HOW WAS PHYSICS PAPER 5 PPL ? HOW WAS YOUR EXAM ?

[Jasmine1204]

my answers are almost identical to yours. .

quite similar..though i scred up no.5..

 

[Jasmine1204]

gradient was 144+/-4 wasn't it?
And one of my % uncertainty came like 17.5% nd i wrote 18%, it it correct?
anyone plz write the answers of the second question

Yea..correct..i got like 16%

 

[aaditya menon]

The gradient was positive, and in Q1, it WAS passing through the origin.

oh yes. right coz it's 1/v^2! Phew thank god
but er, it can't pass through the origin, coz when wind speed is 0, what are u gonna plot on the x-axis which reads 1/v^2 ? 1/0 is undefined..
also theta is 90 degrees, when v is 0, so tan 90 is again undefined. (0,0) cannot be a point, hence doesn't pass through the origin.

 

[Sannikutti]

oh yes. right coz it's 1/v^2! Phew thank god
but er, it can't pass through the origin, coz when wind speed is 0, what are u gonna plot on the x-axis which reads 1/v^2 ? 1/0 is undefined..
also theta is 90 degrees, when v is 0, so tan 90 is again undefined. (0,0) cannot be a point, hence doesn't pass through the origin.

Ya no origin but proportional

i think saying straight line is enough

 

[Ali31a]

My fan was revolutionary, I attached it to a computer system.
From the computer you input the speed of the wind V.
The fan revolves according to that speed and then you measure the angle using a protractor.

Cool xD

 

[Ahmed Tariq]

oh yes. right coz it's 1/v^2! Phew thank god
but er, it can't pass through the origin, coz when wind speed is 0, what are u gonna plot on the x-axis which reads 1/v^2 ? 1/0 is undefined..
also theta is 90 degrees, when v is 0, so tan 90 is again undefined. (0,0) cannot be a point, hence doesn't pass through the origin.

TanQ was inversely proportional to v^2 so the equation we'd get would be tanQ=k/v^2. Comparing it with y=mx+c would give c and hence, the y-intercept to be 0. Doesn't that mean it should be passing through the origin? :/

 

[ying27kia]

if we plot tan theta vs 1/v^2, the suggested relation is correct ONLY if it is straight line and pass through origin with tanQ=m/v^2 where m is the gradient. In fact it is not correct because when v=0, theta is 90 deg, in other words it has y intercept. I think that is all the essence of this experiment.

 

[Gooners]

I think that a direct propotional graph is ONLY true if it is a STRAIGHT LINE PASSING THRU ORIGN! it is what we used throughout our practicals in As last year and also in O LEVELS too, lol!
And i looked up on the internet and it states tht if a graph has a y intercept it cannot b direct propotion :/!
Nd as tanQ was inverse prop. To v^2 hence for any value tanQ must be direct propotion to 1/v^2 hence a straight line passing thru orign shd b rite!! :/ otherwise tanQ cannot be inverse prop. To v^2 As was stated in the Q!

 

[Sannikutti]

I think that a direct propotional graph is ONLY true if it is a STRAIGHT LINE PASSING THRU ORIGN! it is what we used throughout our practicals in As last year and also in O LEVELS too, lol!
And i looked up on the internet and it states tht if a graph has a y intercept it cannot b direct propotion :/!
Nd as tanQ was inverse prop. To v^2 hence for any value tanQ must be direct propotion to 1/v^2 hence a straight line passing thru orign shd b rite!! :/ otherwise tanQ cannot be inverse prop. To v^2 As was stated in the Q!

Well lets do it mathematically :
tan(thetha)=k/v^2 rite? imagine V is speed of air due to the fan.
Assuming no air from other source when V=0 ,k=1,2,3,4,5,etc:-
tan(tetha)=5/0,60,7/0 and balllon will make angle of 90 degree.
Which means that it will get a intersept. U can refer to online graph plotter and do it also

Boils law have this type of proportionality but dont forget that when pressure is 0 volume is zero because it is at absolute zero but here the Q is not zero when v=o hence u cant take same relation

 

[ousamah112]

WAIT people ! we just have to write that if such garph was plotted and IF that was straight line passing thru origin thn the given relationship was CORRECT. Thats it.

 

[Sannikutti]

WAIT people ! we just have to write that if such garph was plotted and IF that was straight line passing thru origin thn the given relationship was CORRECT. Thats it.

Exactly thts what i wrote but actually it wont be like tht thts what im trying to xplain

 

[aliya_zad]

Exactly thts what i wrote but actually it wont be like tht thts what im trying to xplain

Its all about telling to plot a graph of tan@ verses (1/v2)...and if u get a straight line through the origin the relationship's valid!
Why wont it be like that?

 

[Jasmine1204]

TanQ was inversely proportional to v^2 so the equation we'd get would be tanQ=k/v^2. Comparing it with y=mx+c would give c and hence, the y-intercept to be 0. Doesn't that mean it should be passing through the origin? :/

no..it wouldn't pass through origin..it would if it was log smthin against smthin..

 

[Gooners]

no..it wouldn't pass through origin..it would if it was log smthin against smthin..

Wow!!! Log something against something!! U must b a genius to know that!!!

 

[Jasmine1204]

Wow!!! Log something against something!! U must b a genius to know that!!!

Not a genius..but unlike "some people" i have studied before my exam..and yes..if that makes me a genius..I am one..and it is true.. Tan theta against 1/v square would not pass through the origin..but if it was log theta against log v, then it would pass through the origin..

 

[aaditya menon]

well if u think about it, the whole question of passing or not passing revolves around the concept of approximation.

1/v^2 will be o as v tends to infinity
and as v increases, the angle tends to infinity.

But well, u can't achieve a velocity of infinity with a table fan or whatever, hence you can't approximate 1/v^2 to zero, and thus tan theta to 0. It only tends to zero.

I still think it can't pass through origin

 

[thunderingthunder]

It would be difficult for the examiner to stop laughing hard

 

[Gooners]

well if u think about it, the whole question of passing or not passing revolves around the concept of approximation.

1/v^2 will be o as v tends to infinity
and as v increases, the angle tends to infinity.

But well, u can't achieve a velocity of infinity with a table fan or whatever, hence you can't approximate 1/v^2 to zero, and thus tan theta to 0. It only tends to zero.

I still think it can't pass through origin

Dude! Lets finish this chapter! Just clear one thing for me! For a graph to b direct propotion, it MUST pass thru orign or not?
I looked up on the net and it states tht it must but i am still in doubt!! :/

 

[Jasmine1204]

the graph will be a curve NOT passing through the origin since it is not theta against v or theta against 1/v..its 1/v square...

 

[aaditya menon]

the graph will be a curve NOT passing through the origin since it is not theta against v or theta against 1/v..its 1/v square...

Thats something new there. You know what, **** it. Let the examiners decide. I'm outta here, gotta enjoy my holidats till they last