OCR M1 - 31st May 2012 :)

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Hey, Could u helpme out please.
http://www.ocr.org.uk/download/pp_11_jun/ocr_66474_pp_11_jun_gce_4728.pdf

June 11
Q2 part 4
I have got that P will further go up by 0.036 M. I then added this to the 0.36M to get 0.396M.. In order to get the greastest height of P
However the markscheme has said that the greatest height of P is 0.756 M
I don't understand how and why they have got that value...

Just realised its already 0.36 M height... Then U add the 0.36M that it goes up when Q hits the floor and then add the 0.036 ...

Reply 21

raheem94

Original post by Smiley Face :)

Initially it was 0.36m above the ground, when Q hits the floor it has gone a further 0.36m above the ground, the distance travelled while coming to rest is 0.036m, so the answer is 0.36+0.36+0.036=0.756m.

Reply 22

raheem94

Original post by Smiley Face :)

Just realised its already 0.36 M height... Then U add the 0.36M that it goes up when Q hits the floor and then add the 0.036 ...

Just saw you have done it, but remember the unit for metre is lower case 'm' not the upper case 'M'.

Reply 23

Smiley Face :)

Original post by raheem94

Just saw you have done it, but remember the unit for metre is lower case 'm' not the upper case 'M'.

I will remember that, thanks :smile:

Reply 24

Smiley Face :)

http://www.ocr.org.uk/download/pp_08_jan/ocr_18635_pp_08_jan_l_gce.pdf
Hi Can anyone help me out ...
Q1 I used the formula
70g-F=(0.3)*70
Thus F = 665N
But the answer is 707N
So then I did 70 g + (0.3*70)
... But I don't know why I added the 0.3*70
Any explanations would be helpful, thanks :smile:

Reply 25

raheem94

Original post by Smiley Face :)

See the below diagram(the box represents the man):

Now resolve the forces,

\displaystyle (\uparrow) \ \ R - 70g = 70 \times 0.3 \implies R = 21+70g \implies \boxed{R=707N}

Reply 26

Smiley Face :)

Original post by raheem94

See the below diagram(the box represents the man):

Now resolve the forces,

\displaystyle (\uparrow) \ \ R - 70g = 70 \times 0.3 \implies R = 21+70g \implies \boxed{R=707N}

Thanks so much!! :smile:

Reply 27

Smiley Face :)

Jan 2007 Q6
http://www.xtremepapers.com/papers/OCR/Maths/2007%20Jan/AS_GCE_Mathematics_4728_01_January_2007_Question_Paper.pdf
Im really stuck
This is my working out ...
Part 1 I got 0.64 m/s which is right...
After this I fail to gain any marks ...
Please can I have some help, thanks
Jan 07 Q6.jpg

Reply 28

megank

ew im dreading mechanics! cannot do them at all!

Reply 29

Smiley Face :)

Original post by megank

ew im dreading mechanics! cannot do them at all!

Haha, me too!! can u help with the above q??

Reply 30

raheem94

Original post by Smiley Face :)

See the diagram below:

The area of the velocity time graph is the distance.

The distance travelled in the region, 0 to 25s, is 8.

So the distance travelled in the region, 65 to 135s, is 32(=40-8)

We know acceleration in the region, 65 to 105s, is 0.02.
To find 'V', use the formula, v=u+at

v= u+at \implies V = 0 + 0.02(40) \implies \boxed{V=0.8}

Hopefully you will be able to do the other parts yourself.

Remember, part (iii) will require the use of area of trapezium.

Reply 31

Smiley Face :)

Original post by raheem94

See the diagram below:

v= u+at \implies V = 0 + 0.02(40) \implies \boxed{V=0.8}

Hopefully you will be able to do the other parts yourself.

Remember, part (iii) will require the use of area of trapezium.

Thanks, again :smile:

Reply 32

raheem94

Original post by Smiley Face :)

Thanks, again :smile:

No problem.

It looks that now i am the only person left to answer on this thread, why no one answers here?

Reply 33

Smiley Face :)

Original post by raheem94

No problem.

It looks that now i am the only person left to answer on this thread, why no one answers here?

I don't know ... But btw I am deffo getting better at mechanics now :smile:

.. Starting to finally get nearer my aim! :smile:

Reply 34

raheem94

Original post by Smiley Face :)

I don't know ... But btw I am deffo getting better at mechanics now :smile:

.. Starting to finally get nearer my aim! :smile:

What's your aim?

Reply 35

Smiley Face :)

Original post by raheem94

What's your aim?

An A but if possible a 100UMS

Reply 36

raheem94

Original post by Smiley Face :)

An A but if possible a 100UMS

I got 100UMS in M1 :smile:

You need to work a lot harder than you are working now.

Reply 37

Smiley Face :)

Original post by raheem94

I got 100UMS in M1 :smile:

You need to work a lot harder than you are working now.

Ok, What would u suggest i need to do?? Ive started getting in the 60's range (out of 72) Really would appreciate the advice :smile:

Reply 38

raheem94

Original post by Smiley Face :)

Ok, What would u suggest i need to do?? Ive started getting in the 60's range (out of 72) Really would appreciate the advice :smile:

I just solved the whole book, watched a few examsolutions videos, and did around 5 edexcel past papers.

NB: I did edexcel M1.

You just need to work harder, ensure that you know all the concepts, have a quick review of your book, attempt a few questions from each exercise to ensure that you know everything required for M1.