Impulse applied to a bouncing ball

In summary: However, the difference between the original velocity (of going from 1 m/s to 0) and the new velocity (of going from 0 to 1 m/s) is still 1 m/s.
  • #1
Np14
27
2

Homework Statement


A 0.500 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.700 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

Homework Equations


FNETt = mv - mvo
k = 1/2mv2
Ug = magh

The Attempt at a Solution


I first solved Ug = K as the ball's gravitational energy will be converted to kinetic energy in an elastic collision. Point A is the highest point (1.2 m); point B is the highest point the ball bounces back up (0.7m).

magh = 1/2mv2
(0.5)(9.8)(1.2) = (1/2)(0.5)(VB)2

That gives VB = 4.85 m/s.

FNETt = mv - mvo
FNETt = (0.5)(4.85) - (0.5)(0)

I tried to solve for Impulse using the first equation (the impulse momentum theorem) and got +2.42 N(s) upward. The answer is +4.28 N(s) upward. What did I do incorrectly?
 
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  • #2
Np14 said:
What did I do incorrectly?
You only found the impulse necessary to stop the ball. You need to add the impulse that gets it back up to the new height.
 
  • #3
What do you mean by impulse to stop the ball? Please elaborate.
 
  • #4
Np14 said:
What do you mean by impulse to stop the ball? Please elaborate.
That's how I interpreted this:
Np14 said:
Fnett = (0.5)(4.85) - (0.5)(0)
It looks like you set the final velocity to 0. But I now think perhaps you meant this is the momentum gained in falling, so the zero is the initial velocity. Either way, comes to the same thing; the momentum gained in falling is equal and opposite to the impulse necessary to stop it.
As I posted, you need to add the impulse necessary to raise it to the new height.
 
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  • #5
Np14 said:
What do you mean by impulse to stop the ball? Please elaborate.

"Impulse to stop the ball" = "difference in momentum between moving (your first term) and stopped (your second term)". It's the change in momentum to go from falling to not moving.

But there's more of a change than that. It doesn't stop dead with velocity 0.
 
  • #6
Okay so I tried to find second impulse but it did not work out...

KA = Ug,B + KB
1/2(0.5)(4.85)2 = (0.5)(9.8)(0.7) + 1/2(0.5)(VB)2
VB = 5.55 m/s

ImpulseB = (0.5)(5.55 - 4.85)
ImpulseB = 0.35 N(s)

ImpulseA + ImpulseB = 2.42 + 0.35 = 2.77 N(s)
 
  • #7
What second impulse? There's only one impulse. It hit the ground at one velocity (downward), it leaves the ground at another velocity (upward). Find out what those two velocities are and the difference between mv is your impulse, the impulse provided by the ground.

Np14 said:
##1/2(0.5)(4.85)^2 = (0.5)(9.8)(0.7) + 1/2(0.5)(VB)^2##
This seems to be saying that the kinetic energy when it hit the ground has turned into the energy to rise 0.7 m, plus some additional kinetic energy when it's done rising 0.7 m.

But that doesn't happen. 0.7 m is as high as it rises. The kinetic energy it has when it leaves the floor is just enough to reach 0.7 m. There's no leftover KE at the top of the bounce.

Kinetic energy is not conserved here. Since it doesn't bounce as high as its initial height, it has obviously lost some energy.

Np14 said:
VB = 5.55 m/s
Does that make sense to you given the above (incorrect) equation? That when you subtract off 0.5 * 9.8 * 0.7 from the KE of moving at 4.85 m/s, you end up with even MORE kinetic energy than when you started?

Look. Go back to your first reasoning. How did you figure out the velocity you get from falling a given height? You said the PE turns into KE, right?
Now it goes back up. The new KE turns back into PE. So what's the new KE?

I see another misunderstanding in your last post also. Ask yourself this. Suppose you found that something went from 1 m/s to the right, to 1 m/s to the left. How much did the velocity change? Is it 0, meaning those are exactly the same velocity? Remember that velocity is a vector. Is 1 m/s right the same vector as 1 m/s left? Is the difference between them 0, or non-zero?
 
  • #8
RPinPA said:
The new KE turns back into PE. So what's the new KE?
I am not sure what you mean by asking what the new kinetic energy is.
It would just be K = Ug, right??

1/2mv2 = magh
1/2(0.5)(VB)2 = (0.5)(9.8)(0.7)
VB = 3.7 m/s

RPinPA said:
I see another misunderstanding in your last post also. Ask yourself this. Suppose you found that something went from 1 m/s to the right, to 1 m/s to the left. How much did the velocity change? Is it 0, meaning those are exactly the same velocity? Remember that velocity is a vector. Is 1 m/s right the same vector as 1 m/s left? Is the difference between them 0, or non-zero?

I understand your point that velocity is a vector, and the answer would be a nonzero ±2 (depends on which direction is specified as positive). But I don't understand how it relates to my problem. I subtracted 0 from Vf because the object is initially at rest, not because it didn't have any velocity when it was falling. At the lowest point (instant it touches the ground), the velocity would be 4.85 m/s, because this is an elastic collision.
 
  • #9
Np14 said:
At the lowest point (instant it touches the ground), the velocity would be 4.85 m/s, because this is an elastic collision.
Yes, it hits the ground at 4.85 m/s, but not because it is an elastic collision. The collision hasn't happened at that stage.
You have now found that it rebounds at 3.7m/s. So what is the total change in velocity?
 
  • #10
OK, I see a very basic point of confusion we need to clear up.

Np14 said:
I subtracted 0 from Vf because the object is initially at rest, not because it didn't have any velocity when it was falling. At the lowest point (instant it touches the ground), the velocity would be 4.85 m/s, because this is an elastic collision.

The problem is asking about how the floor affects it. Where in this motion of "drop, hit the floor, bounce up from the floor" is it that the floor is exerting force on the ball?

Was it the floor that caused it to go from 0 m/s at the top to 4.85 m/s? Is that really the part of the motion where the floor is affecting it? If the floor wasn't there, would it not fall?

Np14 said:
At the lowest point (instant it touches the ground), the velocity would be 4.85 m/s, because this is an elastic collision.
What is an elastic collision? Before it hits the floor, when it's still in the air, what are you calling a collision?
 
Last edited:
  • #11
haruspex said:
So what is the total change in velocity?

Ft = m(v - vo)
Ft = (0.5)(3.7 -4.85)
Ft = - 0.575 Ns, which is incorrect

RPinPA said:
What is an elastic collision? Before it hits the floor, when it's still in the air, what are you calling a collision?

I misspoke. The ball falls due to gravity and at the instant of collision changes velocity.
 
  • #12
Np14 said:
Ft = m(v - vo)
Ft = (0.5)(3.7 -4.85)
Ft = - 0.575 Ns, which is incorrect
I misspoke. The ball falls due to gravity and at the instant of collision changes velocity.

From what to what? What direction is the 4.85 m/s? What direction is the 3.7 m/s? This is precisely why I said "I see another misunderstanding in your last post" and why I asked you to think about the right / left issue.
 
  • #13
Np14 said:
3.7 -4.85)
That's the change in speed. It is not the change in velocity.
 
  • #14
Thank you, now it makes so much more sense.

Ft = 0.5(3.7 - (-4.85))
Ft = 4.28 Ns
 
Last edited:

Related to Impulse applied to a bouncing ball

What is impulse?

Impulse is a measure of the change in momentum of an object. It is equal to the force applied to the object multiplied by the time over which the force is applied.

How is impulse applied to a bouncing ball?

When a ball bounces, it experiences a change in momentum due to the force of the ground pushing against it. This change in momentum is equal to the impulse applied to the ball.

What factors affect the impulse applied to a bouncing ball?

The three main factors that affect the impulse applied to a bouncing ball are the force of the bounce, the time of contact between the ball and the ground, and the elasticity of the ball and ground.

How does the elasticity of a ball and ground affect the impulse?

The more elastic the ball and ground are, the longer the time of contact and the smaller the force of the bounce will be. This results in a smaller impulse being applied to the ball.

Why is understanding impulse important for studying the behavior of bouncing balls?

Understanding impulse is important for studying bouncing balls because it helps us understand how they interact with the ground and how they change direction and speed. It also helps us predict the height of a bounce and the overall behavior of the ball.

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