Just before a bouncy ball hits a wall, it is traveling parallel to the ground at 18 m/s, and it then bounces directly backwards at 17 m/s. The interaction time with the wall was 0.40 ms (4.0 x 10-4 s). If the ball's mass is 0.17 kg, what is the average magnitude of the force of the wall on the ball during their interaction?
V = Vo + a*t = 17 m/s,
18 + a*4*10^-4 = 17,
a = ?. (It will be negative.)
F = M*a.
To find the average magnitude of the force of the wall on the ball during their interaction, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a).
1. First, we need to find the change in velocity (acceleration) of the ball during the interaction. Since the ball bounces directly backward at 17 m/s, and it was initially traveling forward at 18 m/s, the change in velocity is the difference between these two speeds: Δv = 18 m/s - (-17 m/s) = 35 m/s.
2. Next, we need to calculate the acceleration. Acceleration is defined as the change in velocity divided by the time: a = Δv / Δt. In this case, the change in velocity is 35 m/s, and the time of interaction is given as 0.40 ms (or 4.0 x 10^-4 s): a = 35 m/s / (4.0 x 10^-4 s) = 87,500 m/s^2.
3. Now, we can calculate the force using Newton's second law: F = m * a. The mass of the ball is given as 0.17 kg, and the acceleration is 87,500 m/s^2: F = 0.17 kg * 87,500 m/s^2 = 14,875 N.
Therefore, the average magnitude of the force of the wall on the ball during their interaction is 14,875 Newtons.
To find the average magnitude of the force of the wall on the ball during their interaction, we can use the impulse-momentum principle.
The impulse-momentum principle states that the change in momentum of an object is equal to the force applied to it multiplied by the time over which the force is applied.
First, let's find the initial momentum of the ball before it hits the wall. Momentum is calculated by multiplying mass and velocity.
Initial momentum (before collision) = mass (m) × velocity (v)
= 0.17 kg × 18 m/s
= 3.06 kg·m/s
Next, we need to find the final momentum of the ball after it bounces off the wall. Since the ball bounces directly backwards at 17 m/s, the final velocity is negative.
Final momentum (after collision) = mass (m) × velocity (v)
= 0.17 kg × (-17 m/s)
= -2.89 kg·m/s
The change in momentum is given by the difference between the final and initial momentum:
Change in momentum = Final momentum - Initial momentum
= (-2.89 kg·m/s) - (3.06 kg·m/s)
= -5.95 kg·m/s
Now, we can calculate the average force using the impulse-momentum principle:
Average force = Change in momentum / Time
= -5.95 kg·m/s / (4.0 × 10^-4 s)
= -1.49 × 10^4 N
Since force is a vector quantity and we are interested in the magnitude of the force, we can take the absolute value:
Average magnitude of force = | Average force |
= |-1.49 × 10^4 N|
= 1.49 × 10^4 N
Therefore, the average magnitude of the force of the wall on the ball during their interaction is 1.49 × 10^4 N.