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

The answer is as follows in the photo