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.
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?
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.
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.
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?
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.
@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
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
Yep, that's correct. Basically don't worry - what you did is fine (and what I would do, coming to think of it).
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).
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?