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# Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread watch

1. (Original post by KloppOClock)
[Trailer] <--------> [Car]

the arrow is the direction of thrust
also it says the magnitude of thrust is 100N

if you dont get that I can draw a diagram for you
thanks very much! i finally got the right answer!
2. (Original post by 99joey)
oh ok, however if they asked us present it in the format i and j, would I then be penalised if I didn't underline the vectors?
As far as Im aware they cant penalise you...but to be safe you can figure it out in bracket method then just stick the answer down in the format they asked for, but usually in a mark scheme it says accept bracket method (or alternative)
3. (Original post by 83457)
Can I theoretically do them with Edexcel m1 and m2 knowledge? And do they have 3d vectors like with OCR MEI?
Have you done all of the real Edexcel papers from 2001 onwards? The specification for M1 and M2 hasn't changed since then.
4. Is my working out too OCD compared to some of your working out?

Obviously, I won't do them like this in the real exam I.e free hand drawing but I'd rather
perfect everything neatly drawn so I can get into a perfect pattern etc.

Also, how are some of your working outs when you answer questions compared to mine?

5. (Original post by KloppOClock)
would there be tension when A and B are back to there original position, or if A and B reach a height of 2L? Also if A gets tension before B does after reaching its original position first, does it rebound?

(Original post by KloppOClock)
if A and B reach a height of 2L?
1. The question states that . This inequality is here such that the particle with speed does not collide with the pulley. Hence neither particle goes above a height from their original position.

(Original post by KloppOClock)
Also if A gets tension before B does after reaching its original position first, does it rebound?
2. Both particles must get tension at the same time as they are connected by a single string. Tension occurs will the length of the string is of definite length with no slack. In reality there would be rebound but this does not have to be considered for the question.

(Original post by KloppOClock)
would there be tension when A and B are back to there original position
3. There would be tension if, and only if both particles were in their original position.

Hints for question
Split the problem into the 3 parts:
- When both A and B are going upwards.
- When A is going upwards and B goes downwards.
- When both A and B are going downwards.

Hope this helps,
Cryptokyo
6. (Original post by XxKingSniprxX)
Is my working out too OCD compared to some of your working out?

Obviously, I won't do them like this in the real exam I.e free hand drawing but I'd rather
perfect everything neatly drawn so I can get into a perfect pattern etc.

Also, how are some of your working outs when you answer questions compared to mine?

I pretty much do the same but I make the diagrams as simplistic as possible while still ensuring all the information is there. But a slight difference may be that I:
-represent vector quantities with solid lines and use a little arrow (or arrows!) - but more on that later.
-represent solid surfaces with a solid line with stripes on the opposite side to the surface.
-always represent an object with a single blob of ink.
-use dashed lines for measurements.
Just on arrows...
-arrows halfway along the line for velocity.
-use bracket endings on an arrow for displacement.

But that is much overkill for one to do. But it makes the drawings only have about 4 lines.
7. How do I do part A?
8. (Original post by Sam1999)
How do I do part A?
just from looking at it I would guess that you resolve the weight of P and apply F=ma to it
9. (Original post by KloppOClock)
just from looking at it I would guess that you resolve the weight of P and apply F=ma to it
But how do I resolve it as I don't know the hypotenuse
10. (Original post by Sam1999)
But how do I resolve it as I don't know the hypotenuse
im just doing the question now so ill have a look
11. (Original post by Sam1999)
But how do I resolve it as I don't know the hypotenuse
just set the hypotenuse to equal a mass of "P" times G to get a weight. Put that into F=MA and the P should cancel out i think

Edit: just checked the mark scheme, P's cancel out and final answer is 8.49N
12. (Original post by KloppOClock)
just set the hypotenuse to equal a mass of "P" times G to get a weight. Put that into F=MA and the P should cancel out i think
But doesn't the weight act vertically down?
13. (Original post by Sam1999)
But doesn't the weight act vertically down?
Never mind I'm being stupid 😤 I've been stuck on that for 20 mins
14. (Original post by Sam1999)
But doesn't the weight act vertically down?
yes but your trying to find acceleration which isnt vertical.

First resolve the force of P. Use a weight of P*G (mass times gravity). Then taje the horizontal component and sub that into F=MA
15. (Original post by Cryptokyo)
-use bracket endings on an arrow for displacement.
what do you mean by that?
16. Stuck on 1b from iygb paper v (dont think there is ms)

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17. Can someone please explain to me why for part d we use 2s1 instead of just s1 + s2, taken from Jan 10.
Spoiler:
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18. (Original post by _Priyesh_)
Stuck on 1b from iygb paper v (dont think there is ms)

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Which paper? Whats iygb

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19. Just marking the M1 Solomon Paper D that I did and I have a question about the mark scheme:

Surely it would also depend upon the size of the stone as well? How can you say the tennis ball will have more wind resistance if they don't state the size or shape of the stone?

Would I get any marks for my answer?

"The minimum speed required would be different depending upon the mass of either object as the resultant force of 'F' in F=MA would change which would effect the acceleration as they have different masses, therefore more or less speed may be required."
20. (Original post by KloppOClock)
just from looking at it I would guess that you resolve the weight of P and apply F=ma to it
I would have just resolved 9.8 with the angle they gave, would it have been the same answer?

Edit yeah same answer, its because the ball is acellerating vertically down due to gravity so you can resolve

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