Posted by **Michael** on Sunday, April 12, 2009 at 11:36pm.

A rubber ball is thrown against the wall. The ball has a mass of 4 g and strikes the wall perpendicularly with a speed of 12 m/s. The acceleration is constant while the ball is touching the wall. After touching the wall, the center of mass moves .5 cm toward the wall, and back out from the wall.

what is the magnitude of the time average force?

>>First I found the impulse delivered by the wall which is Px'-Px (momentum) = .096 N/s. Then I tried to find the change in time by taking (.5*2*(1/1000))/12. For this change in time i got 8.33E-4. However, the answer .096/8.33E-4 was incorrect. Help!

- Physics -
**drwls**, Sunday, April 12, 2009 at 11:55pm
The average velocity from first impact to turnaround (maximum compression) is 6 m/s. The time required to reach maximum compression (at constant acceleration) is T = 0.5 cm/(600 cm/s)= 8.3*10^-4 s. Acceleration is

a = (delta V)/T

a = (12 m/s)/(8.3*10^-4s) = 14,400 m/s^2

M = m a = (0.004 kg)(14,400 m/s^2) = 57.6 N

I understand how you got the impulse. The units of momentum are not N/s. The time agrees with mine, but mine is half the total interval of contact, and yours is the total time of contact. Something is wrong there

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