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

`F_(a)` = 0.0009866 N

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)` =
`(1.776)/(1800)`

`F_(a)` = 0.0009866 N