OCR A 2021 Paper 1 - A-level Physics Past Paper

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0:00 Multiple choice
12:30 Q16 - g from freefall - impulse
17:33 Q17 - Uncertainties - oscillating strip experiment
22:48 Q18 - SHC & SLH - internal energy
27:03 Q19 - Moments - wheelie bin
30:34 Q20 - Gravitational fields & satellites
35:00 Q21 - Spring loaded car - SHM - projectile motion
38:58 Q22 - Particle accelerator - magnetic fields
42:13 Q23 - Lumonisity & g for planets - uncertainties - Stefan's & Wien's law -
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i have watched all the past paper videos. idk what was going on in this one but the energy at times was unmatched 😂

sparks
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Thank you for making these, they're really useful ❤️
Also your sarcasm makes it so much less boring to watch

missy
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These videos are super helpful! Got my Alevel physics exam in 2 weeks and I’ve been using your videos to mark my papers so I fully understand the answers which has been really useful- thankyou!!😄

fleur
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I would just like to add a comment about qu20 as one of my 6th form students asked me about this. I calculated total energy as - 1.18 x10 ^10 J. so i am in agreement with OCR. If the satellite had net positive energy it would not be in a bound system and it would break free from the earths gravitational field. The total energy has to be negative as it is in a bound system.
In bound systems potential energy is negative!! and total energy is negative. To take it a bit further you can show that the magnitude of the potential energy is twice the magnitude of the kinetic energy (this comes from equating Newtons law of grav to mv^/r) So the total energy actually turns out to be minus the kinetic energy!!
Well thats my take on it, I hope this helps someone!

brianwardle
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I did the exact same for q17, it should be clearer that mass is changing.

thomassmith
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Question 19 - what about the vertical component of F. The handle is displaced from the wheel in the diagram but the distance is not given. The vertical component of F will also contribute to the overall turning moment. The question as it stands is not actually possible without assuming the handle is directly above the wheel.

martynwheeler
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Q11 you don't need to do all those calculations because 1AU is that similar to that between the Earth and sun so the orbital period would be the same technically. And I would assume everyone knows how many days there are in a year.

saeth
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(Q19) Why isn't the downward component of F considered as it will also contribute a clockwise moment.... Simply an oversight/bad diagram from OCR??

josephheldenhammer
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I was a physics A level teacher for 34 years and really do not like question 19b i. I am currently coaching my grandson and came across this question in his mock exam. As several people have commented, what about the vertical component of F (Fsinθ). Its line of action clearly has a perpendicular distance to the pivot point giving rise to another clockwise moment. The only way to get an answer is to assume that this moment is negligible. It is certainly a lot smaller than the moment from F cos20 because sin20 is small and the perpendicular distance is much smaller than that of Fcos20. The author of this video has clearly done the solution in line with the official answer ignoring Fsinθ.
I am appalled that an exam board could set such a question. Strong candidates could have wasted a lot of time trying to solve this impossible problem not wanting to assume that they can ignore the clockwise moment produced by Fsinθ. Weak candidates could just ignore it without knowing why and get the marks

johnbrid
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35:52 .. where does T=2pisrt(m/k) come from?

samayahone
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In Q20: the minus value means the satellite is bound within the Earth's gravitational field :)

ohmreggienius
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hi will you do the 2022 and 2023 papers?

sofiaalonss
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Can't wait for ur reaction this years paper next year 😭

jkstr
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These past paper videos were very helpful, thanks :)

RY
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For question 21 can you explain why it is 1/4 of the full oscillation please

bredli
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i got the same answer as you for q20 so i'm glad that you said the marksheme was wrong because i was so confused

honeyvirdi
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These are so helpful I appreciate you uploading these

JwNineNineNine
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For Q20, I also got 3.54x10^10J, Would examiners mark me down for this

ggx
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For 21(a): I used x=Acos(wt) with x = 0 (since the extension is 0 when it reaches the equilibrium position) and solved for t and got the same answer. Is this method valid?

AK-enib
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Hey, would it be possible to do more edexcel papers? They are so helpful

cutzaky