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Infinite series Geometric series

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    hello
    i have a question to this solved problem in the book

    " Mathematical Methods for Physics and Engineering Third Edition K. F. RILEY, M. P. HOBSON and S.J. BENCE "
    page 118

    Consider a ball that drops from a height of 27 m and on each bounce retains only a third
    of its kinetic energy; thus after one bounce it will return to a height of 9 m, after two
    bounces to 3 m, and so on. Find the total distance travelled between the first bounce and
    the Mth bounce.

    2. Relevant equations
    i have to use this equation
    [tex]S_M=a \frac{1-r^M}{1-r}[/tex]
    so [tex]a=9[/tex] because he wants the distance from the first bounce
    if he just said " find the total distance " i have to put a=27 right?
    the common ration is [tex]d=\frac{1}{3}[/tex]
    the ball goes up and down so i have to multiply the sum by 2 to get the total distance
    now
    [tex]S_M=2*9 [\frac{1-(\frac{1}{3})^{M-1}}{1-\frac{1}{3}}][/tex]
    [tex]S_M=27 [1-(\frac{1}{3})^{M-1}][/tex]

    i only don't understand why he put (M-1) not just (M)


    3. The attempt at a solution
    i thought maybe if I just put M then the ball will bounce for last time and be hanging in the air.
    but if i put M-1 then the ball will hit the ground and stop
    right?
     
  2. jcsd
  3. Oct 13, 2015 #2

    andrewkirk

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    How many return trips from ground to an apex up in the air are there between the 1st bounce and the M-th bounce?
     
  4. Oct 13, 2015 #3
    i think until the factor [tex](\frac{1}{3})^{M-1}[/tex] approachs zero
    so im not sure but may i put this factor to zero
    [tex](\frac{1}{3})^{M-1}=0[/tex]
    then i can calculate M
    right?
     
  5. Oct 13, 2015 #4
    i think until the factor [tex](\frac{1}{3})^{M-1}[/tex] approachs zero
    so im not sure but may i put this factor to zero
    [tex](\frac{1}{3})^{M-1}=0[/tex]
    then i can calculate M
    right?
     
  6. Oct 13, 2015 #5

    Ray Vickson

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    That equation is very easy to solve: its solution is ##M = + \infty##.

    However, that equation has absolutely nothing to do with the original question. I suggest you re-read the problem.
     
    Last edited: Oct 13, 2015
  7. Oct 13, 2015 #6

    vela

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    Forget about the formula for a minute, and try working out what distance the ball travels between the first bounce and the second (M=2) bounce, the first bounce and the third (M=3) bounce, and so on. Once you see the pattern, write down the appropriate series for the Mth bounce, and then apply the sum formula to that series.

    You should see your guesses about M vs. M-1 don't really make sense.
     
  8. Oct 14, 2015 #7
    the series is not difficult
    [tex]S_M=a+ar^2+ar^3+...ar^M[/tex]
    substituting my information
    [tex]S_M=2[9+9\frac{1}{3}+9(\frac{1}{3})^2+....+9(\frac{1}{3})^M][/tex]

    if i want the distance between the first bounce and the second bounce i have to calculate the first 2 terms of the series.
     
  9. Oct 14, 2015 #8

    vela

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    Really? After the first bounce, it reaches a height of 9 m, so the total distance is travels until the second bounce is 18 m. That's not equal to the sum of the first two terms of your series.
     
  10. Oct 14, 2015 #9
    Hey i get it. The distance traveled between the first bounce and the second bounce is just the first term.

    The distance traveled between the first bounce and the third bounce is the first two terms.

    Therefore the distance between the first bounce and the Mth bounce is adding all terms until (Mth-1)
    Right?

    But if the question just said ( find the total distance ?)
    I think i have to add 27m to the final sum and replace (M-1) by (M)
     
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