A 47.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 15.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval?
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(29.5+15.5)/(5*10^-3)
To find the magnitude of the average acceleration of the ball during the time interval it is in contact with the wall, we can use the equation:
average acceleration = (change in velocity) / (time interval)
First, we need to find the change in velocity of the ball. The initial velocity (vi) of the ball is 29.5 m/s and the final velocity (vf) is -15.5 m/s (since the ball rebounds in the opposite direction). Therefore, the change in velocity is:
change in velocity = vf - vi = -15.5 m/s - 29.5 m/s = -45.0 m/s
The negative sign indicates that the ball's velocity changed direction.
Next, we need to convert the time interval from milliseconds to seconds. The given time interval is 5.00 ms. To convert it to seconds, divide by 1000:
time interval = 5.00 ms / 1000 = 0.005 s
Now we can calculate the average acceleration:
average acceleration = (change in velocity) / (time interval) = -45.0 m/s / 0.005 s = -9000 m/s²
The negative sign indicates that the acceleration is in the opposite direction of the ball's initial velocity.
Therefore, the magnitude of the average acceleration of the ball during the time interval it is in contact with the wall is 9000 m/s².