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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: /0 on June 29, 2008, 02:08:49 pm

Title: Recreational Problems (Physics)
Post by: /0 on June 29, 2008, 02:08:49 pm: Yay! A Physics problem thread!

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 $m, \Delta m <<M \Rightarrow MV = \Delta mv$
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 $\Rightarrow X_{max} = \frac{m^2v}{4RM}$
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 03, 2008, 03:15:10 pm: 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
Title: Re: Recreational Problems (Physics)
Post by: phagist_ on July 03, 2008, 03:50:59 pm: 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).
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 03, 2008, 04:53:02 pm: 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.

Answer given by bigtick :)
Title: Re: Recreational Problems (Physics)
Post by: Mao on July 03, 2008, 07:51:06 pm: (http://imgs.xkcd.com/comics/nerd_sniping.png)
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 03, 2008, 08:14:18 pm: Quote from: Mao on July 03, 2008, 07:51:06 pm
(http://imgs.xkcd.com/comics/nerd_sniping.png)

LMAO! Nerd sniping ;D
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 03, 2008, 08:59:03 pm: Quote from: DivideBy0 on July 03, 2008, 04:53:02 pm
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)
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 03, 2008, 09:10:50 pm: That is correct bigtick, how did you get the solution though?
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 03, 2008, 09:15:33 pm: Quote from: DivideBy0 on July 03, 2008, 09:10:50 pm
That is correct bigtick, how did you get the solution though?
Ask Fibonacci
Title: Re: Recreational Problems (Physics)
Post by: enwiabe on July 03, 2008, 09:43:05 pm: Fibonacci said to stop beating around the bush and show your <expletive deleted> working.
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 03, 2008, 10:05:36 pm: Quote from: enwiabe on July 03, 2008, 09:43:05 pm
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.
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 04, 2008, 01:10:14 am: Don't worry, the solution to the problem is here (It is IPhO '67 Problem 2)

http://www.jyu.fi/kastdk/olympiads/

I was just wondering if you had an alternative way of solving (I suspected you might because of the unrationalized denominator).
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 04, 2008, 09:11:49 am: [IMG]http://img161.imageshack.us/img161/8776/effectiveresistance2sb5.gif[/img]
[URL=http://g.imageshack.us/g.php?h=161&i=effectiveresistance2sb5.gif]
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 04, 2008, 03:18:24 pm: Let r be 1 unit of resistance.
Resultant resistance is a continuous fraction 1+1/(1+1/(1+1/....)) = (1+rt5)/2
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 04, 2008, 04:32:06 pm: Wow very nice mathematical solution ^^
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 05, 2008, 10:37:35 pm: Here is one about resistance.
http://itute.com/board/viewtopic.php?t=321
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 05, 2008, 11:24:34 pm: Here are another ~~two~~ three (aren't resistors fun):

5. A speeding motorbike passes an unmarked police car on the highway. The bike is traveling at a speed of $v$ , and the police car at a speed of $V$ . After a delay of 1 second, the police car accelerates constantly at $a\ ms^{-2}$ until he catches the bike. The bike does not accelerate, but travels on at the same speed.
Calculate the distance it takes the police car to overtake the bike in terms of a, V and v.

6. In the (first) diagram, all resistors have resistance $4 \ \Omega$ and all the ideal batteries have emf $4 \ V$ . What is the amount of current and direction of current through resistor R?

7. In the (second) diagram, $r_1=2 \Omega$ , $r_2=4 \Omega$ , $r_3=7 \Omega$ , $r_4=14 \Omega$ , $r_5=10 \Omega$ , $emf = 4V$ . Find the current through the battery.
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 06, 2008, 12:44:07 am: 7. 6/7 A

5. From the moment the bike passed the police. (v/a){v-V+a+sqrt[(v-V)(v-V+2a)]}

6.
(http://img352.imageshack.us/img352/6854/vcenotesforum2ag7.gif)
I=8/4=2A down
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 06, 2008, 10:12:25 pm: 6 and 7 are correct, but for 5 I got

$x = \frac{v\left(v-V+a+\sqrt{v^2+2v(a-V)+V^2}\right)}{a}$

Quote from: bigtick on July 05, 2008, 10:37:35 pm
Here is one about resistance.
http://itute.com/board/viewtopic.php?t=321

If the current just before P is $i$ , then the current will split equally at the junction and $\frac{i}{3}$ of current will go through each $r \ \Omega$ resistor. Similarly, a currrent of $\frac{i}{3}$ will go through each $3r \ \Omega$ resistor. $\frac{\frac{i}{3}}{2}=\frac{i}{6}$ of current goes through the $2r \ \Omega$ resistors because it must split again.

So the equivalent resistance is:

$R_{eq} = \frac{V}{I}=\frac{\frac{i}{3} \times r + \frac{i}{3} \times 3r + \frac{i}{6} \times 2r}{i}=\frac{5r}{3}$
Title: Re: Recreational Problems (Physics)
Post by: bigtick on July 07, 2008, 10:52:07 am: For Q5
Try v=4, V=2 and a=1 to find whether the two displacements are the same.
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 07, 2008, 10:59:52 pm: Quote from: bigtick on July 07, 2008, 10:52:07 am
For Q5
Try v=4, V=2 and a=1 to find whether the two displacements are the same.

Sorry, yeah that's correct

I missed the $V$ in $x = Vt_p+\frac{1}{2}at_p^2+V$
Title: Re: Recreational Problems (Physics)
Post by: /0 on June 30, 2009, 11:48:47 pm: You are in a car holding a helium balloon. The car suddenly swerves to the right. In which direction will the balloon move?
Title: Re: Recreational Problems (Physics)
Post by: evaporade on July 01, 2009, 12:09:52 am: to the right
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 01, 2009, 12:11:51 am: correct!
Title: Re: Recreational Problems (Physics)
Post by: TrueTears on July 01, 2009, 12:16:08 am: A baseball with mass 0.14 kilogram and speed 40 m/s is caught. If all of its kinetic energy is converted to heat as it is caught, and all of the heat is absorbed by the ball, what is its temperature change? (Assume the baseball's specific heat capacity is 1,000 joules per kilogram degree Celsius.)
Title: Re: Recreational Problems (Physics)
Post by: Mao on July 01, 2009, 12:37:20 am: +0.8 K (assuming no other forms of energy, no energy loss)
Title: Re: Recreational Problems (Physics)
Post by: TrueTears on July 01, 2009, 12:50:45 am: Correct.
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 01, 2009, 12:56:04 am: Another few (I haven't tried solving these yet so I hope they work but it's still ROCKET SCIENCE fo shizle)

a) The Snobbits of planet Kazak wonder what is beyond the sky, so they launch a rocket vertically upwards from the ground to take photos.
The mass of planet Kazak is $10^{24}kg$ and its radius is $10^6$ m. The material to build the rocket weighs $500kg$ and the rocket contains $100kg$ of potassium used for thrust. If the potassium is exhausted at a constant rate of $1kg/s$ and ~~each kilogram~~ produces $10^5N$ of thrust, find the maximum height reached by the rocket.

b) The inhabitants of planet Australia already know the mass of their planet ( $10^{24}kg$ ) but they don't know the radius. Being the practically-minded geniuses they are, they decide to send up several rockets into space and measure the speed of the rocket at infinity. Eventually they find that a rocket with the same specifications as the Snobbits' rocket is stationary at infinity. What is the radius of Australia?
Title: Re: Recreational Problems (Physics)
Post by: kamil9876 on July 01, 2009, 01:40:19 am: a.) Why did I get some fucked up differential equation? :-\

Maybe there is a move clever way of doing it, it's too late :P
Title: Re: Recreational Problems (Physics)
Post by: TrueTears on July 01, 2009, 03:00:07 am: I just have to lol at "planet Kazak"
Title: Re: Recreational Problems (Physics)
Post by: zzdfa on July 02, 2009, 10:51:38 am: you are trapped in the centre of a circular pen, radius R.

there is a dog on the outside of the pen, it always runs towards the point on the fence that you are closest to.

The dog has a speed 4v, you have speed v. How can you reach the edge of the pen before the dog?
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 03, 2009, 03:01:45 pm: Let the pen have radius $R$ .

You have to run in a small circle around the middle post and then when the dog is exactly opposite for you, dash for the fence.

If you're running straight for the fence and want to just outrun the dog then you must leave a distance r from the circle, where:

$\frac{R-r}{v} < \frac{\pi R}{4v}$

$\therefore r > \left(1-\frac{\pi}{4}\right)R=0.2146R$

However, as long as you keep within a radius of $0.25R$ your angular velocity will be greater than the dog's, allowing you to position yourself opposite the dog.

So circle between $0.2146R$ and $0.25R$ until the dog is opposite you, then run to the edge.
Title: Re: Recreational Problems (Physics)
Post by: /0 on July 03, 2009, 03:33:50 pm: 1. You are on a bumpy train ride, drinking mineral water out of a cylindrical cup. How much should you drink so that, after placing your beverage on a rattling table, it is most stable?

2. An astronaut in space turns on two flashlights simultaneously and pointing in opposite directions, sending out two beams that both move away from the astronaut at the speed of light. How fast does the front of one beam move away from the front of the other beam?

3. What is the shape of an object that has the greatest free-fall acceleration at a particular point on its surface?

4. Object A is moving at velocity $v$ and collides elastically with stationary object B, what is the greatest speed object B can have after the collision and why?
Title: Re: Recreational Problems (Physics)
Post by: /0 on August 07, 2009, 04:24:55 pm: A helicopter is hovering stationary above the ground. It has a propeller overhead of radius $L$ m that rotates with a frequency of $f$ Hz. A uniform magnetic field of strength $B$ T goes into the ground. Find the EMF between the edge of the propeller and the centre.

(Sorry fixed the problem: the helicopter is not moving)