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MEI M1 exam discussion - Friday 1st June

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Reply 160
Considering I really don't like mechanics, that went surprisingly well. :smile:
Reply 161
Original post by firebolt
Thats what I put :smile: but my friend put that she assumed gravity was 10


Reckon the assumption that there is no horizontal acceleration will be ok?
Anyone got an estimate for and A/B/C. Thought section A was quite good but my god section B killed me!! The fact that it was in tonnes threw me off and I forgot how to do some of the stuff, really got me down :/ don't want to dwell though, just got to prepare for C4!
Original post by firebolt
Thats what I put :smile: but my friend put that she assumed gravity was 10


it should be air resistance ... g shoud be quoted as 9.8. it says on the front of the paper
Original post by TimetoSucceed
guys i did the projectile differently

i did s = ut + o.5at^2

set s as 2.44, used quadratic formula to get t, got 2 values

used the highest one and that was when it was descending

used that t to work out the horizonal distance at that time which was 55m

therefore as 55>50, and 2.44 is the vertical distance off the ground at 55m, then at 50m itd be above 2.44 hence going over the pole.


EXACTLY what i did, Im sure we are right :smile:! HI FIVE
Reply 165
Original post by xiyangliu
it should be air resistance ... g shoud be quoted as 9.8. it says on the front of the paper


Will I get the mark for the assumption horizontal acceleration = 0?
Reply 166
Original post by xiyangliu
truck is 5900N
locomotive is 100300N

i got taht


sorry but do you remember how you worked it out :biggrin:?
Reply 167
Original post by cloudhang
sorry but do you remember how you worked it out :biggrin:?


I got the same answers. You had to consider the tension, acceleration (1.1 or whatever was given) and the mass of individual parts of the train and apply newton's second law.
Oh and the value of R.
(edited 11 years ago)
Reply 168
Anyone get a copy of the paper to do a markscheme?
Reply 169
For question 7, could anyone clear up for me how they got the value of beta? i got a round 4.5 degrees.. must have been a bad moment or something!

It was an odd paper, quite different to pretty much every past paper ive done, but the only questions that threw up problems were those in section B, and it was only like the last part so i shouldn't have dropped too many marks!
Reply 170
Original post by NJam
For question 7, could anyone clear up for me how they got the value of beta? i got a round 4.5 degrees.. must have been a bad moment or something!

It was an odd paper, quite different to pretty much every past paper ive done, but the only questions that threw up problems were those in section B, and it was only like the last part so i shouldn't have dropped too many marks!


You had no driving force and the train was travelling at a constant speed so it was in equilibrium. You had the resistance acting up the slope and the 800000gsintheta acting down the slope. You simply equated these and solved for sin theta and got 0.24 to 2.d.p. Can't remember if this is exactly right.
Original post by NJam
For question 7, could anyone clear up for me how they got the value of beta? i got a round 4.5 degrees.. must have been a bad moment or something!

It was an odd paper, quite different to pretty much every past paper ive done, but the only questions that threw up problems were those in section B, and it was only like the last part so i shouldn't have dropped too many marks!


Well it said speed was constant so its in equilibrium. The frictional force was 22R for the whole train. Gravity down (parallel) a slope is mgsin(theta). Basically treat it as 1 big block of mass 800 tonnes and where Friction=mgsin(theta) to give it a force of zero.

so (800*10^3)*9.8*sin(theta)=22*1500

Theta=arcsin((22*1500)/((800*10^3)*9.8))=0.2412 degrees.
(edited 11 years ago)
Reply 172
If i got 100% in sec A, and around 50-75% in sec B, wat grade wud tht giv me ovrall?

This was posted from The Student Room's Android App on my HTC Desire
Reply 173
Original post by ngnav
If i got 100% in sec A, and around 50-75% in sec B, wat grade wud tht giv me ovrall?

This was posted from The Student Room's Android App on my HTC Desire


36 plus 18-27 = 54-63 so probably an A or a high B depending on boundaries.
Reply 174
Original post by Gilsenan
You had no driving force and the train was travelling at a constant speed so it was in equilibrium. You had the resistance acting up the slope and the 800000gsintheta acting down the slope. You simply equated these and solved for sin theta and got 0.24 to 2.d.p. Can't remember if this is exactly right.


was the mass not 120000? which would make it 120000gsintheta?

Edit: crap, just realised what i did. Nevermind!!
(edited 11 years ago)
Original post by Gilsenan
I got the same answers. You had to consider the tension, acceleration (1.1 or whatever was given) and the mass of individual parts of the train and apply newton's second law.
Oh and the value of R.


why does the mass matter? its horizontal forces not vertical?
Reply 176
Original post by NJam
was the mass not 120000? which would make it 120000gsintheta?

Edit: crap, just realised what i did. Nevermind!!


No problem, was gonna say I'm pty sure there was an 800,000 in there somewhere :P
Reply 177
Original post by firebolt
why does the mass matter? its horizontal forces not vertical?


So you can apply F = MA, it wasn't in equilibrium remember.
Anyone want to hazard a guess for the grade boundaries?:smile: kicking myself over section B ow
Original post by Gilsenan
So you can apply F = MA, it wasn't in equilibrium remember.


yes it was it said it was at constant speed so there was no acceleration and all forces equal

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