a 0.2kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds up at 25 m/s. If the ball is in contact with the sidewalk for 0.025s, what is the magnitude of the average force applied by the sidewalk on the ball?
See previous post: Sun,10-21-12,2:44pm.
40N
To find the magnitude of the average force applied by the sidewalk on the ball, we can use the impulse-momentum principle.
The impulse-momentum principle states that the change in momentum of an object is equal to the impulse acting on it. Impulse is defined as the product of force and time.
Given:
Mass of the ball (m) = 0.2 kg
Initial velocity (vi) = 30 m/s
Final velocity (vf) = -25 m/s (negative because the ball is rebounding upwards)
Contact time (Δt) = 0.025 s
First, let's calculate the change in momentum (Δp) of the ball:
Δp = m * (vf - vi)
Δp = 0.2 kg * (-25 m/s - 30 m/s)
Δp = -0.2 kg * 55 m/s
Δp = -11 Ns
The negative sign indicates a change in direction.
Next, let's calculate the impulse (J) acting on the ball:
J = Δp
J = -11 Ns
Since impulse is equal to the force multiplied by the contact time, we can rearrange the equation to find the force:
J = F * Δt
F = J / Δt
F = -11 Ns / 0.025 s
Finally, let's calculate the magnitude of the average force applied by the sidewalk on the ball:
|F| = |-11 Ns / 0.025 s|
|F| = 440 N
Therefore, the magnitude of the average force applied by the sidewalk on the ball is 440 Newtons.