Higher Level Physics

errlloyd

-ilikeshorts wrote: »

The dimples on golf balls are to keep them straight and to do with spin

Yeah that does actually seem to be the case.

Would have you got marks for suggesting the use of a raindrop shaped bearing. (As a rain drop is of course the most aerodynamic shape)

ironclaw

-ilikeshorts wrote: »

The dimples on golf balls are to keep them straight and to do with spin

Thats true. But a ball bearing is completely smooth and wouldn't spin after such a small distance. But a valid point all the same. If I was correcting, I wouldn't give marks for polishing the ball.

errlloyd wrote: »

Yeah that does actually seem to be the case.

Would have you got marks for suggesting the use of a raindrop shaped bearing. (As a rain drop is of course the most aerodynamic shape)

Suppose you would. Good answer I got to say. However, it isn't the most aerodynamic shape.



The base is still a sphere with great surface area. This causes drag. Hence why the nose of concorde and supersonic jets is a cone. The best aerodynmic shape occurs are the volume and diametre of the cyclinder approaches Zero. Thus at zero it can pass between molecules and achieve zero drag. Apparently it comes down to the speed the object is travelling.

I'd say the most efficent shape would be a some sort of leading edge, similar to a plane wing.

Fringe

I'm hoping heavy ball and place it in a vacuum is the best. Some people I was talking to had other ideas. Make 's' small, do it at a high altitude, use a sphere...

timmywex

Right, lets sort out this reducing air resistance business.

• Taking multiple values of t for each s and then using the smallest value
• Any reference to tmin
• ensure no external forces act on it(draughts etc)
• Use LARGE values for s as it reduces percentage error.

They may well accept other answers aswell however.

celtic723

timmywex wrote: »

Right, lets sort out this reducing air resistance business.

• Taking multiple values of t for each s and then using the smallest value
• Any reference to tmin
• ensure no external forces act on it(draughts etc)
• Use LARGE values for s as it reduces percentage error.

thank you!. that's the correct answer and your like the only one i've seen say it.

WolfForager

I had put it in a vacum and use a cone :S Looking back on it now the cone was a poor choice...

smndly

Large values of s does make the experiment more accurate because it gives smaller percentage errors when measuring distance. But the question was referring to the air resistance. So even though making s smaller will in fact have a detrimental effect on the accuracy of the experiment, it still reduces the time exposed to air resistance.

timmywex

celtic723 wrote: »

thank you!. that's the correct answer and your like the only one i've seen say it.

The bigger s value lets the particle accelerate to closer to the value for g. And if anyone doesnt believe me check 2004 marking scheme, similar question was asked for the last 7 marks.(but it was for general precautions)

It was only 4 marks for 2 precautions anyways!!

Also snmdly, see above about acceleration towards g

Pixiesticks

timmywex wrote: »

Right, lets sort out this reducing air resistance business.

• Taking multiple values of t for each s and then using the smallest value
• Any reference to tmin
• ensure no external forces act on it(draughts etc)
• Use LARGE values for s as it reduces percentage error.

They may well accept other answers aswell however.

I know using small valus of t and big values of s reduce percentage error but they were asking about air resistance so surely small s?

errlloyd

ironclaw wrote: »

Thats true. But a ball bearing is completely smooth and wouldn't spin after such a small distance. But a valid point all the same. If I was correcting, I wouldn't give marks for polishing the ball.



Suppose you would. Good answer I got to say. However, it isn't the most aerodynamic shape.



The base is still a sphere with great surface area. This causes drag. Hence why the nose of concorde and supersonic jets is a cone. The best aerodynmic shape occurs are the volume and diametre of the cyclinder approaches Zero. Thus at zero it can pass between molecules and achieve zero drag. Apparently it comes down to the speed the object is travelling.

I'd say the most efficent shape would be a some sort of leading edge, similar to a plane wing.

Just spit balling here, but isn't a plan wing based on a raindrop.



Like obviously double that and you would get something that looks like a valentines heart, but in essence a rain drop.

Like the reason a rain drop is so perfect is simply because it shapes itself to the path of least resistance. In fact how cool would it be to do this experiment with mercury to get that effect.

timmywex

Pixiesticks wrote: »

I know using small valus of t and big values of s reduce percentage error but they were asking about air resistance so surely small s?

Ah im not arsed arguing it, its worth 2 marks outta 400!!

I can see both sides of the argument though

I'm not so sure about keeping s large.

Typically that's the answer, but typically they ask you what you can do to minimise timing errors etc. or they just ask you for general precautions.

This specified minimising the effects of air drag. Having a greater distance s would result in a higher velocity, and [latex]\mbox{F_{drag}} \ \propto \ \mbox{v}^2[/latex]. So, increasing the distance would result in a higher force of drag. I think if you can justify your answer for keeping s small with respect to minimising air drag you'll get the marks.

artemus

RandomUserName! wrote: »

Yeah I got 22 for that.
Then you multiply it by two for images on each side.
I forgot the central image though, so left it as 44 when it should have been 45.

god help you when your results come out if your using that logic!the first order image ; n1 consists of two images! therefore 22 images is the max...work backwards put 22 in and solve for tee-ta(excuse the spelling) it will be less than 90...test for 23 and your result is greater than 90...this is impossible...therefoer you are wrong...light caanot be difracted more than 90 degrees!!you are wrong...wrong

That question about the number of images was too vague: it could have meant one of two things:

How many order images were there. If that's what it meant, then the answer was 22.

How many images were there. If it meant that, then it would have been 44/45.

I personally wrote 22, but I can see the logic in saying 44/45; I wouldn't be suprised if 22 was wrong.

Risteard

I said 45. Maximum of 22 bright fringes either side. That's 44 and central is 45.

dermo1990

For the acceleration on the ramp in Q6 I used some applied maths technique where you break up the acceleration into seperate components and ended up with 70(9.8)Sin20=70a. Ended up getting 3.35 m/s^2. Not sure if its right though

likely_lass

dermo1990 wrote: »

For the acceleration on the ramp in Q6 I used some applied maths technique where you break up the acceleration into seperate components and ended up with 70(9.8)Sin20=70a. Ended up getting 3.35 m/s^2. Not sure if its right though

i did that too just used cos70 instead dont see how that method could be wrong

Fince

-JammyDodger- wrote: »

I did it in a far too complicated way.

I used the conservation of mechanical energy ([latex]\mbox{mgh} = \frac{1}{2}\mbox{m}\mbox{v}^2[/latex]) to find what speed it would be doing at the top of the ramp, then I just used [latex]\mbox{v}^2 = \mbox{u}^2 %2B \mbox{2as}[/latex] (letting a = g) to find its vertical height from the top of the ramp. It worked out at something like 5.2 - 5.4 metres above the ground.

I just had to try the new Latex add-on for boards for equations. Works pretty well!

if my memory serves me correct you had the mass and the velocity. bang that into 1/2mv^2 get a value for energy. leave this value equal to mgh. you get h= first value for energy divided by mg. maybe i over simplified, i tend to that that in applied maths, but in physics its normally 'if it seems too obvious, it probably is the answer'.

Delta Kilo

-ilikeshorts wrote: »

The dimples on golf balls are to keep them straight and to do with spin

errlloyd wrote: »

Just spit balling here, but isn't a plan wing based on a raindrop.



Like obviously double that and you would get something that looks like a valentines heart, but in essence a rain drop.

Like the reason a rain drop is so perfect is simply because it shapes itself to the path of least resistance. In fact how cool would it be to do this experiment with mercury to get that effect.

Firstly, the dimples on golf balls aren't really to keep it straight. The idea of the dimples is to disturb the laminar flow of the air around it, reducing drag and allowing the ball to go further and faster. But your right it is to do with spin, it needs to be spinning to work and dropping it from a small height will not generate the necessary spin so it wouldn't work.

The wing is out of context here too. The shape of the wing gives it lift, it does not reduce the drag on the wing. It is to do with Bernuolli's theorem about pressure diferences. Therefore having a wing shaped or tear shaped ball would not help.


People on about the diffraction grating. For n to be max doesnt sintheta = 1. Therefore you end up with d/lambda. this gives you 11.7 which is the 12th order image?

ZorbaTehZ

Numerical answers to question 6:

(i.) 2.9768 m/s^2
(ii.) 23.94 kg
(iii.) 26.6 N

Initial Centripetal Force = 771.75 N
Maximum Vertical Height = 5.619 m

B0X

Risteard wrote: »

I said 45. Maximum of 22 bright fringes either side. That's 44 and central is 45.

I also got 22, i never multiplied it by 2, ah sure that barley a loss

randylonghorn

Artemus, whether he's wrong or right, there's no need to be so OTT about it!

Indeed, others seem to think that he may have a point, depending on how the question is interpreted.

AndyWhite

ZorbaTehZ wrote: »

Numerical answers to question 6:

(i.) 2.9768 m/s^2
(ii.) 23.94 kg
(iii.) 26.6 N

Initial Centripetal Force = 771.75 N
Maximum Vertical Height = 5.619 m

Got the same answer for part i but for ii I think they asked for weight, not mass.

errlloyd

randylonghorn wrote: »

Artemus, whether he's wrong or right, there's no need to be so OTT about it!

Indeed, others seem to think that he may have a point, depending on how the question is interpreted.

He is actually right though, when calculating fridges you can't round from 89.99999 to 90

Jack Sheehan

B0X wrote: »

I also got 22, i never multiplied it by 2, ah sure that barley a loss

Automatic failure.

shanethemofo

dermo1990 wrote: »

For the acceleration on the ramp in Q6 I used some applied maths technique where you break up the acceleration into seperate components and ended up with 70(9.8)Sin20=70a. Ended up getting 3.35 m/s^2. Not sure if its right though

By doing it this way you dont count for external forces such as friction, your just presuming that g is the only force acting on him. i did it this way first and got 3.35 but then went back and changed it.

You use that to get the force of friction. the actual acceleration was 2. something. you subtract them from eachother and slot in the acceleration into f=ma.

UnknownSpecies

artemus you are a dick, so what if hes wrong, i suppose you got/will get 100% in your leaving cert physics!

AndyWhite

haha, i didnt do that question but the other peoples answer seems more logical to me than artemis's one! I bet hes feelin just a tiny bit silly right about now!