# Homework Help: Find the time it takes for bouncing ball to come to rest

Tags:
1. May 9, 2015

### freutel

1. The problem statement, all variables and given/known data
A ball, which is launched in the air with velocity V, has inelastic collisions with the floor: the kinetic energy after each collision is k times the kinetic energy before the collision, where k<1. Assume that the gravitational acceleration is constant: g [m/s^2]. I was asked to show that the time interval between the nth and the (n+1)th bounce is tn=(2V/g)*kn/2. This was pretty simple using conservation of energy and motion along straight-line equations. The second question asks me to find the total time T the bouncing ball takes to come to rest. This is where I am stuck.

2. Relevant equations
• Conservation of energy -> mgh=½mV2
• after the nth bounce Kinetic Energyn = kn(½mV2) which is equal to ½m(vn)2.
• height the ball reaches after the nth bounce -> hn=knV2/(2g)
• Motion along straight-line equations -> v=at, x=x0+v0t+½at2
• Time interval between the nth and the (nth+1)th bounce -> tn=(2V/g)*kn/2

3. The attempt at a solution

I honestly did not know where to start but I started with the thought that if the ball stops bouncing it means that hn would be zero. I splitted the time interval equation in (V/g)*kn/2 + (V/g)*kn/2) = tn. With the height of the ball equation i subbed one part of the time interval equation which results in tn=(V/g)*kn/2 + √(2hn/g). I moved everything except the height part to the left and when removing the square root I got (tn)2 - (V2/g2)*kn = 2hn/g. It is clear that for hn to be zero then (tn)2 has to be equal to (V2/g2)*kn. I have a feeling I'm drifting towards a wrong answer with this because I really do not know what to do now. Did I approach this problem correctly? If so, what do I have to do to finalize it and if not, I would really appreciate some help to push me in the right direction!

2. May 9, 2015

### PeroK

Can you think of something from your maths classes that might help to find the total time?

3. May 9, 2015

### freutel

Now i came up with that the total time it will take is T=t0 + Σtn with n=1, 2, 3,...,. The energy that is lost after every bounce is also ½mv2(1-kn). The number of bounces needed is when kn approaches zero. That's all I know.
Now that I think of it, it may be impossible to know what n is with the data given. So maybe the proper answer is T=t0 + Σ(2V0/g)*kn/2 with n starting from 1 till the value of n when kn is close to zero so I guess till the value of n when n=klog(0,0001)

4. May 9, 2015

### PeroK

Yes, but think mathematically rather than physically. What kind of sum might you be dealing with?

5. May 9, 2015

### freutel

It's power series right? With the power of k constantly increasing. I honestly do not know and I'm just saying stuff.

6. May 9, 2015

### PeroK

It's a "geometric" series. Remember them?

7. May 9, 2015

### freutel

Wow, I would seriously have never come up with that. Should have paid more attention in class. So I looked it up how to solve a geometric series (because I never quite learned how to do it in school) so i came up with this

T=t0 + t1 + t2 + t3 +... + tn.

T=(2V/g) + (2V/g)*√k + (2V/g)*√k2 + (2V/g)*√k3 + ... + (2V/g)*√kn

So the common ratio is √k

T - (√k)T=2V/g

T=2V/(g*(1-√k))

Thanks a lot PeroK!