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Edexcel AS/A2 Mathematics M2 - 17th June 2016 - Official Thread

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can someone plz confirm whether we shuld write mass per unit area in centre of mass wuestions as k and use it in our calculationsare marks allocated fr datif so wat is da correct way of writin it i find it a bit confusing
Original post by saitama kun
can someone plz confirm whether we shuld write mass per unit area in centre of mass wuestions as k and use it in our calculationsare marks allocated fr datif so wat is da correct way of writin it i find it a bit confusing


Nearly had a stroke reading this, but..

I'm pretty sure, if I understand what you're asking, that the answer is no. In centre of mass questions you don't need to consider mass per unit area as such, just the mass of each individual part of the lamina/body and the area of that part when summing different parts together.
(edited 7 years ago)
In which way does the thrust act?
Just did the June 2015 paper... died... Exam is tomorrow :s


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Original post by target21859
In which way does the thrust act?


up towards the rod
Original post by TheMarshmallows
up towards the rod


Ok thanks so do you just consider it as force like this?
can someone explain question 2 b) http://qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2013/Exam-materials/6678_01_que_20150609.pdf
Original post by target21859
Ok thanks so do you just consider it as force like this?


yeah thats fine.
Reply 128
Avatar for alvosm
alvosm
7
Original post by somevirtualguy
The tension in the string acts through the point D, so when taking moments will equal 0. It works the same for the normal reaction at B, you don't need to include it when taking moments about D because as you can see it would pass straight through D meaning it would equal 0.

For part b), taking moments about C will eliminate the tension when taking moments again, leaving you with the reaction force, the weight of the rod and the frictional force which you can solve for the magnitude.


understand it now!! thank youuu!!
Can someone help explain q4b of june 2012.
i know how to do it and did it right but messed up on b because i assumed centre of mass would be on the left of O as it didnt specify if k could be less than 1 or not but in the markscheme it says the centre shifts to the right of O (for this k has to be larger than 1).
In these type of questions do you just assume k is larger than one if no other info is given as otherwise the centre of mass can end up being on the wrong side due to the assumption k is very small????temp 1.png
temp 2.pngAlso can someone please help me with part a, its 2 marks but i dont get where they get the 30 from??
Original post by cjlh
Nearly had a stroke reading this, but..

I'm pretty sure, if I understand what you're asking, that the answer is no. In centre of mass questions you don't need to consider mass per unit area as such, just the mass of each individual part of the lamina/body and the area of that part when summing different parts together.

r u sure bro coz they have mentioned mass per unit area = m/a2 in da text book as first statement n den have carried out using m not with a2 in da calculation
Original post by rm761
temp 2.pngAlso can someone please help me with part a, its 2 marks but i dont get where they get the 30 from??


60/2 = 30
its a circle theorem (two tangents to a circle that intersect)
Original post by saitama kun
r u sure bro coz they have mentioned mass per unit area = m/a2 in da text book as first statement n den have carried out using m not with a2 in da calculation


Have you got a page reference for that? Flicked through the book just now and didn't see it.

For a uniform lamina I'm certain you don't have to mention the pass per unit area.
Original post by milogillot
60/2 = 30
its a circle theorem (two tangents to a circle that intersect)


Ah ok, i didnt think about it like that
thanks
Original post by saitama kun
r u sure bro coz they have mentioned mass per unit area = m/a2 in da text book as first statement n den have carried out using m not with a2 in da calculation


just do area. as long as answer is correct it should be fine as cuz its uniform, the mass per unit area is the same through all parts so cancels out anyway?
When you do distance = speed x time when looking in the horizontal sense,

Do you use the initial speed of projection for this calculation?


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Avatar for LLk1
LLk1
1
I just wanted to confirm that if an object is pulled or a car’s engine has a force, the energy of these forces is lost by friction and kinetic energy once it has moved a distance x.
So an initial kinetic energy + work by engine = final kinetic energy + friction
but when resolving forces its ( P - weight component - resistance to motion = mass * acceleration ) where the forces are all acting simultaneously.
Original post by Don Joiner
When you do distance = speed x time when looking in the horizontal sense,

Do you use the initial speed of projection for this calculation?


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you use the horizontal component of the initial speed
I had a couple of questions about collisions questions if anyone can answer them:
1. In the coefficient of restitution equation, if you're finding the relative speed of two particles a and b moving in the same direction, does it matter whether you do say, V(a) - V(b) or V(b) - V(a)?
2. If you were doing a question and you modelled a particle A as going to the left following a collision (calling right positive), but you found that because its speed came out negative it was actually moving to the right, would you correct this and make it positive before using it in any more calculations, or keep it negative?
I thought you would keep it negative, but I've been thinking about it and I'm not sure.

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