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# Edexcel Mathematics: Mechanics M3 6679 01 - 17 May 2017 [Exam Discussion] watch

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1. Hey guys, got a question.

June 2009 Q7b)

The particle is moving towards B. So the particle was displaced from closer to A. Shouldn't the resolving equation be : 8(x-1.2) - 12(x+0.8)= 0.5a ?

The mark scheme says the particle was displaced towards B..

2. Hi guys,

How would I go about resolving the reaction force in question 8? I am confused as to where alpha would go and why. Initially I did Rcos(alpha) for vertical and Rsin(alpha) for horizontal but apparently it's the other way around.

http://pmt.physicsandmathstutor.com/...%20circles.pdf
3. (Original post by physicalgraffiti)
Hi guys,

How would I go about resolving the reaction force in question 8? I am confused as to where alpha would go and why. Initially I did Rcos(alpha) for vertical and Rsin(alpha) for horizontal but apparently it's the other way around.

http://pmt.physicsandmathstutor.com/...%20circles.pdf

Try page 19.

4. http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

how would I go about 2b?
5. (Original post by physicalgraffiti)
Hi guys,

How would I go about resolving the reaction force in question 8? I am confused as to where alpha would go and why. Initially I did Rcos(alpha) for vertical and Rsin(alpha) for horizontal but apparently it's the other way around.

http://pmt.physicsandmathstutor.com/...%20circles.pdf
Alpha would be the angle between the reaction force and the horizontal. This is useful as this angle is then the same as the angle between the vertical and the slant height of the cone so you have values for tanalpha which would be 3a/4a
6. Anyone done Q.2 in the exam style paper in the back of the Edexcel M3 textbook? I resolved vertically to get 2Tsintheta=mg and got an equation from Hooke's law but I don't know how I should move on from there
7. (Original post by physicalgraffiti)
Anyone done Q.2 in the exam style paper in the back of the Edexcel M3 textbook? I resolved vertically to get 2Tsintheta=mg and got an equation from Hooke's law but I don't know how I should move on from there
sin theta= 0.6/ 0.3+x

substitute T and sin theta in the expression and solve
8. (Original post by saira.p)
sin theta= 0.6/ 0.3+x

substitute T and sin theta in the expression and solve

I tried that but I keep getting a negative answer :/ Is the T=0.7x/0.6 and 2Tsintheta=0.2g bit correct?
9. (Original post by physicalgraffiti)
I tried that but I keep getting a negative answer :/ Is the T=0.7x/0.6 and 2Tsintheta=0.2g bit correct?
t=0.7x/0.3 as you consider half the string
10. When is the tension in one string the same as the tension in the other string and when is it different? As I see both being used in questions. Also, when is the tension in two different strings the same.
11. I think the mark scheme used ratios, but i didn't really understand it. My way was to make the moments about the point G (in mark scheme diagram) equal to each other
12. (Original post by Akkiakki98)
Hey guys, got a question.

June 2009 Q7b)

The particle is moving towards B. So the particle was displaced from closer to A. Shouldn't the resolving equation be : 8(x-1.2) - 12(x+0.8)= 0.5a ?

The mark scheme says the particle was displaced towards B..

I'm guessing that since it does not reach A or B, it is decelerating, which means that acceleration is in the negative direction.
13. (Original post by -Phoenix-)
I think the mark scheme used ratios, but i didn't really understand it. My way was to make the moments about the point G (in mark scheme diagram) equal to each other
how would you take moments?
14. (Original post by saira.p)
how would you take moments?
(Taking diagram from mark scheme)
You know that r*tan(30) = OG, so OG = r/(root3). And you know that the angle between OG and the vertical line (7mg) is 60 degrees. Therefore, (r/(root3))*sin(60)= r/2 = perpendicular distance between O & G. So, 7mg*(r/2) = moments from O to G. Then, do the same on the other side.
Distance from G to λmg = (2r-(r/(root3))). Angle between horizontal and (G to λmg) = 30 degrees, so for perpendicular distance on this side, do ((2r-(r/(root3)))*cos(30))). And for moments on this side, do λmg*((2r-(r/(root3)))*cos(30))).
Then, make the two sides equal to each other, so you get:
(7mgr/2) = ((6-(root3)/3)*((root3)/2))* λmg
When you rearrange, you will get: λ = (7 +14(root3))/11) = 2.84

Sorry for the long explanation. It's much simpler than what my explanation makes it look like.
15. less than a day left!!!:/
16. (Original post by dididid)
less than a day left!!!:/
Less than a day before I lose my virginity!
17. whats everyone going to be doing today? Im gonna have a chilled day tbh - just perfect my methods for some first principle derivations so I can get really quick at them if they come up and just do some madasmaths SHM stuff then that is my M3 revision done! :O
18. (Original post by kavindajd)
Less than a day before I lose my virginity!
Shouldn't you be happy about that?
19. We need to take the direction AB as positive as it's moving away from the equilibrium in that direction. With these type of questions we take F=ma in the direction of x increasing. Hence F=ma towards B. The tension in BP will be positive but in AP it will be negative due to acting in the direction x decreasing. I hope this helps
20. What do you guys think is the hardest Solomon paper?

They feel much easier than Edexcel's past papers.

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