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

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

you cant move and you feel like someones pushing you down onto your bed it can feel like you want to get up but you just can't it can be extremely scary

what is the question because that is not a question.