What Is the Average Acceleration of a Super Ball Bouncing Off a Wall?

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To calculate the average acceleration of a Super Ball bouncing off a wall, the initial velocity is 28.0 m/s and the final velocity is -19.0 m/s due to the change in direction. The time of contact with the wall is 3.50 ms, which converts to 0.0035 seconds. The average acceleration can be determined using the formula a = (v_final - v_initial) / t, resulting in a negative value that reflects the change in direction. The correct magnitude of acceleration must account for this directional change, leading to a final calculation of approximately 2571 m/s². Understanding the direction of velocities is crucial for accurate results in such problems.
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Homework Statement


A 50.0 g Super Ball traveling at 28.0 m/s bounces off a brick wall and rebounds at 19.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)


Homework Equations


i used Vxf = Vxi + at


The Attempt at a Solution



I put vxf = 19, vxi = 28 and t = 3.50 ms and .0035, i get an answer of - 2.57 and -2571 (respectively) and made it positive since its the magnitude, but its wrong. Any ideas?
 
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You must account for the fact that the ball changes direction, and so the velocities before and after have different directions.
 
F=ma=dp/dt
 
Impulse = F * t = change in momentum = (mv1 - (- mv2)

F = average force
a = F/m

= (v1 +v2)/t
 
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