# AQA MM1B and MPC3 exams January 23 2013Watch

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Poll: After these 2 exams, how would you rate them?
Mechanics: Easy (4)
8%
Mechanics: Okay (10)
20%
Mechanics: Could have been better (7)
14%
Mechanics: I've failed (2)
4%
Core 3: Easy (12)
24%
Core 3: Okay (4)
8%
Core 3: Could have been bettwe (8)
16%
Core 3: I've failed (3)
6%
6 years ago
#41
(Original post by SherlockHolmes)
I would guess 2 marks would be lost because one for the getting the wrong time and one for an incorrect answer.

I think the boundary for an A will be the same as last year (57) plus/minus 2. More likely to be minus though, as generally, people found the exam harder than the previous year's papers.
Thank you! You are my favourite new friend of the day
2
6 years ago
#42
(Original post by SherlockHolmes)
It should have been 1.25 and 2.75

To get 1.25 you assumed that the A and B would travel in the same direction after the collision with a positive velocity (to the right).

To get 2.75 you had to assume A and B would rebound and travel in the opposite direction to their initial direction. Velocity of A would now be negative and velocity of B would be positive.

Here are the numbers if anyone was looking for them.

Before the Collision

Particle A
Mass = 5kg
velocity = 4ms-1

Particle B
Mass = 4kg
Velocity = -3ms-1

After the Collision

Particle A
Mass = 5kg
velocity = 0.6ms-1

Particle B
Mass = 4kg
Velocity = Vms-1

Question asked for the two possible velocities of B after the collision.
Yeah I thought theA particle could be going in the minus direction and the B particle could be going in the minus direction so I worked that out when each was in the minus direction, okay this might not make sensee but I feel like it was right haha
0
6 years ago
#43
I taught myself mech in year 12 so I only did core 3 and fp2 today, core 3 was easyyyyy but fp2 was impossible!
0
6 years ago
#44
(Original post by OliT95)
The integral was pie/4 and 0 btw. But then you make U in terms of sec^2x, so you square both sides so U^2 = tan^2x, then add 1 so U^2 + 1 = tan^2x + 1. With the identity it means U^2 + 1 = sec^2 so then you just subsitute in U^2 + 1.

Then it is (U^2 + 1) root U dU then from there multiply out brackets and integrate
My son found this question hard. I did my a-levels 25 years ago, but I work with Maths so I'm not too rusty. I just gave it a go and I reach the same conclusion as above (only I built my substitution by direct observation rather than reverse manipulation then observation).

Here's the full version:

integral sec^4(x) * sqrt( tan(x) ) dx between 0 and pi/4.

where integral is integral symbol and sqrt is square root symbol and ^ means power of eg x^2 is "x squared".

with substitution u = tan(x)

ok so du/dx = sec^2(x) therefore dx = 1/sec^2(x) substitute that into
integral gives:

= integral sec^4(x) * sqrt( u ) * 1/sec^2(x) du

= integral sec^2(x) * sqrt( u ) du [because sec^4(x)/sec^2(x) = sec^2(x)]

Normally its obvious how to substitute u for tan(x) into the remaining x
part, but in this case it is not. That doesnt mean its gone wrong, dont panic think what is it that you are trying to do. Answer: you need to manipulate the x part somehow to be expressed in terms of u (ie to be some multiple or power of tan(x) then you can get an all u form with no x's ready for actual integration).

Well u is tan(x) and we have sec(x) related things to manipulate. A formula relating sec and tan is tan^2(x)+1 = sec^2(x) so substitute that:
(If you forget that formula just divide sin^2(x) + cos^2(x) = 1 by cos^2(x) throughout).

= integral (tan^2(x)+1) * sqrt(u) du

now replace tan(x) with u:

= integral (u^2 + 1) * sqrt(u) du

expand brackets, replace sqrt(u) by u^(1/2) to make it easier first:

= integral (u^2 + 1) * u^(1/2) du
= integral u^(5/2) + u^(1/2) du

then using general integral rule x^n = 1/(n+1)*x^(n+1) rule on each part:

= [ 2/7*u^(7/2) + 2/3*u^(3/2) ]

substitute u = tan(x):

= [ 2/7*tan^(7/2)(x) + 2/3*tan^(3/2)(x) ]

= [2 sqrt(tan(x))*(1/7*tan^3(x) + 1/3*tan(x)]

put the valus in pi/4 and 0 subtract etc.

Quite a difficult question looks like most of the people didnt get it so
that helps the UMS mark stay low.

btw I am not sure if it would give you any marks but you could've put the
definite integral formula into a calculator and calculated the answer
(though it would give no hint as to how to do the integral!) Maybe they would give one mark only for the correct answer as it would be obvious you used a calculator if there was no working.

You would also get a mark each for writing the first few stages I would think:

[1] du/dx = sec^2(x) therefore dx = 1/sec^2(x) substitute that

[2] = integral sec^4(x) * sqrt( u ) * 1/sec^2(x) du

probably one for the right answer mark [8] (if you got it via a calculator with support for definite integrals) and lose 5 for the unknown middle.

I hope you scavenged those marks or maybe not if you ran out of time, but it shows that you can get marks for parts of a question you dont know.

If you get spare time you might get a mark for explaining what you are trying to do (eg rearrange sec^2(x) in terms of tan(x) so I can substitute it for u).

ps my tip for current students: practice being a speed demon. A guy with 25% time left over at the end can scavenge marks for things he doesnt know or do things the hard way and retrofit a passable facsimile of a correct writeup for something he has a mental block about, and if there is nothing else left to do use remaining time to battle the remaining impossible question.

pps It really does work, I once got 99% on university math test that I knew 75% of the topic and not that well. University rules allowed you to choose 5 of 8 questions. But they would award marks for parts of any 8 questions. I aggressively used speed and error carried forward and guesses etc and managed to get 99%. In the previous math test where I did know what I was talking about I got 156% out of 160% (docked down to 100%). I finished after 2hrs into 3 hr exam and used all spare time checking and proof reading tee hee.
0
6 years ago
#45
(Original post by SherlockHolmes)
The boat travelled back along the same path but from B to A on the return. The velocity of the boat remained the same at 4ms-1 but the question did not state the direction. The current of the river is still 3ms-1 due east.

Using the above, you could draw a triangle which would not be a right angle (a few got the answer as 5 which is incorrect), and then using the sine rule or cosine rule to find the resultant velocity which was 1.40ms-1.

regarding mm1b paper.

for 6b) I get
R
2 = 42 + 32 – 2(4) (3) COS 53.13˚

R = 3.25 m/s
0
6 years ago
#46
Well....holy moly... walked out of the mechanics exam almost trembling at the difficulty, forgot to answer one question, totally messed up another, lots of scribble and mess where I couldn't figure things out - expected a C or D... got an A! Yeeeeeah

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