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# M1 Edexcel Maths 8th June 2016 [MY ANSWERS] [COMPARE]

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1. Hi, here are my answers to the paper sat today, not 100% sure if their entirely correct obviously but its up for comparison and discussion.

1)a) 104degrees
b) i) P : 400i + t(15i+20j)
ii) Q : 800j + t(20i-5j)
c) 640i + 640j

2) a) 20.6N
b) 15.45N

3) 3Ns

4)a) n/a diagram
b) T = 8.75s

5) Coefficient of friction = 0.73

6) i) d=1.2m
ii) m=42kg

7)a) 2.5i+2.5j
b) 13 ms^-1

8)a) T=11.76N
b) 16.6N at a bearing of 225degrees

Hope this helps.

Thanks, Keenan.
2. Exactly the same as myself , hopefully 100% then.
For 1b would you not have to combine the i&j components into one vector?
3. Got the same also
4. is this the edexel foundation 2016 paper 2 answers
5. (Original post by james5665)
is this the edexel foundation 2016 paper 2 answers
No sorry
6. (Original post by JN17)
Exactly the same as myself , hopefully 100% then.
For 1b would you not have to combine the i&j components into one vector?
Haha yeah hopefully, time will tell! & im not too sure what you mean? you mean like in column vector notation? if so it wouldnt make a difference
7. (Original post by keenafied)
Haha yeah hopefully, time will tell! & im not too sure what you mean? you mean like in column vector notation? if so it wouldnt make a difference
b) i) P : 400i + t(15i+20j)
ii) Q : 800j + t(20i-5j)
As in replacing these answers with (400+15t)i + (20t)j and (20t)i + (800-5t)j, so collecting like components into one vector.
8. (Original post by JN17)
b) i) P : 400i + t(15i+20j)
ii) Q : 800j + t(20i-5j)
As in replacing these answers with (400+15t)i + (20t)j and (20t)i + (800-5t)j, so collecting like components into one vector.
Oh I see, I dont think it will make a difference, in mark schemes for similar questions they do the same... you had to collect like terms for part c) to equate j components as it was due west, so they wouldnt mark you down anyway cos you proved you can still do it
9. (Original post by keenafied)
Hi, here are my answers to the paper sat today, not 100% sure if their entirely correct obviously but its up for comparison and discussion.

1)a) 104degrees
b) i) P : 400i + t(15i+20j)
ii) Q : 800j + t(20i-5j)
c) 640i + 640j

2) a) 20.6N
b) 15.45N

3) 3Ns

4)a) n/a diagram
b) T = 8.75s

5) Coefficient of friction = 0.73

6) i) d=1.2m
ii) m=42kg

7)a) 2.5i+2.5j
b) 13 ms^-1

8)a) T=11.76N
b) 16.6N at a bearing of 225degrees

Hope this helps.

Thanks, Keenan.
2b 5.15
10. (Original post by youreanutter)
2b 5.15
you didnt take into acount acceleration? its still accelerating with the scale pan. If you stand in a lift and it doesnt move the force exerted on you is lower but when it accelerates upwards its greater..
11. You may have got the first vector question slightly wrong. If it were (640i + 640j), Q would be East of P not West. So the i coords have to be in the negative direction - Thus (-640i+640j) would have been the answer. Well thats what I put anyway.
12. (Original post by RKM21)
You may have got the first vector question slightly wrong. If it were (640i + 640j), Q would be East of P not West. So the i coords have to be in the negative direction - Thus (-640i+640j) would have been the answer. Well thats what I put anyway.
t was equal to 32 I believe, at this time Q is at (640, 640) and P is at (880,640) if you plot these, P is further to the right than Q and hence Q is west of P.
13. (Original post by JN17)
t was equal to 32 I believe, at this time Q is at (640, 640) and P is at (880,640) if you plot these, P is further to the right than Q and hence Q is west of P.
Yeah were right, he thinks its relative to the origin
14. (Original post by youreanutter)
2b 5.15
I got that, because when I did F=ma I used the mass of both instead of one. You probably did the same

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15. hi does anybody know the answers for the maths edexel foundation paper 2 2016 or know anyone that does?
16. In number 3 would i gain at least 1 or 2 marks for finding out what the friction force acting on the particle was. i couldn't answer the question but just found out the friction
17. Paper anyone?
18. (Original post by Rafeerahman)
In number 3 would i gain at least 1 or 2 marks for finding out what the friction force acting on the particle was. i couldn't answer the question but just found out the friction
It was out of 7, you might get around 2 marks because their was quite a bit you had to do after, including finding the acceleration, velocity after the collision.. and then the impulse calculation
19. (Original post by keenafied)
Hi, here are my answers to the paper sat today, not 100% sure if their entirely correct obviously but its up for comparison and discussion.

1)a) 104degrees
b) i) P : 400i + t(15i+20j)
ii) Q : 800j + t(20i-5j)
c) 640i + 640j

2) a) 20.6N
b) 15.45N

3) 3Ns

4)a) n/a diagram
b) T = 8.75s

5) Coefficient of friction = 0.73

6) i) d=1.2m
ii) m=42kg

7)a) 2.5i+2.5j
b) 13 ms^-1

8)a) T=11.76N
b) 16.6N at a bearing of 225degrees

Hope this helps.

Thanks, Keenan.

Although I left 5 as 3sf and rounded 8a to 3sf meaning that b was 16.7N, which I said was at 45 degrees (with a diagram) Hopefully shouldn't get penalised for that...
20. Does anyone know how many marks I'd lose if I got the method correct but values wrong? E.g for moments question I got d=1.2m but to find m I accidentally used 1.2 when doing moments about A, the method was correct but due to accidentally using this my answer was wrong, I also did the same for implies question, got method correct but my value for deceleration was wrong as I did a=-1/8*0.4*g, completely forgot its 0.4a is equal to muR but everything else was fine on these examples, so how many marks would I roughly lose? Thanks

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