Edexcel - M3 - 18th May 2016 Watch

Mihael_Keehl
Badges: 13
Rep:
?
#1
Report Thread starter 3 years ago
#1
Here we go! We can discuss exam questions or general content if you need help.

I won't be revising for a while but if I can help, I'll try, means less revision for me in the summer if I help people earlier on (or if people explain to me!)


http://www.examsolutions.net/maths-r.../M3/module.php

https://www.youtube.com/watch?v=7pzW...e0WVwW&index=4

http://www.physicsandmathstutor.com/...ths-papers/#M3
2
reply
Krollo
Badges: 17
Rep:
?
#2
Report 3 years ago
#2
First :ahee:

Posted from TSR Mobile
0
reply
雷尼克
Badges: 2
Rep:
?
#3
Report 3 years ago
#3
2nd

this is gonna be my first exam
1
reply
Krollo
Badges: 17
Rep:
?
#4
Report 3 years ago
#4
(Original post by 雷尼克)
2nd

this is gonna be my first exam
Ditto I should think.

Just got my mock result back for M3 yesterday, managed an A* so quite happy. Nevertheless I bet we get a right arsy paper in May

Posted from TSR Mobile
0
reply
Student403
Badges: 18
Rep:
?
#5
Report 3 years ago
#5
Same my first exam for June series
0
reply
Ayman!
Badges: 12
Rep:
?
#6
Report 3 years ago
#6
I thought there was an M3 thread already?

Anyway, subbing! I'll be sitting the IAL paper.
0
reply
TheFarmerLad
Badges: 21
Rep:
?
#7
Report 3 years ago
#7
Joining
0
reply
Mihael_Keehl
Badges: 13
Rep:
?
#8
Report Thread starter 3 years ago
#8
yeah my first too

hf
0
reply
Louisb19
Badges: 14
Rep:
?
#9
Report 3 years ago
#9
Is it safe to say that whenever there is an object (such as a pole for instance) which is in contact (at an angle) with the ground, that the reaction force always acts perpendicular to the ground. It makes sense but I don't fully understand why it would be perpendicular to the ground and not perpendicular to the object.

This is what I'm talking about, I'm never quite sure where the reaction force is supposed to point.
0
reply
Louisb19
Badges: 14
Rep:
?
#10
Report 3 years ago
#10
(Original post by KFazza)
Yes, and because weight also acts perpendicular to the ground


Posted from TSR Mobile
That was my thinking too! Good old Newton's 3rd law.
0
reply
tiny hobbit
Badges: 15
Rep:
?
#11
Report 3 years ago
#11
(Original post by KFazza)
Yes, and because weight also acts perpendicular to the ground


Posted from TSR Mobile
(Original post by Louisb19)
That was my thinking too! Good old Newton's 3rd law.
It's nothing to do with the direction of the weight.

When a curved end meets a flat thing, the normal reaction is perpendicular to the flat thing. That also explains the direction of the force N on your diagram.
0
reply
Ayman!
Badges: 12
Rep:
?
#12
Report 3 years ago
#12
(Original post by Louisb19)
Is it safe to say that whenever there is an object (such as a pole for instance) which is in contact (at an angle) with the ground, that the reaction force always acts perpendicular to the ground. It makes sense but I don't fully understand why it would be perpendicular to the ground and not perpendicular to the object.



This is what I'm talking about, I'm never quite sure where the reaction force is supposed to point.
(Original post by KFazza)
Yes, and because weight also acts perpendicular to the ground


Posted from TSR Mobile
The reaction force is always perpendicular to the surface. Remember M1 - inclined slopes?
0
reply
CLevels
Badges: 2
Rep:
?
#13
Report 3 years ago
#13
Could someone please help me understand what's going on in question 4a here? I really don't get the method. I've self-taught myself M3 for the most part so I'm not the best yet at recognising which methods to use.
0
reply
ombtom
Badges: 16
Rep:
?
#14
Report 3 years ago
#14
(Original post by caaleb)
Could someone please help me understand what's going on in question 4a here? I really don't get the method. I've self-taught myself M3 for the most part so I'm not the best yet at recognising which methods to use.
Haven't looked at the question, but it looks like you were given v = 4/(x+2). You know v = dx/dt, so just solve the differential equation (FP2 I think?)

I assume the question said for the first 2 seconds, hence the integral from 0 to 2 with respect to time, and 0 displacement to X displacement for dx. Then just integrate and solve the quadratic.
0
reply
CLevels
Badges: 2
Rep:
?
#15
Report 3 years ago
#15
(Original post by ombtom)
Haven't looked at the question, but it looks like you were given v = 4/(x+2). You know v = dx/dt, so just solve the differential equation (FP2 I think?)

I assume the question said for the first 2 seconds, hence the integral from 0 to 2 with respect to time, and 0 displacement to X displacement for dx. Then just integrate and solve the quadratic.
Oh well that's handy, I haven't started that bit of FP2 yet, doing complex numbers at the moment :rolleyes: Thanks though haha, appreciate it.

I feel like I'm alone in finding M3 really hard, surely other people have felt like that?

Edit: It actually looks to be C4, but funnily enough I haven't finished the Integration chapter yet there
1
reply
ombtom
Badges: 16
Rep:
?
#16
Report 3 years ago
#16
(Original post by caaleb)
Oh well that's handy, I haven't started that bit of FP2 yet, doing complex numbers at the moment :rolleyes: Thanks though haha, appreciate it.

I feel like I'm alone in finding M3 really hard, surely other people have felt like that?
I thought there were a few hard bits (particularly one question on simple harmonic motion), but then I learnt M4 to make M3 easier. You should definitely learn first order differential equations before continuing with M3.

Edit: I have no idea how far C4 differential equations goes.
0
reply
Student403
Badges: 18
Rep:
?
#17
Report 3 years ago
#17
(Original post by ombtom)
I thought there were a few hard bits (particularly one question on simple harmonic motion), but then I learnt M4 to make M3 easier. I strongly suggest you learn FP2 differential equations before continuing with M3. (Don't think you need second order for M3 but there's definitely a lot of first order.)
Wait really? i didn't need any FP2 for M3
0
reply
ombtom
Badges: 16
Rep:
?
#18
Report 3 years ago
#18
(Original post by Student403)
Wait really? i didn't need any FP2 for M3
Ah, I must be getting my units confused then.
0
reply
Student403
Badges: 18
Rep:
?
#19
Report 3 years ago
#19
(Original post by ombtom)
Ah, I must be getting my units confused then.
According to a flowchart I saw, FP2 is needed for M4 and FP2 and FP3 are needed for M5
0
reply
ombtom
Badges: 16
Rep:
?
#20
Report 3 years ago
#20
(Original post by Student403)
According to a flowchart I saw, FP2 is needed for M4 and FP2 and FP3 are needed for M5
Definitely FP2 for M4. I struggled through M5 without needing FP3. Maybe that's for the old specification, or I learnt FP3 without realising?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Birmingham City University
    Undergraduate Open Day Undergraduate
    Sat, 23 Mar '19
  • University of Bolton
    Undergraduate Open Day Undergraduate
    Sat, 23 Mar '19
  • Harper Adams University
    Animals and Veterinary Sciences Open Day Undergraduate
    Sat, 23 Mar '19

Where do you need more help?

Which Uni should I go to? (119)
17.68%
How successful will I become if I take my planned subjects? (71)
10.55%
How happy will I be if I take this career? (119)
17.68%
How do I achieve my dream Uni placement? (99)
14.71%
What should I study to achieve my dream career? (65)
9.66%
How can I be the best version of myself? (200)
29.72%

Watched Threads

View All
Latest
My Feed