33.6k views
2 votes
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.

1 Answer

7 votes
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
User Dwenaus
by
5.7k points