A 52.5g super ball traveling at 28.0m/s bounces off a brick wall and rebounds at 18.5m/s. A high speed camera records this event. If the ball is in contact with the wall for 4.10ms, what is the magnitude of the average acceleration of the ball during this time interval?
acceleration=changevelocity/time=(18.5+28.5)/.0041 m/s^2
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 formula:
Average acceleration = change in velocity / time interval
First, we need to find the change in velocity of the ball. The ball's velocity changes from 28.0 m/s to -18.5 m/s (since it rebounds in the opposite direction). So, the change in velocity is:
Change in velocity = -18.5 m/s - 28.0 m/s = -46.5 m/s
Next, we need to convert the time interval from milliseconds (ms) to seconds (s):
Time interval = 4.10 ms ÷ 1000 = 0.00410 s
Now, we can plug in the values into the formula to calculate the average acceleration:
Average acceleration = (-46.5 m/s) / (0.00410 s)
Calculating this value gives us the magnitude of the average acceleration of the ball during the time interval.