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# AQA MM1B - Mechanics 1 -Tuesday 21st June 2016

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1. (Original post by wil_is_he)
If you got all the rest exactly right then i reckon 6 maybe... But you might be lucky and get 7 its very hard to say as there hasnt been a 10 marker before
ok thanks, I'm a further maths student and managed to do 6x4 = 32 instead of 24 :-(
2. (Original post by Olsmarto)
I did what a lot of people did and made the velocities equal and getting 5 and 25 then using the R equations to get the position and doing Rb -Ra and then pythag to find the distance for each t=T and t=25, which was like 80.5m and 160.9m, I know this is wrong but will a get any marks here and there?

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It would be extremely harsh if they didn't award marks for this method. If the A grade boundary is very low, this will be why.
But there will be at least 6M1 marks at a guess, but really depends on the mark scheme. It could be really linear :/
3. (Original post by OturuDansay)
It would be extremely harsh if they didn't award marks for this method. If the A grade boundary is very low, this will be why.
But there will be at least 6M1 marks at a guess, but really depends on the mark scheme. It could be really linear :/
But the velocities were not equal, only parallel. You cannot find a value of t such that .
However you can find value of k such that where k is a scalar constant.
4. you need to think about the components of the velocities.

I component/divided by J component leads directly to direction

so this is the same when the velocities are parallel

Still very hard for M1 and the AS students
5. (Original post by dm999)
you need to think about the components of the velocities.

I component/divided by J component leads directly to direction

so this is the same when the velocities are parallel

Still very hard for M1 and the AS students
Maybe that's what AQA are pushing towards, harder exams.
6. (Original post by jtebbbs)
SAME horizontal plane only means that they are at the same height, i.e. don't move up or down relative to the ground. The question specifically said "instant" at which the velocities are parallel, which straight away means there's only one value for t - which is 12. The big misconception is really that most people didn't realise that when two velocities are parallel, one is a multiple of the other but they are not necessarily the same magnitude. With your method, if you put t=5 into the expressions for both components of both velocities, you would find that the velocities are neither the same, nor multiples of each other, so could not possibly be parallel.

The key knowledge that AQA were looking for here was that multiples of vectors are parallel, which is GCSE knowledge, not Core 4 (although it is used briefly again in Core 4 with vectors in i, j and k) and that is what made this a 10 mark question when everything else it was asking had come up in 6-8 markers in previous years.

I'm not saying it wasn't a tough question, it was, but that is what they were testing for.
On which exam board for GCSE is this on because I certainly didn't learn it 😂
7. I would think its intentional, it seems the exams are getting harder each year,

The thing that stands out about making this question difficult, is that it does not lead you into the solution at all.

It could say, find the velocity of each particle at time t.

And the position at time t.

Then the distance between the particles when the velocities are parallel.

More candidates would then score at least some of the marks.

Some candidates might even score full marks, that didnt have a clue where to start.

Its certainly aimed at "sorting" out the grade As
8. (Original post by rainbowtwist)
On which exam board for GCSE is this on because I certainly didn't learn it 😂
Its how you interpret it,

All exams boards do tangents at gcse.

The direction is purely that, the i component divided bu the j component.
9. yeah i did
10. How did you guys get 1.76 for question 7a. I keep trying and i got 1.9 something... :/
11. (Original post by the_malis)
How did you guys get 1.76 for question 7a. I keep trying and i got 1.9 something... :/
you have to make s=1 and a= -9.8 I think?
12. (Original post by kiiten)
you have to make s=1 and a= -9.8 I think?
I'm an idiot, i made s=-1 for some reason. How many marks would I lose for thatc question, do you think?
13. (Original post by the_malis)
I'm an idiot, i made s=-1 for some reason. How many marks would I lose for thatc question, do you think?
I think that would still work if you put a= 9.8 (not sure though). It depends really (how many marks was that ques i forgot xD).

Youll probably get some method marks and possibly error carried forward for later ques. Maybe half marks in total?
14. For the 10 marker I did everything right until working out the position of B where I did not add it's original position vector so I got a distance much smaller then 106 - how many lost?
With the 7 marker I also did everything right until right at the end where I forgot to square the cos50 so I only divided by cos50 not cos^2(50) and so got 10.6 instead of 13.2 - how many lost again?
15. (Original post by Colsop50)
ok thanks, I'm a further maths student and managed to do 6x4 = 32 instead of 24 :-(
haha I did 3+7=11 in one of them, so I know how you feel!
16. (Original post by SM-)
For the 10 marker I did everything right until working out the position of B where I did not add it's original position vector so I got a distance much smaller then 106 - how many lost?
With the 7 marker I also did everything right until right at the end where I forgot to square the cos50 so I only divided by cos50 not cos^2(50) and so got 10.6 instead of 13.2 - how many lost again?
Maybe 3 marks? For the 7 marker possibly 2 marks you might want someone else to double check as im not too sure.
17. (Original post by Sam Webb)
You can't do 16/1.7 because 1.7 was the time taken for the ball to hit the front edge, you had to rework out the time taken to be 1.9 as it would take longer to go further then do 16/1.9 etc
I said the intiaI veIcoity was xsin(theta) and worked at an expression for t with x as verticaI distance is the same. I added 3 to previous horizontaI distance and then times my verticaI time by 2 and subbed that t into a suvat. got around 13.
18. (Original post by That bloke 2016)
I'm a bit late but I'm pretty sure it was 081 degrees as it seemed too big so I flipped the opp/adj fraction to ensure it was the most sensible answer and about 7.something degrees came out, confirming it was actually about 081

note: there will still be a fair few method marks for getting that far.
Wait sorry, I'm confused so what was the actual correct answer?

A 51/75
b 45/75

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