Question

A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 24 m/s downward to 11 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 in units of N.

Answer 1

Correct question:

A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 24 m/s downward to 11 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 in units of N.

Answer:

The magnitude of the average force exerted on the superball by the sidewalk is 592.8 N

Explanation:

Given;

mass of the superball, m = 57 g = 0.057 kg

initial velocity of the superball, u = 24 m/s

final velocity of the superball, v = 11  m/s

contact time with the sidewalk, t =  1 / 800 s

To determine the magnitude of the average force exerted on the superball by the sidewalk, we apply Newton's second law of motion;

F = ma

$But, a = \frac{v-u}{t} \\\\Thus, F = m(\frac{v-u}{t} )\\\\F = 0.057(\frac{11-24}{1/800} )\\\\F = 0.057(\frac{-13}{1/800})\\\\F = -0.057(\frac{800*13}{1})\\\\ F = -592.8 \ N\\\\Magnitude \ of \ the \ force \ is \ 592.8 \ N$

Therefore, the magnitude of the average force exerted on the superball by the sidewalk is 592.8 N