** HL Physics Before / after **

Kingkumar

jamo2oo9 wrote: »

When I came into the exam today this morning, I was genuinely bricking it but now when I finished it, I was so happy. That may be the nicest paper I've come across in Physics. The STS was simply beautiful with plenty of easy marks for everyone to get. Section A was handy enough and had no problems. For the metallic conductor question on the graph, I just explained that the graph contradicts the theory that was given to us on the course. I wouldn't worry much on that part. The mobile phone charger was very straight forward although I couldn't figure how find how much heat energy would be left in 10mins..? Any ideas?

Power dissapated is = I^2 R so heat energy = I^2RT u have R the resistance from the previous q and u have I and its 10 mins. so thats all. what did people put for the part labelled F was it a laminated iron core?

KildareKing

jamo2oo9 wrote: »

When I came into the exam today this morning, I was genuinely bricking it but now when I finished it, I was so happy. That may be the nicest paper I've come across in Physics. The STS was simply beautiful with plenty of easy marks for everyone to get. Section A was handy enough and had no problems. For the metallic conductor question on the graph, I just explained that the graph contradicts the theory that was given to us on the course. I wouldn't worry much on that part. The mobile phone charger was very straight forward although I couldn't figure how find how much heat energy would be left in 10mins..? Any ideas?

why did it contradict it, I think you had to draw a line of best fit, and for the energy one it was W=I^2 R t .

Copper_pipe

That was a lot nicer than the DEB pre!

Kingkumar

Jabama wrote: »

For question 7, did anyone know how to find the permittivity of the string?? i just guessed and used mass/length :P

ehh for what do u need the permittivity of the string for. the formula is F = 1/2L sqroot(T/mew) and mew is mass/length

dorcan

Leaving Cert Student wrote: »

q5 was tough ya what was the principle of storage heater??

i wrote heat capacity using bricks which are heated during the night and the energy is released during the day. liked that question because that's how our science room is heated.

KildareKing

Kingkumar wrote: »

Power dissapated is = I^2 R so heat energy = I^2RT u have R the resistance from the previous q and u have I and its 10 mins. so thats all. what did people put for the part labelled F was it a laminated iron core?

F was a transformer, the diagram is in the tables.

Jackobyte

Leaving Cert Student wrote: »

q5 was tough ya what was the principle of storage heater??

I talked about heat capacity and how the bricks could taken in and hold heat for long periods of time due to their high heat capacity.

jamo2oo9 wrote: »

For the metallic conductor question on the graph, I just explained that the graph contradicts the theory that was given to us on the course. I wouldn't worry much on that part.

Hopefully they'll accept curve or straight line because of this.

Jabama wrote: »

For question 7, did anyone know how to find the permittivity of the string?? i just guessed and used mass/length :P

μ is the mass per unit length, not permittivity in this case. I got tension to be 193N or so?

xJEx

**** i explained fusion instead of fision. Kill me.. I was so tired cause i got no sleep

Leaving Cert Student

xJEx wrote: »

**** i explained fusion instead of fision. Kill me.. I was so tired cause i got no sleep

Eh what q??

jamo2oo9

KildareKing wrote: »

why did it contradict it, I think you had to draw a line of best fit, and for the energy one it was W=I^2 R t .

Because it was supposed to be directly proportional through the origin. I got mine as a curve like everyone else.

Glee_GG

Not too bad, 7 out of 8 questions were grand, not so sure about my last one though but hopefully managed the A1, really need it for points at this stage! Q6 was pretty similiar to one they had a few years ago so that was grand as was Q5 and Q10 on the particle physics! For the experiment for boyles did anyone else assume the length they gave was in cm? I was sitting there for ages thinking that someone would never be measuring 52 metres in a lab so I changed it before doing the 1/L, anyone else?!

Jabama

Kingkumar wrote: »

ehh for what do u need the permittivity of the string for. the formula is F = 1/2L sqroot(T/mew) and mew is mass/length

Oh right cool, i thought that mew was the permittivity of the wire but i just guessed it should be mass/length, did you get 5N so?

Leaving Cert Student

Glee_GG wrote: »

Not too bad, 7 out of 8 questions were grand, not so sure about my last one though but hopefully managed the A1, really need it for points at this stage! Q6 was pretty similiar to one they had a few years ago so that was grand as was Q5 and Q10 on the particle physics! For the experiment for boyles did anyone else assume the length they gave was in cm? I was sitting there for ages thinking that someone would never be measuring 52 metres in a lab so I changed it before doing the 1/L, anyone else?!

ya the units were kPa and cm cubed had to be changed and handled that way I would say

Undeadfred

Jabama wrote: »

For question 7, did anyone know how to find the permittivity of the string?? i just guessed and used mass/length :P

Yeah that's it im pretty sure cause mew is mass per unit length so divide the total mass by the length gives you the mass per unit length

Leaving Cert Student

Define volt??

Glee_GG

Leaving Cert Student wrote: »

ya the units were kPa and cm cubed had to be changed and handled that way I would say

Grand! I left the pressure the way it was which is a bit stupid thinking back but sure it'll only be one or two marks!

curly135

Glee_GG wrote: »

Not too bad, 7 out of 8 questions were grand, not so sure about my last one though but hopefully managed the A1, really need it for points at this stage! Q6 was pretty similiar to one they had a few years ago so that was grand as was Q5 and Q10 on the particle physics! For the experiment for boyles did anyone else assume the length they gave was in cm? I was sitting there for ages thinking that someone would never be measuring 52 metres in a lab so I changed it before doing the 1/L, anyone else?!

Yeah, even thinking about the scale for the volume on the Boyle apparatus, it would be unrealistic to have it marked in meters... I got a straight line through the origin using centimeters anyway.

Overall nice enough paper, section A was grand, a few bits that caught me here and there in section B, but I'll definitely be going up from 64 in the pres :P

xJEx

Leaving Cert Student wrote: »

Eh what q??

9 the one about radioactivity

KildareKing

jamo2oo9 wrote: »

Because it was supposed to be directly proportional through the origin. I got mine as a curve like everyone else.

but it was directly porprotional, you just had to draw a line of best fit, you probably joined the points together.

maughantourig

Q11:
Given in log tables:

• T=2Pi/w
• a=-(w^2)s

w=root of (a/s) (leaving out - )

->T=(2Pi/(root of a/s))

->T(root of a/s)=2Pi

->a/s=(4Pi^2)/(T^2)

->a=((4Pi^2)s)/(T^2)

Anyone else get this?

KildareKing

curly135 wrote: »

Yeah, even thinking about the scale for the volume on the Boyle apparatus, it would be unrealistic to have it marked in meters... I got a straight line through the origin using centimeters anyway.

Overall nice enough paper, section A was grand, a few bits that caught me here and there in section B, but I'll definitely be going up from 64 in the pres :P

cm^3 not cm, I hope you didnt ,iss that.

Leaving Cert Student

maughantourig wrote: »

Q11:
Given in log tables:

• T=2Pi/w
• a=-(w^2)s

w=root of (a/s) (leaving out - )

->T=(2Pi/(root of a/s))

->T(root of a/s)=2Pi

->a/s=(4Pi^2)/(T^2)

->a=((4Pi^2)s)/T

Anyone else get this?

T was squared for me

mulciber

Leaving Cert Student wrote: »

ya the units were kPa and cm cubed had to be changed and handled that way I would say

I said cm^3 and Pascals. Are you sure it was kilopascals?

maughantourig

Leaving Cert Student wrote: »

T was squared for me

Me too, I left it out there by accident

xJEx

So if everyone did their graph differently what did yous get for the resistance when I = 0.7 ?

Leaving Cert Student

maughantourig wrote: »

Me too, I left it out there by accident

Q11 was a handy question. I threw it in at the end as an extra q after finishing the rest of the paper!

curly135

KildareKing wrote: »

cm^3 not cm, I hope you didnt ,iss that.

Luckily in the exam I didn't :P

Jackobyte

KildareKing wrote: »

why did it contradict it, I think you had to draw a line of best fit, and for the energy one it was W=I^2 R t .

KildareKing wrote: »

but it was directly porprotional, you just had to draw a line of best fit, you probably joined the points together.

jamo2oo9 wrote: »

Because it was supposed to be directly proportional through the origin. I got mine as a curve like everyone else.

Current at 1V was 0.17A. Therefore, if relationship was purely linear, at 6V, current should have been about 0.17 x 6 = 1.02A
However, on graph it was only 0.88, showing that it didn't remain linear at higher voltages.

I came out and asked my teacher straight away and she said that she couldn't be sure whether they wanted straight line or curve until the marking scheme came out, but that it didn't appear directly proportional based off the points given.

Leaving Cert Student

q7 was tricky I thought? For drawing out the diagram of the sound waves, what did ye do?? I did the first 4 harmonics but then scribbled out the second!

xJEx

Jackobyte wrote: »

Current at 1V was 0.17A. Therefore, if relationship was purely linear, at 6V, current should have been about 0.17 x 6 = 1.02A
However, on graph it was only 0.88, showing that it didn't remain linear at higher voltages.

I came out and asked my teacher straight away and she said that she couldn't be sure whether they wanted straight line or curve until the marking scheme came out, but that it didn't appear directly proportional based off the points given.

I looked up the diagram in the book and the points are slightly off there too. To be honest, mine looked fairly similiar to the straight line in the book