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OCR MEI AS Mathematics M1 10/06/2013

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Reply 40

Geronimo!

Help me ahhhh!! ****ing ****ty mechanics driving me ****ing nuts ****ing ****ing ****ing ****!!! :mob:

Reply 41

rohan11

Can someone explain to me what Q7 part v C) from june 2009 asking for. And how to do it please.

Reply 42

Axion

Original post by rohan11

Can someone explain to me what Q7 part v C) from june 2009 asking for. And how to do it please.

Find how long it takes until the sphere is back at point P using g as -9.8 now as the string has broken

Reply 43

rohan11

Original post by Axion

Find how long it takes until the sphere is back at point P using g as -9.8 now as the string has broken

thanks i got it now :smile:

Reply 44

bluekipper

Original post by JG1027

Hey does anyone know what cpt means on the mark schemes?

cpt = component (vertical or horizontal)

Reply 45

loua96

Original post by mathsftww

Can someone please help me, I need help on the june 2008 paper, Q7) iv). Thanks :frown:

Have you tried drawing a triangle with all the forces and angles labelled? The forces in PC and QC are T1 and T2 respectively, you know the resultant force downwards is 8g N and the horizontal force is 10N (as the box is in equilibrium) :smile:

Reply 46

Kbatch

I just did a paper and went to mark it, if I've used a different method than stated, but one thats still valid and got the correct answer would I still achieve all the marks?

Reply 47

Axion

Original post by Kbatch

I just did a paper and went to mark it, if I've used a different method than stated, but one thats still valid and got the correct answer would I still achieve all the marks?

yep

Reply 48

Matt_dragon

Sorry for being rude, but I am genuinely ****TING myself. Not doing well in mocks though :frown:

Reply 49

CWE

Have no Idea how to do Question 7i on june 2010. It doesn't seem to make any sense, I have checked an explanation on the last page but it does not seem to make work, surely when t=0 is only one point on the curve?

http://www.mei.org.uk/files/papers/m110ju_lmkm.pdf

would be grateful if someone could take a look!

Reply 50

Axion

Original post by CWE

Lol, Kbatch asked me the same question earlier. Here was my PM

Ok its a displacement time graph. I.e. distance time graph that deals with positives and negatives. The y axis is metres up the string so you can visualize that over the course of four seconds, the object moves up a little from its 12m high starting place, then moves down to below the middle and then reaches the origin where y =0

A) greatest displacement ABOVE. It starts at 12m high and reaches a maximum height of 16m (at 1 second), hence the greatest displacement above is 4m

B) Greatest distance of P from 12m. This is just the lowest point reached in this case. The lowest point reached is -4m. Therefore the greatest distance is 16m

C) It's moving downwards when the y values (or height) starts moving downwards. This occurs between 1 second < t < 3.5 seconds

D) It's at rest when the gradient = zero. This occurs at t = 1 and t = 3.5