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Urgent a level maths help

Hi guys, my a level maths exams start on friday (14th June) and i’m kind of in a sticky situation right now. Since i’m self teaching myself M1, C3 and C4, it’s difficult when I get stuck on a question and have no one to ask for help.

If there’s anyone who sees this, whose studied/studying OCR MEI a level maths (M1/C3/C4) or can help me with some questions regarding these modules, please let me know! I’d really appreciate it.

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Reply 1
Ask away.
Original post by alevel.student99
Hi guys, my a level maths exams start on friday (14th June) and i’m kind of in a sticky situation right now. Since i’m self teaching myself M1, C3 and C4, it’s difficult when I get stuck on a question and have no one to ask for help.

If there’s anyone who sees this, whose studied/studying OCR MEI a level maths (M1/C3/C4) or can help me with some questions regarding these modules, please let me know! I’d really appreciate it.
Thank you! This is a C3 question I was struggling on, I don’t know what i’m doing wrong
(iii)
CFEFB732-1468-4BDF-A25E-9A008F43E150.jpeg
this is what i did so farimage.jpg
Original post by mqb2766
Ask away.
Reply 3
That looks right continue doing it
Reply 4
Hopefully you should find the x term cancels at some point. But the way you are doing it is correct
Reply 5
Looks like the dx = ... du expression should be the inverse (for the x term)
Original post by alevel.student99
Thank you! This is a C3 question I was struggling on, I don’t know what i’m doing wrong
(iii)
CFEFB732-1468-4BDF-A25E-9A008F43E150.jpeg
this is what i did so farimage.jpg
(edited 4 years ago)
Anyone do a level further maths?
ahhhh that’s why I kept messing it up, thank you!!
Original post by mqb2766
Looks like the dx = ... du expression should be the inverse (for the x term)
image.jpgfor question 5ii, why do we not consider the tension in the string, 55N when resolving upwards to find the reaction of the man of the lift floor, but we consider the weight of the parcel?
Reply 9
There are quite a few explanations about this online, just search for lift or elevator reaction force.
But the weight of the parcel is merged with the weight of the man (fairly obvious, just think if the parcel was very heavy or light). It would be affected by the lift falling (its acceleration), but the tension in thd string produces a force on the lift which determines its acceleration. When considerimg the acceleration of the lift, the lift, person and parcel is considered as a lump.
Original post by alevel.student99
image.jpgfor question 5ii, why do we not consider the tension in the string, 55N when resolving upwards to find the reaction of the man of the lift floor, but we consider the weight of the parcel?
Who knows someone doing a level further maths, further pure (srsrly)
Have done.
Original post by Kirthana.b
Who knows someone doing a level further maths, further pure (srsrly)
Physics and maths tutor has some good explanation of modules for our exams. If you have any questions, I'm happy to help. Should we share our answers after the exams on this thread?
I’ll check that out, thank you! Also are you sitting ocr mei exams?
Original post by BioChemDave
Physics and maths tutor has some good explanation of modules for our exams. If you have any questions, I'm happy to help. Should we share our answers after the exams on this thread?
Original post by alevel.student99
Hi guys, my a level maths exams start on friday (14th June) and i’m kind of in a sticky situation right now. Since i’m self teaching myself M1, C3 and C4, it’s difficult when I get stuck on a question and have no one to ask for help.

If there’s anyone who sees this, whose studied/studying OCR MEI a level maths (M1/C3/C4) or can help me with some questions regarding these modules, please let me know! I’d really appreciate it.


Yo I’m gonna start making videos on a level maths soon what do you find the most difficult
image.jpgfor part (ii) how do you work out the vertical component of velocity for the speed? I thought it’ll be zero, but the mark scheme says something else
The initial vertical speed is zero (stone thrown horizontally), but gravity accelerates the stone downwards, so just use the suvat rules to calculate the (vertical) speed when its descended from the initial height to zero.
Original post by alevel.student99
image.jpgfor part (ii) how do you work out the vertical component of velocity for the speed? I thought it’ll be zero, but the mark scheme says something else
E917BD17-AA18-4B01-9DC7-1310A2BF569F.pngahh that makes sense, gravity pulls it downwards, okay I just used suvat to work out initial velocity and also got 4.9ms^-1, I was wondering why in the mark scheme they used a quicker method and just multiplied the time taken to reach the water by acceleration?
Original post by mqb2766
The initial vertical speed is zero (stone thrown horizontally), but gravity accelerates the stone downwards, so just use the suvat rules to calculate the (vertical) speed when its descended from the initial height to zero.
vertical speed:
v = u + at
= 0 + 9.8*0.5
Its the usual / correct formula?
Original post by alevel.student99
E917BD17-AA18-4B01-9DC7-1310A2BF569F.pngahh that makes sense, gravity pulls it downwards, okay I just used suvat to work out initial velocity and also got 4.9ms^-1, I was wondering why in the mark scheme they used a quicker method and just multiplied the time taken to reach the water by acceleration?
ohhh that’s what they used, I was using s=ut+0.5at^2 (I did s=1.225m) and worked out ‘u’
Original post by mqb2766
vertical speed:
v = u + at
= 0 + 9.8*0.5
Its the usual / correct formula?

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