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