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3 years ago

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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

1 answer:

fgiga [73]3 years ago

4 0

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

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