# Calculating the Impulse of a dropped ball

• LukasG
In summary: From what I gather, you mean some kind of constant that we can use to find the force while the ball is being compressed. If so, I do not know of any.In summary, the conversation discusses the calculation of impulse and time for a ball dropped from a certain height onto a powder-covered ground. The ball bounces back to a certain height and the print it leaves can be used to calculate the distance it was compressed during the bounce. There is some confusion about the direction of the impulse and the use of an experimentally determined constant to calculate the force during compression.
LukasG
I would like to start off by excusing myself in advance for not knowing all of the correct terminology, since I am not studying physics in English.

1. Homework Statement

A ball is dropped from a certain height, it then falls down towards the ground and consequently bounces up again to a certain height. The ball is also dropped from the same height but on a powder covered ground, which allows measurement of the print the ball has left. Calculate the Impulse and the time the ball touches the ground.

##h_1 =## Height ball is dropped from: 0.5 m
##m =## Mass of the ball: 0.0544 kg
##h_2 =## Height the ball reaches after bounce: 0.31 m
##r =## Radius of the ball: 0.0153 m
##l = ## Radius of the print: 0.01 m

## Homework Equations

##I = F /Δ t = Δp = mv_2 - mv_1##
##W = F s##
##E_p = m g h##
##E_\text{k} =\tfrac{1}{2} mv^2 ##

## The Attempt at a Solution

I start off by attempting to calculate the Impulse. I see that I can calculate it by using ##I = mv_2 - mv_1##, although I need the two different velocities. Since the change in momentum occurs during the bounce, the two different velocities should be just before, and just after the bounce.

##v_1 = ## Velocity before contact
##v_2 = ## Velocity after contact

Since we know the height of which the ball was dropped, we should be able to calculate the potential energy it has at that height. We also know that just before the bounce, all of that energy is transformed into kinetic energy. Therefore:

##mgh_1 = \tfrac{1}{2} mv_{1}^2##

The desired variable is ##v_1##

##v_1 = \sqrt{2gh_1}##

The values are put in and we get a value of ##v_1 = 3.13368\, m/s##

The same procedure is used, but this time with ##h_2## to calculate ##v_2##.

##v_2 = \sqrt{2gh_2} = 2.46746\, m/s##

Here I assign the positive direction as downwards, resulting in ##v_2## becoming negative (it moves upwards, in the opposite direction of the positive).

##v_2 = -2.46746\, m/s##

We can now calculate the impulse of the ball.

##I = mv_2 - mv_1 = m(v_2-v_1) = -0.304\, Ns##

Here is where my first problem arises. The impulse becomes negative, which I don't fully understand. The force of the impulse would in turn have to be negative to give a negative answer. The other option would be to calculate ##I = m(v_1-v_2)## instead to not give a negative value. I suppose we can do this since it is still the difference in momentum.

Edit: The value of the impulse is negative because of the set positive direction downwards.

##I = m(v_2-v_1) = -0.304\, Ns##

Since I am not quite sure about the value of ##I##, I am sceptical to continuing my calculation.

If we assume the value of ##I## is correct, we now need the force of the impulse to calculate the time the ball touches the ground.

Since there is a change in energy during the bounce, we can assume that the ball does some kind of work.

##W = F s##

Although we now see that we need a distance. I assume this distance is the distance the ball is pressed together during the bounce. The work done is some kind of "braking" work. This is where the calculation we did with powder is useful, since we can use that distance ##(l)## to calculate ##s##.

Keep in mind this might be hard to imagine. The part of the ball that is pushed together can be seen as an arc. I have found this relevant article about it to explain it better: http://en.wikipedia.org/wiki/Sagitta_(geometry)

##s## is calculated using the pythagorean theorem, where ##r## is the hypotenuse, ##l## one catheter and ##(r-s)## another.

##r^2=l^2+(r-s)^2##

##s = r - \sqrt{r^2 - l^2}##

##s_1 = 0.00372027\,m##
##s_2= 0.131098237\, m##

I do not think that ##s_2## is a reasonable answer, because it is greater than the diameter of the ball.

To now calculate the ##Δt## we need ##F##. This is the average force during the impulse. I think that we can calculate this by using ##F=W/s##. The ##W## can be substituted with some form of energy, although which one I am not sure of.

One idea is to use ##E_{kb} / s = F##, where ##E_{kb}## is the kinetic energy before the bounce. Although I would only get the force before it started to go up again, so we would need to do another calculation.

##E_{kb} / s = F_b##
##E_{ka} / s = F_a##

These two forces would then together be the average force, which we could use in ##t=I/F##

I do not have a clear answer nor do I know what is correct. All help and input is welcome! I will be checking this thread as often as I can to answer any thoughts you may have.

Last edited:
You got a negative value for the impulse because you assigned the positive direction to be down. Assigning the positive direction to be up would result in a positive value for the impulse. As for the second part, the ball would be elastic, yes? So wouldn't the force that the ball exerts on the ground be proportional to the distance by which the ball is compressed? Wouldn't there also be an experimentally determined constant involved?

AlephNumbers said:
You got a negative value for the impulse because you assigned the positive direction to be down. Assigning the positive direction to be up would result in a positive value for the impulse. As for the second part, the ball would be elastic, yes? So wouldn't the force that the ball exerts on the ground be proportional to the distance by which the ball is compressed? Wouldn't there also be an experimentally determined constant involved?

Thanks for taking the time to reply, and thanks to your explanation of why the impulse became negative I now understand. I do not know if the ball would be elastic. That means that it doesn't lose energy and momentum during the collision? But I think it does, since it doesn't reach the same height. I do not understand what you mean by a experimentally determined constant.

I want you to take your finger and push your finger into your belly (lightly), and keep pushing. Your finger stops, even though you are exerting a constant force on your finger. This means that as you push your finger further into your belly, your belly exerts a larger force in response. The same thing happens if you punch a concrete wall, but the concrete is really, really elastic (resistant to deformation), so your hand only pushes into the concrete a very small amount. The difference between these two scenarios can be accounted for via an experimentally determined constant. The truth is, however, that even the function modeling the force should be determined experimentally.

Don't punch a concrete wall, by the way.

I think the reason that the ball does not reach its original height is because the ball becomes permanently deformed to some extent. You are right that part of the collision is inelastic.

I think I understand what you are saying, and the part about the response force I recognize from my lessons. Although in this particular experiment I do not think that you need the constant you are talking about to solve the problem, atleast from my understanding of the lesson and the information we were given. Do you think it is solvable without it? Can we find some kind of connection between the average force during the impulse and the vaules which we have?

Is it correct to assume that ##F_b + F_a = F## ?

Or is ##W## equal to the lost energy during the bounce? in that case ##W = E_{kb} - E_{ka} = Fs##

Another thing I thought about was that since we now know that the impulse is negative, that means that the average force also has to be negative, since it isn't reasonable to use a negative value for ##t## in ##I = Ft## ?

Yes, the average force must be negative. I agree that the difference between the final mechanical energy of the ball and the initial mechanical energy of the ball is the amount of energy that goes into the inelastic part of the collision.

However, the print left by the ball is not useful unless you have a function expressing the force exerted by the ball on the ground as a function of the distance the ball is compressed.

Hmm... yes that certainly would make things a lot easier. Although because I was not given any information regarding that I do not think that is how I'm supposed to solve it.

It would seem you are attempting to solve for the amount of time the ball is in contact with the ground. I am almost 100% sure that cannot be solved for without a force function like the one I described above.

Do you have any example of such a function? I am getting kind of frustrated here over this experiment... :P

Yes! I think the best thing to do would just be to assume the force is proportional to the distance the ball is compressed, or F(x)= kx. In any real situation the function would be something like F(x)= akx where a is an experimentally determined constant.

Oh ok, it's a shame our physics teacher didn't say anything about that. I suppose you would complete a number of different drops from different heights to determine it?

Do you have a slow motion camera and/or photogates and a decent set up for capturing and modeling data? This seems considerably more involved than any solo assignment I ever received in intro physics. What exactly is it that you have been told to find? Or is this more of an exploratory experiment you are doing on your own time?

We did this experiment in class, and we were told to calculate the impulse and the time the ball touches the ground. The equipment we received was: a large ruler, stopwatch, powder, ball, way of measuring the radius (smaller, more exact instrument), and a scale. Although now that I think of it, the stopwatch wasn't needed... Anyways, I find it quite absurd that our teacher would give us an experiment that isn't possible to solve with the provided tools. Also, like you say, this assignment seems quite intricate for our class, since we are only studying our first year in physics.

Well, I am at a loss. I have never met nor heard of an incompetent physics teacher (who was actually qualified to teach physics), so I feel like I must be wrong. Can we get a second opinion?

I will be going over the experiment tomorrow afternoon, so I will make sure to check back with the complete calculation and answer :)

LukasG said:
I would like to start off by excusing myself in advance for not knowing all of the correct terminology, since I am not studying physics in English.

1. Homework Statement

A ball is dropped from a certain height, it then falls down towards the ground and consequently bounces up again to a certain height. The ball is also dropped from the same height but on a powder covered ground, which allows measurement of the print the ball has left. Calculate the Impulse and the time the ball touches the ground.

##h_1 =## Height ball is dropped from: 0.5 m
##m =## Mass of the ball: 0.0544 kg
##h_2 =## Height the ball reaches after bounce: 0.31 m
##r =## Radius of the ball: 0.0153 m
##l = ## Radius of the print: 0.01 m

## Homework Equations

##I = F /Δ t = Δp = mv_2 - mv_1##
##W = F s##
##E_p = m g h##
##E_\text{k} =\tfrac{1}{2} mv^2 ##

## The Attempt at a Solution

I start off by attempting to calculate the Impulse. I see that I can calculate it by using ##I = mv_2 - mv_1##, although I need the two different velocities. Since the change in momentum occurs during the bounce, the two different velocities should be just before, and just after the bounce.

##v_1 = ## Velocity before contact
##v_2 = ## Velocity after contact

Since we know the height of which the ball was dropped, we should be able to calculate the potential energy it has at that height. We also know that just before the bounce, all of that energy is transformed into kinetic energy. Therefore:

##mgh_1 = \tfrac{1}{2} mv_{1}^2##

The desired variable is ##v_1##

##v_1 = \sqrt{2gh_1}##

The values are put in and we get a value of ##v_1 = 3.13368\, m/s##

The same procedure is used, but this time with ##h_2## to calculate ##v_2##.

##v_2 = \sqrt{2gh_2} = 2.46746\, m/s##

Here I assign the positive direction as downwards, resulting in ##v_2## becoming negative (it moves upwards, in the opposite direction of the positive).

##v_2 = -2.46746\, m/s##

We can now calculate the impulse of the ball.

##I = mv_2 - mv_1 = m(v_2-v_1) = -0.304\, Ns##

Here is where my first problem arises. The impulse becomes negative, which I don't fully understand. The force of the impulse would in turn have to be negative to give a negative answer. The other option would be to calculate ##I = m(v_1-v_2)## instead to not give a negative value. I suppose we can do this since it is still the difference in momentum.

Edit: The value of the impulse is negative because of the set positive direction downwards.

##I = m(v_2-v_1) = -0.304\, Ns##

Since I am not quite sure about the value of ##I##, I am sceptical to continuing my calculation.

If we assume the value of ##I## is correct, we now need the force of the impulse to calculate the time the ball touches the ground.

Since there is a change in energy during the bounce, we can assume that the ball does some kind of work.

##W = F s##

Although we now see that we need a distance. I assume this distance is the distance the ball is pressed together during the bounce. The work done is some kind of "braking" work. This is where the calculation we did with powder is useful, since we can use that distance ##(l)## to calculate ##s##.

Keep in mind this might be hard to imagine. The part of the ball that is pushed together can be seen as an arc. I have found this relevant article about it to explain it better: http://en.wikipedia.org/wiki/Sagitta_(geometry)

##s## is calculated using the pythagorean theorem, where ##r## is the hypotenuse, ##l## one catheter and ##(r-s)## another.

##r^2=l^2+(r-s)^2##

##s = r - \sqrt{r^2 - l^2}##

##s_1 = 0.00372027\,m##
##s_2= 0.131098237\, m##

I do not think that ##s_2## is a reasonable answer, because it is greater than the diameter of the ball.

To now calculate the ##Δt## we need ##F##. This is the average force during the impulse. I think that we can calculate this by using ##F=W/s##. The ##W## can be substituted with some form of energy, although which one I am not sure of.

One idea is to use ##E_{kb} / s = F##, where ##E_{kb}## is the kinetic energy before the bounce. Although I would only get the force before it started to go up again, so we would need to do another calculation.

##E_{kb} / s = F_b##
##E_{ka} / s = F_a##

These two forces would then together be the average force, which we could use in ##t=I/F##

I do not have a clear answer nor do I know what is correct. All help and input is welcome! I will be checking this thread as often as I can to answer any thoughts you may have.
No doubt the force may vary in a fairly complicated way during the short impact time. That's quite common in a problem that lends itself to being analyzed by use of impulse.

It's often the case that the force is treated as being constant for the short time duration of the impact.

Therefore assume uniform acceleration for the ball coming to rest - average speed is ##\displaystyle\ \frac{| v_1 |}{2}## and uniform acceleration for the ball rebounding from rest gives an average speed of ##\displaystyle\ \frac{| v_2 |}{2}## .

You could simply average those two and find the time it takes the ball to travel a distance of ##\ 2s\ ##.

This can also be used to find the average force exerted on the ball or equivalently the average force exerted on the ground.

Last edited:
Thank you so much for the answer! I really appreciate you taking your time to help me. If I understand you correctly, ##\frac{\mid v_1 \mid}{2}## is the average speed going from touching the ground to being still? (And the same with the other average speed). Is the average speed calculated by ##\frac{\mid v_1 -v_0\mid}{2}##, where ##v_0 = 0## ?

AlephNumbers said:
the reason that the ball does not reach its original height is because the ball becomes permanently deformed to some extent
No. It can fully recover its original shape, but the force it exerts as it expands, at each stage in the expansion, is less than the force needed to compress it at the same point during compression.
SammyS said:
It's often the case that the force is treated as being constant for the short time duration of the impact.
Sadly, that's true. In practice, the profile would depend on e.g. whether the ball is solid or gas-filled.
For gas-filled, taking the pressure as constant, the force is proportional to the contact area. Taking the ball to be approximately a sphere with a small cap removed, the contact area can be found as a function of the compression distance, etc.
The graphs at http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/5. BallBounce.pdf suggest that in all cases it would be more accurate to treat it as SHM than as a constant force.

## 1. How do you calculate the impulse of a dropped ball?

To calculate the impulse of a dropped ball, you need to know the mass of the ball, the initial velocity at which it was dropped, and the time it takes to hit the ground. The formula for impulse is impulse = mass x change in velocity.

## 2. What is impulse in physics?

In physics, impulse is defined as the change in momentum of an object. It is a vector quantity and is measured in units of kg*m/s.

## 3. What factors affect the impulse of a dropped ball?

The impulse of a dropped ball is affected by the mass of the ball, the initial velocity at which it was dropped, and the duration of contact with the ground. The greater the mass and initial velocity, and the longer the duration of contact, the greater the impulse will be.

## 4. How does impulse relate to force?

Impulse and force are related through Newton's Second Law, which states that force is equal to the rate of change of momentum. Therefore, a larger impulse will result in a larger force.

## 5. Can the impulse of a dropped ball be negative?

Yes, the impulse of a dropped ball can be negative if the ball bounces back up after hitting the ground. This would result in a decrease in momentum and a negative change in velocity, leading to a negative impulse.

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