M3 28/01/2011 Edexcel

I put in the wrong conditions into question 1 so I got C as 12 instead of 18 so got the wrong quadratic

I didn't manage to derive the inequality in the circular motion question (Thankfully it was only two marks)

I think I got the wrong value of lambda in the question in which you had to hang the toy from A and then take moments. I got 0.35 but I think it should of been the reciprical of that to give 2.56 of something...

Apart from that it went ok. I just managed to derive the very last inequality (the BIG 9marker). I also got 0.222s for the time it takes to get directly from A-->B.

The inequality came from T in BP being something like $\frac{m}{2}({\omega}^2l - \sqrt{2}g)$. As there was a tension so

I think I got lambda = 2.84 or something, I know my friend got it too...

Also, for the question involved to strings, I got the speed at C to be $v^2 = \frac{232}{5}gl$ (Something like that).

For circular motion question again, I got the tensions to be:
$\frac{1}{2}m(w^2l + \sqrt2g)$
$\frac{1}{2}m(w^2l - \sqrt2g)$

For the vertical circle question I said the max and min values of "T" occur at the bottom and top of the circle so when theta = pi and theta = 0.

For the strings I think I got

$v^2 = \frac{184}{15}gl$

For the strings I think I got

$v^2 = \frac{184}{15}gl$

For the strings I think I got

$v^2 = \frac{184}{15}gl$

I think I know how you got your answer. The particle had mass 3m I believe, but your answer uses mass = m

This.


$v^2 = \frac{232}{5}gl$ is what one would get if they ignored the fact the mass was 3m (relative to the 10mg $\lambda$) in the second part. Edit: See above.

FFS

Another stupid mistake...

This.


$v^2 = \frac{232}{5}gl$ is what one would get if they ignored the fact the mass was 3m (relative to the 10mg $\lambda$) in the second part. Edit: See above.


Personally, I didn't feel this was a particularly difficult M3. Sure, the above, and one of the COM questions came out with rather nasty looking answers, but a lot of the other stuff was all 'show that...'. I'd expect reasonably high boundaries, but I don't really know.

Also what do people think about difficulty/grade boundaries for this paper?

The paper was straightforward for the first bit until it got to Question 6 more or less and Question 7 was a complete b*tch

I think the boundaries will be like Jan '10's

I got the same tensions as you guys but my teacher said the tension in A was 0

I made the silly mistake of not reading the part of 'A and B are on the same positive axis' SHM question

I messed up. I worked so hard for it! How much UMS would you think 57 raw would be?

Does anyone know if you're allowed to use M2 in your A2 modules to contribute to the A*? And then use M3 in your AS further maths? (Incase I screw up M3) So I'll have the following modules:

AS:

FP1 M1 M3

A2:

FP2 FP3 M2

They count your best 3 A2 results when awarding the A* so any 3 from FP2 FP3 M2 M3

EPE = KE + GPE

$2(\frac{hx^2}{2l}) = \frac{1}{2}mv^2 + mgh$

$\frac{hx^2}{l} = \frac{1}{2}mv^2 + mgh$

$\frac{10mgx^2}{l} = \frac{1}{2}mv^2 + mg(\frac{12}{5}l)$

$\frac{20gx^2}{l} = v^2 + \frac{24}{5}gl$

$\frac{20(\frac{8}{5}l)^2}{l} = v^2 + \frac{24}{5}gl$

$\frac{1280}{25}l = v^2 + \frac{24}{5}gl$

$v^2 = (\frac{256}{5} - \frac{24}{5})gl$

$v^2 = \frac{232}{5}gl$

That's what I did....

Surely you have forgotten about Final EPE as the position you want the speed at was where the strings are still stretched, unless i have missed something.

Well it said find the speed when the particle is at "C". "C" is the mid-point of AB where AB = 2m. The natural lengths of the two strings combined is 2m so there is no E.P.E stored at C.

(I used the wrong mass so ignore my final answer anyway )

For the strings I think I got

$v^2 = \frac{184}{15}gl$

I think I know how you got your answer. The particle had mass 3m I believe, but your answer uses mass = m