Physics - How'd it go?

stainluss

Someone please tell me that the star was moving away for the doppler effect?!:eek:

validusername

stainluss wrote: »

Someone please tell me that the star was moving away for the doppler effect?!:eek:

Yes.

H2student

johnmcdnl wrote: »

james joyce that was such an easy couple of marks.. surely every teacher told their students that james joyce named the quarks :rolleyes::rolleyes:
good job mine did anyways lol :P

LMFAO, once yes. However, I guessed his name because everytime I flippin see the last page of the particle physics chapter of Exam Edge, the name James Joyce jumps out at me.

validusername

H2student wrote: »

LMFAO, once yes. However, I guessed his name because everytime I flippin see the last page of the particle physics chapter of Exam Edge, the name James Joyce jumps out at me.

Yeah! lol. I was reading it this morning thinking do we honestly need to know this? Guess we did! :pac:

UglyFuc

validusername wrote: »

Someone please tell me that the star was moving away for the doppler effect?!
Yes.

thank christ, and the red shift was the the star moving towards earth because wavelenght is shorter and at red end of spectrum

hmmmm52

UglyFuc wrote: »

thank christ, and the red shift was the the star moving towards earth because wavelenght is shorter and at red end of spectrum

think blue shift is when its movin towards earthe.....

johnmcdnl

IronGirl92 wrote: »

We didn't have a book :rolleyes: I said a beeper emitting a sound of known frequency attached to a string and swirled around in the air. The person swinging hears the same frequency but then two observers either side hear variations???

that's actually how you do it isn't it... or it is.. that swinging a yoke around in a circle rings a bell... hmmm

i'd say all our experiment are ok tbh.... they all showed the object or observer moving so... can't see how they'd dock marks like

validusername

UglyFuc wrote: »

thank christ, and the red shift was the the star moving towards earth because wavelenght is shorter and at red end of spectrum

em, I think for the red shift, the star was also moving away. red has the longest wavelenth of the spectrum. I'm afraid you mixed it up.

I thought it went well for me but it was a strange paper. Q6 looks horrid, I attempted it at the end for the craic and gave up when they didn't give us the masses! The James Joyce Q was a bit harsh, luckily my teacher mentioned it in class but I would have regarded it as trivia moreso than exam material...

Also, data logger graphs in Q12? Never used one before in my life, thought that was a bit unfair, even though the choice in the question meant I didn't have to do it.

Doppler Q & X rays were lovely though. Happy overall.

johnmcdnl

The James Joyce Q was a bit harsh, luckily my teacher mentioned it in class but I would have regarded it as trivia moreso than exam material...[

why does everyone think this?? we have to know who named the electron and all sorts of other names and stuff that i couldn't be bothered even thinkin about now... why is knowing who named the quark any different... just becasue joyce wasn't a scientist?? it was a fair question.. it's in our books..

and besides it's only gonna be worth 3 marks overall anyways so... not worth worrying over... exact same as if you left out a unit of measurement so...

Blerdiii

validusername wrote: »

oooo, I think I got that too!

howd you work it out?
mine is worng id say i got the real frequency and apparent frequency ( a la relative velocity in applied maths XD)
then subbed into doppler eqtn ..think i used the wrong sign though

Fizzical

Taco Chips wrote: »

Yeah 6 was tricky. You were given the value for g at the earth's surface so I use g = GM/r^2 to get M for earth. Then they told you that the moon mass was 81 times the earth mass so just multiply and then stick it back into the g=GM/4R^2. It was tricky though. Experiments were nice I thought.

That would work - except it should be 9R^2 (twice the radius above the surface = 3R total)

Also, since g = GM/d^2 then g is proportional to 1/d^2, so if you triple the distance the answer is 1/9 of g i.e. 9.81/9. No M required.

Similarly for part (iii). F of earth = F of moon. So GM(e)M(astronaut)/D(e)^2 = GM(m)M(astronaut)/(1-D(e))^2. Substitute 81M(e) for the mass of the moon, and cancel G, M(e) and M(astronaut). Don't forget to subtract the radius of the earth from your final answer for height above the earth's surface. No G or M(e) required.

validusername

Blerdiii wrote: »

howd you work it out?
mine is worng id say i got the real frequency and apparent frequency ( a la relative velocity in applied maths XD)
then subbed into doppler eqtn ..think i used the wrong sign though

well I found the frequency of both using c=(lamda)f .
the laboratory one is the actual frequency f and the star is f '
..then I used the doppler effect formula.
f ' = fc / c+u

I'm not certain if it's right either but I think the figures you got are similar to mine.
and ew, I don't like relative velocity! but kudos to you for being able to do those questions!

Blerdiii

validusername wrote: »

well I found the frequency of both using c=(lamda)f .
the laboratory one is the actual frequency f and the star is f '
..then I used the doppler effect formula.
f ' = fc / c+u

I'm not certain if it's right either but I think the figures you got are similar to mine.
and ew, I don't like relative velocity! but kudos to you for being able to do those questions!

yup exactly what i did well if you thought in the same way as me maybe we are ok ( im terrified i did c - u)

bahaha i can barely do them just remembered it XD

Blerdiii

UglyFuc wrote: »

thank christ, and the red shift was the the star moving towards earth because wavelenght is shorter and at red end of spectrum

nope other way round red shift is the farthest away becaus eof reds wavelength sorry!
plus my reasoning is with the expansion of the universe everything is getting farther and farther awy XD

validusername

Blerdiii wrote: »

yup exactly what i did well if you thought in the same way as me maybe we are ok ( im terrified i did c - u)

bahaha i can barely do them just remembered it XD

yeah we're ok.

god, I'm dreading app. maths on friday!

johnmcdnl

Fizzical wrote: »

That would work - except it should be 9R^2 (twice the radius above the surface = 3R total)

Also, since g = GM/d^2 then g is proportional to 1/d^2, so if you triple the distance the answer is 1/9 of g i.e. 9.81/9. No M required.

Similarly for part (iii). F of earth = F of moon. So GM(e)M(astronaut)/D(e)^2 = GM(m)M(astronaut)/(1-D(e))^2. Substitute 81M(e) for the mass of the moon, and cancel G, M(e) and M(astronaut). Don't forget to subtract the radius of the earth from your final answer for height above the earth's surface. No G or M(e) required.

that doesn't sound very fair that your allowed used applied maths in the physics paper.. are you sure it's in the marking scheme.. hmmm :rolleyes: yes i'm jsut jealous that you got to study applied maths and my school is too small to offer it

CosyAcorn

johnmcdnl wrote: »

we're not supposed to have to look at the tables though this year because we're doing the 2008 leaving cert which is suppsoed to supply us with those details... i tried doing it with and with the value for G but didn't work out eitehr way... how are you supposed to do it when there's no mass of the earth or the moon.. ugh 81moon = mass of earth...

like in fairness...

other than that it was pretty handy.. experiments were grand.. question 5 was really easy too...
q11 looked sooooo tricky but one you read the passage it was really esay

q12 - part A - wtf my teacher said that their not supposed to ask that question so i can see that causing a bit of bother perhaps..


delighted it's over anyways B2 for me me thinks.. spent the last 2 minutes running through the paper marking myself and being hard on myself so...

hopefully I'll get it anyways

you were supposed to do the accel.due to gravity at a height 2 times the radius of the earth above the surface bit in Q6 using the relationship between force and distance (force of gravity is inversely proportional to the distance between the objects). you did not need a value for G or the mass of the earth to work it out. the acceleration was 1.0888888... m/s^2

Blerdiii

validusername wrote: »

yeah we're ok.

god, I'm dreading app. maths on friday!

oh god ......:(:(:(:(:(:(:(

johnmcdnl

CosyAcorn wrote: »

you were supposed to do the accel.due to gravity at a height 2 times the radius of the earth above the surface bit in Q6 using the relationship between force and distance (force of gravity is inversely proportional to the distance between the objects). you did not need a value for G or the mass of the earth to work it out. the acceleration was 1.0888888... m/s^2

i can see where your coming from - god that question involved way to much thinking and playing with numbers to be fun... why couldn't they have asked the dervation of that formula which would have been soooo much nicer but still a decent test

twas an extra question for me anways so i'm not too worried tbh

brianorourke

Was disappointed at the lack of a half-life question but it was a pretty do-able paper

johnmcdnl

brianorourke wrote: »

Was disappointed at the lack of a half-life question but it was a pretty do-able paper

12B was the only nuclear question on the whole paper and i didn't do 12... i like nuclear question... twas a bit unfair that is right

FreeT

Hey boarders, did anyone get any of these answers?:

Q1: Mass = 3.51 kg
Q2: Latent heat = 2,338,477 J/kg
Q3: n = 1.492
Q6: a = 1.09 m/s/s
Height above surface = 3.392x10^8 m
Velocity = 1022.9 m/s
Q7: Speed = 1.48x10^6 m/s
Q11b: 0.2 W/kg
g: 6.25 cm


Thanks in advance.

Holysock

FreeT wrote: »

Hey boarders, did anyone get any of these answers?:

Q1: Mass = 3.51 kg
Q2: Latent heat = 2,338,477 J/kg
Q3: n = 1.492
Q6: a = 1.09 m/s/s
Height above surface = 3.392x10^8 m
Velocity = 1022.9 m/s
Q7: Speed = 1.48x10^6 m/s
Q11b: 0.2 W/kg
g: 6.25 cm


Thanks in advance.

I got the same answers as you for Q 2,3,6 and the velocity of the moon one.Can't remember the rest sorry!:D

c0nor

delighted! an identical question with the doppler effect a few years ago came up- that was a stroll

citizenerased1

Whats done is done guys

it was kinda a naggy paper though..

jawileth

for Q8(ii) did anybody else use P=RI^2 and change it to P=V/R? i think thats right!

and for Q10 (ii) did anyone else get 0.725J?

Q10 (iii)!! as scabby a question as i've ever seen! your dont have to know the make up specific of mesons!!

ihavequestions

for the wiightlessness Q I said 36,000km.... no workings out or anything :P remembererd that is the height for a geostationary orbit for some strange reason!
Did 6 as an extra Q !

Fizzical

johnmcdnl wrote: »

that doesn't sound very fair that your allowed used applied maths in the physics paper.. are you sure it's in the marking scheme.. hmmm :rolleyes: yes i'm jsut jealous that you got to study applied maths and my school is too small to offer it

That's ordinary maths and physics! The inverse square law comes up a couple of times in the course - Newton's law and Coulomb's law. And the rest is just substitution...

calnand

Fizzical wrote: »

Similarly for part (iii). F of earth = F of moon. So GM(e)M(astronaut)/D(e)^2 = GM(m)M(astronaut)/(1-D(e))^2. Substitute 81M(e) for the mass of the moon, and cancel G, M(e) and M(astronaut). Don't forget to subtract the radius of the earth from your final answer for height above the earth's surface. No G or M(e) required.

the mass of the earth is 81 times the mass of the moon not the other way around, I got around 2.7*10^7m for the height above the surface, was so happy when it worked out. i thought the paper was very fair experiments were so easy i was expecting the f=ma and latent heat q so i had no bother and the snells law is really easy. i managed to do 7 sectionb questions so hopefully ill do well. also people were saying we wernt given the mass of the earth all we had to do was use g=GM/r^2 and then use M to calculate the new acceleration due to gravity. just took some logical thinking