complete question:
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1776 N
Explanation:
mass of ball = 60 g = 0.06 kg
velocity of downward direction = 22 m/s = v1
velocity of upward direction = 15 m/s = v2
Δt = 1/800 = 0.00125 s
Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.
p = mv
When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .
I = pf − pi = ∆p
F = ∆p/∆t = I/∆t
let the upward velocity be the positive
Δp = mv2 - m(-v1)
Δp = mv2 - m(-v1)
Δp = m (v2 + v1)
Δp = 0.06( 15 + 22)
Δp = 0.06(37)
Δp = 2.22 kg m/s
∆t = 0.00125
F = ∆p/∆t
F = 2.22/0.00125
F = 1776 N
