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

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But on the question it specifies that it is perpendicular?

Yes, so the point on the path that you are thinking of is not the one you want.

Options:

Draw a couple of triangles with the same angle in them, using the perpendicular idea then use trig.

Write down 2 column vectors for the velocities, with an unknown (downwards) component for the one you want and say that the dot product is 0.

Use the equation of the path and use perpendicular gradients: m1 x m2 = - 1

Reply 21

Alby1234

Original post by tiny hobbit

Thank you, using the dot product worked for me. I guess I was visualising it in a different way

Reply 22

maddywald

Original post by LLk1

Can all questions be solved using moments and resolving up/down ? Or is it necessary to sometimes use triangle of forces?

The way i have been taught it is possible to solve without triangle of forces - i have never come across a question where i have had to do that :smile:

Reply 23

fpmaniac

Anyone can help me with a few questions???

1. A ball is dropped from 0 velocity and after falling for 1s under gravity meets another equal ball which is moving upwards at 7ms-1
a. Calculate the velocity of each ball immediately after the impact given that the coefficient of restitution is 0.25.

Reply 24

fpmaniac

Another one
http://postimg.org/image/na45v0uiz

the angle(a) is represented by tan a = 0.5

AB weighs 400N. The weight (circle) weights 200N. The apparatus remains in equlibirum. AB is modelled as a road and BC as light inextensible cable.
Find the tension in AB.

Reply 25

candol

Original post by fpmaniac

2 eqns, solve simultaneously.
1. conservation of momentum
eg 7m - 9.8m = mv1 + mv2
2. 0.25 = v2-v1/(16.8)
Then solve

Reply 26

fpmaniac

Original post by candol

2 eqns, solve simultaneously.
1. conservation of momentum
eg 7m - 9.8m = mv1 + mv2
2. 0.25 = v2-v1/(16.8)
Then solve

what answer did you get

Reply 27

candol

Original post by fpmaniac

what answer did you get

v1 = -3.5 and v2 = 0.7

Reply 28

fpmaniac

Original post by candol

v1 = -3.5 and v2 = 0.7

oh. Figured it out what i did wrong. Thanks

Reply 29

MaxWalker1

I constantly find myself spending too long on center of mass questions, any one have any tips or shortcuts when answering those types themselves?

Reply 30

fpmaniac

how to do june 2005, question 7b?

Reply 31

jamestg

Original post by maddywald

Edexcel M2 - 17th June 2016 - Official Thread
I hope you find this very useful as a directory for resources and a help desk for any questions about M2 homework, the exam and last minute tips! There are some extremely able people on here, who will be very happy to help - use this to your advantage!

Exam Date
Friday 17th June 2016 pm - 90mins

Exam Info/Guidance
M2 is a step up from M1 - although some of the topics are the same, it is more complicated and it's easy to make small mistakes that can cost you marks that you may not be able to afford to lose.
The topics are listed as follows:
Kinematics
Centres of Mass
Work, Energy and Power
Collisions
Statics of rigid bodies

Resources
Exam solutions videos for M2
Edexcel A-level Maths section
Various M2 resources
MyMaths
MrBartonMaths Applied Module stuff
MadAsMaths M2 Booklets

Past Papers/Practice Papers
Edexcel M2 Past Papers
OCR M2 Past Papers
OCR MEI M2 Past Papers
M2 Solomon Papers
Bronze/Silver/Gold Practice Papers (Bronze = easy, Silver = intermediate, Gold = hard)
Chapter Practice Questions for M2
Try an International A-level Past Paper

Hopefully this thread has enough resources to help you gain top marks, and enough questions to last a lifetime. If you ever (unlikely) or think you will run out of questions, buy some really old textbooks off Amazon for 1p and you will hopefully never run out of questions.

Good Luck!

I don't like being a groan and/or being salty, but you could've credited me for the OP format :colonhash:

Reply 32

candol

Original post by fpmaniac

how to do june 2005, question 7b?

The gain in K.E + work done against friction = Loss in P.E
(I have written it in these terms, because if you think about it, the KE has not gained as much as it would do on a smooth surface, so that difference must be the work done against friction)
The number crunching is now straightforward.
K.E = 1/2mv^2 = 0.5*3*5^2 = 37.5
P.E = 12g from part a)
Work done = F*d = 8F
so 12g - 37.5 = 8F (rearranging above)
F = 10N

Reply 33

LLk1

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Getting really confused with this question, any help would be sick!

When in the mark scheme it says Dist of G from DC, I get a positive answer meaning that G must be below DC. Yet I dont I dont understand why doubling it results in O from DC? Surely thats double the required distance??

Screen Shot 2016-05-30 at 09.18.24.png36.8KB

Reply 34

candol

Original post by LLk1

Attachment not found

Not sure I like the mark scheme
if you consider the vertical components of the centres of mass of the 3 shapes you get:
Square + rest of triangle = Whole triangle
16 * y component of O + (32 * 6.43) = 48 * 6
Note 6.43 comes from tan 25 in mark scheme
Then rearrange to find y comp of ) = 5.14
So centre is 0.86 from DC