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Derivation of a cinematic formula

  1. Apr 18, 2015 #1
    I'm really happy about every hint!

    1. The problem statement, all variables and given/known data
    There is a ball on hight h which is dropped on a table.
    With every Impact the ball loses velocity v by a factor a<1.
    I Need to Show the following:
    The time T after the ball stopps bouncing is:
    [tex] T = \frac{1+a}{1-a} (\frac{2h}{g})^\frac{1}{2} [/tex]

    There is a hint to start to calculate the time t_k between the takeoff and the next Impact
    and that I should use the geometric series.
    We dont need to think about air resistance.

    2. The attempt at a solution
    My Problem is that if I calculate t_k, I dont include the first time t_0 when the ball gets dropped.
    [tex] t_k = 2\frac{h_k}{v_k}=2\frac{h_k}{a^k v_0}=a^{-k} (\frac{2h_k}{\sqrt{(2 h_0 g)}}) = a^{-k} (\sqrt{(\frac{4h_k^2}{2 h_0 g}}) [/tex]
    with [tex] v=\sqrt{2gh}[/tex] which I got from E_pot=E_kin.
    But now I cant shorten h_0 with h_k and a^-k <1 so I cant use the geometric series?!
    Is it wrong to use v=s/t ⇔t=s/v since we have a accelarated movement.
     
  2. jcsd
  3. Apr 18, 2015 #2

    Delta²

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    Yes its wrong to use that. You should take instead [itex]t_k=2\frac{v_k}{g}, k\geq 1[/itex], [itex]t_0=\sqrt\frac{2h}{g}[/itex].
     
    Last edited: Apr 18, 2015
  4. Apr 19, 2015 #3
    How does the v_k looks like?
    [itex] t_k=2 \frac{a^k v_0}{g} = 2 a^k \frac{\sqrt{2gh}}{g}= 2 a^k \sqrt{\frac{2h}{g}} [/itex] ?
    What's next?
     
  5. Apr 19, 2015 #4

    Delta²

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    Next is geometric series of ##t_k##. ##T=t_0+\sum\limits_{k=1}^{\infty}t_k=\sqrt\frac{2h}{g}+\sum\limits_{k=1}^{\infty}t_k##. Notice that the actual ##t_0## is half of what you get from the formula of ##t_k## for k=0(because we drop the ball from above the table), thats why i put it outside the summation which summation is done from k=1 for this reason.

    or you can do the series from k=0 but then subtract ##\sqrt\frac{2h}{g}##.
     
    Last edited: Apr 19, 2015
  6. Apr 20, 2015 #5
    Thank you. To finish that threat I conclude:

    t_k means the time between the k and k+1 impact.

    for t_0
    we get
    [tex]t_0=\sqrt{\frac{2h}{g}}[/tex] with [tex]s=ut+\frac{1}{2}gt^2[/tex]

    for t_k with k>0
    [tex]\Sigma{t_k}=\Sigma_1^n{2 \frac{a^k v_0}{g}}=\Sigma_0^n{2 \frac{a^k v_0}{g}}-\frac{2v_0}{g}[/tex]
    now we let n go to infinity and use the geometric series:
    [tex]\Sigma_{n=0}^{\infty}{t_k}=\Sigma_{n=0}^{\infty}{2 \frac{a^k v_0}{g}}-\frac{2v_0}{g}=2\frac{v_0}{g}(\frac{1}{1-a}-1)=\frac{v_0}{g}\frac{2a}{1-a}=\sqrt{\frac{2h}{g}}\frac{2a}{1-a}[/tex]
    by using [tex]E_{kin}=E_{pot}[/tex]

    Adding 1. plus 2. we get: [tex]\sqrt{\frac{2h}{g}}\frac{1+a}{1-a}[/tex]
     
  7. Apr 20, 2015 #6

    haruspex

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    There is a slightly easier way.
    Consider the ball starting by rising from the ground at speed v. The time to first bounce is 2v/g. If the time to coming to rest is t(v) then t(v) = 2v/g + t(av). Guess a linear solution, t(v) = kv, and find that ##k=\frac{2}{g(1-a)}##. Then it is a matter of converting to the given initial conditions.
     
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