A 54.0-g Super Ball traveling at 29.0 m/s bounces off a brick wall and rebounds at 18.0 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 in m/s^2? Please show work
Vo = 29 m/s.
V = -18 m/s.(opposite direction of Vo).
t = 4 mS = 0.004 s.
a = (V-Vo)/t = (-18-29)/0.004=-11,750 m/s^2.
To find the magnitude of the average acceleration of the ball during the contact with the wall, we can use the formula:
average acceleration = (final velocity - initial velocity) / time
First, let's convert the given masses to kilograms (since SI units are standard in physics equations):
Mass of the ball = 54.0 g = 0.054 kg
Next, we will substitute the given values into the formula:
Initial velocity (u) = 29.0 m/s
Final velocity (v) = -18.0 m/s (since the ball rebounds and changes direction)
Time (t) = 4.00 ms = 0.004 seconds
Now let's calculate the average acceleration:
average acceleration = (v - u) / t
average acceleration = (-18.0 m/s - 29.0 m/s) / 0.004 s
To simplify the equation, let's calculate the numerator first:
Numerator = (-18.0 m/s - 29.0 m/s) = -47.0 m/s
Now, we can substitute the numerator and time values into the formula:
average acceleration = (-47.0 m/s) / 0.004 s
average acceleration = -11750 m/s^2
Therefore, the magnitude of the average acceleration of the ball during the contact with the wall is 11750 m/s^2.