x Turn on thread page Beta
 You are Here: Home

# Getting to Cambridge: STEP by STEP! watch

1. (Original post by Zacken)
Thanks for that! I will be working through the rest of the pure and might even try a stats question out, it looked like S2 integration; if you want to give it a try?
Do let me know the question number and I'll look through it tomorrow morning! I occasionally look at some STEP I and II mechanics from time to time but I usually don't attempt them because they're usually slogs of algebra.
2. (Original post by aymanzayedmannan)
Do let me know the question number and I'll look through it tomorrow morning! I occasionally look at some STEP I and II mechanics from time to time but I usually don't attempt them because they're usually slogs of algebra.
III, 2005, Q14. Let me know what you think of it.
3. Did my first STEP III stats question, absolute blast!
4. (Original post by Zacken)
Did my first STEP III stats question, absolute blast!
Awesome job man
5. (Original post by Student403)
[awesome physics dude]
I have le physicz questionzzz.

Q7(i) and (iii) - Is my reasoning correct here?

(i) Because and are obviously attracted to one another, this means they have opposite charges and hence the particle is neutral? Or is that complete out of whack? Even if I'm correct, I'm not entirely sure why the two X and Y particles having different polarities mean that the particle is neutral.

(iii) So obviously we're going to have the particle decaying into the two particles, and we know that a property of radiation is that the total energy is conserved, be it in the form of mass or not which would make the obvious choice, but what's wrong with (how do you know that that's not correct?), option is definitely ruled out. What's the subtle caveat with option - what does the difference from adding 'energy' in front of the 'mass' make a difference? I thought that mess and energy were equivalent, so either option seemed fine to me; but that's just a misunderstanding on my part.
6. (Original post by Zacken)
I have le physicz questionzzz.

Q7(i) and (iii) - Is my reasoning correct here?

(i) Because and are obviously attracted to one another, this means they have opposite charges and hence the particle is neutral? Or is that complete out of whack? Even if I'm correct, I'm not entirely sure why the two X and Y particles having different polarities mean that the particle is neutral.

(iii) So obviously we're going to have the particle decaying into the two particles, and we know that a property of radiation is that the total energy is conserved, be it in the form of mass or not which would make the obvious choice, but what's wrong with (how do you know that that's not correct?), option is definitely ruled out. What's the subtle caveat with option - what does the difference from adding 'energy' in front of the 'mass' make a difference? I thought that mess and energy were equivalent, so either option seemed fine to me; but that's just a misunderstanding on my part.
Not quite!

i) It's a lot more simple - you just can't see the lambda particle before point O because it's in a particle detector (i.e. magnetic field) as it doesn't leave a trail. This lack of a trail means it's neutral

iii) A - we can't be conclusive about this one because X has a smaller radius than Y. r = mv/BQ and we do not know the masses or the velocities
B - established why that's incorrect
C - because mass =/= energy. If a 1 tonne rocket ship splits up in to two components with energies 100 KJ and 200KJ.. Well... 100kg =/= 100 + 200 KJ
D - You've got it

EDIT: Just another quick thing, you can't really say they're attracted to each other. It's true that they have opposite charges (evidenced by curving in opposite directions in a magnetic field), but the reason they experience circular motion in such arcs is because the Magnetic force due to the magnetic field acts perpendicular to their velocity and so causes circular motion
7. (Original post by Student403)
Not quite!

i) It's a lot more simple - you just can't see the lambda particle before point O because it's in a particle detector (i.e. magnetic field) as it doesn't leave a trail. This lack of a trail means it's neutral
I literally just read about neutral particles not leaving a trail. Thanks for that! Very obvious now.

iii) A - we can't be conclusive about this one because X has a smaller radius than Y. r = mv/BQ and we do not know the masses or the velocities
B - established why that's incorrect
C - because mass =/= energy. If a 1 tonne rocket ship splits up in to two components with energies 100 KJ and 200KJ.. Well... 100kg =/= 100 + 200 KJ
D - You've got it
Ah, the A bit is a bit annoying, I was half-suspecting that it'd be more "inconclusive" instead of just "definitely wrong". If the 1 tonne rocket ship splits into those two components, the energy released in the splitting thing is what gives the two components more energy than the rocket ship by itself, riiiight? But we can say that the "mass energy" (what does this term even mean tbh?) is the sum of the total energy of the two constituents, yeah?
8. (Original post by Zacken)
I literally just read about neutral particles not leaving a trail. Thanks for that! Very obvious now.

Ah, the A bit is a bit annoying, I was half-suspecting that it'd be more "inconclusive" instead of just "definitely wrong". If the 1 tonne rocket ship splits into those two components, the energy released in the splitting thing is what gives the two components more energy than the rocket ship by itself, riiiight? But we can say that the "mass energy" (what does this term even mean tbh?) is the sum of the total energy of the two constituents, yeah?
Yeah I agree. It's just a thing you have to be able to do I guess.. Work out of you haven't got enough information to say "This is a conclusion".

Well it doesn't have more energy. The splitting would require chemical energy from an explosion or whatever so truly all energy would be conserved (assuming no relativistic effects) (I know this seems pedantic bringing up chemical energy but you have to be like that with energies in U4 because there are so many transfers - this mention of chemical energy was actually a mark in a past paper question)

Mass energy is the sum of the energy due to rest mass + any other energy possessed by the particle. So in most cases of unit 4 motion Emass-energy = (mrest mass x c2) + Ek. We use e = mc2 to give us the energy due to rest mass. Then we use the Ek = 1/2 mv2 ... And add the two to give us mass-energy. Mass-energy = truly ALL of the energy possibly possessed by the object, whether in the form of actual energy or mass. This is measured in Joules.

We could probably say that, but this assumes it experiences no relativistic effects (i.e. no increase in mass at the cost of energy). What this implies is that we can use the terms mass-energy and total energy interchangeably (I'm not quite sure if this is true past the 0.8c threshold when it experiences relativistic effects), but personally I'd always choose to use the former term as it sounds a lot more correct and is more likely to be a mark criterion.
9. (Original post by Student403)
[explanationz]
Awesome, thanks for that - very well explained. :-)
10. (Original post by Zacken)
Awesome, thanks for that - very well explained. :-)
My pleasure
11. (Original post by Student403)
...
If only you were my Physics teacher and Zacken was my Maths teacher...
12. (Original post by aymanzayedmannan)
If only you were my Physics teacher and Zacken was my Maths teacher...
Oh man thanks for that :'D But a maths teacher like Zain is goals.. I'm not even exaggerating he teaches AND knows more maths than half the maths department in our school.. And our school was ranked in the top 50 British schools outside the UK That's how good he'd be!
13. (Original post by Student403)
My pleasure
(Original post by aymanzayedmannan)
...
Quick check - Jan 2010, U4 seems to have 63 as the full UMS boundary, I'm assuming this isn't the usual thing for U4 papers?

Edit: Is U5 much more lenient?
14. (Original post by Zacken)
Quick check - Jan 2010, U4 seems to have 63 as the full UMS boundary, I'm assuming this isn't the usual thing for U4 papers?
Once you start doing IAL papers you'll find them a lot easier than normal AL (this only applies to U4) so the boundaries are a lot higher, yes, but they're just as doable for full.

In fact I found full easier on IAL than normal AL

Also with that in mind, U5 isn't like U4 with the higher boundaries. I think U4 is the hardest to get full UMS in A2 compared to U5/6 if you know the content for each equally well
15. (Original post by Student403)
Oh man thanks for that :'D But a maths teacher like Zain is goals.. I'm not even exaggerating he teaches AND knows more maths than half the maths department in our school.. And our school was ranked in the top 50 British schools outside the UK That's how good he'd be!
He could teach our Y12 FP1 teacher a thing or two, that's for sure. I got into an argument with that guy about how my calculator isn't "illegal" in the middle of my U4 mock while he was invigilating.

(Original post by Zacken)
Quick check - Jan 2010, U4 seems to have 63 as the full UMS boundary, I'm assuming this isn't the usual thing for U4 papers?
It was the first paper of the new syllabus. The boundaries are a bit steeper, but having said that, I've only done 2 U4 papers.
16. (Original post by Student403)
Once you start doing IAL papers you'll find them a lot easier than normal AL (this only applies to U4) so the boundaries are a lot higher, yes, but they're just as doable for full.

In fact I found full easier on IAL than normal AL
Oh, that sounds weird. How come?

What's the boundaries like compared to AS? Unit 1 was super forgiving, Unit 2 was quite forgiving but not as much, how do U4/5 stack up to that?
17. (Original post by Student403)
Once you start doing IAL papers you'll find them a lot easier than normal AL (this only applies to U4) so the boundaries are a lot higher, yes, but they're just as doable for full.
IAL papers are easier??? That's the first time I've heard that.
18. (Original post by Student403)
Oh man thanks for that :'D But a maths teacher like Zain is goals.. I'm not even exaggerating he teaches AND knows more maths than half the maths department in our school.. And our school was ranked in the top 50 British schools outside the UK That's how good he'd be!
So your maths department has no one with a maths degree? (Not denying Zacken's brilliance)
19. (Original post by Zacken)
Oh, that sounds weird. How come?

What's the boundaries like compared to AS? Unit 1 was super forgiving, Unit 2 was quite forgiving but not as much, how do U4/5 stack up to that?
Unit 1 boundaries are a fu**ing joke. It's like the easiest physics unit out of ALL of them and yet it has the lowest boundaries wtf? U5 compares to U2 well, I think. U4 is the most harsh probably on par with U3 for full. And U6 is quite nice. Maybe on par with U5/U2 where full is doable (I think it's 35 give or take a couple each year)
20. (Original post by zetamcfc)
So your maths department has no one with a maths degree? (Not denying Zacken's brilliance)
Many of them do but they're veteran teachers. So they have a good teaching style and some even have maths degrees but since they've been teaching for so long, they've forgotten the degree content and we often can't ask them about undergrad level stuff.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 3, 2017
Today on TSR

### Boyfriend slept with someone else

...we were on a break

### Mum says she'll curse me if I don't go to uni

Useful resources
Don't miss out on a place at uni - get clearing email alerts

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE