Physics !!!! (H)

vote4pedro

Very fair paper, but I'm annoyed at myself for making stupid mistakes (wrote Photoelectric instead of Thermionic effect in q.5 :mad: ) Was stumped by height of the ball bouncing in q 12. a, and I also messed up the first 3 questions of Particle Physics bit but still should have got a B3 at least. Any Physics, ahem, 'enthusiasts' want to go through the paper again and write up results to maths probs/calculations?

stevoxbx

I just found 1value for sin i and sinr from the graph and put them into the thingy sin i/sin r.

Sandals

Did anyone else think that the maths questions were incredibly easy, in comparison to book and other years?


Even the first question in q.5 on pressure was so easy i didnt fu=igure it out for while it was just force/area as in from junior cert, i just remembered my teacher describing it in terms of a woman wearing stilletos!

The question 6 was much easier than normal circular motion questions!
I am glad that my one gues "The doppler effect"for the last part appears to be right

Q.7 I didnt like although it was easy 'cause i mixed up the youngs slits in the diagram for a diffraction grating!

q.8 was a very easy radioactivity question, the maths at the end was a joke
for ionisation i said
Where particles become charged into ions and accelerate into more ions causing the ionisation of more particles
is that in anyway right?

q.9 was fine except for the last 2 q's i bluffed my way through them getting attempt marks,
1. I just drew the graph of res. vs. temp and said they were not proportional
2. I said the total r has gone down v=ixr therefore v has gone down


q.10
i think i did well, for the 2nd last q was it e=q/d

and for the last one f=q1q2/4pied^2?

q.11(a)
was simple!
except for circular acellerators i said
(a)they can have accelerate to higher velocitys
(b) they can cause particles to collide
and (c) they are safer?? are any of those right??

q.12 i did a c and d, all ok b was too easy to be true!

Overall handy out I thought

colhol wrote:

What i did was draw the graph and all correct, then took two points of the x-axis, drew a line up (as you can see, and used the corresponding y-values to calculate the slope using the formula. I also put in the the slope of the line would give sin i/sin r. Thats what your supposed to do yeah?

yes that is perfect., most people dont do it that way though!
was your answer near 1.5?

sci0x

Thats the right way to do it, find the slope of the line to get the refractive index. I got 1.45

stevoxbx

can it be done by just taking th sin i value and the sin r value and put them over each other to get 1.5?

CoffeeFreak

It was a really easy paper...I think the easiest of all my exams so far...weird. I hope I get an A1

Richard_Fonzie

q.11(a)
was simple!
except for circular acellerators i said
(a)they can have accelerate to higher velocitys
(b) they can cause particles to collide
and (c) they are safer?? are any of those right??

I'm not sure if that's right.
a) might be correct, because particles can go around once, and then go around again (increasing speed)? But I think that for any circular acclerator, you could match it with a linear one by just making the linear one longer and increasing the voltage across it.
b) was something like what I had. I reasoned that in the linear, if two particles miss each other, they can't have another chance to collide without starting the whole thing again (i.e. you'd need to accelerate more particles). Whereas in the circular accelerator if they miss, the same two particles can just go around again and have another go.

Richard_Fonzie

can it be done by just taking th sin i value and the sin r value and put them over each other to get 1.5?

Yeah, but you'll probably lose marks, because I think you're supposed to use the graph (and hence the average of ALL the values) to find it.

*Angel*

stevoxbx wrote:

can it be done by just taking th sin i value and the sin r value and put them over each other to get 1.5?

It says 'from your graph'.

Richard_Fonzie

Yes but when you use the graph to find it, you typically find the slope of the graph, and in this case the slope is sini/sinr=n. If you didn't find the slope accurately, then you could have just picked a sini & a sinr value from the table and found n without using the graph at all.

edit: nevermind

Moira

overall very happy! exactly like my pre!
q1 conservation of momentum, great that we didnt have to describe the big long experiment cos thats just wicked long. just hope my calculations were right.

q2. please let my maths be right, had to do it so many times cos each time forgot to convert something in standard international units. 24 marks is alot gone if its wrong.

q3. lovely, knew it had to come up.

q4. didnt do it cos the end of it looked a bit dodgey for me.

q5. best ever, such a beauty.

q6. nice, good to see that keplers law coming up but couldnt think of doppler effect but did say something about speed if i remember right.

q7. nice, everything except the last bit cos we had never written that exp up prop.

q.8 didnt do, hate half life and that stuff

q.9 didnt do, but looked ok

q10. when i saw this i panicked, like where was the option q. but did this, was grand, again calculations i wouldnt b the best at.

q11a thought it was really unfair to mix up the theme q and the option q cos i find the theme q a bitch anyway, but did it away turned out alright, anyone the answer to part iii?? prob simple but couldnt think.
b didnt do that option

q12. really nice part a, and in b is it ok to write down the formula of magnetic flux?? just couldnt think of the definition.

prob will mark it harder now but still they cant change it that much cos well i usually base my choice of q on the distribution of mark, they cant just throw that guide out the window. id say it will be for things like forgetting units we'll be shafted for etc.

nobodythere

This is gonna be one HELL of a marking scheme

Sandals

Moira wrote:

q11a thought it was really unfair to mix up the theme q and the option q cos i find the theme q a bitch anyway, but did it away turned out alright, anyone the answer to part iii??

I just GUESSED
because neutrons dont have a charge - are stable

OTliddy

Richard_Fonzie wrote:

(and hence the average of ALL the values) to find it.

No. Pick two values, fairly far apart and use the slope formula to get the resistance, thats if u have sin i on the y-axis. Not sure if picking the point (0,0) would count....they could use it as an excuse to harden the marking scheme.

Here are some answers I got that I noted down:
Section A
Q2: Latent heat=2.285x10^6
Q3: Refractive index=1.44
Q4: change in resistance=17(or 19).69cant read if i wrote 17 or 19, mite b 17.

Section B
5.(d) 15 cm
9.(i) 600 ohms (ii) 4 milliamps i think
10(last bit) 7.153x10^-13
12(a) kinetic energy=17.934 J new height=2.03 m
12(c) 384.6 Hz 769.2Hz
12(b)(i)0.01 Wb
(ii)0.05 V

Other notes
For the radio signal part of Q6, there are prob 3 points going for c=f(lambda)
3 for substition, and 3 for answer.
For the last part, i'd say you'd get 5 marks if u just wrote down "doppler effect"

Webmonkey

Hmm I got the height in 12a as 1.02m i think asfar as i remember.
Kenetic Energy lost = Potential energy gained.
Ep = mgh so lost 6 Jules wasn't it?
6 = (.6)(9.8)x
x = 6/5.88 = 1.02m?

I probably took wrong figures knowing me. I dont' have the paper here so please tell me im right :S

OTliddy

Webmonkey wrote:

Hmm I got the height in 12a as 1.02m i think asfar as i remember.
Kenetic Energy lost = Potential energy gained.
Ep = mgh so lost 6 Jules wasn't it?
6 = (.6)(9.8)x
x = 6/5.88 = 1.02m?

I probably took wrong figures knowing me. I dont' have the paper here so please tell me im right :S

U got the right figures. my method was different(and much longer), but ur method is right aswell.I said the potential energy is the old energy minus the loss in energy, thus:
Ep=17.934-6
11.934=5.88x
x=2.02 m

If there was no loss in energy, the ball would bounce back up 3.05 m again, so i figured that the height decrease must be proportioanal to the energy loss

Sandals

What is ionisation?

defiantshrimp

Sandals wrote:

What is ionisation?

Far as I know it is when a neutral molecule/element/atom is ionised (in other words given a charge) when electrons are knocked out of its orbit. Usually by radiation (alpha, beta or gamma) or other ions.

StupidLikeAFox

Sandals wrote:

yes that is perfect., most people dont do it that way though!
was your answer near 1.5?

1.5 on the button baby! Happy days!

Sandals

ColHol wrote:

1.5 on the button baby! Happy days!

ok thats good!

whenthepawn

for the frequency part of q6 does it make sense to mention that the densities of the air in space and in the earths atmosphere are different thus the frequency inceases as it enters earth's atmosphere?

all in all delighted with the paper although i had learnt of the alternate method for the PCM exp which threw me initially. got to ignore electricity and waves etc completely!

Sandals

one word and nother
doppler effect

JackKelly

whenthepawn wrote:

of the air in space

:eek:

Sandals

TimAy wrote:

:eek:

Nice!

Richard_Fonzie

No. Pick two values, fairly far apart and use the slope formula to get the resistance, thats if u have sin i on the y-axis. Not sure if picking the point (0,0) would count....they could use it as an excuse to harden the marking scheme.

The slope is the average of all the values (that's what I meant), since it's based on the line you drew.

Richard_Fonzie

Ep = mgh so lost 6 Jules wasn't it?
6 = (.6)(9.8)x
x = 6/5.88 = 1.02m

hmmm, that's not what I did. I used Ep = mgh, where Ep is the energy needed to bring the ball to a height h. Kinetic energy was 17.934 when it reached the ground, but it lost 6, so then it had 11.934 on the way up:
(11.934)=(.6)(9.8)h
h=2.03m

10(last bit) 7.153x10^-13

That was the same number as I had, except the power was ^-9 .

To be honest, i thought the Q.11s were sweet. Just about everything they wanted you to say was indicated in each question. Better than having 18 mark questions that simply translate as "write all you know about". Instead the types of questions today helped you associate questions to information.

whenthepawn

TimAy wrote:

:eek:

lol, you know what i mean

sixdraw

ya that 6J is lost to sound and heat. you take it away from the kinetic energy to get what the potential will be so i think 2.03m is right

upmeath

Right, here's the notes I took down on my exam paper...

Q1...Momenta work out to be 2.57 and 2.62, before and after interaction, respectively

Q2...Got 2.3X10^6 for Lsteam, was delighted because I detest heat experiments and equations and omitted them from all revision!

Q3...Refractive index worked out at 1.4875 on my graph...
Said that as the sine function decreases as angle decreases that values of i smaller than 20' would give greater percentage error...Is that right?

Q4...Change in resistance was about 36Ohms if I recall correctly?

Q5
d...Obtained 15cm for the image position
h...Lesser voltages produce eddy currents, or cannot overcome resistance of wires, producing heat instead and waste energy?
i...Thermionic emission?
j...Strong nuclear force

Q6...Got 1.22X10^9m for the satellite's orbital radius... That sound familiar to anyone else?
Got 4000s, 66.7mins or 1.1hrs for the time taken by radio signals to reach earth?

Q8...Co-60's decay constant was 4.17X10^-9
Rate of decay of Co-60 with 2.5X10^21atoms was 1.044X10^13

sixdraw

yeah they look about right. for the high voltage one, i said larger voltage means smaller current and less heat loss from cables as p=i^2R