OCR M1 - 31st May 2012 :)

hi does anyone have the 2005 and 2006 papers and mark schemes please?

also does anyone know any website that shows all the worked out solutions for the papers please?

good luck everyone!

In this question: http://d.pr/i/oFKP, when you resolve parallel and perpendicular to the plane, what do you get?

I am getting so far like 66+ out of 72, personally I want an easy paper, there are 2 topics which I know I will slit my wrists over, the (t,x) graphs and the big questions on vertical motion, which require to conjure up 2 or 3 suvat equations.

how about you?

yep, same. but for me it's more in the vertical motion form when particles are thrown up in the air and come back down, when do they pass eachother. etcetc. I just hope the paper is a nice... simple... paper. please.

Go to Xtremepapers.com there you can find mark schemes, but its going to be hard to find working if there is particular question you need help in I can lend you a hand.

It doesn't help that I've also got FP1 on friday too so I'm kinda having to split the revision for that too ://

morning guys

..... i still couldn't do that question from last night wa wa waaa

morning guys

..... i still couldn't do that question from last night wa wa waaa

What question?

Ok this is what I got:
Tcos45-Fr=massX0
so Tcos45=Fr

and
R=10cos20+Tsin45

you know the coefficient of friction and you put it into the equation of
Fr=uR
Tcos45=0.364x(10cos20+Tsin45)
re-arrange and factorise to get T=(3.64cos20)/(cos45-0.364sin45)
T=7.6...
might have some accuracy errors as I used 0.364 for u instead of the value worked out in 5 i) b
Edit: sorry forgot that you wanted components. Parallel is Tcos45-Fr=0 and perpendicular is 10cos20+Tsin45.

P.s what paper is that?

few pages back, june 2007 question 5 iii or something. I could get T, and get 2 simultaneous equations but i don't understand how they got the final answer.

might just void it haha

I'll have a look in a minute

It depends on the paper really. I really liked Jun 11 and Jan 12, and got 65 + or something, but i HATE the 2010 papers. I REALLLLLLLLY hope the exam is easy! Its my last exam!

lad

morning guys

..... i still couldn't do that question from last night wa wa waaa

morning guys

..... i still couldn't do that question from last night wa wa waaa

Here you go:

From part ii, we know the the projection speed of Q is 5.6.

iii) P: (taking up to be positive)
s=s
u=8.4
v=
a=-9.8
t=t

Q:
s=s-2
u=5.6
v=
a=-9.8
t=t

use s=ut + 0.5at^2 for both of these.

Ie p: s=8.4t-4.9t^2
q: s-2 = 5.6t - 4.9t^2
q: s= 5.6t - 4.9t^2 +2

equate the s as you know they are at the same height:
8.4t-4.9t^2 = 5.6t - 4.9t^2 +2

cancel the -4.9t^2:
8.4t=5.6t+2
get t=5/7...
then you have this:
P:
s=
u=8.4
v=
a=-9.8
t=5/7

Q:
s=
u=5.6
v=
a=-9.8
t=5/7

Then use v=u+at for P to get v=1.4 and then use it for Q to get -1.4. Hence, they both have the same speeds, but are travelling in opposite directions, with Q travelling downwards and P travelling upwards.

I hope that's the question you meant anyways xD

Hope it helps!