Question

g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 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 1

The average force exerted on the superball by the sidewalk is 0.00122 N.

Step-by-step explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)

final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)

time of motion, t = 1800 s

The average force exerted on the superball by the sidewalk is given by;

`F = ma\\\\F = (m(v-u))/(t) \\\\F = (0.05(17-(-27)))/(1800)\\\\ F = (0.05(44))/(1800)\\\\F = 0.00122 \ N`

Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.