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### AuthorTopic: Recreational Problems (Physics)  (Read 6055 times) Tweet Share

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##### Recreational Problems (Physics)
« on: June 29, 2008, 02:08:49 pm »
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An astronaut is marooned from his spaceship in the Dagobah system. He is at a distance x from it. He has a mass of M He has a tank of oxygen of mass m and he breaths at a rate R kg/second.

He can use the oxygen to get himself back to the spaceship. Oxygen will squirt out of his tank into space with a speed of v. If he uses too much oxygen he will move quicker, but may not be able to breathe; if he uses too little he will get back too slowly and may also run out of oxygen to breathe. He is only allowed one squirt which expels a mass Δm of oxygen.

Assume that m << M (i.e. M + m = M)

a) Find the possible values for m in terms of the other variables so the astronaut will make it back to the ship.

b) In terms of the variables, what is the greatest distance (X) he can be from the spaceship and still survive?

Hint: Conservation of Momentum

Solution

By conservation of momentum: $(M+m-\Delta m)V=\Delta m v$     But $m, \Delta m <
To have enough oxygen left: $m- \Delta m \geq Rt$
And also: $X = Vt$

After playing around with those three equations you reach
$v (\Delta m)^2-mv \Delta m +RXM \leq 0$ (at this stage graphing $\Delta m$ is helpful in visualising the inequality)
For this to be satisfied, we must have the discriminant $m^2v^2-4RMXv \geq 0$
So the values of m are (as $m \geq 0$): $\Rightarrow m \geq 2\sqrt{\frac{RMX}{v}}$

For the maximum distance, we require the discriminant to be equal to zero $\Rightarrow X_{max} = \frac{m^2v}{4RM}$
« Last Edit: July 04, 2008, 07:30:51 pm by DivideBy0 »

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##### Re: Recreational Problems (Physics)
« Reply #1 on: July 03, 2008, 03:15:10 pm »
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A cyclist starts from rest at a height h and rides his bike down a frictionless loop-the-loop track as shown in the diagram.

a) Find the minimum height h in terms of r so that he doesn't fall off.

Part a) solved by phagist_

b) Hence, if the rider starts at the minimum possible height, find the acceleration at point B in terms of h and g
« Last Edit: July 03, 2008, 04:18:54 pm by DivideBy0 »

#### phagist_

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##### Re: Recreational Problems (Physics)
« Reply #2 on: July 03, 2008, 03:50:59 pm »
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Ok, so we need to find the conditions that are required for the bicycle to stay on the track at $B$.
They are that the centripetal force supplied by the loop ($\frac{mv^2}{r}$) needs to be greater than $mg$.

The initial energy of the system is all gravitational potential which is

$mgh$

and at part $B$ this energy is transformed into some kinetic (note that some potential still remains)

$Energy ~at~ B$;

$mg2r + \frac{1}{2}mv^2$ (where $v$ is the minimum speed required to stay on the loop)

solving $\frac{mv^2}{r}=mg$

=> $v^2=gr$

subbing in $v^2$ in terms of $g~and~r$;
Total Energy required to complete loop in terms of $g,~r~and~m$;
$mg2r+\frac{1}{2}mgr=\frac{5}{2}mgr$

Since energy is conserved we can equate the initial gravitational potential ($mgh$) with this new expression.

$mgh=\frac{5}{2}mgr$

cancelling $m~and~g$

$h=\frac{5}{2}r$
but since h needs to be greater than this height, otherwise it would give a 0 reaction force at point B.

so $h>\frac{5}{2}$

I think it's right (had a similar problem at uni).

#### /0

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##### Re: Recreational Problems (Physics)
« Reply #3 on: July 03, 2008, 04:53:02 pm »
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3. Consider an infinite network consisting of resistors (resistance of each of them is r) shown in the diagram. Find the resultant resistance $R_{AB}$ between the points A and B.

« Last Edit: July 03, 2008, 09:11:23 pm by DivideBy0 »

#### Mao

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##### Re: Recreational Problems (Physics)
« Reply #4 on: July 03, 2008, 07:51:06 pm »
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Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

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##### Re: Recreational Problems (Physics)
« Reply #5 on: July 03, 2008, 08:14:18 pm »
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#### bigtick

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##### Re: Recreational Problems (Physics)
« Reply #6 on: July 03, 2008, 08:59:03 pm »
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3. Consider an infinite network consisting of resistors (resistance of each of them is r) shown in the diagram. Find the resultant resistance $R_{AB}$ between the points A and B.
(3+sqrt5)r/(1+sqrt5)

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##### Re: Recreational Problems (Physics)
« Reply #7 on: July 03, 2008, 09:10:50 pm »
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That is correct bigtick, how did you get the solution though?

#### bigtick

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##### Re: Recreational Problems (Physics)
« Reply #8 on: July 03, 2008, 09:15:33 pm »
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That is correct bigtick, how did you get the solution though?

#### enwiabe

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##### Re: Recreational Problems (Physics)
« Reply #9 on: July 03, 2008, 09:43:05 pm »
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Fibonacci said to stop beating around the bush and show your <expletive deleted> working.

#### bigtick

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##### Re: Recreational Problems (Physics)
« Reply #10 on: July 03, 2008, 10:05:36 pm »
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Fibonacci said to stop beating around the bush and show your <expletive deleted> working.
<expletive deleted> off, you prick

I withdraw the comment. It was too gentle.
« Last Edit: July 03, 2008, 10:13:07 pm by bigtick »

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##### Re: Recreational Problems (Physics)
« Reply #11 on: July 04, 2008, 01:10:14 am »
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Don't worry, the solution to the problem is here (It is IPhO '67 Problem 2)

I was just wondering if you had an alternative way of solving (I suspected you might because of the unrationalized denominator).

#### bigtick

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##### Re: Recreational Problems (Physics)
« Reply #12 on: July 04, 2008, 09:11:49 am »
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[IMG]http://img161.imageshack.us/img161/8776/effectiveresistance2sb5.gif[/img]
[URL=http://g.imageshack.us/g.php?h=161&i=effectiveresistance2sb5.gif]
« Last Edit: July 04, 2008, 09:14:34 am by bigtick »

#### bigtick

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##### Re: Recreational Problems (Physics)
« Reply #13 on: July 04, 2008, 03:18:24 pm »
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Let r be 1 unit of resistance.
Resultant resistance is a continuous fraction 1+1/(1+1/(1+1/....)) = (1+rt5)/2
« Last Edit: July 04, 2008, 03:25:24 pm by bigtick »

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##### Re: Recreational Problems (Physics)
« Reply #14 on: July 04, 2008, 04:32:06 pm »
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Wow very nice mathematical solution ^^