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A child bounces a 48 g superball on the sidewalk. The velocity change of the super bowl is from 23 m/s downward to 14 m/s upward. If the contact time with the sidewalk is 1800 s, what is the magnitude of the average force exerted on the superball by the sidewalk?

Respuesta :

Answer:

[tex]F_{a}[/tex] = 0.0009866 N

Step-by-step explanation:

For resolving this problem we can use the impulse-momentum theorem; this theorem establishes that  the change in momentum of an object equals the impulse applied to it

The impulse is defined how:

I = [tex]F_{a}[/tex]Δt

    [tex]F_{a}[/tex]:   Average force

    Δt =  Time interval

Then:

[tex]mv_{f}-mv{o} = F{a}[/tex]Δt

(0.048)(14) - (0.048)(-23) = [tex]F_{a}[/tex](1800)

0.672 + 1.104 = [tex]F_{a}[/tex]1800

1.776 = [tex]F_{a}[/tex]1800

[tex]F_{a}[/tex] = [tex]\frac{1.776}{1800}[/tex]

[tex]F_{a}[/tex] = 0.0009866 N

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