A 47.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 15.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval?
change in velocity/change in time
= (15.5 +29.5)/5*10^-3
if the second part asks for the average force, multiply by the mass
To find the magnitude of the average acceleration of the ball, we can use the equation:
Average acceleration = (Change in velocity) / (Time interval)
The change in velocity can be calculated by subtracting the initial velocity from the final velocity:
Change in velocity = Final velocity - Initial velocity
Given:
Mass of the ball (m) = 47.0 g = 0.0470 kg
Initial velocity (u) = 29.5 m/s
Final velocity (v) = 15.5 m/s
Time interval (t) = 5.00 ms = 5.00 × 10^(-3) s
First, let's calculate the change in velocity:
Change in velocity = 15.5 m/s - 29.5 m/s
Change in velocity = -14.0 m/s (Note: The negative sign indicates that the velocity is in the opposite direction)
Now, let's calculate the magnitude of the average acceleration:
Average acceleration = (-14.0 m/s) / (5.00 × 10^(-3) s)
Average acceleration = -2800 m/s^2 (Note: The negative sign indicates acceleration in the opposite direction)
Therefore, the magnitude of the average acceleration of the ball during this time interval is 2800 m/s^2.
To calculate the magnitude of the average acceleration of the ball during the given time interval, we can use the formula:
average acceleration (a) = change in velocity (Δv) / time interval (Δt)
First, we need to determine the change in velocity (Δv) of the ball. The initial velocity (v1) of the ball before the bounce is 29.5 m/s, and the final velocity (v2) after the bounce is 15.5 m/s. The change in velocity can be calculated as:
Δv = v2 - v1
Δv = 15.5 m/s - 29.5 m/s
Δv = -14.0 m/s
As the ball rebounds, its velocity changes in the opposite direction, leading to a negative change in velocity.
Next, we need to convert the time interval (Δt) from milliseconds to seconds. Given that the time interval is 5.00 ms:
Δt = 5.00 ms ÷ 1000 (to convert ms to seconds)
Δt = 0.005 s
Now, we can calculate the magnitude of the average acceleration (a) by dividing the change in velocity (Δv) by the time interval (Δt):
a = Δv / Δt
a = -14.0 m/s / 0.005 s
a = -2800 m/s²
Since the magnitude of acceleration is always positive, we take the absolute value of the calculated result:
|a| = |-2800 m/s²|
|a| = 2800 m/s²
Therefore, the magnitude of the average acceleration of the ball during the 5.00 ms time interval is 2800 m/s².