Different option topics.Guys, what do you think a 90 raw in this exam will scale to? I looked at the rawmarks.info page for physics. The table shows that a 90 in 2013 got scales to 95 but a 90 in the same year also got scaled to 94. How is this possible?
I've just read through this and seen all your arguments about that motion trolley question and I don't agree with any of your drawings (not that I'm an expert by any means, but I think I at least understand this.)
Firstly, as many people have already explained, it definitely doesn't accelerate as it leaves the field or whatever, that violates the Law of Conservation of Energy. While the Eddy Currents do switch direction as they pass B, that is so that they expel a magnetic field which is attracted to the permanent magnet, instead of repelled. Either way, acceleration is negative.
Next, nobody seems to have considered the fact that a magnetic field coming from a point like that doesn't have a "boundary" per say and actually goes on forever, although its only significant up close, (similar to the Earth's gravitational field, except obviously much smaller range of significance." This implies the curve should have 2 horizontal asymptotes (1 at v, or slightly higher I guess, and one lower, which it approaches as it experiences negligible acceleration leaving the field)
Another thing to note is clearly entering and leaving the curve should be the same, so the curve will be symmetrical, save for the decrease in gradient upon slowing down, and should have maximum gradient right under the magnet (as there is highest change in flux there, assuming velocity doesn't decrease too significantly.) Overall it should look like an arctan(-x) graph but slightly squashed down the bottom.
That's just my two cents
I derived it using integration but basically you use 1/2 mv^2 =FdHow did everyone do the find the velocity of the electron question? I tried finding acceleration, then finding time by subbing it into s=ut+1/2at^2 then subbing into a=delta v/delta t, but i was getting values way over the speed of light.
Why do you let kinetic energy equal to torque?I derived it using integration but basically you use 1/2 mv^2 =Fd
Work =FD and energy = 1/2 mv^2 so they're both functions of energy. Or you can just derive it like I didWhy do you let kinetic energy equal to torque?
Actually the method i said gave the same answer, i was just being a fuckwit and used 9.8 for a.Work =FD and energy = 1/2 mv^2 so they're both functions of energy. Or you can just derive it like I did
i didn't even extrapolate my LOFB just drew the one that suited the pointsI got 2.6m for the height of the platform. What's the degree of error BOS usually allows? Then again at the same time I didn't force the LOBF through the centre as I read a few people here did, which might be the 0.1 difference. (you're not supposed to FYI, even thought theoretically there is zero range at zero height, the whole point of graphing is it allows you to determine the systematic error and by forcing it through the origin it distorts the original purpose of graphing to produce a more accurate answer)
How do you propose the graph looks then, your logic says it should start flat (at each asymptote), get less flat and then suddenly go flat again that doesn't ring with me. I understand your logic about the trolley sticks out on each side of the field though, but I can't mathematically accept it. Maybe two different Eddy Currents form on each side, attracting either way? I'm not sure. I guess we'll just have to wait for the guidelines. I imagine they would have to be quite lenient seeing as so maybe variables were undefined, for all we know velocity was pretty much zero at D, in which case the curve would look drastically differentI don't agree with this part. The graph when the trolley is right under the magnet should have zero gradient, as the net force on the trolley would be zero. Remember Emf (induced) = -delta (phi)/delta (t). And phi = B (magnetic field strength) x A (area of the trolley affected by the magnetic field). When the trolley is right under the magnet, there is no change in B or A, and hence no change in phi, and so delta (phi)/delta (t) = 0. So the graph here would have zero gradient.
Also, if the trolley is long enough so that some part of the trolley sticks out of the field on either side of the magnet even if it's right under it, the rear part of the trolley would be repelled, and the front part of the trolley would be attracted, meaning that the net force on the trolley at this point would be zero anyway, hence, as acceleration = 0, there is no change in v and the graph is still flat (i.e. gradient is zero). I agree with the rest of your stuff though about the graph having two asymptotes and the velocity decreasing over time all, but I think the velocity only decreases when the trolley is entering and leaving the magnet's field...
the hard part was finding the right value for a, since you have to use f=ma with force from your first answer and m from the back of the sheet, dont think you'll get any marks for that q tbhActually the method i said gave the same answer, i was just being a fuckwit and used 9.8 for a.
I think the graph would look something like this. I don't 100% remember where A, B, C and D was labelled and the graph's shape may be a bit inaccurate (so take this with a grain of salt) but you get the idea...How do you propose the graph looks then, your logic says it should start flat (at each asymptote), get less flat and then suddenly go flat again that doesn't ring with me. I understand your logic about the trolley sticks out on each side of the field though, but I can't mathematically accept it. Maybe two different Eddy Currents form on each side, attracting either way? I'm not sure. I guess we'll just have to wait for the guidelines. I imagine they would have to be quite lenient seeing as so maybe variables were undefined, for all we know velocity was pretty much zero at D, in which case the curve would look drastically different
YEAH THATS WHAT I PUT WOOOOI think the graph would look something like this. I don't 100% remember where A, B, C and D was labelled and the graph's shape may be a bit inaccurate (so take this with a grain of salt) but you get the idea...
View attachment 32617
Yeah me tooYEAH THATS WHAT I PUT WOOOO
I feel this is legit because in between B and C, you can approximate it as a constant field (and moving through a constant field will yield no change in emf)
or alternatively, the force is a continuous function
it switches from being repulsive to being attractive, so at some point it should pass through zero (dodgy logic i know)
i really liked that q actually
You think so? I think that the hard part was knowing what to do with a, i found a using f=ma but didnt know what to do with it, hopefully i get 1 mark.the hard part was finding the right value for a, since you have to use f=ma with force from your first answer and m from the back of the sheet, dont think you'll get any marks for that q tbh