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MEI Mechanics 2 (M2) - 18 May 2016

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is there anything you need explaining? :smile:
How do you decide which point to choose when working out internal forces?
Original post by Mr Moon Man
How do you decide which point to choose when working out internal forces?

I tend to start vertically at a point with just two opposing forces, but it makes no difference provided there's only 1 unknown at the point along the axis you're resolving.

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(edited 7 years ago)
Reply 23
If a particle is stuck right at the end of the rod. Is there a reaction force
can someone explain question 3iv on Jan 13 paper. I think it's the only question of this type to have ever come up, where you have to specifiy the stable area for the centre of mass. I don't really understand the mark scheme but am I right in thinking you can just connect all of the points in contact with the ground, then shade the shape you create?

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Reply 25
Original post by 11234
If a particle is stuck right at the end of the rod. Is there a reaction force


There is, but it doesn't affect anything because it's the force the rod is exerting and hence plays no part in moments or resolving (i.e: doesn't affect equilibria) since we are only concerned with the forces that is being exerted on the rod.
Reply 26
3 iv was weird, my teacher went through it and thought it was a pretty bad question, will probably not come up again - examiner's report described it quite negatively. You were right in that you do just connect the points and shade it in, although for the second part you were meant to draw on the centre of mass and say it's stable because is within the shaded region.
3iv.jpg

Probably too late for anyone to respond, but for part 1 b iii of the same jan 2013 paper you needed to find the frictional force which the markscheme allowed you to either consider impulse or find acceleration. I did it by stating the work done by friction was reponsible for the change in kinetic energy (friction was the only force involved). I belive this is a perfectly valid method and it got the right answer but it wasn't one of the methods listed in the markscheme and there was no comment saying alternate methods allowed, do you know if I would not get the mark or would they allow this?
Reply 27
Original post by bash!


Probably too late for anyone to respond, but for part 1 b iii of the same jan 2013 paper you needed to find the frictional force which the markscheme allowed you to either consider impulse or find acceleration. I did it by stating the work done by friction was reponsible for the change in kinetic energy (friction was the only force involved). I belive this is a perfectly valid method and it got the right answer but it wasn't one of the methods listed in the markscheme and there was no comment saying alternate methods allowed, do you know if I would not get the mark or would they allow this?


@Alex621 tagging to let you know your question was answered.

Also, bash! - I think you'd get the marks for that, was does the examiner report say? The examiners receive a more detailed copy of the markscheme you see that outlines more alternative methods, so I'm fairly sure yours is fine.
Reply 28
Original post by Zacken
@Alex621 tagging to let you know your question was answered.

Also, bash! - I think you'd get the marks for that, was does the examiner report say? The examiners receive a more detailed copy of the markscheme you see that outlines more alternative methods, so I'm fairly sure yours is fine.


Ah yeah, sorry I forgot to reply properly (don't post much).
The examiner's report doesn't really say much about this question, it was only worth 1 mark (was only the first step of iii), but I'm worried the same situation could apply to a larger question. I believe I've seen similar questions that allow considering the change in KE, so it was weird that this one doesn't. Didn't know the examiners got a different markscheme, seems strange they'd put effort into creating a different one to upload publicly.
I thought I'd heard before that using a correct method not listed would still get the mark but just wanted to check so I'm not worrying so much tomorrow :tongue:
Reply 29
Original post by bash!

I thought I'd heard before that using a correct method not listed would still get the mark but just wanted to check so I'm not worrying so much tomorrow :tongue:


Yep, that's correct. Basically don't worry - what you did is fine (and what I would do, coming to think of it).
Good luck everyone!
Nearly everything was easy enough, but then there's Q4 parts one and two... o boi
(edited 7 years ago)
Reply 32
Did anyone get. 0.198 m for the energy question
Original post by 11234
Did anyone get. 0.198 m for the energy question


Yeah, I also know someone else who did too, so we're probably right.
Reply 34
For the 2 marker involving Q in question 1 when an object is dropped off Q did the speed of Q change or stay the same?
Reply 35
how did you guys prove that friction was present
Reply 36
Original post by 11234
how did you guys prove that friction was present

If it were a smooth plane, the velocity of the ball parallel to the plane would be unchanged by the collision. Therefore to show that it is not smooth (and therefore that there is friction) you just need to show that the horizontal elocity is not the same before as it is after. 10cos(60) is not equal to 6cos(40).
(edited 7 years ago)
Original post by Maths77
For the 2 marker involving Q in question 1 when an object is dropped off Q did the speed of Q change or stay the same?


I said it stayed the same because no force acted on Q to change it's speed
Reply 38
Original post by Rawsonj
If it were a smooth plane, the velocity of the ball perpendicular to the plane would be unchanged by the collision. Therefore to show that it is not smooth (and therefore that that is friction) you just need to show that the horizontal elocity is not the same before as it is after. 10cos(60) is not equal to 6cos(40).

I did that!!!! Thanks. I said that 10cos60 does not equal 6cos40 and then said something about impulse due to friction. Could I get 3 marks?
Was the coefficient of friction 1.57 and was the tension in one of the rods 394?

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