.
b) F (av) (floor on ball) - W (weight force) = F (av) (net)
F (av) (floor on ball) = N (normal force)
Therefore, N - mg = 28.8 (from part a)
N = 28.8 + (0.080 x 10) = 28.8 + 0.8 = 29.6 N
Therefore F (av) (floor on ball) = 29.6 N in the upwards direction.
c) F (av) (floor on ball) and F (av) (ball on floor) are a Newton's 3rd Law action reaction pair.
Hence F (av) (ball on floor) = - 29.6 N
Therefore F (av) (ball on floor) = 29.6 N in the downwards direction.
This should be correct! Hope it helps!