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Arsey's M2 Edexcel revision and resources thread

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Original post by R2D2.
Okay and when rods are stuck on a wall (hinges) does it matter which direction my X and Y force goes? Like in the equations and diagrams, will I still get right answer regardless? There was one question where this rod was about to slip from this hinge question. It had a mass at the very end so I assumed that the bit near the wall will try to slide up but it was going to slide down? I don't understand how :frown:

Have you tried the june 2010 paper? I found that so easy. Almost got full marks. I keep doing good in some papers and bad in others. I have always got over or around 55 but I need to get solid 60s


Yeah the maths will work itself out, say if you put Y upwards and it's actually downwards, through resolving vertically upwards you would get a negative value, thus telling you it's actually in the other direction and vice versa
Original post by R2D2.
Does anyone know why the boundaries for jun 11 and jan 12 were extremely high? I haven't attempted them but assuming they are really easy?


Yep they were almost C2 difficulty
Pretty nervous for this paper as M1 was my worst module last year. Hope it's gonna be okay.

Thanks for all the revision materials Arsey! You're the best :biggrin:


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Original post by SummerPi
Pretty nervous for this paper as M1 was my worst module last year. Hope it's gonna be okay.

Thanks for all the revision materials Arsey! You're the best :biggrin:


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Hi, I was wondering do you just so happen to know which the hardest M2 papers are?
Reply 64
Original post by sneakbo2
Yeah the maths will work itself out, say if you put Y upwards and it's actually downwards, through resolving vertically upwards you would get a negative value, thus telling you it's actually in the other direction and vice versa


nice. I tried the same thing with the last impulse question. Getting it the wrong way around. But it gave me a negative value which meant that the direction of motion is now reversed and because I had arrows on my diagram, it helped me understand which way it will go. In impulse questions with restitution, where you don't know final velocities of either... same thing applies right? It doesn't matter which way they are going? But if they are going the same way, I have to do sub values in so that the final velocities show that they don't collide?

i.e. if I have A and B colliding and their final speeds are X and Y respectively. I draw the final velocities in --> way Y has to be bigger than X so that it doesn't collide again. And if for some reason X actually reverses after collision, I would get negative answer?
Reply 65
Original post by Boy_wonder_95
Yep they were almost C2 difficulty


I am going to try the first one now.

Btw if anyone has attempted the review exercises, can you tell me how much time I would need to roughly go through most of the questions? I have done a good 10 papers and after about 4-5 more I will probably just work off the book questions. I am feeling a bit more confident but does anyone have any tips to handle the tricky questions? By tricky I mean questions like Jan 2011 6.b) I know it's only two marks but sometimes they come up as bigger marks :frown: I didn't know I had to change the signs on j component because it's downwards...
Reply 66
Can someone explain the last part of question 2? http://examsolutions.net/a-level-maths-papers/Edexcel/Mechanics/Mechanics-M2/2011-June/paper.php I got 70 in this paper. This was supposedly an easy paper. I found it fairly easy as well. I only lost marks on final answer on q2 and the last part of last question where I didn't think -.-

I got both my velocities and assumed I would need to say velocity for one particle has to be greater than velocity of the other since the travel opposite each other. But the mark scheme and exam solutions guy just used one of the velocities? I don't understand why you can't use any other identities. I guess I should have looked at my two velocities and worked it out but I am curious as to whether there was a logical method behind this. Thanks :smile:
Original post by R2D2.
nice. I tried the same thing with the last impulse question. Getting it the wrong way around. But it gave me a negative value which meant that the direction of motion is now reversed and because I had arrows on my diagram, it helped me understand which way it will go. In impulse questions with restitution, where you don't know final velocities of either... same thing applies right? It doesn't matter which way they are going? But if they are going the same way, I have to do sub values in so that the final velocities show that they don't collide?

i.e. if I have A and B colliding and their final speeds are X and Y respectively. I draw the final velocities in --> way Y has to be bigger than X so that it doesn't collide again. And if for some reason X actually reverses after collision, I would get negative answer?


Yeah the that's the principle behind the whole of collisions. E.g a ball A hits B, B then rebounds off a wall.

If Final velocity of A comes out of negative then it is going in reverse, B obviously changes direction hitting the wall and if it's moving faster than A it will catch up and collide again, all about visualising moving balls and understanding the topic really, the maths isn't too difficult
Reply 68
Original post by sneakbo2
Yeah the that's the principle behind the whole of collisions. E.g a ball A hits B, B then rebounds off a wall.

If Final velocity of A comes out of negative then it is going in reverse, B obviously changes direction hitting the wall and if it's moving faster than A it will catch up and collide again, all about visualising moving balls and understanding the topic really, the maths isn't too difficult


I got 65 in jan 2012 paper :frown: apparently that was an easy paper :frown:(((( really worried now
Reply 69
I have a quick question. Sometimes, I don't follow the method the mark scheme has. And because I use rounded answers (3sf ones) I get different values from mark scheme. Not by a lot. For instance instead of 8.49, I got 8.47

Sometimes they round to 2 sf (idk why) but I round to 3 all the time if I can. And sometimes you are expected to use a formula in a given context but you can get the same answer a different way. How many marks will I lose if I don't follow the mark scheme exactly? It doesn't provide any alternate solutions. I am assuming for the 1 dp difference answers, I will still get full mark?
Reply 70
Original post by R2D2.
I have a quick question. Sometimes, I don't follow the method the mark scheme has. And because I use rounded answers (3sf ones) I get different values from mark scheme. Not by a lot. For instance instead of 8.49, I got 8.47

Sometimes they round to 2 sf (idk why) but I round to 3 all the time if I can. And sometimes you are expected to use a formula in a given context but you can get the same answer a different way. How many marks will I lose if I don't follow the mark scheme exactly? It doesn't provide any alternate solutions. I am assuming for the 1 dp difference answers, I will still get full mark?


Normally they round to 2sf when using g as the value we us is 9.8 which is also to 2sf
Reply 71
Original post by chica111
Normally they round to 2sf when using g as the value we us is 9.8 which is also to 2sf


so if mine is to 3 do I lose final mark?
Reply 72
Original post by R2D2.
so if mine is to 3 do I lose final mark?


Im not sure, but I don't think you would as I give all my answers to 3sf too and have been told to do so
Could anyone shed some light on question 4 on the June 2009 paper. It's the first ever M2 question I really can't get my head around.
Reply 74
Original post by chica111
Im not sure, but I don't think you would as I give all my answers to 3sf too and have been told to do so


I believe you will lose marks if you give an answer with g to more than 3 , I am not 100% sure though

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Does anyone know what freely hinged means in the context of M2. I think that's where the problem lies in this question for me.


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Original post by lebron_23
Could anyone shed some light on question 4 on the June 2009 paper. It's the first ever M2 question I really can't get my head around.

Thrust is basically reverse tension. Take moments about A to remove the forces by the hinge to get part A.

In part B resolve horizontally and vertically for the equations and then once you found the respective vertical and horizontal forces simply do a inverse tan of the triangle you get. You can express your final answer as an angle above the horizontal AB or from the vertical AC (no need for a bearing)
Original post by lebron_23
Does anyone know what freely hinged means in the context of M2. I think that's where the problem lies in this question for me.


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Hinged just means there is a reaction in two directions. Upward and horizontal acting at point A or wherever the wall is connected to the rod. Questions like that you usually obtain magnitude of tension or whatever thrust etc by taking moments, then resolving horizontally and vertically, finally if it asks for the magnitude, just use pythagoras.
Original post by sneakbo2
Hinged just means there is a reaction in two directions. Upward and horizontal acting at point A or wherever the wall is connected to the rod. Questions like that you usually obtain magnitude of tension or whatever thrust etc by taking moments, then resolving horizontally and vertically, finally if it asks for the magnitude, just use pythagoras.


I meant to say freely jointed, but thanks for the explanation anyway. My M2 knowledge is somewhat lacking when it comes to concepts exclusive to M2. I'm basically just running of M3 knowledge lol.


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