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

Maetras
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#141
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#141
(Original post by rm761)
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????Attachment 551219
I really want to know the answer to this as well.
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Nijuuninichi
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#142
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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.
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izofficiallyme
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#143
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#143
(Original post by target21859)
I posted it as an attachment on page 4 I think
Oh sorry should have checked before, thank you!
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Maetras
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#144
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(Original post by izofficiallyme)
Oh sorry should have checked before, thank you!
It's actually page 2 sorry
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econam
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#145
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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
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alevelstresss
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#146
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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?
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TheMarshmallows
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#147
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(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
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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
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Patrick Gekko
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#148
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(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.

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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.
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pineneedles
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#149
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(Original post by Patrick Gekko)
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.
Alright, that makes sense, thank you 😊

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TheMarshmallows
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#150
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(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*.
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Nick-F007
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#151
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(Original post by target21859)
I really want to know the answer to this as well.
Essentially you solve this queston by calculating moments about the centre of mass.

Would you agree that the extra mass km shifts the centre of mass to the right?

Knowing that the verticle which makes an angle of [email protected]=5/6 passes through the new centre of mass, youcan test this by trying the c.o.m. on either side of the centre.

What you find is that the angle woild actually extend beyond the original centre of mass on the left which doesnt make sense so it must go on the right.

You calculate the position of the centre of mass by saying 5/6 = o/4 asthe ratios are the same.

Then calculate moments which are equal about the centre, hope this made sense, good luck tomorrow!
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kkboyk
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#152
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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
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oinkk
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#153
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#153
Not looking forward to this exam. Statics will always be a weakness for me.
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TheMarshmallows
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#154
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(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
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econam
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#155
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(Original post by TheMarshmallows)
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
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?
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oinkk
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#156
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(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).
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Maetras
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#157
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(Original post by Nick-F007)
Essentially you solve this queston by calculating moments about the centre of mass.

Would you agree that the extra mass km shifts the centre of mass to the right?

Knowing that the verticle which makes an angle of [email protected]=5/6 passes through the new centre of mass, youcan test this by trying the c.o.m. on either side of the centre.

What you find is that the angle woild actually extend beyond the original centre of mass on the left which doesnt make sense so it must go on the right.

You calculate the position of the centre of mass by saying 5/6 = o/4 asthe ratios are the same.

Then calculate moments which are equal about the centre, hope this made sense, good luck tomorrow!
Ah yeah that makes sense thanks for clearing that up. Good luck to you too. Hopefully it's a nice paper.
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TheMarshmallows
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#158
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(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:

Name:  ImageUploadedByStudent Room1466108569.385315.jpg
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As for the reaction force, it's acting through D (since D is vertically above B) and so can be ignored.



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eftio.gea
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#159
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Any of you tried the gold paper 4?
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econam
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#160
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(Original post by TheMarshmallows)
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:

Name:  ImageUploadedByStudent Room1466108569.385315.jpg
Views: 154
Size:  118.0 KB

As for the reaction force, it's acting through D (since D is vertically above B) and so can be ignored.
Ahh ok when you draw it like that I can see it now. Thank you!!
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