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OCR (non mei) M1 Friday 17th June 2016

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it's not 97.18 because the triangle you are using isn't the resolved force and you could draw the forces with that angle and resolve them to find it's incorrect, drawing a line connecting the 2 ends and finding it is equal to 6 is not a resolved force
Original post by physicskid123
QQ


QQ
Original post by ajamesg
IMG_0005.jpgthis is the proof that it is 97.18 degrees


You are wrong, put one of the forces due north, and the other with 97.18 and resolve to get the resultant and it won't be 6N.
Original post by sandpitturtle
was P 12 on q5?


Yes
Reply 64
Avatar for ajamesg
ajamesg
1
Original post by Furthermaths100
You are wrong, put one of the forces due north, and the other with 97.18 and resolve to get the resultant and it won't be 6N.

As I said previously, I agree now that I am wrong. Do you remember how many marks the whole question was worth?
Reply 65
Avatar for czj1997
czj1997
8
* unrelated question.. but
IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?
Reply 66
Avatar for ss1529
ss1529
Original post by Parallex
agree w/ 6.76 for the last one

tension was 1.81N or something for the pulleys

can't remember much lol


yes thats what i got too
Original post by czj1997
* unrelated question.. but
IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?


definitely
Original post by czj1997
* unrelated question.. but
IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?


Yes but you will need higher As to make up for the Bs. You just need to average 80 UMS overall to get an A (480+ total UMS points).
Any idea what 68 would be UMS wise?
its angle to the vertical not horizontal that's why its 88 degrees. (180-97)
Original post by Buymoria
Any idea what 68 would be UMS wise?


68 marks could get you 100 ums, depends on how everyone did overall. Doubt that will get you less than 96 ums though.
I also did the cosine rule but youre forgetting to minus the value of 97 whatever from 180
Original post by Arnold12345
its angle to the vertical not horizontal that's why its 88 degrees. (180-97)


82.7 degrees
forgot to work on time on question 1 :unimpressed:, 2 marks off?

(also what was the question on the contact force asking?)
Reply 75
Avatar for G_8
G_8
1
I got the answer right for the 82.8 degrees question, however the method I used was definitely over complicated because when I was checking I realised that I could have done it in a much simpler way i.e. using the cosine rule. Do you think they will mark me down for having used an alternative complex method despite getting the right answer?
Reply 76
Avatar for G_8
G_8
1
also does anyone have any ideas as to what the 90 ums boundary will be? around 65-66 maybe....??
Original post by G_8
I got the answer right for the 82.8 degrees question, however the method I used was definitely over complicated because when I was checking I realised that I could have done it in a much simpler way i.e. using the cosine rule. Do you think they will mark me down for having used an alternative complex method despite getting the right answer?


No chance they will mark you down, I was thinking of doing it the complicated way but realised before I started, both are legitemate ways to do it and will both be on the mark scheme
Original post by G_8
also does anyone have any ideas as to what the 90 ums boundary will be? around 65-66 maybe....??


This was quite hard for an M1 paper I would say - papers like this are usually about 63 for 90 ums and this will be no exception.
(edited 7 years ago)
Original post by Snicherz
forgot to work on time on question 1 :unimpressed:, 2 marks off?

(also what was the question on the contact force asking?)


Yes 2 marks off. Not too sure on the contact force question.

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