- #1

LukasG

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**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

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.

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