Question

A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 26 m/s downward to 16 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? 1. 2183 N 2. 2107 N 3. 1992 N 4. 1877 N 5. 2068 N 6. 1839 N 7. 2030 N 8. 1954 N 9. 2145 N 10. 1915 N

Answer 1

Answer:

10. 1915 N

Explanation:

The impulse exerted on the ball, I, is equal to the product between the force (F) and the contact time ($\Delta t$):

$I=F \Delta t$ (1)

The impulse is also equal to the change in momentum of the ball, $\Delta p$, equal to the product between the mass of the ball (m) and the change in velocity ($\Delta v$):

$I=\Delta p = m\Delta v$ (2)

By merging the two equations (1) and (2) we find:

$F \Delta t = m \Delta v$

Solving it for F, we can find the magnitude of the average force:

$F=\frac{m \Delta v}{\Delta t}=\frac{(0.057 kg)(16 m/s-(-26 m/s))}{\frac{1}{800} s}=1915 N$