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Aqa mm1b!!

I do not understand the AQA MM1B at all! please could i have some help!! exam on thursday and dont get a thing about mechanics.
Reply 1
Avatar for greenfro
greenfro
I am doing the exam and I've pretty much sussed it now. But don't worry, I felt the same like a week ago! What do you need help with?
Reply 2
Avatar for Sagacious
Sagacious
13
Original post by adrian.trueman
I do not understand the AQA MM1B at all! please could i have some help!! exam on thursday and dont get a thing about mechanics.


Well, we can't help you if we don't know what you're stuck with... Pick a question, try it, tell us how far you get and we will help you.

Also, go to the maths forum.
Reply 3
Avatar for kenau
kenau
The best things you can do is to go through some past paper first,and then post some questions which you don't understand,so that we can help you...
The machanic paper is not difficult actually,just practice more you will get an idea how to do it.As all the question type are repeat and repeat.
some propular like:suvat,momentum,connect partical,projectal must be in there.
You don't have so much time,so advise do the pp from 10 jan to 12 jan.
I think this would probably help you to improve.Do pp can always help you to solve some stuff.
so work more hard:smile:
Reply 4
Avatar for tom472
tom472
I've been revising for this like crazy, I think i've done every past paper 4 times :/ I find kinematics in two dimensions quite tricky depending on the question. Good luck for the exam!
Reply 5
Avatar for ZTSR
ZTSR
Do past papers. Literally repeats its self, any questions just post them up.

As easy as it seems :/ I find the bearing the most difficult.
get help.jpg

can someone please help with this question?! exam is tomorrow and i'm panicking!
Reply 7
Avatar for greenfro
greenfro
Original post by shannonmcnamara
get help.jpg

can someone please help with this question?! exam is tomorrow and i'm panicking!


6. (a) t=4 θ= 0 (because its horizontal) and x= 1000

use x=Vcosθt 1000=Vcos(0) x 4

1000=4V V=250ms-1

(b) y=Vsinθt-0.5gt^2 + h

0=(250sin(0) x 4)-(4.9 x (4^2)) + h

h= 78.4

(c) Vcosθ i + Vsinθ-gt j

250cos(0) i + 250sin(0)- (9.8 x 4) j

250 i -39.2 j

magnitude= 253 ms-1

(d) x= Vcosθt

1000=253cosθ x 4
θ= 8.91

if you use the rounded 253 value you get 8.83 which is also accepted as the right answer!

its all using the equations
hope it helps :smile:
ah i get it now, thanks for your help! :smile: x

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