Question

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?

Answer 1

Answer:

$F_{a}$ = 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 = $F_{a}$Δt

    $F_{a}$:   Average force

    Δt =  Time interval

Then:

$mv_{f}-mv{o} = F{a}$Δt

(0.048)(14) - (0.048)(-23) = $F_{a}$(1800)

0.672 + 1.104 = $F_{a}$1800

1.776 = $F_{a}$1800

$F_{a}$ = $\frac{1.776}{1800}$

$F_{a}$ = 0.0009866 N