Calculating Forces When Bouncing a Ball

[Sirsh]

Bouncing a ball!

Hey all, Just wondering if anyone could help me with this handful of questions? thank you.
IF you can't specific ways to figure them out, please tell me the logic/theory. thanks :D

Inital velocity = 10m/s, final velocity = 8m/s. mass = 80grams.

6 The time of contact between the ball and the floor during the bounce was 0.0500 s.

A. Calculate the average net force acting on the ball during its contact with the floor.

F = MΔV / ΔT
F = 0.08*10/0.05
F = 16 N

B. Calculate the average force that the floor exerts on the ball.

F = MΔV/ΔT
= 0.08*18/0.05
= 28.8 N

C. Calculate the average force that the ball exerts on the floor.

Fav = P1(mΔv) + P2(mΔv)
= 0.08*10 + 0.08*8
= 1.44 / 2
= 0.72 kg-1 m/s


Basically my question is, what are these forces and how are they calculated? thanks a lot!

 

[Doc Al]

A. Calculate the average net force acting on the ball during its contact with the floor.

F = MΔV / ΔT
F = 0.08*10/0.05
F = 16 N

For some reason, here you have ΔV = 10 m/s. Why?

B. Calculate the average force that the floor exerts on the ball.

F = MΔV/ΔT
= 0.08*18/0.05
= 28.8 N

This is the answer to A.

Hints: What individual forces act on the ball? What's the net force?

C. Calculate the average force that the ball exerts on the floor.

Fav = P1(mΔv) + P2(mΔv)
= 0.08*10 + 0.08*8
= 1.44 / 2
= 0.72 kg-1 m/s

Just invoke Newton's 3rd law.

 

[Sirsh]

I thought for the first question, it was when it's coming down so at 10m/s. Forces which act on the ball.. do you mean when it collides with the floor or when its coming down before it hits the floor. Is the net force just the initial force - the final force? thank you!

 

[Doc Al]

I thought for the first question, it was when it's coming down so at 10m/s. Forces which act on the ball.. do you mean when it collides with the floor or when its coming down before it hits the floor.

No, they are talking about during the collision, not before it collides. (While it's coming down the only force on it is its weight.)

Is the net force just the initial force - the final force? thank you!

No. Net force means the sum of all individual forces acting. (Don't confuse 'net' with change.)

The only thing you can calculate is the average force during the entire interaction (duration = 0.0500 s).

 

[Sirsh]

Okay well, so far from what you've said I've made these calculations out of it.

A. Calculate the average net force acting on the ball during its contact with the floor.

F = MΔV / ΔT
F = 0.08*18/0.05
F = 28.8 N

B. Calculate the average force that the floor exerts on the ball.

F = MΔV/ΔT
= 0.08*2/0.05
= 3.2 N

C. Calculate the average force that the ball exerts on the floor.

F = P1(mΔv)/T + P2(mΔv)/T
= 0.08*10/0.05 + 0.08*8/0.05
= 16 + 12.8
= 28.8N

With Newtons third law, it's that the force exerted on the floor, should be the same magnitude as the force exerted on the ball by the floor in the opposite direction. Does this mean that the ball exerts a force of 28.8N downwards and the floor exerts a force of 28.8N upwards?

Im confused because the ball goes down, then back up? is it just the dowards for the floor exerts back. so 0.08*10/0.05 = 16N?

thanks a lot!

 

[HallsofIvy]

No, while the ball is "going down" (but already struck the floor) the floor exerts a force to slow the ball to 0 velocity, but then whle the ball is "going upward", the floor exerts a force giving the ball an upward velocity.

Assuming this is an "elastic" collision, the upward speed of the ball after the collision is 10 m/s, the same as the downward speed before. The net change in velocity is 10- (-10)= 20 m/s and so the net change in momentum is (.08)(20)= 16 kg m/s.

 

[Sirsh]

Oh so, the floor exerts a force to slow the ball down to 0m/s but then also exerts a force to accelerate it to 8m/s,

so it's the stopping force + the accelerating force = net force.

 

[Doc Al]

A. Calculate the average net force acting on the ball during its contact with the floor.

F = MΔV / ΔT
F = 0.08*18/0.05
F = 28.8 N

Good. In what direction does that average net force act?

B. Calculate the average force that the floor exerts on the ball.

F = MΔV/ΔT
= 0.08*2/0.05
= 3.2 N

Remember that the net force equals the (vector) sum of the individual forces. What individual forces act on the ball? Hint: Two forces act. One of them is the force that the floor exerts.

C. Calculate the average force that the ball exerts on the floor.

F = P1(mΔv)/T + P2(mΔv)/T
= 0.08*10/0.05 + 0.08*8/0.05
= 16 + 12.8
= 28.8N

No. First solve B correctly, then consider Newton's 3rd law.

With Newtons third law, it's that the force exerted on the floor, should be the same magnitude as the force exerted on the ball by the floor in the opposite direction.

Yes!

Does this mean that the ball exerts a force of 28.8N downwards and the floor exerts a force of 28.8N upwards?

No. 28.8 N is the net force, not the force between ball and floor.

Im confused because the ball goes down, then back up? is it just the dowards for the floor exerts back. so 0.08*10/0.05 = 16N?

No.

 

[Sirsh]

Okay, A) 28.8N upwards.
B)

F(slowing ball to 0m/s) = MΔV/ΔT
= 0.08*10/0.05
= 16 N UP
F(accelerating the ball to 9m/s) = MΔV/ΔT
= 0.08*8/0.05
= 12.8 N Down

C. Calculate the average force that the ball exerts on the floor.

F(slowing to 0m/s) = MΔV/ΔT
= 0.08*10/0.05
= 16 N Down
F(accelerating to 9m/s) = MΔV/ΔT
= 0.08*8/0.05
= 12.8 N UP

is that correct?

 

[Doc Al]

Okay, A) 28.8N upwards.

Right. And if we use + to represent up, the net force = + 28.8 N.

B)

F(slowing ball to 0m/s) = MΔV/ΔT
= 0.08*10/0.05
= 16 N UP
F(accelerating the ball to 9m/s) = MΔV/ΔT
= 0.08*8/0.05
= 12.8 N Down

No. Forget MΔV/ΔT. You needed it for part A, but that's done. Now just use the definition of net force. Reread the hints I gave earlier.

 

[Sirsh]

B) 0.08*18/0.05 = 28.8N?

so the positive force and negative force are used together in the equation to figure out the exertion forces?

 

[Doc Al]

B) 0.08*18/0.05 = 28.8N?

Nope. (Didn't I just say to forget MΔV/ΔT?)

so the positive force and negative force are used together in the equation to figure out the exertion forces?

Not sure what you mean here.

Do this. There are two forces acting on the ball during the collision. List them and their direction. Then add them up and set the sum equal to the net force.

 

[Sirsh]

Force accelerating the ball towards the floor, then the ball accelerating after hitting the floor. so the ball would exert a force of 16N and the floor would exert 28.8N?

 

[Doc Al]

Force accelerating the ball towards the floor, then the ball accelerating after hitting the floor.

I want you to actually name the forces that act on the ball (or at least name the body exerting the force on the ball). There are only two forces acting. One acts down; the other acts up.

 

[Sirsh]

Im so confused, I've come to a conclusion that the floor exerts a force of +28.8N on the ball while the ball only exerts a force of -16N. this therefore allows the ball to acceralate to 8m/s i think.

 

[Doc Al]

Im so confused, I've come to a conclusion that the floor exerts a force of +28.8N on the ball while the ball only exerts a force of -16N. this therefore allows the ball to acceralate to 8m/s i think.

Forget all that. What forces act on the ball during the collision?

As we've already determined, +28.8 N is the average net force acting on the ball during the collision. That's the sum of all forces acting.

 

[Sirsh]

Um, gravity is acting on the ball as well as the normal force.

 

[Doc Al]

Um, gravity is acting on the ball as well as the normal force.

Yes! Gravity acts down (how do you calculate that force?); the normal force acts up.

That normal force is the force exerted by the floor on the ball. That's what we need to solve for in part B.

You know the net force (from part A), so set up an equation (add up the forces and set them equal to the net force) and solve for the normal force.