A 51.5-g Super Ball traveling at 29.0 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 3.60 ms, what is the magnitude of the average acceleration of the ball during this time interval?
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To find the magnitude of the average acceleration of the ball during the time it is in contact with the wall, we can use the following formula:
average acceleration (a) = change in velocity (Δv) / time interval (Δt)
First, we need to calculate the change in velocity of the ball. The initial velocity (u) is 29.0 m/s, and the final velocity (v) is 19.0 m/s. The change in velocity (Δv) is then:
Δv = v - u
= 19.0 m/s - 29.0 m/s
= -10.0 m/s
Next, we need to convert the time interval from milliseconds (ms) to seconds (s). The time interval (Δt) is 3.60 ms, which is equivalent to 0.00360 seconds.
Now, we can calculate the magnitude of the average acceleration using the formula:
average acceleration (a) = Δv / Δt
Plugging in the values:
a = -10.0 m/s / 0.00360 s
= -2777.78 m/s^2
The magnitude of the average acceleration of the ball during the time it is in contact with the wall is approximately 2777.78 m/s^2.