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# Edexcel M2/M3 June 6th/10th 2013 Watch

1. (Original post by cant_think_of_name)
It depends whether the velocity is given in terms of t or x. If in terms of t you differentiate with respect to t, if in terms of x, use a = v(dv/dx), and separate the variables
thank you so so much!! makes so much more sense now.
2. Hey could somebody help me with question 6c on m3 jan 2013, why is the displacement a/2?
3. (Original post by RYRK)
how you finding it??
i dont know how i feel
4. i dont know how moments for work for M3 chapter 5, someone please tell me how its done
5. (Original post by *mike)
Hey could somebody help me with question 6c on m3 jan 2013, why is the displacement a/2?
its the height distance?
6. (Original post by JenniS)
6a is bad enough, I didn't know how to f****** integrate a triangle
It's two symmetrical lines through the origin, so you take the axis of symmetry of the triangle as the x axis and split the shape into rectangles of length 2y and width DX and then use integration of 2y*x dx. Area is easy to find as integral of 2y dx or I reckon the formula of the area of triangle should do. And then you divide them to get the answer. Bu it's easier said than done hope it helped. BTW, any advice for me guys? I just don't know what else to do.
7. (Original post by study beats)
i dont know how moments for work for M3 chapter 5, someone please tell me how its done
You first find the centre of mass of the solid or the object given, then as you are given its mass, and using trigonometry you can find the normal distance to the vertical through the point of suspension. Also you have the mass and the position of the added object so you can find the normal distance from the vertical through the suspension point. Then as usual with moments, multiply the masses and the distances and if it's in equilibrium the moments are equal. Hope it's easy to follow.
8. (Original post by study beats)
its the height distance?
Thanks for the reply, I understand that you need to know the height to model it as a projectile, but I do not understand why it is a/2. How do you know the particle only travels a/2 ?
9. (Original post by *mike)
Thanks for the reply, I understand that you need to know the height to model it as a projectile, but I do not understand why it is a/2. How do you know the particle only travels a/2 ?
IIRC, the a/2 is the initial height above the horizontal through OA, i.e. asin(30) = a/2.
10. If anyone could upload a photo of their working to 4b on june 2012 please please please do, I really can't do it
11. (Original post by *mike)
Hey could somebody help me with question 6c on m3 jan 2013, why is the displacement a/2?
What you do is to find the speed just before loosing contact, then knowing the direction to be at 90 degrees to the radius of the circle at that point, you can find out the vertical and the horizontal component of velocity because you have already found the angle through which it has turned. Now all you need to do is to consider it as a projectile ie the horizontal velocity is constant but the vertical component changes. Use suvat to find the final vertical velocity ( you have the displacement to be -a/2, the acceleration to be -g and the u you have already found out.) Knowing that the horizontal speed is constant, use arctan to find the answer.
12. (Original post by JenniS)
If anyone could upload a photo of their working to 4b on june 2012 please please please do, I really can't do it
I'll try to do that.
13. (Original post by JenniS)
If anyone could upload a photo of their working to 4b on june 2012 please please please do, I really can't do it

Hopefully my working is a little clearer than my explanation? I took V to be the origin.

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14. (Original post by StUdEnTIGCSE)
Agreed.

But minimum velocity was -4.5 not 0 but minimum speed is.
You're right, sorry. Slip of the brain. I've edited my post.

15. a less neat and more angle based version of doing it.
16. (Original post by JenniS)
If anyone could upload a photo of their working to 4b on june 2012 please please please do, I really can't do it
the post above. hope it helps.
17. (Original post by LShirley95)

Hopefully my working is a little clearer than my explanation? I took V to be the origin.

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a less neat and more angle based version of doing it.
Thank you so so much both of you, I think I understand it now, mashmammad's way was how I was trying and failing to do it because I couldn't get theta! thanks!
18. Anyone got Jan 2001???? Can't find it anywhere. Its the only one I have not done. Please share if you have it.
Anyone got Jan 2001???? Can't find it anywhere. Its the only one I have not done. Please share if you have it.
http://www.freeexampapers.com/#A%20L...ths/Edexcel/M3 First paper
What you do is to find the speed just before loosing contact, then knowing the direction to be at 90 degrees to the radius of the circle at that point, you can find out the vertical and the horizontal component of velocity because you have already found the angle through which it has turned. Now all you need to do is to consider it as a projectile ie the horizontal velocity is constant but the vertical component changes. Use suvat to find the final vertical velocity ( you have the displacement to be -a/2, the acceleration to be -g and the u you have already found out.) Knowing that the horizontal speed is constant, use arctan to find the answer.

(Original post by Miken Moose)
IIRC, the a/2 is the initial height above the horizontal through OA, i.e. asin(30) = a/2.

Thanks to both of you, I think I didnt fully understand the qesution, so does the particle literally go through point A. which is the only way displacement could be a/2, when I first read the question I thought he went below point A to a point D.

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