05 Mar 2008

Puzzle: Will the ladybugs meet? If yes, when? If not, why?

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Puzzle Taken From : www.archimedes-labs.org
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© Archimedes Labs

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[FONT=Verdana, Arial, Helvetica, sans-serif]Infinite beetle path[/FONT]
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A high quality rubber band is fastened and hung from a horizontal pole with a cannonball at its end. Two facing ladybugs are crawling along this rubber band toward each other.
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[FONT=Verdana, Arial, Helvetica, sans-serif] From their respective starting positions (8 cm apart -- see image), each small beetle crawls toward the other at a speed of 1 cm per second. However, in the length of time each beetle crawls 1 cm, the cannonball, thanks to the force of gravity, stretches the rubber band an additional 8 cm. Will the poor ladybugs ever meet? And, if yes, when? If not, why?![/FONT]


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gohm

gohm

Branch Unspecified
08 Mar 2008
OK, probably will fail miserbly but here's my thought-

Assuming the band is strong enough to handle the weight of the ball/gravity force without breaking, yes they will meet. If the band breaks the poor bugs are in trouble! =) They will meet because eventually the band will either reach equilibrium with the force of gravity on the ball or the ball will come to rest on a surface which absorbs the force (like the ground) also halting the stretching of the band. Thus the bugs will continue to move, closing the gap & eventually meeting once the band no longer stretches. We also need to know the mass of the cannonball so as to calculate the total stretching distance to then calculate when the bugs will meet.
gohm

gohm

Branch Unspecified
16 Mar 2008
Ok, I am wallowing in curiosity as to the answer to this!
16 Mar 2008
Oops. Anyone wants to scratch brain over this problem?

I'll post the answer soon.
satheesh27887

satheesh27887

Branch Unspecified
28 Jul 2008
yes... the bugs meet... but it take very long time.....
since the band is stretching 8cm when the bug moves 1cm,the stiffness of the band is very low.....
KSHIRABDHI

KSHIRABDHI

Branch Unspecified
28 Jul 2008
yes the bugs will meet after a certain duration when the band reaches the limit of its elasticity after which the distance to be travelled by bugs will become a constant distance.........bt it is only possible if the band doesn't break
01 Sep 2008
[FONT=Verdana, Arial, Helvetica, sans-serif]Source: Previous monthly puzzles: October-November 2007

Yes, the beetles will meet! Even if D0 were 8 cm (one ladybug on the pole, the other on the cannonball), the change is thus:
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[FONT=Verdana, Arial, Helvetica, sans-serif]Sec.
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]Band Length[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]Bugs Gap[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]1)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D1 = 16 [cm];[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A1 = 12 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]2)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D2 = 24 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A2 = 15 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]3)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D3 = 32 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A3 = 17 1/3 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]4)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D4 = 40 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A4 = 19 1/6 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]5)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D5 = 48 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A5 = 20.6 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]6)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D6 = 56 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A6 = 21.7 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]7)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D7 = 64 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A7 = 22 4/7 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]8)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D8 = 72 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A8 = 23 3/35 [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]9)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D9 = 80 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A9 = 23.422... [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]10)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D10 = 88 [cm];
[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]A10 = 23.564... [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]11)[/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]D11 = 96 [cm][/FONT]; [FONT=Verdana, Arial, Helvetica, sans-serif]A11 = 23.524... [cm][/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]The beetles are now closing faster than the distance between them grows. After 30 seconds, the rubber band is 248 cm long and the bugs are 0.316... [cm] apart. They will meet just before the next second.[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]But... There is a more mathematical way to solve this puzzle without ITERATIVE calculations![/FONT]​
[FONT=Verdana, Arial, Helvetica, sans-serif]Consider that each time the cannonball stretches the rubber band, the bug comes with it, which means that the FRACTION of band left to cross stays the same.[/FONT]​
[FONT=Verdana, Arial, Helvetica, sans-serif]Each beetle moves specularly to its counterpart toward the central point of the rubber band. Their relative position being after each move specular to each other, to solve this puzzle we just need then to take into consideration the moves of one ladybug on a HALF section of the rubber band.[/FONT]​
[FONT=Verdana, Arial, Helvetica, sans-serif]OK. After one crawl, the beetle has done 1 cm, that is 1/4 of the rubber band (fig. a.1), while the cannonball stretches the band to 8/2 = 4 cm (and the ladybug is now 2 cm along). The beetle crawls 1 cm again, which is now 1/(4 + 4) = 1/8 of the total band (fig. a.2). His next crawl will only be 1/(8 + 4) = 1/12 of the band (fig. a.3). And so on...
The fraction of band that the ladybug has crawled at 6 steps is:
1/4 + 1/8 + 1/12 + 1/16 + 1/20 + 1/24
or
(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6)/4
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[FONT=Georgia, Times New Roman, Times, serif]fig. a)[/FONT][​IMG]
[FONT=Verdana, Arial, Helvetica, sans-serif]This is more evident when we analyze the beetle's route until step 6. As shown in fig. b) below, the ladybug has covered m + m/2 + m/3 + m/4 + m/5 + m/6, but the total route it must crawl to meet its counterpart is 4m long. Then the fractions of band that both ladybugs have to crawl respectively until the meeting are:
m + m/2 + m/3 + m/4 + m/5 + m/6 + ... + m/n = 4m
or
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/n = 4
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[FONT=Georgia, Times New Roman, Times, serif]fig.b)[/FONT][​IMG]
[FONT=Verdana, Arial, Helvetica, sans-serif]The sum 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/n represents the first n terms of consecutive reciprocals of natural numbers called "harmonic series"[/FONT].​
[FONT=Verdana, Arial, Helvetica, sans-serif]There is a relatively 'simple' formula to estimate the sum H of n terms of an harmonic series:
Hn ≈ ln(n) + 0.5772156649 + 1/2n (it gets closer, the larger n is)
Since in our case Hn = 4, therefore
ln(n) + 0.5772156649 + 1/2n ≈ 4
ln(n) + 1/2n ≈ 3.422784336
[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]We can omit here 1/2n, then the equation becomes:
ln(n) ≈ 3.422784336
n ≈ e3.422784336 ≈ 30.65464915...
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[FONT=Verdana, Arial, Helvetica, sans-serif]Approximatively one half second after 30 seconds, the ladybugs will meet.[/FONT]​

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