I thought these multiple choice questions were pretty hard, potentially harder than the real papers. Hope the very last one makes sense! Thank YOU for using my channel!!
For the last question, I calculated KE using 0.5mv^2, where v = (2 x pi x radius) / (24 x 60^2) but this doesn't work. Could you please explain why? Great video btw, thank you!
@@user-ru5sw3ss2u Hi, the time period may not be 24 hours. To do this this way you'd need to calculate the time period using kepler's third law. Easiest way to calculate orbital velocity is sqrt(GM/r) . Hope this helps!
Zphysics, however I feel coming out of the exam tomorrow. Just know that it would he alot worse if I had not found your channel. Thank you for everything thus far
For Q7 you can go straight to using the v^2 equation but just put s = 1.3m in to find the vertical velocity component upon hitting the ground, no need for 2 suvats.
I agree! I was almost solving them "live" in a free period, so pretty similar to what I would have spotted first in a real exam. I completely agree though, this would be more efficient : )
@@zhelyo_physics Also for the pressure question you made a pretty big assumption that the thin outer strip can be approximated well by a rectangle of length 2piR and width t, this worked for that one example but a better way that works more generally is to relate t to the inner and outer radii and expand, then say that t^2 is negligible etc.
Quick Question about 6. Would the driving force be 4000 X 7 or 4000 X 6 ? Because in the question they specify a clear difference between the trucks and the locomotive. And the 4000N resisting force is for each truck
@@ali-bt1cp Yh I thought what he did was a bit dodgy but ultimately his way just looks at the change in resultant force so is wrong physics, because it should be assumed the loco will have a different resistive force as muR=frictional force and R=mg and m is larger and the wheels will have roughly the same mu, but his way will lead to the correct answer every time
For question 3 you could probably figure out that its C much quicker by thinking that watts is joules per second and if its dissipating 6 joules in one second then its dissipating 12 joules in 2 seconds. Might be helpful if ur under time pressure.
For 7, you do not have to do SUVAT twice if you use the distance as -1.3 and U as positive 7. I know both methods work, but when under time pressure you should probably use the most efficient method
I suspect it doesn't matter since the resultant force is the force from the loco minus the resistive force which initially is 0 and removing 4 carriages will lead to a reduction in the resistive force by 4x the resistive force of a single carriage leaving the a resultant force of 16000N
I agree, I think the definition of a truck is somewhat dubious in this question. I just assumed the locomotive will have the same drag as all the other ones (pretty good approximation, a real locomotive will have very similar resistive force). Hope this helps!
for question 5 would it not be more accurate to say that the area which the force is applied over is pi(R+t)^2-piR^2 which yields, when expanded, 2piRt+2Pit^2 but t^2 is negligible since t
@@abrahamtharakan Originally t is much smaller than R so t^2 will be even smaller in comparison to R but in the question is only looking for an approximate value for the pressure so using the most significant term is sufficient though if they wanted a fully accurate value then the t^2 should be included. As an example of the squared term being negligible, the difference between 1cm to a meter is 100 fold but from 1cm^2 to 1m^2 is 10000 fold, there is also the case of small angle approximations where the higher terms of the Maclaurin series are ignored because of there large exponents.
in my printed version of the last question, the multiple choice answers different and include are 6x10^5 which is the initial answer without including the orbital speed.
Because the pressure is only applied to the circumference of the cylinder (I suspect technically the area is 2pit^2 since is the difference in areas of a circle with radius R+t and R which leads to the extra term but as t is negligible in comparison to R i.e. R>>t then that term can be ignored)
For 6, the question says the resistive forces are on the trucks, I would've thought this means the total resistive forces is 6 x 4000N not 7 x 4000N as the locomotive is not a truck. Is this incorrect?
I suspect it doesn't matter since the resultant force is the force from the loco minus the resistive force which initially is 0 and removing 4 carriages will lead to a reduction in the resistive force by 4x the resistive force of a single carriage leaving the a resultant force of 16000N
I think for this one I assumed the locomotive is a truck, otherwise we are kinda saying the locomotive has no resistive force which is very much not possible. Hope this helps!
I also found that generally the specimen was way harder than any of the past papers. I don't think tomorrow's one will be as difficult though, hopefully
For question 13, why is statement 2 not correct when the diffraction grating is on the earth, Do the atoms in the atmosphere not absorb any photoenergy at all?
excellent question, for physics we assume not, in practice there would be some however not a significant amount and also we can compare the solar specturm to other stars and differentiate what's from the atmosphere and what's from the sun. The amount absorbed by the atmosphere was pretty insignificant if I remember correctly.
for question 3 although electricity isn't tested work, energy and power is isn't it? and the question only uses the power and not the p.d. given in the question
@@zhelyo_physics it is, it has the same paper code at the top as OCR A (H556) instead of the OCR B code (H557). And it literally says "physics A" at the top
I thought these multiple choice questions were pretty hard, potentially harder than the real papers. Hope the very last one makes sense! Thank YOU for using my channel!!
And thank you for creating quality and accessible learning materials
I'm glad you said this cuz I thought they were pretty tough!
For the last question, I calculated KE using 0.5mv^2, where v = (2 x pi x radius) / (24 x 60^2) but this doesn't work. Could you please explain why? Great video btw, thank you!
@@user-ru5sw3ss2u Hi, the time period may not be 24 hours. To do this this way you'd need to calculate the time period using kepler's third law. Easiest way to calculate orbital velocity is sqrt(GM/r) . Hope this helps!
@@zhelyo_physics Ah I see thank you!
What a legend. Good luck to everyone on their physics exams.
thank you for the comment, good luck!
Zphysics, however I feel coming out of the exam tomorrow. Just know that it would he alot worse if I had not found your channel. Thank you for everything thus far
wow thank you so much for this comment. GOOD luck!
good luck to us all tomorrow, hopefully the exam boards are kind
good luck everyone. I'm cooked 💀
we all are, especially those doing AQA
Same 💀💀
Bro Ive got a few hours to go through the whole Spec and I’m gonna be learning not revising 💀
Same
If its anything like aqa 2023 we’re ashes icl 😭
you’re a god sent! the last question was driving me insane yesterday cuz i just kept getting 6.0 x 10^9 and i couldn’t find anyone solving it
No problem! It's tricky, even more so, I don't think the real answer is that as it presumes the initial KE is 0 but the earth itself is rotating.
thank you so much, been watching since the start of a levels you're an absolute legend mate
thank you so much for your comment! very much appreciated, good luck tomorrow!!
thank god these are specimen questions they require so much working especially the last question
Are specimen papers generally harder?
@@fpsa_ yes for some reason
@@fpsa_ the one he just did is, idk about other exam boards
@@fpsa_ yes they try to include as much content as possible, usually.
agreed, they are trickier
thank you very much and goodluck to all of us .
Good luck guys!!!
Im cooked for tomorrow. Regardless thanks zphysics you are the goat
you got it! Good luck tomorrow!! Thanks for the comment.
Guys there is an easy tip if you struggle an 1 mark Q
Skip it (and maybe come back)
For Q7 you can go straight to using the v^2 equation but just put s = 1.3m in to find the vertical velocity component upon hitting the ground, no need for 2 suvats.
I agree! I was almost solving them "live" in a free period, so pretty similar to what I would have spotted first in a real exam. I completely agree though, this would be more efficient : )
@@zhelyo_physics Also for the pressure question you made a pretty big assumption that the thin outer strip can be approximated well by a rectangle of length 2piR and width t, this worked for that one example but a better way that works more generally is to relate t to the inner and outer radii and expand, then say that t^2 is negligible etc.
Wow i was actually doing this paper printed right in front of me and then i get this video recommended. Excellent.
great minds think alike!! : ) Good luck!
Thank you!
anytime!
thanks so much, truly.
Anytime!
Prayers for anyone doing Aqa
Good luck!
7:30 i dint get why you have to use area= 2PiRt if its hollow but if it is not hollow area=PiR^2t
Quick Question about 6.
Would the driving force be 4000 X 7 or 4000 X 6 ?
Because in the question they specify a clear difference between the trucks and the locomotive. And the 4000N resisting force is for each truck
It doesn't matter either as the change in resisting force is the same i.e. 16000N
that’s what i thought too and surprisingly it gives the same answer
I think the definition of a "truck" in this question is a little dubious
@@ali-bt1cp Yh I thought what he did was a bit dodgy but ultimately his way just looks at the change in resultant force so is wrong physics, because it should be assumed the loco will have a different resistive force as muR=frictional force and R=mg and m is larger and the wheels will have roughly the same mu, but his way will lead to the correct answer every time
last question is crazy
Agreed
For question 3 you could probably figure out that its C much quicker by thinking that watts is joules per second and if its dissipating 6 joules in one second then its dissipating 12 joules in 2 seconds. Might be helpful if ur under time pressure.
For 7, you do not have to do SUVAT twice if you use the distance as -1.3 and U as positive 7. I know both methods work, but when under time pressure you should probably use the most efficient method
I completely agree : ) it was the first method that came to mind when I filmed this : )
thank you for your continued service to spreading physics sir. and good luck to all the soldiers entering no mans land tomorrow. salute o7
Do you have a video compilation of all of the one mark definitions for OCR? (e.g. what is the principle of moments)
sadly not but they are within my all of revision videos
in question 6 dont you do 4000 x 6 since there is 6 trucks and not 7?
exactly my question. Why do you count a locomotive as a truck?
I suspect it doesn't matter since the resultant force is the force from the loco minus the resistive force which initially is 0 and removing 4 carriages will lead to a reduction in the resistive force by 4x the resistive force of a single carriage leaving the a resultant force of 16000N
I agree, I think the definition of a truck is somewhat dubious in this question. I just assumed the locomotive will have the same drag as all the other ones (pretty good approximation, a real locomotive will have very similar resistive force). Hope this helps!
@@teacupcakes2739 Hes cooked
for question 5 would it not be more accurate to say that the area which the force is applied over is pi(R+t)^2-piR^2 which yields, when expanded, 2piRt+2Pit^2 but t^2 is negligible since t
Yeah this is the correct way, physicists usually avoid expanding brackets at all cost!
I'm a little confused as to why t^2 would be negligible, could you perhaps explain that?
@@abrahamtharakan Originally t is much smaller than R so t^2 will be even smaller in comparison to R but in the question is only looking for an approximate value for the pressure so using the most significant term is sufficient though if they wanted a fully accurate value then the t^2 should be included. As an example of the squared term being negligible, the difference between 1cm to a meter is 100 fold but from 1cm^2 to 1m^2 is 10000 fold, there is also the case of small angle approximations where the higher terms of the Maclaurin series are ignored because of there large exponents.
@@abrahamtharakan How big is 0.001^2?
in my printed version of the last question, the multiple choice answers different and include are 6x10^5 which is the initial answer without including the orbital speed.
hmmm I wonder if it was changed after and this is an older version? Never seen this version I think.
why is the circumference multiplied by the thickness when working out the pressure on the hollow tube in q5?
Because the pressure is only applied to the circumference of the cylinder (I suspect technically the area is 2pit^2 since is the difference in areas of a circle with radius R+t and R which leads to the extra term but as t is negligible in comparison to R i.e. R>>t then that term can be ignored)
For 6, the question says the resistive forces are on the trucks, I would've thought this means the total resistive forces is 6 x 4000N not 7 x 4000N as the locomotive is not a truck. Is this incorrect?
Yes it should be 6x4000
no youre right, it said 6 trucks and a locomotive. i guess we are assuming locomotive is perfectly aerodynamic- ie drag coefficient is zero
i think also because of the diagram he drew which only had trucks 1,2,3 so he took 4000*3 that made the answer correct
I suspect it doesn't matter since the resultant force is the force from the loco minus the resistive force which initially is 0 and removing 4 carriages will lead to a reduction in the resistive force by 4x the resistive force of a single carriage leaving the a resultant force of 16000N
I think for this one I assumed the locomotive is a truck, otherwise we are kinda saying the locomotive has no resistive force which is very much not possible. Hope this helps!
I also found that generally the specimen was way harder than any of the past papers. I don't think tomorrow's one will be as difficult though, hopefully
yeah i went though the specimen and it was def harder than any past paper i’ve gone through
Could you do the same for paper 2?
on it!
Goat
Hopefully one more video before the end of today...stay tuned! : )
@@zhelyo_physicsEven more goated 🐐
Thank you Zhelyo, however i fear to suggest that doing every single past paper still didnt help for OCR A paper 1
grade boundaries are likely to reflect this. Good luck in preparation for the next one!
For question 13, why is statement 2 not correct when the diffraction grating is on the earth, Do the atoms in the atmosphere not absorb any photoenergy at all?
excellent question, for physics we assume not, in practice there would be some however not a significant amount and also we can compare the solar specturm to other stars and differentiate what's from the atmosphere and what's from the sun. The amount absorbed by the atmosphere was pretty insignificant if I remember correctly.
for question 3 although electricity isn't tested work, energy and power is isn't it? and the question only uses the power and not the p.d. given in the question
It is not for OCR A. I have no idea what that question is doing in the specimen paper.
@@zhelyo_physics it is, it has the same paper code at the top as OCR A (H556) instead of the OCR B code (H557). And it literally says "physics A" at the top
OCR B doesn't have a paper called "Modelling physics"
@@stanb5840 Hi, I meant electricity is not tested for OCR A on Paper 1 normally
ITS OVER
how is question 15 1 mark lol.. in 2021 the 6 marker was asking this
I agree.
my goat
wohoo, thanks
Ngl this is hopeless for me I’m beyond cooked
It's so over