Achild bounces a 56 g superball on the sidewalk. the velocity change of the superball is from 20 m/s downward to 13 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.

Using the impulse-momentum theorem and taking this down as positive, we will get: 
FΔt = Δp 

F = Δp / Δt 

= m(v - v₀) / t 

= 0.056kg [13m/s - (- 20m/s) / 0.00125s 

= 1478.4 N when properly rounded off is the answer we are looking for in this problem.

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