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# OCR (non mei) M1 Friday 17th June 2016

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1. (Original post by physicskid123)
look it's cos a = (4^2 + 4^2 - 6^2) / 2*4*4 so cos a = -.125 and so a is 97.18
Resultant of the 2 forces was 6N..

2. (Original post by Jackandrew1)
Just a quick question for question 4 part iii. In the unofficial mark scheme it mentioned that it had already been verified. However in the question it said that after B collides with CD that there are no more collisions after this point. This means that surely B can not have a velocity of zero as A would then collide with B. Therefore shouldn't the velocity of B be at least greater than 4ms^-1 (which was the previous speed of A).
Where's the unofficial markscheme??
3. (Original post by swagmister)
Where's the unofficial markscheme??
Theres a link for it on the first page

4. Hey, I don't think it matter but I think questions 2 and 3 on the mark scheme were the other way round, in case there is any referencing confusion correct me if wrong
5. (Original post by Parallex)
Resultant of the 2 forces was 6N..

6. (Original post by Parallex)
Resultant of the 2 forces was 6N..

here u go my friend.
7. (Original post by Jackandrew1)
Just a quick question for question 4 part iii. In the unofficial mark scheme it mentioned that it had already been verified. However in the question it said that after B collides with CD that there are no more collisions after this point. This means that surely B can not have a velocity of zero as A would then collide with B. Therefore shouldn't the velocity of B be at least greater than 4ms^-1 (which was the previous speed of A).
Yes I agree. I can't remember my answer but I know that I had momentum before = 0.2 x 4 + 0.4v because the minimum possible velocity of B after the collision would have to be 4ms-1 to avoid further collisions.
8. where is the unofficial mark scheme???????????
9. (Original post by physicskid123)
here u go my friend.
What's that trying to show? If you're going to say I'm wrong at least post your working that leads you to a resultant force of 6N, instead of putting numbers in to an online calculator without even knowing what they're showing.
10. (Original post by Parallex)
What's that trying to show? If you're going to say I'm wrong at least post your working that leads you to a resultant force of 6N, instead of putting numbers in to an online calculator without even knowing what they're showing.
it's to show the resultant aka if you put in 2 sides of 4cm and 4cm (which are our 2 forces of 4N) and you put in the resultant force of 6N you get the angles shown

I'm done here mate bye seeu soon xxx
11. (Original post by physicskid123)
it's to show the resultant aka if you put in 2 sides of 4cm and 4cm (which are our 2 forces of 4N) and you put in the resultant force of 6N you get the angles shown

I'm done here mate bye seeu soon xxx
Like I said, draw me a triangle with the angle between the 2 4N forces being 97.18 degrees and show me a resultant of 6N.
12. (Original post by Parallex)
Like I said, draw me a triangle with the angle between the 2 4N forces being 97.18 degrees and show me a resultant of 6N.
I'm sorry to say that he is right:

if you use the sine rule, you get the same answer (97.181 degrees)

as you get x = 4sin(97.181)/sin(41.41) and that gives you 6
13. (Original post by Parallex)
Like I said, draw me a triangle with the angle between the 2 4N forces being 97.18 degrees and show me a resultant of 6N.
QQ
15. here is my solution
16. this is the proof that it is 97.18 degrees
17. (Original post by daniellon)
For question 5 I said the direction is to the right. Can I get away with it
Yes almost certainly
18. was P 12 on q5?
19. (Original post by ajamesg)
this is the proof that it is 97.18 degrees
You don't get the resultant by drawing a line between the 2, use the parallelogram rule for adding vectors.
20. (Original post by Parallex)
You don't get the resultant by drawing a line between the 2, use the parallelogram rule for adding vectors.
yes you are right. how many marks was the whole question worth?

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