AQA A2 Mathematics MM2B Mechanics 2 - Monday 22nd June 2015 [Exam Discussion Thread]

Thanks a lot!

Just done Jan 2013 and the last part of question 9 took forever to do I'm worried I'm not going to have enough time for a tricky 'show that' question

That question is horrid. I still can't do it despite trying it twice. It was in my opinion the hardest ever M2 question


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I can post my working if you want? I gave up, marked my paper and then just as I found a video to explain it I had an 'omg I think I know how to do this' moment. I'm glad it was only 5 marks though because it meant I still got an A* UMS-wise

Oh what video? if you wouldnt mind posting I'd appreciate it.Posted from TSR Mobile

https://www.youtube.com/watch?v=kA4F-p5Hj_4

My method:




Sorry about them being sideways

Thanks! That makes so much more sense. I still wouldn't have got that in the exam haha.


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Can someone please help me understand how to find the angle in no (c).. Mark scheme is also attached

I've just done this paper so I'll post my working out:

For simplicity I've drawn the lamina the same way up as the diagram gives it, but the dashed line represents the horizontal direction and the solid line represents the vertical.
You know that when the lamina is suspended from point H that the vertical will pass directly through the centre of mass. You're given the distance of the COM from AH and you've just worked out the distance from AB. The distance of the COM from HC is just 30cm - 13.6cm and so you can form a triangle and work out the angle required.
Edit: don't be thrown off by the COM not actually being where the lamina 'exists' as per say
Another edit: Theta is the angle between HG and the horizontal and x is the other angle in the right angled triangle. I did it a long winded way and you could actually just do the tan of theta straight away but apparently I make things harder for myself

Neither would I. I was panicking as I couldn't do that one and my answer for the epe question seemed odd (but fortunately it was right) and M2 grade boundaries are unforgiving.
The examiner's report said this question was 'designed to discriminate between grades A and A*' and 'There was much inventive algebra shown in part (c) to obtain the printed result.'
I can't stand how cutting and sarcastic the examiners reports are (although sometimes it is quite funny)

I've just done this paper so I'll post my working out:

For simplicity I've drawn the lamina the same way up as the diagram gives it, but the dashed line represents the horizontal direction and the solid line represents the vertical.
You know that when the lamina is suspended from point H that the vertical will pass directly through the centre of mass. You're given the distance of the COM from AH and you've just worked out the distance from AB. The distance of the COM from HC is just 30cm - 13.6cm and so you can form a triangle and work out the angle required.
Edit: don't be thrown off by the COM not actually being where the lamina 'exists' as per say
Another edit: Theta is the angle between HG and the horizontal and x is the other angle in the right angled triangle. I did it a long winded way and you could actually just do the tan of theta straight away but apparently I make things harder for myself

can someone explain why for jan 13 q 3 you have to work in radians?

Sorry, what do you mean?

They give you what $\sin \theta$ equals so that you can just use the numerical value when multiplying by the weight to find the component of the weight acting down the slope.

This means the driving force at maximum power must equal the resistance force plus the component of the weight down the slope. You don't have to find $\theta$ at any stage.




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Well when i first did the calculation in degrees i came to the wrong answer, redid it in radians and got the right answer?

Which are the hardest past papers?

Not sure about the hardest paper in itself, but if you haven't already then try Q9 January 2013.


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