Average force the floor exerted on the ball

[Sinusoidal]

Homework Statement

A ball of mass 200 g is released from rest at a height of 2.00 m above the floor and it rebounds straight up to a height of 0.900 m. (a) Determine the ball's change in momentum due to its contact with the floor. (b) If the contact time with the floor was 0.0950 seconds, what was the average force the floor exerted on the ball, and in what direction?

Homework Equations

Δp=m(v-v_o)
F_avg=Δp/Δt
w=mg

The Attempt at a Solution

The answer to a) is 2.09 kg x m/s, and the answer to b) is 24.0 N upward.

I understood everything except for a portion of b. I used F_avg=Δp/Δt=2.09 kg x m/s/.0950s to calculate 22.0 N, but you're supposed to add weight to that as well. That is where I get stuck. I don't understand why you're supposed to add w=(.200kg)(9.8 m/s) to the 22 N.

Can someone draw a picture of this and also explain why you add g/why it's upward?

My book says "The floor also needs to support the weight of the ball during impact."

I just don't get that

Thanks so much!

 

[PhanthomJay]

When you solve for F_avg in your rate of momentum change equation, which is Newton's 2nd law, don't forget that what you calculate for F_avg , 22N, is the NET force acting on the ball. The net force on the ball consists of both its weight acting down on it, and the normal average contact force of the floor acting up on it. Draw a FBD of the ball in contact with the floor and note the direction of the forces.

 

[baba89]

As a scientist, it is important to understand the principles and concepts behind the equations and calculations used to solve a problem. In this case, we are dealing with the concept of momentum and the forces acting on an object during a collision.

First, let's look at the concept of momentum. Momentum is defined as the product of an object's mass and its velocity. In this problem, the ball has a mass of 200 g (0.200 kg) and is initially at rest, so its initial momentum (p0) is 0 kg·m/s. After the collision with the floor, the ball rebounds with a velocity of 0.900 m/s, so its final momentum (pf) is 0.200 kg x 0.900 m/s = 0.180 kg·m/s. Therefore, the change in momentum (Δp) is equal to the final momentum minus the initial momentum, or Δp = pf - p0 = 0.180 kg·m/s - 0 kg·m/s = 0.180 kg·m/s.

Now, let's look at the concept of force. Force is defined as the rate of change of an object's momentum, or F = Δp/Δt. In this problem, we are given the change in momentum (Δp = 0.180 kg·m/s) and the contact time with the floor (Δt = 0.0950 s), so we can calculate the average force exerted by the floor on the ball using the equation Favg = Δp/Δt = 0.180 kg·m/s / 0.0950 s = 1.89 N.

But wait, what about the weight of the ball? The weight of an object is the force exerted on it by gravity, and it is given by the equation w = mg, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2 on Earth). In this problem, the weight of the ball is w = (0.200 kg)(9.8 m/s^2) = 1.96 N. This weight is always acting on the ball, whether it is in motion or at rest. During the collision, the floor exerts a force on the ball (Favg) to change its momentum, but it also needs to support the weight of the ball (