A tennis ball with a speed of 5.0 m/s is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 3.0 m/s. If the ball is in contact with the wall for 0.013 s, what is the average acceleration of the ball while it is in contact with the wall?
Ft = m(final velocity) - m(initial velocity)
F = ma so equation changes to
mat = m(final velocity) - m(initial velocity)
plug in values and solve for 'a'
hturtu
To find the average acceleration of the ball while it is in contact with the wall, we can use the formula:
Average acceleration (a) = Change in velocity (Δv) / Time taken (Δt)
First, we need to calculate the change in velocity of the ball during its contact with the wall. The initial velocity of the ball before hitting the wall is 5.0 m/s, and the final velocity after rebounding is -3.0 m/s (opposite direction).
Δv = final velocity - initial velocity
Δv = -3.0 m/s - 5.0 m/s
Δv = -8.0 m/s
Next, we need to calculate the time taken for the ball to rebound. The problem states that the ball is in contact with the wall for 0.013 s.
Therefore:
Δt = 0.013 s
Now we can calculate the average acceleration:
a = Δv / Δt
a = -8.0 m/s / 0.013 s
a ≈ -615.4 m/s²
So, the average acceleration of the ball while it is in contact with the wall is approximately -615.4 m/s². The negative sign indicates that the acceleration is in the opposite direction to the initial motion of the ball.