A child bounces a 48 g superball on the sidewalk. the velocity change of the superball is from 26 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 in units of n.

By definition we have the momentum is:

P = m * v

Where,

m = mass

v = speed

Before the impact:

P1 = (0.048) * (26) = 1.248 kg * m / s

After the impact:

P2 = (0.048) * ( - 17) = - 0.816 Kg * m / s.

Then we have that deltaP is:

deltaP = P2-P1

deltaP = ( - 0.816) - (1,248)

deltaP = - 2,064 kg * m / s.

Then, by definition:

deltaP = F * delta t

Clearing F:

F = (deltaP) / (delta t)

Substituting the values

F = ( - 2.064) / (1/800) = - 1651.2N

answer:

the magnitude of the average force exerted on the superball by the sidewalk is 1651.2N