Find the time it takes for bouncing ball to come to rest

[freutel]

Homework Statement

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 t_n=(2V/g)*k^n/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.

Homework Equations

• Conservation of energy -> mgh=½mV²

• after the nth bounce Kinetic Energy_n = kⁿ(½mV²) which is equal to ½m(v_n)².

• height the ball reaches after the nth bounce -> h_n=kⁿV²/(2g)

• Motion along straight-line equations -> v=at, x=x₀+v₀t+½at²

• Time interval between the nth and the (nth+1)th bounce -> t_n=(2V/g)*k^n/2

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 h_n would be zero. I splitted the time interval equation in (V/g)*k^n/2 + (V/g)*k^n/2) = t_n. With the height of the ball equation i subbed one part of the time interval equation which results in t_n=(V/g)*k^n/2 + √(2h_n/g). I moved everything except the height part to the left and when removing the square root I got (t_n)² - (V²/g²)*kⁿ = 2h_n/g. It is clear that for h_n to be zero then (t_n)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!

 

[PeroK]

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

 

[freutel]

Now i came up with that the total time it will take is T=t₀ + Σt_n 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)*k^n/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)

 

[PeroK]

Now i came up with that the total time it will take is T=t₀ + Σt_n 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)*k^n/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)

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

 

[freutel]

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

 

[PeroK]

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

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

 

[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 +... + t_n.

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!