A 45 g superball traveling at 28.0 m/s bounces off a brick wall and rebounds at 24.5 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? (Note: 1 ms = 10-3 s.)
Please show the work and the answer. Thanks so much!!
a = (V-Vo)/t.
a = (24.5-0) / 0.0036 = 6806 m/s^2.
Nope, Henry. Vo=-28
a= (24.5+28)/1E-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 formula:
average acceleration = (change in velocity) / (time interval)
First, let's calculate the change in velocity. The initial velocity of the ball before hitting the wall is 28.0 m/s, and the final velocity after rebounding is 24.5 m/s.
change in velocity = final velocity - initial velocity
= 24.5 m/s - 28.0 m/s
= -3.5 m/s
The negative sign indicates that the ball's velocity is changing in the opposite direction during the rebound.
Next, we need to convert the time interval from milliseconds to seconds. Given that 1 ms = 10^-3 s,
time interval = 3.60 ms * (10^-3 s/ms)
= 3.60 * 10^-3 s
Now, we can substitute the values into the formula for average acceleration:
average acceleration = (-3.5 m/s) / (3.60 * 10^-3 s)
= -972.22 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity of the ball.
Therefore, the magnitude of the average acceleration of the ball during the time interval it is in contact with the wall is 972.22 m/s^2.