Edexcel M2/M3 June 6th/10th 2013

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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 :smile:

thank you so so much!! :smile:

makes so much more sense now.

Reply 621

*mike

Hey could somebody help me with question 6c on m3 jan 2013, why is the displacement a/2?

Reply 622

study beats

Original post by RYRK

how you finding it??

i dont know how i feel

Reply 623

study beats

i dont know how moments for work for M3 chapter 5, someone please tell me how its done

Reply 624

study beats

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?

Reply 625

username916760

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 :biggrin:

hope it helped. BTW, any advice for me guys? I just don't know what else to do.

Reply 626

username916760

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. :biggrin:

Reply 627

*mike

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 ?

Reply 628

Miken Moose

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.

Reply 629

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

Reply 630

username916760

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.

Reply 631

username916760

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.

Reply 632

LShirley95

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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Reply 633

tiny hobbit

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.

Reply 634

username916760

a less neat and more angle based version of doing it.

Reply 635

username916760

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.

Reply 636

JenniS

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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Original post by mashmammad

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!

Reply 637

username916760

Anyone got Jan 2001???? Can't find it anywhere. Its the only one I have not done. Please share if you have it.

Reply 638

Whitetiger757

Original post by mashmammad

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%20Level/Maths/Edexcel/M3 First paper

Reply 639

*mike

Original post by mashmammad

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.