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

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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

I really want to know the answer to this as well.

Reply 141

Nijuuninichi

Don't know if this has been mentioned on the thread yet, but which paper have you guys found the hardest?

I remember June 2015 being quite hard but I had just finished learning the specification when I did it so I would probably find it a lot easier now.

Reply 142

izofficiallyme

Original post by target21859

I posted it as an attachment on page 4 I think

Oh sorry should have checked before, thank you!

Reply 143

Maetras

Original post by izofficiallyme

Oh sorry should have checked before, thank you!

It's actually page 2 sorry

Reply 144

econam

Could anyone explain how to do this question? I get taking moments about D but cant seem to get the answer. Its the jan 2014 IAL paper. Thanks :smile:

Reply 145

alevelstresss

Can I just double double check with you guys

I got 275/300 last year, and if I get 90,90 on C3 and C4, I should have 455/600, meaning I only need 25 in M2 for an A* still right?

Reply 146

TheMarshmallows

Original post by econam

Could anyone explain how to do this question? I get taking moments about D but cant seem to get the answer. Its the jan 2014 IAL paper. Thanks :smile:

draw diagram with forces and such (mass, friction and reaction at B, tension in string)
note that the horizontal distance from D to the mass is a*sin(theta)

part a:
Moments about D:
2aFr = mg * asin(theta)
2Fr = mgSin(theta)

Fr = mgsin(theta)/2 as required

(edited 7 years ago)

Reply 147

Patrick Gekko

Original post by pineneedles

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.

Posted from TSR Mobile

1- yes, use Vb-Va / Ua-Ub
2- No, if you've done everything right just leave it how it is. but just think carefully about the question, sometimes you may get rid of the negative sign- for example if it is asking whether/ to show that there will be a second collision between particles.

Reply 148

pineneedles

Original post by Patrick Gekko

Alright, that makes sense, thank you 😊

Posted from TSR Mobile

Reply 149

TheMarshmallows

Original post by alevelstresss

Can I just double double check with you guys

I got 275/300 last year, and if I get 90,90 on C3 and C4, I should have 455/600, meaning I only need 25 in M2 for an A* still right?

For A-Level maths yes, that is correct. >80UMS average across all papers, plus >90UMS average across c3 and c4 gets you an A*.

Reply 150

kkboyk

For working out the coefficient of restitution of two particle, should the bottom part of the fraction so you add both speed of approach; or the difference? :s

Reply 151

oinkk

Not looking forward to this exam. Statics will always be a weakness for me.

Reply 152

TheMarshmallows

Original post by kkboyk

For working out the coefficient of restitution of two particle, should the bottom part of the fraction so you add both speed of approach; or the difference? :s

Speed of approach / speed of separation

Reply 153

econam

Original post by TheMarshmallows

Thanks! Sorry if im being stupid but how come you use the verticle distance for friction and the horizontal distance for the weight? Also what happens to the normal reaction at B?

Reply 154

oinkk

Original post by kkboyk

For working out the coefficient of restitution of two particle, should the bottom part of the fraction so you add both speed of approach; or the difference? :s

It's the difference in the speed of approach. Take care with signs. Two particles may be heading in opposite directions so one will have negative velocity (this turns it into an addition which you may see in certain mark schemes).

Reply 155

Maetras

Ah yeah that makes sense thanks for clearing that up. Good luck to you too. Hopefully it's a nice paper.

Reply 156

TheMarshmallows

Original post by econam

Thanks! Sorry if im being stupid but how come you use the verticle distance for friction and the horizontal distance for the weight? Also what happens to the normal reaction at B?

Both the frictional force and weight force are already perpendicular to our point, D. We're given that the distance from D to B (where the frictional force is) is 2a. So we just need the distance from the weight force to the point. I drew a picture to illustrate the idea and where the distance comes from:

ImageUploadedByStudent Room1466108569.385315.jpg