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# OCR Mechanics 1 25th Jan watch

1. (Original post by jj314d)
Mr M, on Q4 I found the T=3.36N and a=1.4 for the pulley. I then used them to find R=0.28 and F=0.49 (apparently wrong). On dividing to find mu, I got 1.75. Was there a quick way to find R and F without having to find T and a?
I don't know what you did there. The system is in equilbrium so clearly the tension must be the same as the weight of Q (2.94 N).
2. on Question 4 :

if the mass of the particle was 0.4kg and the angle was 60degrees.....surely you would do 90-60 = 30 .....then for the perp and parrallely forces it would be 0.4gsin(30) and 0.4gcos(30) ???????

Mr.M I would be really grateful if you could post up the full solution to question 4....I will rep aswell .....thanks
3. (Original post by Mr M)
I don't know what you did there. The system is in equilbrium so clearly the tension must be the same as the weight of Q (2.94 N).
read post above....sorry i thought id quote you to grab your attention haha.....thanks again
4. (Original post by Goldfishy)
http://en.wikibooks.org/wiki/A-level..._of_a_Particle Under friction, it has a limit on mu. Dammit.

I think the website is wrong, or worded badly or I didn't read it properly. Either way, I read the article the night before the exam and thought that mu had max value of 1 - spent a good 15mins thinking that I had done something wrong in that q4 when instead I could have have checked my q9...
That "article" contains a number of errors I'm afraid. I suppose that is the risk associated with Wiki entries.
5. (Original post by Mr M)
I don't know what you did there. The system is in equilbrium so clearly the tension must be the same as the weight of Q (2.94 N).
Now it makes sense - I thought I had to find the accel and tension in the pulley, like past pulley questions. It does make it a lot easier now. I liked the rest of the paper which I thought was quite easy compared to past exams.Thanks
6. (Original post by SK-mar)
read post above....sorry i thought id quote you to grab your attention haha.....thanks again
Ok very briefly.

Look at Q. It has weight 2.94 N and is in equibrium so the tension in the string is also 2.94 N.

The weight of P is 3.92 N.

Perpendicular to plane:

R + 2.94 cos 60 = 3.92 cos 60

R = 0.49 Newtons

Parallel to plane:

3.92 cos 30 = 2.94 cos 30 + Friction acting up slope

Fric = 0.49 sqrt 3 Newtons

mu = Fric / R = sqrt 3

(Or you can take a short cut and just type tan 60 into your calculator!)
7. (Original post by Mr M)
Ok very briefly.

Look at Q. It has weight 2.94 N and is in equibrium so the tension in the string is also 2.94 N.

The weight of P is 3.92 N.

Perpendicular to plane:

R + 2.94 cos 60 = 3.92 cos 60

R = 0.49 Newtons

Parallel to plane:

3.92 cos 30 = 2.94 cos 30 + Friction acting up slope

Fric = 0.49 sqrt 3 Newtons

mu = Fric / R = sqrt 3

(Or you can take a short cut and just type tan 60 into your calculator!)
oh god I didnt think you needed to take Q into account........would i possibly still get method marks for resolving P only...and also would i get a mark on the next bit for stating coeff of friction = F/R but because I had the wrong values I got the wrong answer.....orr will they be nice and carry the marks forward so whatever u got for F and R then if u plugged it in it works? thanks bit confusing sry
8. (Original post by Mr M)
OCR (not OCR MEI) answers January 2010

All non-exact answer given to 3 s.f.

1. (i) Speed of P at ground = 18.9 m/s (2 marks)

(ii) Height = 17.3 m (2 marks)

(iii) Speed of P when 5 m above ground = 16.1 m/s (2 marks)

2. (i) Magnitude of resultant = 27.1 N (5 marks)

(ii) Angle = 37.4 degrees (3 marks)

3. (i) m = 0.096 kg (4 marks)

(ii) Change in momentum = 1.2 Ns (2 marks)

(iii) Common speed = 3.25 m/s (3 marks)

4. (i) (a) R perpendicular = 0.49 N

R parallel = 0.849 N (4 marks)

(b) Coefficient of friction = 1.73 (2 marks)

(ii) T = 4.36 N a = 1.09 m/s² (4 marks)

5. (i) t = 12.5 seconds (3 marks)

(ii) t = 26.7 s (5 marks)

(iii) Sketch graph (3 marks)

6. (i) a = 0.012t - 0.18 (2 marks)

(ii) Show k = 2 (5 marks)

(iii) x = 0.002t³ - 0.09t² + 2t

x = 30.0 m (5 marks)

7. (i) T = 1960 N (5 marks)

(ii) (a) Show 3.15 (2 marks)

(b) t = 9.50 seconds (9 marks)

Edit: question 7 fixed.

Mr M,, Thanks as always for the answers!

I thought this paper was slightly above standard difficulty, but it was fair enough.

I was very annoyed about question 4) I have been led to believe that the co-efficient of friction never exceeds 1, and i had never seen a question where mu > 1. Therefore on reaching my answer of 1.73, i crossed it out, and made up a load of rubbish to get what i thought was a reasonable answer of .43ish.

Do you not think that it was very unfair to throw in a question with a mu > 1? I do

Most other question were fine!

2 quick queries:
1) I really cant remember the change in momentum question. Any chance you could show your working for this one. Obviously it's simple enough, but i can't remember the values used and I don't regonise the answer.

2) For 4ii Why was tension 4.36N. Now that q = 0.5kg, does T not equal 0.5g N = 4.9N?

Thanks for all the great help
x
9. (Original post by Mr. Cricket)
Mr M,, Thanks as always for the answers!

I thought this paper was slightly above standard difficulty, but it was fair enough.

I was very annoyed about question 4) I have been led to believe that the co-efficient of friction never exceeds 1, and i had never seen a question where mu > 1. Therefore on reaching my answer of 1.73, i crossed it out, and made up a load of rubbish to get what i thought was a reasonable answer of .43ish.

Do you not think that it was very unfair to throw in a question with a mu > 1? I do

Most other question were fine!

2 quick queries:
1) I really cant remember the change in momentum question. Any chance you could show your working for this one. Obviously it's simple enough, but i can't remember the values used and I don't regonise the answer.

2) For 4ii Why was tension 4.36N. Now that q = 0.5kg, does T not equal 0.5g N = 4.9N?

Thanks for all the great help
x
The change in momentum was 9 x 0.096 - - 3.5 x 0.096 = 1.2 Ns

For the tension question, the system is not longer in equilibrium (it is accelerating) so the tension does not equal the weight (the weight must be greater than the tension for it to accelerate downwards).
10. (Original post by SK-mar)
oh god I didnt think you needed to take Q into account........would i possibly still get method marks for resolving P only...and also would i get a mark on the next bit for stating coeff of friction = F/R but because I had the wrong values I got the wrong answer.....orr will they be nice and carry the marks forward so whatever u got for F and R then if u plugged it in it works? thanks bit confusing sry
probably yes and yes
11. What marks did people get in this. Everyone was complaining including me i think about the wording and the coefficient of friction, but somehow i managed 86!
Post what you expected and what you got. I think the mark scheme must of been quite low!
12. (Original post by Will1692)
What marks did people get in this. Everyone was complaining including me i think about the wording and the coefficient of friction, but somehow i managed 86!
Post what you expected and what you got. I think the mark scheme must of been quite low!
I got 99/100 UMS :O
13. (Original post by Josh-H)
I got 99/100 UMS :O
Congrats, what did you truthfully expect?
14. (Original post by Will1692)
Congrats, what did you truthfully expect?
Honestly, I put a lot of work into mechanics, and found the paper ok, so I did expect a high A.. just not that high :P!
15. (Original post by Josh-H)
Honestly, I put a lot of work into mechanics, and found the paper ok, so I did expect a high A.. just not that high :P!
great stuff, its so weird to be told at my school noone has ever surpassed 91 in c4, then i did well for 86. Then you come on tsr and feel like a right failure! So many clever people on here haha.
Nice isnt it when the hard work pays off! I only need three Cs now for AAB to manchester so im happy! Well done!
16. (Original post by Will1692)
great stuff, its so weird to be told at my school noone has ever surpassed 91 in c4, then i did well for 86. Then you come on tsr and feel like a right failure! So many clever people on here haha.
Nice isnt it when the hard work pays off! I only need three Cs now for AAB to manchester so im happy! Well done!
Mate, I wish all my grades were like my mechanics one :')
I'd be chuffed with 86 in C4!
My other maths grades were: C1: 81, C2: 74, C3: 84, S1: 97 M1: 99

So as you can see, they're not all as high as M1, but I'm still v.pleased :P
17. (Original post by Josh-H)
Mate, I wish all my grades were like my mechanics one :')
I'd be chuffed with 86 in C4!
My other maths grades were: C1: 81, C2: 74, C3: 84, S1: 97 M1: 99

So as you can see, they're not all as high as M1, but I'm still v.pleased :P
Well yeah i worded it wrong i got 91 in C4, 86 in M1 lol. I see your point, but it just proves there is always someone better academically.
Go on the C4 thread and everyone got more than 95! Its crazy.
I do think some become a little too obsessed though with academics, you seem to have a good balance and really enjoy/understand the applied stuff?
An A is great, but all this crazy 6 A* stuff is just obsessive!
18. Well, I did get a 100 in one of my exams though M1 was my tied worst
19. (Original post by JChoudhry)
Well, I did get a 100 in one of my exams though M1 was my tied worst
Which was what 99? :P
20. (Original post by Will1692)
Which was what 99? :P
73 But for me, that is still an achievement, self-taught 5 modules in 3 months with 4 hours of maths a week while in Year 11

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