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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: dekoyl on March 18, 2009, 09:54:25 pm

Title: Dekoyl's Questions
Post by: dekoyl on March 18, 2009, 09:54:25 pm: I just encountered a centre of mass question which I'm unfamiliar with and has never been mentioned by my teacher.
Are these type of questions in the course? The question also requires the equation:

$x_{cm}=\frac{m_1x_1 + m_2x_2}{m_1+m_2}$

Thank you
Title: Re: Centre of mass?
Post by: TrueTears on March 18, 2009, 09:56:18 pm: Hmmm I met one of these questions in the Jacaranda book, and only ONE lol

I don't think centre of mass is in the course, but it's good to know, not hard to remember the equation.
Title: Re: Centre of mass?
Post by: /0 on March 18, 2009, 09:56:29 pm: Haven't read the 2009 study design but I've seen questions on them in 2009 Jacaranda and also I think you're quite likely to use it in Structures and Materials.
Title: Re: Centre of mass?
Post by: dekoyl on March 18, 2009, 09:57:20 pm: Quote from: TrueTears on March 18, 2009, 09:56:18 pm
Hmmm I met one of these questions in the Jacaranda book, and only ONE lol

I don't think centre of mass is in the course, but it's good to know, not hard to remember the equation.
Yeah I'm referring to that ONE. =P

Quote from: /0 on March 18, 2009, 09:56:29 pm
Haven't read the 2009 study design but I've seen questions on them in 2009 Jacaranda and also I think you're quite likely to use it in Structures and Materials.
Ah okay we have a different DS =]

Thanks TT and /0.
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 28, 2009, 03:16:39 pm: Hmm.

We are given a scenario with a car towing a trailer (with a pull bar).
The car has a constant driving force of 2400N. The car experiences 400N friction and the trailer experiences 200N friction.

We're told to calculate the work done by the car over a distance.
Using $W=Fx$ , for F I used 2400 - 400 - 200 = 1800N. However in the solutions, they used F = 2400

Am I wrong? I thought it was $\mbox{W=F_{\mbox{net}}d}$ . Was $\mbox{F_{\mbox{driving force}}=F_{net}}$ ?

Thanks
Title: Re: Dekoyl's Questions
Post by: TrueTears on March 28, 2009, 03:32:26 pm: Yes, answers are right, you gotta becareful here because it says the work DONE BY the CAR. It does not matter whether there's friction or anything else acting on the car, but only the forces that the CAR ITSELF exerts.
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 28, 2009, 03:38:57 pm: Quote from: TrueTears on March 28, 2009, 03:32:26 pm
Yes, answers are right, you gotta becareful here because it says the work DONE BY the CAR. It does not matter whether there's friction or anything else acting on the car, but only the forces that the CAR ITSELF exerts.
Argh the language.

Thanks man. TrueTears to the rescue again :P
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 28, 2009, 05:05:32 pm: In physics, if you don't use the method assessors expect, would you lose marks even if you get the solution right?

I encountered a question where the knowledge of $GPE\rightarrow KE$ needed to be applied to calculate velocity but I used $v^{2} = u^{2} + 2as$ and got the correct answer. As long as the correct answer is obtained, any method is fine right?

Thanks
Title: Re: Dekoyl's Questions
Post by: TrueTears on March 28, 2009, 05:12:07 pm: Quote from: dekoyl on March 28, 2009, 05:05:32 pm
In physics, if you don't use the method assessors expect, would you lose marks even if you get the solution right?

I encountered a question where the knowledge of $GPE\rightarrow KE$ needed to be applied to calculate velocity but I used $v^{2} = u^{2} + 2as$ and got the correct answer. As long as the correct answer is obtained, any method is fine right?

Thanks
yeap as long as your answer is correct, give your working method makes sense and is also logical, then you still get the marks.

If you get the answer but your working was not logically, then you won't get the method marks.
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 28, 2009, 05:14:11 pm: Alrighty =] Thanks again TT.

I'm asking all these questions because I have a SAC on Monday. Yep =|
Title: Re: Dekoyl's Questions
Post by: Over9000 on March 28, 2009, 08:17:25 pm: Impulse = $F\triangle t=m\triangle v$ , so you can use both (whichever one you have more values for)
Title: Re: Dekoyl's Questions
Post by: TonyHem on March 28, 2009, 08:17:31 pm: I thought its the same as the formulas u mentioned above, it just depends on the information given/known.
I think..
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 02:18:34 am: First one I'm not sure what I've done wrong.
A 20kg wagon is pulled along the level ground by a rope inclined at $30^{o}$ above the horizontal. A friction force of 30 N opposes the motion. How large is the pulling force if the wagon is moving with an acceleration of $0.4 ms^{-2}$ ?

Okay - after taking the friction and angle of the rope into account, I end up with:
$F_{pulling force}=34.64 + 20(0.4)$
$= 42.64N$
The answer, however, is $43.9 N$ .

Second one: I'm just not sure what to do as I can't picture the question =S
An inclined plane making an angle of $25^{o}$ with the horizontal has a pulley at its top. A 30kg block on the plane is connected to a freely hanging 20kg block by means of a cord passing over the pulley. Compute the distance the 20kg block will fall in 2s starting from rest. Neglect friction.

Thanks everyone.
Title: Re: Dekoyl's Questions
Post by: Over9000 on March 29, 2009, 02:36:53 am: For q.1
F=ma
F= $20 \times 0.4$
F= 8N
so the total net force is 8N
if friction force is 30N, there must be a force of 38N acting in the opposite direction (i.e direction of pulling) horizontally, however we must find the pulling force at $30^{\circ}$ to horizontal
Therefore the force is $\frac{38}{cos(30)}$ (draw a triangle if you dont understand this part)
This gives 43.8786N (4 d.p)
Which is 43.9N (1 d.p)
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 02:40:48 am: ^Ah crap. My answer to the first question was a fluke then :(

Thanks Over9000 :D Didn't expect anyone to help at this hour
Title: Re: Dekoyl's Questions
Post by: Over9000 on March 29, 2009, 02:50:09 am: btw, wats the answer to the second question, I have 7.3m, not sure if its right?? (ill post my working if it is)
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 02:51:47 am: Quote from: Over9000 on March 29, 2009, 02:50:09 am
btw, wats the answer to the second question, I have 7.3m, not sure if its right?? (ill post my working if it is)
The answer is 2.87m.
Title: Re: Dekoyl's Questions
Post by: Over9000 on March 29, 2009, 02:52:46 am: oh, kk, way off, ill have to check where I made an error ;D
Title: Re: Dekoyl's Questions
Post by: Over9000 on March 29, 2009, 03:08:49 am: Ok, I shall try to explain what the picture is like
Inclined plane $25^{\circ}$ to horizontal, Imagine a block halfway (doesnt have to be) along the incline plane and a pulley (simple one pulley) at the very top of the plane, now, the rope connects to the block which is on the inclined plane goes over the single pulley and then downwards connected to another lighter block. So you can imagine a block on an inclined plane, connected to a block which is hanging freely in mid air via the pulley system (I dont know if you will understand this, a bit hard to explain).

So $T +mgSin\theta$ (block on inclined plane) = $mg$ (block hanging freely)
$T= mg- mgSin\theta$
$T= 20\times 9.8 - 30 \times 9.8 sin(25)$
$T= 196 - 124.25$
$T=71.75N$
$F=ma$
$71.25=50a$
$a=1.435ms^{-2}$
use constant acceleration forumla $x=ut + \frac{1}{2}at^{2}$
$x= 0 + \frac{1}{2} \times 1.435 \times 4$
$x= 2.87m$

If you still cant picture it (which may make the working out seem hard to comprehend), send me a pm and ill scan my working out of the question by hand.
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 03:18:04 am: No it's okay ;D Your explanation was great so I got it.
Thanks heaps again Over9000 =]
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 03:23:03 pm: Momentum: A heavy truck collides head on with a car one fifth its mass. The car and truck were both travelling at 20 $kmh^{-1}$ above the speed limit. During the collision, what is the ratio? Answer: 1

Thanks!
Title: Re: Dekoyl's Questions
Post by: dcc on March 29, 2009, 03:41:46 pm: Quote from: dekoyl on March 29, 2009, 03:23:03 pm
Momentum: A heavy truck collides head on with a car one fifth its mass. The car and truck were both travelling at 20 $kmh^{-1}$ above the speed limit. During the collision, what is the ratio? Answer: 1

What is the ratio? Which ratio?
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 03:44:38 pm: Quote from: dcc on March 29, 2009, 03:41:46 pm
What is the ratio? Which ratio?
That's the exact question. =(
It's meant to be in relation to momentum and impulse. And the exact answer is 1. (Ratio of 1:1?)
Title: Re: Dekoyl's Questions
Post by: TrueTears on March 29, 2009, 03:49:56 pm: Quote from: dcc on March 29, 2009, 03:41:46 pm
Quote from: dekoyl on March 29, 2009, 03:23:03 pm
Momentum: A heavy truck collides head on with a car one fifth its mass. The car and truck were both travelling at 20 $kmh^{-1}$ above the speed limit. During the collision, what is the ratio? Answer: 1

What is the ratio? Which ratio?
yeah, what ratio does it ask? ratio of the speed? or...?
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 03:56:23 pm: Quote from: TrueTears on March 29, 2009, 03:49:56 pm
yeah, what ratio does it ask? ratio of the speed? or...?
This is from page 11 of "Motion question booklet.doc" in the thread you created with motion notes + practice SACs.
I'm not sure; This is the exact question (copy + paste):
Quote
20. A heavy truck collides head on with a car one fifth of its mass. The car and the truck were both travelling at 20 kmh-1 above the speed limit.
During the collision what is the ratio ?

Ans: 20) 1

Yeah looks like there might be something missing judging from the space between 'ratio' and '?'.

Thanks anyway dcc and TT
Title: Re: Dekoyl's Questions
Post by: TrueTears on March 29, 2009, 04:08:25 pm: hmm yeah, might be a faulty question lol
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 09:30:26 pm: A simple one:

A rubber ball bounces from a height of 1.0m onto a wooden table and rebounds to a height 0.8m.
Explain how the momentum is conserved during the collision between the ball and the table.
Worth 2 marks and I didn't get any marks. :(
Title: Re: Dekoyl's Questions
Post by: TrueTears on March 29, 2009, 09:36:07 pm: Momentum is always conserved, here the ball does not rebound to its original height because some momentum is transferred into the table.

Considered the table as a car and the ball as a much lighter car.

The table car is stationary, where as the ball car crashes head on into the table car.

Assuming they both rebound, the table car rebounds a tiny bit (most likely none here cos the ball is so light) and the ball car bounces back in the direction where it came from, but it's lost some speed (because its transferred some momentum into the table car) and so it does not go back fully where it started.
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 09:38:42 pm: Thanks TT - I guess my explanation wasn't good enough :P But your explanation should please my marker.
Title: Re: Dekoyl's Questions
Post by: Flaming_Arrow on March 29, 2009, 09:41:01 pm: Quote from: TrueTears on March 29, 2009, 09:36:07 pm
Momentum is always conserved, here the ball does not rebound to its original height because some momentum is transferred into the table.

Considered the table as a car and the ball as a much lighter car.

The table car is stationary, where as the ball car crashes head on into the table car.

Assuming they both rebound, the table car rebounds a tiny bit (most likely none here cos the ball is so light) and the ball car bounces back in the direction where it came from, but it's lost some speed (because its transferred some momentum into the table car) and so it does not go back fully where it started.

And the Normal Reaction Force> Weight, otherwise it wouldn't bounce.
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 11:03:25 pm: Hmm the last one for tonight hopefully.

How is the answer to part b that?
(http://i42.tinypic.com/qx1heg.png)
Title: Re: Dekoyl's Questions
Post by: Mao on March 29, 2009, 11:20:24 pm: net acceleration is -a

that means, $F_{up} - F_{down} = -aM$

if you want to deal with the entire system: then the driving force is the only uphill force, the downhill forces are friction (on both objects) and pull due to gravity (on both objects)
$\implies F_{traction} - (21000\sin 23^o + 17000\sin 23^o + 890 + 920) = -a \times (1700 + 2100)$

if you want to deal with the car only (gives the same answer), then the driving force is the only uphill force, the downhill forces are friction (on car only), pull by gravity (on car only), and tension
$\implies F_{traction} - (7500 + 17000\sin 23^o + 890) = -a \times (1700)$

to calculate net acceleration, find the forces on the trailer, going uphill is the tension force, going downhill is friction and gravity
$\implies -a = \frac{7500 - (21000\sin 23^o + 920)}{2100}$
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 29, 2009, 11:37:53 pm: Ah awesome. Thanks Mao =]
Title: Last-minute-before-SAC question
Post by: dekoyl on March 30, 2009, 12:11:00 am: Gah.. another one.

Part b again. Should it be $m\triangle v = 0.156(26 - 40)$ because it's force on bat BY ball? Solutions has 1.2(12.7-4.12) ie. the bat
(http://i44.tinypic.com/14ui3xy.png)
Title: Re: Dekoyl's Questions
Post by: /0 on March 30, 2009, 12:27:19 am: Force on bat by ball = Force on ball by bat, just in opposite directions (Newton 3).

Also, if you want the impulse on the ball, it should be $0.156(26-(-40)) = 10.296kgms^{-1}$ , since you have to take into account the direction of the initial and final velocity vectors.
Title: Re: Dekoyl's Questions
Post by: dekoyl on March 30, 2009, 12:50:14 am: Quote from: /0 on March 30, 2009, 12:27:19 am
Force on bat by ball = Force on ball by bat, just in opposite directions (Newton 3).

Also, if you want the impulse on the ball, it should be $0.156(26-(-40)) = 10.296kgms^{-1}$ , since you have to take into account the direction of the initial and final velocity vectors.
Yeesh :( Forgot all about that. Not good.

Anyway thanks saviour /0. ;D
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 16, 2009, 08:49:03 pm: Let $d$ =greatest horizontal distance of ball. Find the maximum vertical height to which the ball can be thrown in terms of $d$ . Assume the initial speed is the same in either case.

Thanks
Title: Re: Dekoyl's Questions
Post by: TrueTears on April 16, 2009, 08:57:44 pm: Assuming parabolic path.

$d = \frac{u^2sin(2\theta)}{g}$

solving for $u^2$ leads to $\frac{dg}{sin(2\theta)}$

sub this in $h_{max} = \frac{u^2sin^2(\theta)}{2g}$ yields

$h_{max} = \frac{\frac{dg}{sin(2\theta)}sin^2(\theta)}{2g}$

$sin(2\theta) = 2sin(\theta)cos(\theta)$

subbing this in : $h_{max} = \frac{d}{4}tan(\theta)$
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 19, 2009, 06:51:22 pm: ^Thanks TT

How would one prove that the maximum range of a projectile is launched at $45^{o}$ ?
Title: Re: Dekoyl's Questions
Post by: Flaming_Arrow on April 19, 2009, 06:57:51 pm: we know that range is $\frac{u^2 \sin (2\theta)}{g}$

subbing in $\theta = 45^o$

we get

$\frac{u^2 \sin (90)}{g}$

$\sin(90) = 1$ which is the largest sin value

then we get $\frac{u^2}{g}$

$\therefore$ $45^o$ is the angle that yields the maximum range
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 19, 2009, 07:10:54 pm: Quote from: Flaming_Arrow on April 19, 2009, 06:57:51 pm
we know that range is $\frac{u^2 \sin (2\theta)}{g}$
The book, if I remember correctly, does not use the formula $\frac{u^2 \sin (2\theta)}{g}$ :(

They derive it and I figured how they did it :P They did some implications as well, though.

Thanks FA
Title: Re: Dekoyl's Questions
Post by: Mao on April 19, 2009, 09:31:42 pm: vertical velocity: $u_v = u\sin \theta$
horizontal velocity: $u_h = u\cos \theta$

time to take to get to the top: $t_{1/2} = \frac{u_v}{g} = \frac{u\sin \theta}{g}$

$R = 2\cdot t_{1/2}\cdot u_h = \frac{u^2 (2\sin \theta \cos \theta)}{g} = \frac{u^2 \sin 2\theta}{g}$

$\frac{dR}{d\theta} = \frac{2u^2}{g}\cos 2\theta$ , $\frac{d^2R}{d\theta^2} = -\frac{4u^2}{g}\sin 2\theta$

at maximum range, $\frac{dR}{d\theta} = 0, \; \frac{d^2R}{d\theta^2} < 0,\; 0\le \theta \le \frac{\pi}{2}$

$\implies \theta = \frac{\pi}{4} = 45^o$
Title: Re: Dekoyl's Questions
Post by: Flaming_Arrow on April 20, 2009, 01:54:10 pm: Quote from: Mao on April 19, 2009, 09:31:42 pm
vertical velocity: $u_v = u\sin \theta$
horizontal velocity: $u_h = u\cos \theta$

time to take to get to the top: $t_{1/2} = \frac{u_v}{g} = \frac{u\sin \theta}{g}$

$R = 2\cdot t_{1/2}\cdot u_h = \frac{u^2 (2\sin \theta \cos \theta)}{g} = \frac{u^2 \sin 2\theta}{g}$

$\frac{dR}{d\theta} = \frac{2u^2}{g}\cos 2\theta$ , $\frac{d^2R}{d\theta^2} = -\frac{4u^2}{g}\sin 2\theta$

at maximum range, $\frac{dR}{d\theta} = 0, \; \frac{d^2R}{d\theta^2} < 0,\; 0\le \theta \le \frac{\pi}{2}$

$\implies \theta = \frac{\pi}{4} = 90^o$

i think u mean $\implies \theta = \frac{\pi}{4} = 45^o$
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 20, 2009, 03:58:41 pm: ^Heh didn't catch that.

Q30 from book.
A road is to be banked so that any vehicle can take the bend at a speed of $30ms^{-1}$ without having to rely on sideways friction. The radius of the curvature is 12m. At what angle should it be banked?

I think the book made a mistake in the answers but I'm not that familiar with this so I don't want to make any attempts of correcting the book :P

Thanks :)
Title: Re: Dekoyl's Questions
Post by: Flaming_Arrow on April 20, 2009, 04:08:25 pm: use $\frac{v^2}{rg} = \tan \theta$
Title: Re: Dekoyl's Questions
Post by: TrueTears on April 20, 2009, 04:12:47 pm: Form two simultaneous equations

$N\sin\theta = \frac{mv^2}{r}$ ...[1]

$N\cos\theta = mg$ ...[2]

Divide $\frac{[1]}{[2]}$ yields:

$\tan\theta = \frac{v^2}{gr}$

To get an idea of how the 2 equations were formed, take a look @ bottom of pg 64 right picture :P
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 20, 2009, 04:17:33 pm: Quote from: TrueTears on April 20, 2009, 04:12:47 pm
Form two simultaneous equations

$N\sin\theta = \frac{mv^2}{r}$ ...[1]

$N\cos\theta = mg$ ...[2]

Divide $\frac{[1]}{[2]}$ yields:

$\tan\theta = \frac{v^2}{gr}$

To get an idea of how the 2 equations were formed, take a look @ bottom of pg 64 right picture :P

Ah I thought they made a mistake with the $N\cos\theta = mg$ :-[

Yeah I understand now. Thanks FA and TT

(Thanks for the $\frac{v^2}{rg}=tan(\theta)$ FA and TT for showing how to get it :P)
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 22, 2009, 08:50:24 pm: Hey everyone. The notes I have aren't very elaborate and I'm not sure how the equation:
$\mbox{gravitational field strength}=\frac{GM}{r^2}$
is derived.

I kind of see it, kind of don't. :(

Thanks =]
Title: Re: Dekoyl's Questions
Post by: TrueTears on April 22, 2009, 08:53:48 pm: We know that $F = \frac{GmM}{r^2}$

now $F = mg$

$mg = \frac{GmM}{r^2}$

$g = \frac{GM}{r^2}$
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 22, 2009, 09:13:22 pm: *Doh =_=

Thanks TT.

Also, Chobits!
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 23, 2009, 11:44:31 pm: Just another one:

Why isn't it possible for a geostationary satellite do be over Melbourne?

Thanks:(
Title: Re: Dekoyl's Questions
Post by: TrueTears on April 23, 2009, 11:48:24 pm: A geostationary satellite cannot stay above Melbourne, it must be stationary over the equator. This is because it must orbit Earth's centre of mass (because that's where gravity is acting from) and a satellite over Melbourne would be too low to orbit Earth's centre of mass.
Title: Re: Dekoyl's Questions
Post by: dejan91 on April 24, 2009, 12:59:34 am: Also, a geostationary satellite must have the same orbital period as the Earth, and above the equator is the only location where this is possible.
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 26, 2009, 02:30:22 pm: ^Thanks guys.

I'd printscreen but I don't have the e-book version. Jacaranda Physics 2 Page 82 Question 48.

A disabled satellite of mass 2400kg is in orbit around Earth at a height of 2000km above sea level. It falls to a height of 800km befoe its built in rocket system can be activated to stop the fall continuing.
(There's a graph attached).
Q (c) If the speed of the satellite during its intial orbit is 6900 $ms^{-1}$ , what is its speed when the rocket system is activated?

The solution uses $K_e = \frac{1}{2}mv^2$ for before the built-in rocket and after the built-in rocket.
Could I use $v=\sqrt{\frac{GM}{R}}$ ?
My answer was $7.4$ x $10^{3} ms^{-1}$ whilst the book was $7.9$ x $10^{3} ms^{-1}$ .

Thanks
Title: Re: Dekoyl's Questions
Post by: TrueTears on April 26, 2009, 02:36:07 pm: Equate loss in $U_g$ to change in $E_k$

From b) loss in $U_g = 1.68 \times 10^{10}$

now $U_g = \triangle E_k$

$1.68 \times 10^{10} = \frac{1}{2}(2000)(v^2-6900^2)$

solving for v yields $v = 7.9 \times 10^3 ms^{-1}$ (1d.p)

Note: If you use 1.68 you would get something like $7.83 \times 10^3$ . However I used the exact value in my calc ie, 1.68... and you can round this number up to $7.9 \times 10^3$
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 26, 2009, 07:24:41 pm: Again - lack of understanding and knowledge :(

Why does the area under a gravitational force-distance graph give the energy needed to launch a satellite but the area under a gravitational field strength-distance graph give the energy per kilogram needed to launch a satellite?
(Jacaranda 2)

The answer gives something about $\frac{Fx}{m}$ = $\frac{W}{m}$ . I see that if you multiply $m$ to both sides you get W=Fx which is okay but how did they come up with this? They also mention the equation has "same units".

Thanks!
Title: Re: Dekoyl's Questions
Post by: TrueTears on April 26, 2009, 07:28:19 pm: Gravitational force - distance graph:

gravitational force is measured in Newtons

distance in metres

overall we have $Nm$ , which is the unit for work

Now consider gravitational field STRENGTH - distance graph:

gravitational field STRENGTH is measured in $Nkg^{-1}$ (Newtons PER kilogram) This comes from $E = \frac{F}{m}$ Where F is force in N and m is mass in kg

distance is measured in metres

so overall we have $Nkg^{-1}m$ ie $Nmkg^{-1}$ meaning Work per kilogram, also known as, energy per kilogram.

Therefore to find the energy to launch the satellite you must multiply by the mass.

Eg, $\frac{Nm}{kg} \times kg = Nm$ (energy)
Title: Re: Dekoyl's Questions
Post by: dekoyl on April 26, 2009, 07:32:23 pm: ^You should get paid to write the solutions for them :P
Thanks TT - I got it now =]
Title: Re: Dekoyl's Questions
Post by: dekoyl on May 11, 2009, 10:05:13 pm: Q26
(http://i43.tinypic.com/oqj12g.png)
In the answer, they had:
$U_E=\frac{1}{2}(0.30m)(35N) + 0.2m(35N)+\frac{1}{2}(0.2m)(5N)$

Where did they get the two 35N and 5N from? :(

Thanks.
Title: Re: Dekoyl's Questions
Post by: Mao on May 11, 2009, 10:28:22 pm: No idea, it should just be area under the graph...
Title: Re: Dekoyl's Questions
Post by: dekoyl on May 11, 2009, 10:45:55 pm: ^Yeah I thought so too. I used like $700N$ but they used $35N$ :S
Title: Re: Dekoyl's Questions
Post by: TrueTears on May 11, 2009, 10:50:35 pm: Yea I used 700N as well dekoyl, like Mao said, it's just the area under the graph...
Title: Re: Dekoyl's Questions
Post by: dekoyl on May 11, 2009, 11:03:53 pm: Alright =] Thanks for the confirmations
Title: Re: Dekoyl's Questions
Post by: dekoyl on May 12, 2009, 08:27:23 pm: I've got a few discrepancies between a few sets of notes.

For questions regarding orbits/satellites, do we always use the given radius of the satellite's orbit?

In one of the questions uploaded here, there is:
Radius of satellite: $2.6 \times 10^6 m$
and Radius of satellite's orbit: $1.22 \times 10^9m$

Which one do we use and in what cases?

Thanks! (I have a SAC tomorrow)
Title: Re: Dekoyl's Questions
Post by: TrueTears on May 12, 2009, 08:31:43 pm: Yes here's the trick examiners might use, if they say "orbit" that's the distance you use for "r"

But if they say "altitude" or "height" you must add the earth's radius, or whatever planet the satellite is orbiting.
Title: Re: Dekoyl's Questions
Post by: dekoyl on May 12, 2009, 08:34:46 pm: Thanks TT!

So in the example I gave above, the "Radius of satellite" is pretty much a useless figure?
Title: Re: Dekoyl's Questions
Post by: TrueTears on May 12, 2009, 08:41:41 pm: Yeah, I would use the "Radius of satellite's orbit" as the value for "r"
Title: Re: Dekoyl's Questions
Post by: dekoyl on June 06, 2009, 04:47:09 pm: The magnitude of the gravitational field strength on the surface of the Earth has a value of approximately 10 N/kg.
Calculate the magnitude of the gravitational field strength at a distance of four Earth radii above the surface.

I used $g=\frac{GM}{r^2}$ which got the right answer but they wanted something with:
$\frac{g_1}{g_2}=\frac{{R_2}^2}{{R_1}^2}$ . Where did this formula come from?

Thanks
Title: Re: Dekoyl's Questions
Post by: dcc on June 06, 2009, 04:49:57 pm: Basically, they are acknowledging that $g \propto \frac{1}{r^2}$ , from which the answer easily falls out (without needing to do any calculations).
Title: Re: Dekoyl's Questions
Post by: kurrymuncher on June 06, 2009, 07:50:28 pm: Quote from: dekoyl on June 06, 2009, 04:47:09 pm
The magnitude of the gravitational field strength on the surface of the Earth has a value of approximately 10 N/kg.
Calculate the magnitude of the gravitational field strength at a distance of four Earth radii above the surface.

I used $g=\frac{GM}{r^2}$ which got the right answer but they wanted something with:
$\frac{g_1}{g_2}=\frac{{R_2}^2}{{R_1}^2}$ . Where did this formula come from?

Thanks

They got that formula just by using this $g=\frac{GM}{r^2}$ .Just divide g1 by g2. As G and M are the same in each case they will just cancel out so you will be left with $\frac{g_1}{g_2}=\frac{{R_2}^2}{{R_1}^2}$
Title: Re: Dekoyl's Questions
Post by: methodsboy on June 06, 2009, 07:57:31 pm: ^i believe that's one of kepler's laws
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 06, 2009, 08:12:16 pm: from my raw memory, i don't think it is. Johann Kepler was an astronomer and he lived before Newton, hence his work wouldn't have anticipated anything about acceleratation due to gravity being related to celestial motion. His work was more to do with planetary motion like radius of orbits and periods (i think he was the first to prove that the orbits are elliptical). Though he never anticipated anything about forces or acceleration, I think he believed that magnetism or some whirlpool moved the planets at some point.
Title: Re: Dekoyl's Questions
Post by: dekoyl on June 28, 2009, 10:00:12 pm: Q:A vehicle with constant acceleration travels 20m in 2s and 13m in the next second. What is the acceleration?

Thanks =\
Title: Re: Dekoyl's Questions
Post by: Damo17 on June 28, 2009, 11:02:49 pm: Quote from: dekoyl on June 28, 2009, 10:00:12 pm
Q:A vehicle with constant acceleration travels 20m in 2s and 13m in the next second. What is the acceleration?

Thanks =\

let u= initial velocity, a =acceleration

since 3 seconds total:

$u+3a=13$ -----1

$(u+a) + (u+2a)=20$
$2u+3a=20$ -----2

$u+3a=13$ -----1
$2u+3a=20$ -----2

solve simultaneous equations for u gives $u=7ms^{-1}$

$\therefore$ $7+3a=13$ , $a=2ms^{-2}$
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 28, 2009, 11:04:58 pm: Assuming the direction doesn't change:

x=0.5at^2+ut

x=20, when t=2:
20=2a+2u

x=20+13 when t=3:
20+13=4.5a+3u

Simultaenous equations ftw.
=============================================================
Interestingly, say the direction of motion changed at t=2:

we still have the equation:

20=2a+2u

Only this time:

20-13=4.5a+3u

Gotta love unprecise questions.
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 28, 2009, 11:08:08 pm: Quote from: Damo17 on June 28, 2009, 11:02:49 pm
Quote from: dekoyl on June 28, 2009, 10:00:12 pm
Q:A vehicle with constant acceleration travels 20m in 2s and 13m in the next second. What is the acceleration?

Thanks =\

let u= initial velocity, a =acceleration

since 3 seconds total:

$u+3a=13$ -----1

$(u+a) + (u+2a)=20$
$2u+3a=20$ -----2

$u+3a=13$ -----1
$2u+3a=20$ -----2

solve simultaneous equations for u gives $u=7ms^{-1}$

$\therefore$ $7+3a=13$ , $a=2ms^{-2}$

Nah, I think by "travelling 13m in next second" it means that the distance covered in the interval (2,3) is 13m. Hence the overall position from the origin at t=3 is 20+13 (assuming direction has not changed).

Oh and just realising, I think u misentepreted distance as velocity.
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 28, 2009, 11:35:22 pm: Are you saying that:

average velocity=instantaenous velocity?
Title: Re: Dekoyl's Questions
Post by: Damo17 on June 28, 2009, 11:37:44 pm: Quote from: kamil9876 on June 28, 2009, 11:35:22 pm
Are you saying that:

average velocity=instantaenous velocity?

LOL, I'm so stupid. Get it now.
Title: Re: Dekoyl's Questions
Post by: Flaming_Arrow on June 28, 2009, 11:38:39 pm: just wondering, why are you doing unit 3 stuff? or is this from spesh?
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 28, 2009, 11:46:53 pm: Surprisingly you still get the same acceleration. I think doing this exact same incorrect method woudl still give the same acceleration because of funny relationships in constant acceleration:

ie: $v_{average}=\frac{v+u}{2}$

Is true only in constant acceleration, in non-constant acceleration this isn't true. Hence the reason why the average velocity 13 is halfway in between 12 and 14. I think you can prove using this that this exact incorrect method gives the correct acceleration, buit only in constant acceleration.
Title: Re: Dekoyl's Questions
Post by: dekoyl on June 29, 2009, 12:33:15 pm: Q: A lift descends a distance of 30m, from rest to rest, as follows: constant acceleration for the first 10 m, constant velocity for he next 10 m, constant deceleration for the next 10 m. If the total time taken is 5 s, find the greatest speed reached by the lift.

Thanks =]
Title: Re: Dekoyl's Questions
Post by: TrueTears on June 29, 2009, 01:23:28 pm: (http://img10.imageshack.us/img10/2238/dekoyl.jpg)

$\frac{(a+5) \times x}{2} = 30$ ...[1]

$x \times a = 10$ ...[2]

x = 10 a = 1

Therefore greatest speed reached is $10 ms^{-1}$
Title: Re: Dekoyl's Questions
Post by: dekoyl on June 30, 2009, 05:41:40 pm: Q: A particle starting from point O with uniform velocity 4 m/s. 2 seconds later, another particle leaves O in the same direction with a velocity of 5 m/s and with an acceleration of $3 m/s^2$ . Find when and where it will overtake the first particle.

I know there's something wrong with my understanding but what I had constructed so far was this:
(http://i39.tinypic.com/2pt6udv.gif)
Red: First particle - $x=4t$
Blue: Second particle - $x=\frac{3}{2}t^{2}-t+2$ <--- I suspect this is the wrong.

Thanks for the help. I feel very slow in these holidays =(
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 30, 2009, 05:58:23 pm: Diagram way:

you want to find a line t=T such that the area "bounded by the line t=T, the blue line, the line t=2 and the t axis" equals the area bounded by "t=0, t=, the red line, and the t axis".

Reason: They start at same point hence at time T they should have the same displacement hence same area under graph.

$\int_{2}^{T} Blue dt=\int_{0}^{T} Red dt$

Find equations of blue and red and integrate and find T. There are hundreds of way of picturing this, this is just what I came up with now.
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 30, 2009, 06:05:32 pm: Quote from: dekoyl on June 30, 2009, 05:41:40 pm

Blue: Second particle - $x=\frac{3}{2}t^{2}-t+2$ <--- I suspect this is the wrong.

Thanks for the help. I feel very slow in these holidays =(

$x=\frac{3}{2}(t-2)^2+5(t-2)$

ie: $\Delta t=t-2$ for that particle since it travels two seconds less.
Title: Re: Dekoyl's Questions
Post by: dekoyl on June 30, 2009, 06:54:24 pm: Thanks kamil.
For the 'blue' equation, I obtained it from
$v = 3(t-2)+5 = 3t - 1$ so I didn't think that the travelling two seconds less needed to be taken into account.

For some reason I didn't get it after integrating but I'll keep trying.
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 30, 2009, 07:31:59 pm: No it doesn't, the first post i offered my solution, in the second post I fixed up your attempt. You should think of them seperately, I think you confused the two.
Title: Re: Dekoyl's Questions
Post by: kamil9876 on June 30, 2009, 07:36:48 pm: here is the blue one:

$\int_{2}^{T} 3(t-2)+5 dt$
$[\frac{3}{2}(t-2)^2+5t]_{2}^{T}$
$\frac{3}{2}(T-2)^2+5T-10)$
$\frac{3}{2}(T-2)^2+5(T-2)$

So you see it is equivalent to my second post :)
Title: Re: Dekoyl's Questions
Post by: dekoyl on June 30, 2009, 09:20:45 pm: ^Ah yeah. Thanks for that, Kamil.

Anyway I think I've given up on this question. It's either the answers are wrong (it says T=2 seconds, with distance of 16) but I keep getting T=4. I used your approach and another approach and I keep getting T=4 seconds.

It's just the answers to Fitz is rarely wrong so I'm doubting myself at the moment :P
Title: Re: Dekoyl's Questions
Post by: dekoyl on July 01, 2009, 11:04:46 pm: Hopefully the last one on velocity/time graphs... I don't know why but I seem to have quite a bit of trouble with these =\

A train passed a station, A, at 60 km/h, maintained this speed for 12km, and was then uniformly retarded to stop at B, 15 km from A. A second train started from A the instant the first train passed it, was uniformly accelerated, then kept at constant speed over 3 km, and finally retarded to stop at B at the same time as the first train. Find the constant speed of the second train.

I found the retardation of the first train (I think). $a=\frac{-5}{108}$ , time kept at constant speed: 720 seconds, and total time retarded (360 seconds). but I'm not sure how I could use this info for the second train.

Thanks!
Title: Re: Dekoyl's Questions
Post by: kamil9876 on July 02, 2009, 12:29:38 am: I will assume you have found the time it takes for first train to travel to B. Let's call this time t:

Let $\Delta t$ be the time second train goes for a constant speed v.

Diagram attatched is a velocity time graph, you can see that the total area is:
$<br />0.5(\Delta t + t)v=15$

However the area of the rectangle is:

$v\Delta t=3$

Now sub in $\Delta t=3/v$ into the 1st equation and you should get a quadratic in v. Use the value of t you worked out for the first train.
Title: Re: Dekoyl's Questions
Post by: dekoyl on July 04, 2009, 08:55:51 pm: Q:A train passes a station A at 30 km/h, maintains this speed for 7km and is then uniformly retarded to stop at B, 8 km from A.
A second train starts from A at the instant the first train passes it. It is uniformly accelerated for part of the way and uniformly retarded at the same rate for the rest of the way to stop at B at the same time that the first train stops.
Find the greatest speed of the second train.

I keep getting $\frac{200}{3} m/s$ whilst the book gets $\frac{415}{27}$ .
Did I misinterpret the question?
"...uniformly retarded at the same rate..." Does that mean it takes the same time to travel the first half of the journey as it takes the second half?

Thanks!
Title: Re: Dekoyl's Questions
Post by: zzdfa on July 04, 2009, 09:39:30 pm: I got 400/27

time taken for train1 to get from A to B is $\frac{9000}{v_0}$ where $v_0=30\tex{km/h}$
this is equal for the time taken for the 2nd train to get from A to B (8km)

therefore the average velocity of the trains over the 8km is $\frac{8}{9}\cdot v_0$

and also because the acceleration/deceleration is constant, the max velocity will be twice the average velocity.

if im not wrong in my assumptions above (i cbf doing the algebra to check):

$v_{max}=\frac{8}{9}\cdot 2v_0=\frac{400}{27}\text{m/s}$

which is different from your book. either im wrong or the question actually meant 'uniformly retarded at the same rate as the first train'. or the book is wrong. wierd how we have the same denominator but.
Title: Re: Dekoyl's Questions
Post by: Mao on July 04, 2009, 11:45:43 pm: Q:A train passes a station A at 30 km/h, maintains this speed for 7km and is then uniformly retarded to stop at B, 8 km from A.
A second train starts from A at the instant the first train passes it. It is uniformly accelerated for part of the way and uniformly retarded at the same rate for the rest of the way to stop at B at the same time that the first train stops.
Find the greatest speed of the second train.

zzdfa, I am unsure of how you got the time taken for train 1, do you have working?

I have a totally different answer:

Train 1: $v^2 - u^2 = 2ax \implies -900 = 2a \implies a = -450 km/h^2$

$t = \frac{7}{30} + \frac{0-30}{-450} = \frac{7}{30} + \frac{2}{30} = \frac{3}{10} hr$

Train 2: rate of acc = rate of dec, hence it accelerates half way (4km), decelerates 4km. Looking at the first half (acceleration)

$x = ut + \frac{1}{2}at^2 \implies 4 = \frac{1}{2}a \left(\frac{3}{20}\right)^2 \implies a = \frac{3200}{9} km/h^2$

$v_{max} = a t = \frac{3200}{9} \cdot \frac{3}{20} = \frac{160}{3} km/h$
Title: Re: Dekoyl's Questions
Post by: zzdfa on July 05, 2009, 12:11:51 am: your answer is the same as mine. mine is in m/s =)

time = $\frac{9000\text{m}}{30\text{km/h}}=\frac{3}{10}\text{h}$
Title: Re: Dekoyl's Questions
Post by: kamil9876 on July 05, 2009, 12:21:51 am: I got Mao's ans too last night but cbf checking and hence posting haha.
Title: Re: Dekoyl's Questions
Post by: dekoyl on July 06, 2009, 12:07:11 am: A general question:
When we have questions of two particles being launched vertically from the same position, say 5 seconds after the first one, is launched and we want to find the time they collide, why do we have the time of the second particle as t-5 instead of t+5 since it's launched five seconds AFTER?

If I'm not clear, please tell me and I'll provide an example.

(I'm talking about questions similar to "launch a ball at the ground at "v" and 5 seconds later, a ball is let go directly above the position the first ball was launched. Find the time and where they collide)

Thanks!
Title: Re: Dekoyl's Questions
Post by: zzdfa on July 06, 2009, 12:22:50 am: the 2nd particle was launched AFTER the first, which means its been in motion for less time than the 1st particle.

its similar to why you translate a graph in the -x direction when its f(x+somethingpositive).
Title: Re: Dekoyl's Questions
Post by: dekoyl on July 06, 2009, 12:39:33 am: Ah thanks a lot. The function example really helped.
Hmm I think I saw stuff like that in the Essential's book. I'm not working out of that but I think I should now :P
Title: Re: Dekoyl's Questions
Post by: dekoyl on July 06, 2009, 02:17:45 pm: Okay, a question similar to the above one.

An object is projected vertically upwards from a point A, with a velocity of 40 m/s. Five seconds later another object is dropped from a point 5 m vertically above A. Find when and where the two objects meet.

With $g=10 ms^{-2}$ I have these 2 simultaneous equations:
$x_1 = 40t - 5t^{2}$ $\leftarrow$ Now this is okay. Launched at 40 m/s = u.
$x_2 = 5 - 5(t-5)^{2}$ $\leftarrow$ Not okay :( How is this obtained? Isn't initial velocity = 0 as it is dropped from a stationary point? Hence the $ut$ of $x = ut + \frac{1}{2}at^{2}$ should be 0?

Thaaaaanks.
Title: Re: Dekoyl's Questions
Post by: kamil9876 on July 06, 2009, 02:33:56 pm: Say we have two clocks, one for particle 1 and it shows time $t_1$ , one for particle 2 that shows time $t_2$ . (the clocks are set so that when particle 1 is launched than $t_1=0$ , when particle 2 is launched then $t_2=0$ )
E.g: when clock 1 shows $t_1=5$ , clock 2 shows that $t_2=0$ , when clock 1 shows $t_1=10$ then clock 2 shows $t_2=5$ etc.

the equation for particle 2 is:

$s_2=0.5at_2^2+ut_2$
$s_2=0.5at_2^2+0$

however as the example i provided above shows: $t_1=t_2+5$ . and so when u sub that in:
$<br />s_2=0.5a(t_1-5)^2$

i.e: so now you only have to worry about the time of one clock as we have eliminated $t_2$ from the expression

There are probably better ways of explaining this so I encourage further posts hah.
Title: Re: Dekoyl's Questions
Post by: dekoyl on July 06, 2009, 03:09:54 pm: Hey Kamil - Thanks for the great explanation with the two clocks.

However, there's still a slight problem :(
$s_1=40t-5t^{2}$
$s_2=-5(t-5)^{2}$
Solving simultaneously, you get $t=12.5$ But the answer is $t=12$

However, if $s_2=5 - 5(t-5)^2$ , you will get $t=12$ .

Do you know how the "5" in $s_2$ comes in?
Title: Re: Dekoyl's Questions
Post by: kamil9876 on July 06, 2009, 03:33:18 pm: $s=0.5at^2+ut$

The s is not position, but rather displacement (change in position). hence $s=x_{final} - x_{initial}$
$<br />x_{final}=0.5at^2+ut+x_{initial}$

In fact an even more general formula will be this:

$x_{final} - x_{initial}=0.5a(t_{final} - t_{initial})^2 + u(t_{final} - t_{initial})$

This one probably takes the cake though:

$\int_{t_{initial}}^{t_{final}} v dt=x_{final}-x_{initial}$

Which takes into account the possibility of starting at different times that I showed earlier. You just have to be really careful with how you use these formulae.

Remark:
I also could've explained the offset in initial position in terms of different axis (just like i did it in terms of different clocks for time) and just said that the formula $x=0.5at^2+ut$ assumes that $x_{initial}=0$
Title: Re: Dekoyl's Questions
Post by: dekoyl on July 06, 2009, 03:49:34 pm: *Huge head smack* Why of course. I should really think these equations through instead of just stupid number plugging. >:[
Huge thanks for all that kamil! That really helped.
Title: Re: Dekoyl's Questions
Post by: dekoyl on August 16, 2009, 06:45:33 pm: Q: The light from a red LED has a frequency of $4.59 \times 10^{14} Hz$ . What is the energy change of electrons that produce this light?

I don't understand why the answer is negative.
Answer is $\triangle E_{electron} = -1.90 eV$

Big thanks. Got a SAC tomorrow =(
Title: Re: Dekoyl's Questions
Post by: kamil9876 on August 16, 2009, 08:22:49 pm: drops to a lower energy level. So change is negative.

It's simply the conservation of energy:

$E_{i}=E_{f}$

$E_{i light} + E_{i electron}=E_{f light}+E_{f electron}$
$E_{i light}-E_{f light}=E_{f electron} - E_{ i electron }$
$-(E_{f light} - E_{i light})=\Delta E_{electron}$
$-\Delta E_{light}=\Delta E_{electron}$

$\Delta E_{light}$ is positive since it went from 0(wasn't there) to something.
Title: Re: Dekoyl's Questions
Post by: dekoyl on August 26, 2009, 08:51:31 pm: Brain fart: Is it possible to calculate the frequency of electrons?

I'm doing my English language analysis at the moment and this popped into my head. :| Not cool
Title: Re: Dekoyl's Questions
Post by: /0 on August 26, 2009, 08:56:57 pm: Not sure but I'm guessing that you can get it from the de Broglie wavelength

$\lambda = \frac{h}{mv}$

$f = \frac{v}{\lambda} = \frac{v}{\frac{h}{mv}} = \frac{mv^2}{h}$
Title: Re: Dekoyl's Questions
Post by: dekoyl on August 26, 2009, 09:00:55 pm: Oh okay thanks /0. I just flipped through the Jacaranda book and there's a question that just leaves the frequencies of electrons blank (p351)
Title: Re: Dekoyl's Questions
Post by: Mao on August 26, 2009, 10:06:48 pm: Mao weighs 70kg, he is walking at 1m/s. Hence, he has a wavelength of 9.47E-36 m, hence a frequency of 1.06E35 Hz.

TAKE THAT GAMMA RAY, I BEAT YOU BY 10^16 = 10 MILLION BILLION TIMES.
Title: Re: Dekoyl's Questions
Post by: dekoyl on November 10, 2009, 01:23:58 am: A tube of length 0.765m closed at one end, with the speaker at the open end and a microphone is placed 0.085m from the closed end:

[_O___________ <|
^mic ^speaker

The frequency of the sound from the speaker is increased from 0Hz until the microphone detects a maximum intensity of sound.
Q: What number harmonic is presen when the microphone detects maximum amplitude for the first time?

Thanks :(
Title: Re: Dekoyl's Questions
Post by: /0 on November 10, 2009, 01:29:45 am: Wouldn't it be the first harmonic? If you're increasing frequency from 0Hz the first resonance you'll come across is the fundamental.
Title: Re: Dekoyl's Questions
Post by: dekoyl on November 10, 2009, 01:44:00 am: Quote from: /0 on November 10, 2009, 01:29:45 am
Wouldn't it be the first harmonic? If you're increasing frequency from 0Hz the first resonance you'll come across is the fundamental.
The answers says 5th harmonic.

I can't offer any reasons as to why because I'm learning as I do trial exams (ha.)
Title: Re: Dekoyl's Questions
Post by: TrueTears on November 10, 2009, 02:20:29 am: Tricky Q this one:

There must be an antinode at the mic.

Thus the wavelength is $2 \times 0.085$ (Since there's an antinode at the closed end)

So now $\lambda = \frac{2L}{n}$

$2 \times 0.085 = \frac{2 \times 0.765}{n}$

$n = 9$

Now we have n = 1,3,5,7, 9

Which is the 5th resonant freq AKA 9th harmonic. I think answer is wrong if it says 5th harmonic.
Title: Re: Dekoyl's Questions
Post by: dekoyl on November 10, 2009, 02:31:33 am: Thanks TT!

This is the exact solution:

Quote
$\frac{0.765}{0.085}=9$ Therefore there must be $2\frac{1}{4}$ wavelengths present in the tube at this time. This is the 5th harmonic.

That's the exact working out they have (which I don't understand). :(
Title: Re: Dekoyl's Questions
Post by: TrueTears on November 10, 2009, 02:38:28 am: Yeah that's just like my solution, I'm not sure what its talking about 2.25 wavelength present, but I'm pretty sure that is meant to be the 5th resonant freq. Probably a mistake (this was a neap exam right?). So it should be 9th harmonic.
Title: Re: Dekoyl's Questions
Post by: kamil9876 on November 10, 2009, 02:45:31 am: What they did is count how many 1/4 wavelengths are in the pipe for the largest wavelength that satisfies the condition pointed out by TT. ie coz 9*(1/4)=2 and a quarter.

if u look at the patterns formed for this kind of tube you can find the formula:
$<br />L=\frac{1}{4}\lambda (2n-1)$ (prove this lol, you can do so by induction which I know you have learnt from umep :P)

Knowing that 9 quarters of a wavelength fit inside, we solve this as: 2n-1=9 which gives n=5
Title: Re: Dekoyl's Questions
Post by: naved_s9994 on November 10, 2009, 07:48:31 am: Quote from: /0 on August 26, 2009, 08:56:57 pm
Not sure but I'm guessing that you can get it from the de Broglie wavelength

$\lambda = \frac{h}{mv}$

$f = \frac{v}{\lambda} = \frac{v}{\frac{h}{mv}} = \frac{mv^2}{h}$

What H would you use?
Title: Re: Dekoyl's Questions
Post by: crappy on November 10, 2009, 01:26:17 pm: Quote from: naved_s9994 on November 10, 2009, 07:48:31 am
Quote from: /0 on August 26, 2009, 08:56:57 pm
Not sure but I'm guessing that you can get it from the de Broglie wavelength

$\lambda = \frac{h}{mv}$

$f = \frac{v}{\lambda} = \frac{v}{\frac{h}{mv}} = \frac{mv^2}{h}$

What H would you use?

You would use H in joules, it would be illogical to use in in electron volts.
Title: Re: Dekoyl's Questions
Post by: dekoyl on November 10, 2009, 03:47:50 pm: We have a 2 open-ended tube and a 1 open-ended tube (with the other end closed). For a scenario (not necessarily a tube) that has TWO closed ends, are both ends an antinode?
Title: Re: Dekoyl's Questions
Post by: appianway on November 10, 2009, 03:56:22 pm: Both ends are a displacement node and a pressude antinode.
Title: Re: Dekoyl's Questions
Post by: Politik23 on November 10, 2009, 08:39:34 pm: Quote from: appianway on November 10, 2009, 03:56:22 pm
Both ends are a displacement node and a pressude antinode.

Could you perhaps explain this?
Greatly appreciated.
Title: Re: Dekoyl's Questions
Post by: appianway on November 10, 2009, 08:55:15 pm: OK, consider it like this.

It's a solid tube, so the particles at the end can't move "out" of the tube so to speak, so the ends are displacement nodes (no displacement occurs). Pressure antinodes occur where displacement nodes do, so they're also pressure antinodes.
Title: Re: Dekoyl's Questions
Post by: Politik23 on November 10, 2009, 10:07:17 pm: Quote from: appianway on November 10, 2009, 08:55:15 pm
OK, consider it like this.

It's a solid tube, so the particles at the end can't move "out" of the tube so to speak, so the ends are displacement nodes (no displacement occurs). Pressure antinodes occur where displacement nodes do, so they're also pressure antinodes.

Awesome, Cheers :D