A 47.0-g Super Ball traveling at 26.5 m/s bounces off a brick wall and rebounds at 20.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.10 ms, what is the magnitude of the average acceleration of the ball during this time interval?

Question

A 47.0-g Super Ball traveling at 26.5 m/s bounces off a brick wall and rebounds at 20.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.10 ms, what is the magnitude of the average acceleration of the ball during this time interval?

Answer 1

To find the magnitude of the average acceleration of the ball during contact with the wall, we can use the equation:

Average acceleration = (change in velocity)/(time interval)

First, we need to calculate the change in velocity. The initial velocity, vi, is 26.5 m/s, and the final velocity, vf, is -20.5 m/s (since the ball changed direction on rebound).

Change in velocity = vf - vi
Change in velocity = -20.5 m/s - 26.5 m/s
Change in velocity = -47.0 m/s

Now, we can substitute the values into the equation for average acceleration:

Average acceleration = (-47.0 m/s)/(3.10 × 10^-3 s)

We can simplify the equation:

Average acceleration = -47.0 m/s / 0.00310 s
Average acceleration = -15161.29 m/s²

However, the magnitude of the acceleration is always positive, so we take the absolute value:

Magnitude of average acceleration = |-15161.29 m/s²|
Magnitude of average acceleration = 15161.29 m/s²

Therefore, the magnitude of the average acceleration of the ball during contact with the wall is 15161.29 m/s².

Answer 2

To find the magnitude of the average acceleration of the ball during the contact with the wall, we can use the equation:

acceleration (a) = (change in velocity) / (time interval)

First, let's calculate the change in velocity:

change in velocity = final velocity - initial velocity

Given:
Initial velocity (u) = 26.5 m/s
Final velocity (v) = 20.5 m/s

change in velocity = 20.5 m/s - 26.5 m/s = -6.0 m/s

Note: The negative sign indicates that the velocity changes direction.

Next, we need to convert the time interval into seconds:

time interval = 3.10 ms = 3.10 × 10^(-3) s

Now we can calculate the average acceleration:

acceleration = (-6.0 m/s) / (3.10 × 10^(-3) s)

Let's plug in the values and calculate:

acceleration ≈ -1935.48 m/s^2

The magnitude of the average acceleration is simply the absolute value of the negative result. Thus, the magnitude of the average acceleration of the ball during contact with the wall is approximately 1935.48 m/s^2.

Answer 3

a = (V-Vo)/t.

a=(-20.5-26.5)/0.0031=-15,161 m/s^2.