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Official M5 Edexcel Thread June 2016

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Original post by Random1357
A levels are very petty sometimes


How many marks do you think that is? 3?
Reply 201
ive seen a few people who have 1+root5 in their period answer but I think you guys made the mistake of thinking that alpha and theta were the same angles if you see my sketch below

20160621_134956.jpg
Original post by gagafacea1
How many marks do you think that is? 3?


2 you just had to differentiate so M1A1
Easiest way for wire question
image.jpg
Correct solution to MI via integration
Attachment not found
How many marks lost if I did the cylinder with top and bottom? FML
Reply 205
Original post by CharpCharlie
Easiest way for wire question
image.jpg
Correct solution to MI via integration
Attachment not found


Please view my solution. The distance from CoM to pivot is actually root10/2 since CoM lies at 1/2a,1/2a
20160621_133207.jpg
Original post by 8752
ive seen a few people who have 1+root5 in their period answer but I think you guys made the mistake of thinking that alpha and theta were the same angles if you see my sketch below

20160621_134956.jpg


Exactly what I did. Any idea how many marks I'd lose?
Reply 207
Original post by ombtom
Exactly what I did. Any idea how many marks I'd lose?


Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root2 instead of root10/2
(edited 7 years ago)
Original post by 8752
Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root instead of root10/2


How many for finding COM?


Posted from TSR Mobile
Original post by 8752
Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root instead of root10/2


I didn't even see an alpha :lol:
Reply 210
Original post by ombtom
I didn't even see an alpha :lol:


there wasn't an alpha in the question but you had to realise that the masses of the two wires didn't lie on the same line so they made different angles to the downward vertical (like they usually do for M5 pendulum questions - they're normally straight objects) because the wire was kinked. Normally in M5 questions - it's just a normal straight pendulum so the angle both masses make with the downward vertical are the same because they both lie on a straight rod or whatever, however here they aren't.
Reply 211
Original post by physicsmaths
How many for finding COM?


Posted from TSR Mobile


I think probably 2. The rest of the method is all right except M0 on CoM method mark and A0 on accuracy mark of CoM
Original post by physicsmaths
How many for finding COM?


Posted from TSR Mobile


None normally: mark for subbed in eqn
Original post by 8752
The centre of mass of the object lies on 1/2,1/2 not on the line joining the 2 masses. use sum of moments=moments of sum using B as the origin

m(0,a)+m(a,0)=2m(x,y)
(x,y)=1/2a,1/2a----> this does not lie on
the line joining the two masses on your diagram.


Hmmm yes it does actually the only thing incorrect is the use of csq=asq+bsq-abcosC, instead of the correct csq=asq+bsq-2abcosC (=5/2) hmm embarrassing
Reply 214
Original post by CharpCharlie
Hmmm yes it does actually the only thing incorrect is the use of csq=asq+bsq-abcosC, instead of the correct csq=asq+bsq-2abcosC (=5/2) hmm embarrassing


Yes you're right it does lie on the line joining the two masses. The distance should be root10/2 a however. Im curious as to how people got root 2 tho?
(edited 7 years ago)
my answers:
i diddnt do m5
Original post by gagafacea1
My answers:

1) < 6 , 12.5 >

2) < sin t , cos 2t >

3)
a) < 2 , 3 , 0 >
b) < 0 , -2 , -3 >

4) M/6 ( 3r^2 + 2h^2 )

5) Same as Random1357

6) u ln(5) - 19.6

7) don't remember the J thing, but got mg/8 in the lastest part.


They seem to pretty much match mine but I'm pretty sure that the answer to 1 was -6,12.5 as it was (-1,5)+(5/2)*(-2,3)=(-6,12.5) (going from memory)
Original post by Ewanclementson
They seem to pretty much match mine but I'm pretty sure that the answer to 1 was -6,12.5 as it was (-1,5)+(5/2)*(-2,3)=(-6,12.5) (going from memory)


yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r. ie my answer to 2 was incomplete.
(edited 7 years ago)
Original post by gagafacea1
yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r.


Yes I gave them the velocity equation. I cannot remember what it was but it fitted the boundary conditions and when differentiated and subbed back in, gave the required value so I'm confident with it :smile:
Original post by gagafacea1
yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r. ie my answer to 2 was incomplete.


Having checked the answer you agreed with for Q5, I do disagree. The centre of mass was a distance of root(10)/2 from the pivot.
m(2a)+m(a)=2mx leads to x=1.5a then the centre of mass was 1.5a below L and by symmetry, 0.5a from l which gives root(10)/2.

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