A 50.0 g Super Ball traveling at 27.0 m/s bounces off a brick wall and rebounds at 17.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.00 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)
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:
acceleration = (change in velocity) / (change in time)
First, let's find the change in velocity. The ball initially travels at 27.0 m/s before hitting the wall, and after the rebound, it travels at 17.0 m/s. The change in velocity would be:
change in velocity = final velocity - initial velocity
= 17.0 m/s - 27.0 m/s
= (-10.0) m/s
Next, we need to convert the time of contact from milliseconds to seconds. We can use the fact that 1 ms = 10^-3 s. Therefore, the time of contact is:
time of contact = 4.00 ms = 4.00 * 10^-3 s
Now, we can calculate the magnitude of the average acceleration using the formula mentioned earlier:
acceleration = (change in velocity) / (change in time)
= (-10.0 m/s) / (4.00 * 10^-3 s)
To simplify the calculation, let's express the time in seconds:
acceleration = (-10.0 m/s) / (0.004 s)
= -2500 m/s^2
Therefore, the magnitude of the average acceleration of the ball during its contact with the wall is 2500 m/s^2.