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Edexcel M2 - 28/01/11

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Avatar for atofu
atofu
7
In M1 there was the whole thing about losing a mark if you put too many decimal places/significant figures, is this still true for M2? Also what is a good number to go to? I usually go to 3 s.f. but i heard that if you use g (9.8) then you have to round to 2 s.f.
Does anyone know which is correct? xD

also if the question gives you the weight of something do you not multiply it by g when doing forces/moments?
(edited 13 years ago)
I usually have no mood to study, but you guys have made me want to do revision so badly now O_O
Original post by atofu
In M1 there was the whole thing about losing a mark if you put too many decimal places/significant figures, is this still true for M2? Also what is a good number to go to? I usually go to 3 s.f. but i heard that if you use g (9.8) then you have to round to 2 s.f.
Does anyone know which is correct? xD

also if the question gives you the weight of something do you not multiply it by g when doing forces/moments?


You usually follow the question. So if you use 9.8g which is 2 s.f the answer should also be 2.s.f (Answer). But if you use say 9.82g then your answer should be rounded to 3 .s.f.
And no you do not multiply it by g when you are given the weight. Because it is already multiplied by g. Weight(Newtons) is a force, just like the equation F = ma, ( mass times acceleration(Which is g 9.8)
But usually in moments(M2) question, g would cancel out, so you do not need to worry. But M1 can sometimes trick you. So watch out :smile:
(edited 13 years ago)
Can anyone help me with this centre of mass question???
Original post by cazzy-joe
Can anyone help me with this centre of mass question???


Ok, so let the angle O be theta.
By the 'length of arc' formula " L = r x theta " , Theta = (2pi / 6) = (Pi / 3)
Split this 'theta' angle into two equal 'alpha' angles that are (Pi / 6)

Now we only need to find the X value for the C.o.M since we know by symmetry that the 'Y' value will run horizontally through O. The easiest way is to look at each length of the framework individually

For OA, by right angled trig, its C.o.M 'X-coordinate' is 3cos(pi/6). By symmetry OB will also have this same 'X-coordinate' for its C.oM. Finally for AB, by the 'C.o.M for an arc' formula ( r sinalpha / alpha) where alpha = (Pi/6) , it's X-coordinate C.o.M will be ( (6sin(Pi/6) / (pi/6) ) = (18/pi). Now we can sub all of these into the main C.O.M formula.

6 x 3cos(pi/6) + 6 x 3cos(pi/6) + 2pi x (18/pi) = (12+2Pi) X

Rearrange to find X = (36 + 18root3) / (12 + 2pi)

which simplifies to X = 9(2 + root3) / (6 + pi)
I hate collisions in M2, they are so tedious to do!

The amount of time I spend on these question is unthinkable and I still manage to slip up somewhere, especially the inequalities ones. Easiest topic in M1, most frustrating in M2
Original post by CapsLocke
I hate collisions in M2, they are so tedious to do!

The amount of time I spend on these question is unthinkable and I still manage to slip up somewhere, especially the inequalities ones. Easiest topic in M1, most frustrating in M2


Completely agree! What I hate most is when you have to set up those inequalities, and when they decide to change the direction, I get into a muddle with them all the time ¬_¬
I'm fine with collisions, but the inequalities do mess me up some times. I don't like statics, but I'm not too bad with them. Hoping for a simple ladders question, no poles with rings etc (though if you've seen any hard statics questions with answers please link me to them!). C.O.M is really easy, but remembering what to do is annoying, especially the fact that I keep making stupid mistakes. Projectiles are fine, as is the calculus. Work-energy can be tedious and easy to mess up on, still don't understand why I can't use M1 knowledge and get the correct answer for half of them =\ . And well.. that's all of the chapters xD .
I'm generally finding M2 papers to be 'easier' than their younger M1 brothers. The Work-Energy questions, which for some reason are always done the worst (according the examiner reports) just seem like the Dynamics' questions from M1 with an extra formula or two thrown in. And using Calculus instead of the SUVAT equations (at least, for projectiles etc) makes the kinematics questions about 10x easier because you're less likely to make mistakes with signs etc, due to the math cancelling it out. Collisions are relatively straight forward. Not a HUGE increase in difficulty from M1, there's just way more to do in a question. The only part I don't like is the wording a lot of the time. Stop calling 'e' friction damnit!

The only module that really seems like a big increase in difficulty from M1 is the M2 Moments chapter. There just seems like a lot more to do compared to M1.

Am I the only one who's hoping for a hard paper like last January? 57/75 for an A. I generally find that the harder a paper is, the better I do. I think that's because I inevitably make simple mistakes, and the mistakes are more punishing the higher the grade boundary is.

Anyway, back to the grind. Good luck all.
Hmm.. edexcel site is missing some papers and markschemes, please quote if you have them!

Jan/June 03 (paper+ms)
Jan/June 04 (paper+ms)
June 05 (paper+ms)
Jan 07 (ms)
June 07

Thanks!
Original post by ViralRiver
Hmm.. edexcel site is missing some papers and markschemes, please quote if you have them!

Jan/June 03 (paper+ms)
Jan/June 04 (paper+ms)
June 05 (paper+ms)
Jan 07 (ms)
June 07

Thanks!


here
Original post by ViralRiver
Hmm.. edexcel site is missing some papers and markschemes, please quote if you have them!

Jan/June 03 (paper+ms)
Jan/June 04 (paper+ms)
June 05 (paper+ms)
Jan 07 (ms)
June 07

Thanks!


here again
Original post by ViralRiver
Hmm.. edexcel site is missing some papers and markschemes, please quote if you have them!

Jan/June 03 (paper+ms)
Jan/June 04 (paper+ms)
June 05 (paper+ms)
Jan 07 (ms)
June 07

Thanks!


here again ..
>< Thanks sooooo much! I'll pos rep again tomorrow :smile: . Sorry to be a bother... but would you also have the June '07 MS? It seems as if I forgot to mention that xD .
Original post by ViralRiver
>< Thanks sooooo much! I'll pos rep again tomorrow :smile: . Sorry to be a bother... but would you also have the June '07 MS? It seems as if I forgot to mention that xD .


You're most welcome:wink::wink:

It said file 2007 june MS is too big to upload in this forum:frown:
may be you can give me your email in my inbox:tongue:
Oh gawwwd.. that paper must be hard xD . Will do, thanks again :smile: .
Im taking this one. I understand all of the concepts, but in every practice paper I do there's always a really hard question which i just can't get my head around and it loses me loads of marks. I only have three practice papers left as well.
Reply 37
Avatar for patto
patto
Revision for this is going okay, though I've not quite got my head round one thing: In some of the questions about things flying through the air and landing somewhere, the examiners seem happy for you to use the vSquared = uSquared + 2as formula, whilst in others they state that marks are only awarded if you arrive at 'v' using conservation principles (i.e. KE lost = PE gained). SolutionBank points out the same thing in q59 and q60 of the first review exercise. I've looked carefully at the wording of the different questions but they don't seem to differ in any consistent way. Can anyone enlighten me?

By way of examples: June 2009 only lists the conservation approach as getting marks (though both approaches give same answer). June 2008 on the other hand permits both.
Can someone help me with these 2 questions:smile:

June 2007
Q4 part b-----I dont understand what they have put on as potential energy.

Q8 part d-----The mark scheme said after 8 sec direction changed..and I dont know what happened..



Thank youu sooo muchh ....
Probably more questions coming up:frown::frown:
m2.doc174.6KB
Original post by cazzy-joe
Can someone help me with these 2 questions:smile:

June 2007
Q4 part b-----I dont understand what they have put on as potential energy.

Q8 part d-----The mark scheme said after 8 sec direction changed..and I dont know what happened..



Thank youu sooo muchh ....
Probably more questions coming up:frown::frown:


4) I don't like the way round they've put the equation in the mark scheme. Here's how i think of it: Work done against friction = Total loss of energy
W.D against Friction = Friction x distance = (5/8)(4mg/5)h
Total loss of Energy = Loss of P.E - Gain in K.E = (7mgh/5) - (0.5mv^2 + 0.5(2m)v^2)
Set these equal to one another to find v^2 in terms of g and h

8) We know from part (c) that P is "instantaneously at rest" at t=8. What this actually means is that the particle is changing direction.
By subing in t=8 into the displacement equation we get s=48 i.e. after travelling 48m the particle changes direction and starts to move back whe way it came. Since we are asked how far the particle has travelled after "10 seconds" we must now sub in t=10 which gives a displacement of s=44. However whilst this is the particle's "displacement from the origin" after 10 seconds, it is NOT the "total distance" the particle has travelled. The particle has travelled 48m forwards in the first 8 seconds and then 4m backwards in the following 2 seconds in order to get to the point where it is 44m from the origin. Hence it has travelled (48 + (48 - 44)) m = 52m
(edited 13 years ago)

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