A child bounces a 54 g superball on the side-
walk. The velocity change of the superball is
from 23 m/s downward to 19 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?
fal 5mod 3= 150
To find the magnitude of the average force exerted on the superball by the sidewalk, we need to use the impulse-momentum principle. This principle states that the change in momentum of an object is equal to the impulse applied to it.
Impulse (J) can be calculated by multiplying the force (F) applied to an object by the time (Δt) the force is applied:
J = F * Δt
Momentum (p) is the product of an object's mass (m) and velocity (v):
p = m * v
Using the given values:
m = 54 g = 0.054 kg (convert grams to kilograms)
v1 = 23 m/s (initial velocity)
v2 = 19 m/s (final velocity)
Δt = 1/800 s = 0.00125 s
First, let's find the change in momentum (Δp) of the superball:
Δp = p2 - p1
Δp = m * v2 - m * v1
Δp = m * (v2 - v1)
Now, let's calculate the average force (F) using the impulse-momentum principle:
J = F * Δt
Δp = F * Δt
Substituting the values:
Δp = 0.054 kg * (19 m/s - (-23 m/s))
Δp = 0.054 kg * (19 m/s + 23 m/s)
Δp = 0.054 kg * 42 m/s
Δp = 2.268 kg·m/s
Now, we have the equation:
2.268 kg·m/s = F * 0.00125 s
To find F, divide both sides of the equation by 0.00125 s:
F = 2.268 kg·m/s / 0.00125 s
F ≈ 1814.4 N
Therefore, the magnitude of the average force exerted on the superball by the sidewalk is approximately 1814.4 Newtons.