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

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