All solutions here are SUGGESTED. Casey will hold no liability for any errors. Comments are entirely personal opinions.
1.A
2.D
3.C
4.A
5.C
6.A
7.D
8.C
9.A
10.C
11.B
12.A
13.D
14.A
15.A
16.A
17.C
18.A/ C ;I am not sure of Q18 though. Cos in reality, there is a phase change of pi on reflection, in which case, the answer shd be C. But phase change is not in the syllabus, then the answer should be A. so I dun know what Cambridge wants..:-(
19.C
20.B
21.C
22.D
23.B
24.B
25.A
26.B
27.B
28.B
29.D
30.B
31.B
32.D
33.B
34.C
35.C
36.C
37.C
38.A
39.D
40.B
Note to all: Casey will not respond to most of the comments as he is busy. You may contact him by SMS at +65 9474 5005 if you have a burning question.
Feel free to explain the answers, if you are confident. Many thanks.
gg.com this is bloody hard
Hi sorry, may I know is it possible to get the H1 Physics Paper 1 answers up too? I know it will only benefit a small group of us but it’ll be greatly appreciated. Thank you! 🙂
Hi! May I ask if you guys be posting the ans for H1 physics paper 1?
Hi, may I know where is the answer? What is the rough (just estimation will do) mark to get A, B and C based on bell curve?
Tq
Can you upload H1 Paper 1 answer also? Thanks A Lot!
I LOST THE FIRST MARK wasted:’). It’s mol. Not charge. You can search it up
Charge is Current*Time, time already has second in it so this is why it’s not the answer.
I think I can fail this paper 🙁
I think I can get like less than 20?????? Am I the only one who finds it tough cause my friends are saying that they can get full marks…
It is difficult but I dont think you will fail!
I got about 28 based on preliminary answers from classmates…that’s bad 🙁
I got 35!
same i got 35 too 🙂
Anyone wants to post their ans? haha
Can you do for h1 physics too?
GGWP CAMBRIDGEEEEEE MUACKSSSSS
Anyone got a different answer for Q28?
28 was tested in one of the past year alevels so it should be correct
Can do for h1 physics paper 1 too?
How much u guys get?
29, including a for qn 18
Why is 36 C and not B? Doesn’t it have to be the same direction?
Cos the momentum shd occur in the same direction as the velocity.. Which is horizontal.
But diffraction means it spreads vertically..
– Casey
Can apply uncertainty principle with different direction?
hi, isn’t it A though? I did a similar question and the answer was A
Ya I think A is correct but I put B because I thought the arrow before the screen means it has not touched the screen. But A should be the answer because it is definitely correct.
imo I thought x should be at the slit and p at the screen eh but no such option HAHA.
must be simultaneous measurement
Answer is A.
Parallel to the slit, the deltaX is small and limited to the slit width, whereas perp to the slit, the position can vary to a large extent, as there are no obstacles so deltaX is significant. As such momentum uncertainty occurs parallel to the slit, as noticed by the greater spreading in the interference pattern. While the momentum uncertainty occurring perp to the slit would be minute.
were the answers stolen?
No. We aren’t trying to compete to be first, to be honest. I hope you understand that we all have personal commitments and realise that this is not our obligation, nor is it your entitlement. We do not derive any marginal utility doing this, only losing time on our side.
And this set of answers, I helped Casey to upload them. So there is lag time on my side, as I have a full-time job and was held up too. So my bad for the delay.
Also, the reason why your comments were not shown was because you linked a link and copy-pasted it a few times, which is why it picked you up as spam. I “un-spammed” you. We set the comment setting rather high, to remove profanities and external links actually. I hope you understand. Thanks
To all those tutors working at The Culture, thank you so much for putting in the effort to help us with the answers and guide us through our As even though we do not attend classes at your tuition centre. A really big big thank you ;):)
hi, its a pleasure. Most of us are not full-time tutors though we only teach in the newtonapple. The purpose of this was to facilitate discussion, and previous feedbacks from ex-students were that they were able to learn from the mistakes made by us and the students of the year. So it actually helps learning for everybody. This site isn’t really about tuition actually, as we do shall undergraduate & post-graduate stuffs here too. Some students had asked us regarding postgraduate or undergraduate studies, and we are more than willing to share a bit. Lastly, we love your own field of studies and are glad we can help ourselves and you learn from the mistakes. 🙂
Mr Teng why is question 6 A and not C?
This is Casey.. Cos they asked for additional weight.. And the barge was always floating, whether it was over the bridge or not. So, no additional force.. Hope that helps..:-)
Anyone getting >30?
31 only..
I got 31 also :/ But it was expected to be difficult cos paper 3 was quite manageable. Don’t be disheartened! Bellcurve won’t deviate so much I guess? Enjoy guys! It’s the end of As (for most)!
37 I think
35 😀
I got 40 if 36 is A!
q18 is C CONFIRM
Mr Teng why is the answer for Question 4 A and not B?
This is Casey.. Cos the mass decreases continuously . Since propelling force is constant, the acc increases.
Guys how much did you and most of your friends get out of 40 for this paper?
Mostly around 30
Upper tier jc?
This guy called “student” has been trolling on all forums will all the fake bellcurve and stuff so avoid seeking comments from non-professionals..
hey but i only got around 29 for this paper so how is that trolling?
Why is 33 B?
– Casey
Thanks
I feel that the ans for 15 should be B. Why is B wrong?
Cos
.. So the increase in the ePE cannot be equal when the extension doubles..
but the question says from the equilibrium position tho
Ya, so you notice that the change in ePE is 1.9 J from highest point to eqm position.. So from highest point to lowest point, the change in ePE cannot be merely doubled cos of the
u try adding up all the powers together.
by COE, total energy should remain same.
if u do option B, +1.9-3.7 is -1.8 NOT 1.8 so cannot
gpe and epe are given as change from centre position so its not COE
if you say its the change from the top position then i feel A is the correct answer
I agree with him, isn’t it from equilibrium? Then it should be positive change of the same magnitude
I think the spring is not at its natural length at equilibrium point because the mass itself will cause an extension
Lets say at eqm extension is e, and amplitude of SHM is a.
EPE @
top: 0.5k(e-a)^2
eqm: 0.5ke^2
btm: 0.5k(e+a)^2
if you let eqm EPE =0 (aka deducting 0.5ke^2 from top, eqm and btm),
top: 0.5k(e-a)^2 – 0.5ke^2 = kea – 0.5ka^2
eqm: 0
btm: 0.5k(e+a)^2 – 0.5ke^2 = kea + 0.5ka^2
as you can see, EPE@btm is not equal to negative of the EPE@top
hope this helps convince you 🙂
How can 24 be B? I thought it was similar to one of the tys qn
Greater V means greater force and hence acceleration. So vertical speed increases
The tys qns is for magnetic field
Why isn’t 29 answer B?
Cos X is not ohmic, so it’s resistance will change with V.. So need to refer to graph for its specific R at the particular V and I
Why can’t I find the individual resistance of X and Y, then I arrange them in parallel.Then from there I calculated the effective resistance of the circuit. And the emf/I should give the effective resistance.
can you upload h1 physics p1 as well
Sorry.. I dun have the paper.. My son is Also taking his A levels this year and he does H2..
is there a way i can show you the questions? i cannot seem to find answers from anywhere else
you can try whatsapp-ing him and ask if he is free and send the images. His number is above. 🙂
can you upload h1 physics p1 as well
Would recommend looking at
https://books.google.com.sg/books?id=WcAKAAAAQBAJ&pg=PA999&lpg=PA999&dq=uncertainty+electron+diffraction+momentum&source=bl&ots=YLKeRCNa4O&sig=zM7PZDzwX40PV3mR3l33pN-GOH8&hl=en&sa=X&ved=0ahUKEwjmhJzRtLDJAhWOWY4KHS-xCaIQ6AEIRDAG#v=onepage&q&f=false
before typing answer for qn36. Should be A according to this.
Oh.. I din look at the link.. But maybe you are right.. Thanks for the input.. Cheers..:-)
Yup. I believe the answer is A. I have seen this question in Cambridge International A levels before.
If A is the answer then what about B? Aren’t they similar but just along different points of the path of electron?
B cannot be accepted as once the electron arrived at the screen, there will not be any associated uncertainty to it. The reason is simply because any system that is being observed will cause the wave function of the electron to collapse. So in this case, the uncertainty associated with p and x should be along the slit width and not at the screen since the patterns on the screen can be observed.
I thought the arrow is before the screen means it has not reached the screen yet lol
Is 27 a bad score?
No! Enough for an A
Why 36 not D?
Must be simultaneous measurement and same direction
Please consider https://books.google.com.sg/books?id=WcAKAAAAQBAJ&pg=PA999&lpg=PA999&dq=uncertainty+electron+diffraction+momentum&source=bl&ots=YLKeRCNa4O&sig=zM7PZDzwX40PV3mR3l33pN-GOH8&hl=en&sa=X&ved=0ahUKEwjmhJzRtLDJAhWOWY4KHS-xCaIQ6AEIRDAG#v=onepage&q&f=false for qn36! Answer should be A!
What about B?
Agreed. The momentum and postional uncertainity must be in the same direction. This is one of the assumptions of the calculation. Furthermore there needs to be a vertical momentom for diffraction to occur. If not no diffraction will occur and the light is simply a bright spot
How many marks out of 40 did you guys get for this paper?
35 yay
For qn 15, I got B bcos at the bottom most position, mg=kx, from there I calculated k=163.5. Then at the bottom position EPE=(1/2) (k) (x^2)=(1/2) (163.5) (0.15^2) I got B . Is that wrong?
But x shd be the total extension.. So from eqm downwards the total extension shd not be just 0.15m..
What is the more likely answer for 18? A or C?
Most likely it’s C as the physics concept behind it is correct.
I also feel that the answer should be A for Q36!
Why isnt the answer for question 4, C?
It is similar to this qn… http://physics.stackexchange.com/questions/91110/increase-in-velocity-by-loss-of-mass
That question is a bit diff cos there is no external driving force.. So linear momentum is conserved. As in the sand’s X speed while in the trolley and when it dropped out is the same.. So by PCoLM, the
Trolley’s speed remain constant..
Just to clarify…
for Question 5, why isn’t it B?
Taking rightwards as positive , and since both spheres have same mass,
mu1 – mu2 = mv1 + mv2
m(u1-u2) = m(v1+v2)
u1-u2=v1+v2 which is B? What’s wrong with this working ?
Also for Qn 4,
since propelling force horizontally is constant, and the water runs out at a constant rate, shouldn’t it accelerate uniformly? since water is running out at a constant rate
Constant F with decreasing m. Hence a increases. Accelerate uniformly means a constant
I thought accelerate uniformly means like increasing at 2ms-2 constantly! haha thanks though!
For elastic collision of 2 objects with same masses, they will exchange velocities. Hence velocity of v1 should actually be negative, and hence cannot use COM. Ans is C is definitely correct.
There is nothing wrong with your workings for question 5. But, it does not show that the collision above is elastic. It is merely a statement of the conservation of linear momentum. So, the best answer is C as it shows that the collision is indeed elastic (can be derived easily from conservation of KE)
ahh I was contemplating on which to put during the paper, regardless thanks for the explanation!
please upload the h1 paper also thanks!!
Hi, How many overall mark tentatively to get B or C? Roughly only. Based on bell curve. It seems my comments got deleted 🙁
B should be 60-65 range because mcq is difficult
18 should be C bc i rmb got learn before, and can be explained by N3L at the fixed point, and 2010 q21 also got reflected wave qn also (but that one on stationary waves).
my guess wuld be ard 60-70, which is the range for most papers like chemistry or biology. maybe slighly less than 70 cos tats like too xtreme. so ard 60-65??
hi. Thanks for uploading the answers. Rmb guys this mcq is just 20%, so for every 2 wrong answers u lose 1%. So if u get 30 and your friend gets 35, u only lose him by 2.5% which is not that bad. Since its over lets just take a break and rejoice! Enlisting dam soon in 4JAN tho.. Good luck everyone and all the best for your future endeavours
Very true. 🙂
Enjoy your 2 years of break!
Hi, where can I get a hold of all the question papers for H2 physics, paper 1, paper 2 and paper 3? I would like to practice them immediately as I am private candidate.
Thanks for your help
erm… im pretty sure that u can buy the yearly tys frm the Popular. shuld have all 3 papers
I need the 2015 paper for practice, where can I get it immediately?
lol. within such a short period of time. i dun tink u can get it on9, unless u have frens who r taking the A levels
I highly doubt you can find it. Maybe only paper 1 for physics since the rest are collected back. Most scripts have their own bar code for copyright reasons. So you can only have papers that were allowed to bring home like Math P1 & P2, Physics P1, Chem P1 & P3. Pretty much those that we were able to do.
Is 28 good enough for this paper
I got almost the same as you don’t worry this paper is really really hard so ya don’t think too much and enjoy your holidays
hi, what’s the A and B grade overall range for H2 physics?
A ~ 75 B~65
hi what are you guy’s estimated overall score? mine is only 66…pretty worried.
hi sir, please upload H1 physics P1 too 🙂
Is below 70 an A possible for this paper?
not possible A for physics is usually higher than chemistry ~75
I wrote C for qns 4. I was choosing between A and C the whole time. For A to work, the truck had to initially be accelerating, so the loss in mass would lead to an increase in acceleration, Fnet=ma>0. But if it was initially moving at a constant speed, where Resistive force = Propellant force, a change in mass will not increase acceleration as Fnet = ma = 0. I chose C because it says it “continues at a constant speed”. I inferred from the word ‘continued’ and assumed that it was initially at constant speed. I do hope i went wrong somewhere so that i can learn from this. By the way answer for 36 is definitely A, my senior in Cambridge checked with the physics professor already. Momentum has to be in the same direction as the delta X or this conservation law will not hold. Just because vertical velocity is 0 in the question does not mean that the vertical velocity cannot be uncertain. It can be 0 +- 5. That’s why it even diffracts, it’s because when you close the slits, delta X becomes smaller and delta P increases, hence the momentum vertically is more uncertain, which is why divergence even takes place at all. This is universally the reason for diffraction. Check out Veritasium’s video on the uncertainty principle for this! Anyway, what’s more important is that we learn something new everyday! Someone please correct me on qns 4, thanks!
For 4, because water is running out, there is a continual decrease in mass hence acceleration is increasing
But how do you even know that it was accelerating in the first place?
Doesn’t have to be accelerating at first. Even if acceleration is 0 initially it will still increase.
You have to assume whatever resistive force is constant. But they didn’t even mention any resistive force so just assume don’t have. Don’t have to think so much haha
It intuitively makes sense to me that it does not have to be initially accelerating. This is my guess for the reason. Although the equation “Fforward-Fresistance = ma” is not technically wrong. I think that this cannot be explained by such simple maths and we need a more intuitive explanation. Thus let us assume that Fresistance is just constant. Hence we can add this constant directly to Fforward. Now let’s equate Fforward = ma. Since Fforward is constant, m decreases and a increases [Which is the very simple way of doing this question without having my convoluted thoughts XD]. Initially the Fforward was used to overcome the resistance, thus a lower mass lowers the intertia of the system, thus the Fforward can now “afford to” overcome resistance and have a net rate of change of momentum due to the inertia of the body thus it accelerates non uniformly. Thus even if net force was zero initially, it is still possible to accelerate without changing the forward force. This is my guess. But good point, there may not even be resistance. Either way i think i kind of managed to explain how initial constant speed can lead to the answer A. Thanks!
Hi, why is the answer for question 11 B and not C?
It’s okay, I realised where’s my careless mistake already
How many marks out of 40 did you guys get for this paper
Arnd 36/37?
How many marks out of 40 did you guys get for this paper 3?
37
hi sir can you upload H1 physics too? im really worried because it is quite difficult
How many marks out of 40 did you guys get for this paper 1?
Why keep asking same question?
Is it possible for A grade to be arnd 70marks overall?
How many marks out of 40 did you get for this paper?
Can u put H1 physics paper 1?
Mr Casey, can you explain to me why the specific heat capacity for an ideal gas when measured at constant volume of the gas is different from that when measured at constant pressure of the gas?
Cos for constant pressure, there is a change in volume.. Which means work need to be done against intermolecular forces of attraction to separate the molecules and also on the environment in prefer to push back the atmosphere to create room for expansion.. Hence more heat is required..
– Casey
Mr Teng can you explain why the specific heat capacity for an ideal gas when measured at constant volume of the gas is different from that when measured at constant pressure of the gas?
Cos for constant pressure, there is a change in volume.. Which means work need to be done against intermolecular forces of attraction to separate the molecules and also on the environment in prefer to push back the atmosphere to create room for expansion.. Hence more heat is required..
– Casey
Hi Mr Teng,what is the origin of the energy of sound waves, is it the sound signal generator?
Mr Casey what is the origin of the energy of sound waves? Is it the sound signal generator?
Hi Mr Teng and Mr Casey, what is the origin of the energy of sound waves? Is it the sound signal generator?
Hi Mr Teng and Mr Casey what is the origin of the energy of sound waves? Is it the sound signal generator?