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OCR MEI AS Mathematics M1 10/06/2013

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Reply 60
Original post by aishamarie
can anyone help with the qs that I've posted on this threaad?>>>>>>>>>>>>>>> http://www.thestudentroom.co.uk/showthread.php?p=43051970&highlight=

thank you! I will rep!


Q5 solution. Hope it helps!
Reply 61
Original post by Axion
it hangs vertically downwards with the 2.3m length slack. you can get the tensions from that


so one of the 0.5m feels all the tension?
Original post by bluekipper
Q5 solution. Hope it helps!


it really did, thanks a lot! - I've reached my rep limit for today :mad: but I'll defo do it tomorrow ! :smile:
hi does anyone want solutions for june 2010 jan 2012
Reply 64
Original post by rohan11
so one of the 0.5m feels all the tension?


yup!
Original post by Economics247
hi does anyone want solutions for june 2010 jan 2012


yes please!
Reply 66
Original post by Axion
Good luck everyone tomorrow by the way :smile:

Who is AS level and who is A2 Level?


I'm A2. Did S1 at AS which I hated but I don't exactly like M1 either!
Original post by aishamarie
yes please!


they not that great im just scanning them now
Original post by Economics247
they not that great im just scanning them now


all solutions are great solutions :biggrin:
- thanks :smile:
Original post by aishamarie
haha omg sorry, good luck tomorrow! :smile:
& btw, I love your blog, will defo follow after tomorrow is over! :biggrin:


you too! ah just did that paper and I can't believe how much better my mech has gotten after 4 months! :smile: and thanks!
Original post by aishamarie
all solutions are great solutions :biggrin:
- thanks :smile:


ahh man it exceed the amount
Original post by Economics247
ahh man it exceed the amount


uhh could you only send the last question of the june 2010 paper then please? :smile:
Reply 72
http://www.mei.org.uk/files/papers/2012_Jan_m1.pdf

Can anyone explain the first part of the last question to me pleas, the projectile one, it seems impossible!
Reply 73
Original post by CWE
http://www.mei.org.uk/files/papers/2012_Jan_m1.pdf

Can anyone explain the first part of the last question to me pleas, the projectile one, it seems impossible!


Okay to work out T, think about where it will be at it's halfway point, use suvat and consider the vertical bits:

v = 0 ms^-1 as objects are instantaneously at rest at highest point

u = 40 x sin a ms^-1 , vertical component of initial velocity

a = -g , as gravity acts downwards in opposite direction to the motion

t = ?

Now use v= u+at, remembering that this is just for half the time (T/2):

0 = 40sin a - g x (T/2)

T/2 = 40sin a / g

T = 80sin a / g

For R, just think about the horizontal component of the velocity (using cos a) and use suvat, remembering that there isn't any acceleration in this direction as air resistance is negligible. Sub in T = 80sin a / g for time t and you'll get the answer :biggrin:
Original post by aishamarie
uhh could you only send the last question of the june 2010 paper then please? :smile:


Pm me your email
Hi guys need help with question 3 june 07 http://www.mei.org.uk/files/papers/m107ju_8gxe4.pdf
Reply 76
Original post by Economics247
Hi guys need help with question 3 june 07 http://www.mei.org.uk/files/papers/m107ju_8gxe4.pdf


1) The pulley is smooth so there is no friction
2) The system is in equilibrium.
Right side = Left side
20g + rod = 15g
So the force in the rod must be -5g
It is a thrust as the force is -ve :smile:
(edited 10 years ago)
Reply 77
Original post by loua96
Okay to work out T, think about where it will be at it's halfway point, use suvat and consider the vertical bits:

v = 0 ms^-1 as objects are instantaneously at rest at highest point

u = 40 x sin a ms^-1 , vertical component of initial velocity

a = -g , as gravity acts downwards in opposite direction to the motion

t = ?

Now use v= u+at, remembering that this is just for half the time (T/2):

0 = 40sin a - g x (T/2)

T/2 = 40sin a / g

T = 80sin a / g

For R, just think about the horizontal component of the velocity (using cos a) and use suvat, remembering that there isn't any acceleration in this direction as air resistance is negligible. Sub in T = 80sin a / g for time t and you'll get the answer :biggrin:


Perfect, thank you!
Original post by Alotties
1) The pulley is smooth so there is no friction
2) The system is in equilibrium.
Right side = Left side
20g + rod = 15g
So the force in the rod must be -5g
It is a thrust as the force is -ve :smile:


Thanks really appreciated i dont even get question 5
Reply 79
Original post by JG1027
same :frown: so do you think past papers for 5 days would be enough to get at least a C ?



5 days is more than enough! I wish I had that much time! I only started revising yesterday...I'm in A2 and resitting...need an A in mechanics...not had much time to revise though due to synoptic papers for my other subjects! :frown:


have a feeling C4 isn't going to go too well :eek:

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