Post by: DrDusk on September 06, 2019, 01:11:28 am
I've noticed with the new syllabus being rolled out that students have been expressing how questions are tough, especially because they have no resources to study the harder calculation/mathematical type questions. This is why I've created this thread.
I aim in this thread to create a collection of HSC level calculation questions for the new syllabus, especially as there is a LOT more emphasis on the Mathematical aspect of it. I will post questions on here regularly so hopefully at least someone finds them useful. All these questions will be of HSC level, so anyone doing HSC should find them as good practice.
I wish all the best for student's who are and will be taking Physics in high school, and also hope that this thread provides some good questions for you to train your mind with.
Note: I will not be focusing on the essay questions.
You will also need an account to see the questions...
Post by: DrDusk on September 06, 2019, 01:12:16 am
This one is quite easy. Make sure to set your working and equations out clearly.
Post by: Coolmate on September 06, 2019, 07:48:20 am
For the first question is this projectile motion? ???
Thanks again for the thread!
Coolmate 8)
Post by: DrDusk on September 06, 2019, 02:07:33 pm
Hi DrDusk, this is a great thread well done! I'm excited to start answering questions! ;D
For the first question is this projectile motion? ???
Thanks again for the thread!
Coolmate 8)
Very much projectile motion!
A question using using a similar 'type' of approach has been asked in the HSC before, so if you can do this, you can do any question they can ask.
Post by: Coolmate on September 06, 2019, 03:12:25 pm
=1m
Post by: DrDusk on September 06, 2019, 03:39:41 pm
Would you use tan() and sub it in as: 100/100tan(45)?Sorry, could you please elaborate? I don't understand what you mean
=1m
Post by: fun_jirachi on September 06, 2019, 05:04:22 pm
Spoiler
Post by: DrDusk on September 06, 2019, 06:17:08 pm
My solution in the spoiler: (which is hopefully correct :) )Spoiler
Well done! Looks like your all set for Projectile Motion
Post by: DrDusk on September 06, 2019, 09:15:55 pm
This is a multi-part step by step question that requires you to picture the scenario and deduce what is happening. This is a common exam type question that they love asking. If you look in the sample paper provided by NESA, you can see that they like to ask questions like this where you need to use a graph.
This is not a very easy question, you will need to think. Luckily though you would only get one of these types in the actual paper.
Post by: DrDusk on September 11, 2019, 09:40:50 am
Has anyone attempted it? If your stuck I can give hints before I post the solution.
Solution will be posted today
Post by: DrDusk on September 11, 2019, 12:54:33 pm
If you haven't tried the question don't open this first.
Post by: DrDusk on September 14, 2019, 09:38:09 pm
This is a good question. You will need to think!
A very good question for the new syllabus!
Let me add just in case, you may make the assumption that the effect of gravity on the electron is negligible
Also typo. The time spent between the plates should be 2.6 times 10^{-9} in the 2nd last line!
Post by: Coolmate on October 03, 2019, 08:51:16 pm
Would you happen to have any practice questions I could attempt on Projectile Motion, Circular Motion, and Keplers Laws? I am trying to teach myself before the term starts :D
Thanks in advance! :)
Coolmate 8)
Post by: DrDusk on October 03, 2019, 09:01:04 pm
Hey DrDusk! :DSure thing. I'll come up with some =)
Would you happen to have any practice questions I could attempt on Projectile Motion, Circular Motion, and Keplers Laws? I am trying to teach myself before the term starts :D
Thanks in advance! :)
Coolmate 8)
I was actually waiting for this so I know what kind of questions people are looking for
fyi my last question with the electron is a projectile motion one as well ;)
Post by: Coolmate on October 03, 2019, 09:07:43 pm
Sure thing. I'll come up with some =)
I was actually waiting for this so I know what kind of questions people are looking for
fyi my last question with the electron is a projectile motion one as well ;)
Awesome! Thanks DrDusk, I really appreciate this ;D
Cheers,
Coolmate 8)
btw: I may not be correct in some questions as I am still learning 8)
Post by: THSCStudyOnly on October 03, 2019, 10:38:16 pm
Question 3:
This is a good question. You will need to think!
A very good question for the new syllabus!
Let me add just in case, you may make the assumption that the effect of gravity on the electron is negligible
Also typo. The time spent between the plates should be 2.6 times 10^{-9} in the 2nd last line!
Is it 0.10699m = 10.7cm? I'll give working if it's correct (tbh i'm not sure, I am bound to make a silly mistake somewhere even if I have the process right)
Post by: fun_jirachi on October 04, 2019, 11:11:51 am
Question 3:
-snip-
Is it 0.10699m = 10.7cm? I'll give working if it's correct (tbh i'm not sure, I am bound to make a silly mistake somewhere even if I have the process right)
Thanks for bumping this thread, was going to answer Q2 but totally forgot about all it! Now there's a Q3 as well :)
I got a different answer for Q3, but will likewise post if I'm correct? I'm not too sure of myself because of the wordiness of the question, spent quite a bit of time just trying to understand what the question was asking :)
My answer
7.2cm
Post by: THSCStudyOnly on October 04, 2019, 11:15:34 am
Post by: DrDusk on October 04, 2019, 04:06:04 pm
Thanks for bumping this thread, was going to answer Q2 but totally forgot about all it! Now there's a Q3 as well :)
I got a different answer for Q3, but will likewise post if I'm correct? I'm not too sure of myself because of the wordiness of the question, spent quite a bit of time just trying to understand what the question was asking :)My answer7.2cm
Well done!
This is correct
Post by: DrDusk on October 04, 2019, 04:07:26 pm
Is it 0.10699m = 10.7cm? I'll give working if it's correct (tbh i'm not sure, I am bound to make a silly mistake somewhere even if I have the process right)
Could you please post your working out so I can have a look at it
Post by: DrDusk on October 04, 2019, 04:56:28 pm
They wont always provide a diagram. Sometimes there really is a lot of info so you gotta learn to stay calm and work your way through it.
A question like this can definitely appear in a trial or hsc exam.
Post by: fun_jirachi on October 04, 2019, 05:04:52 pm
Question 3 answer.
They wont always provide a diagram. Sometimes there really is a lot of info so you gotta learn to stay calm and work your way through it.
A question like this can definitely appear in a trial or hsc exam.
Yeah, that's what I did. I think a lot of the big words got me confused, and it took me a while to realise what the question was actually asking :) Thanks for the question though, definitely challenging! Hope there's more to come :)
Post by: THSCStudyOnly on October 04, 2019, 05:09:52 pm
I redid it a different way and still got it wrong because I used negative acceleration....
I used qE=ma and solved for acceleration in the first plate (after using V=Ed to find E) then I used it to find the v (using kinematic equations) coming into the second set of plates. After this I found the acceleration in the second set of plates (using same thing I did for the above ones). Then I used s=ut+1/2at^2 to plug those values in, but I used negative acceleration (because a=qE/m gave me a negative value which I forgot to times by negative again because the electron in moving down). My bad.
Post by: DrDusk on October 04, 2019, 05:25:47 pm
Well done for getting it!That's really great as well. You would still get 2/3 for it so good job!
I redid it a different way and still got it wrong because I used negative acceleration....
I used qE=ma and solved for acceleration in the first plate (after using V=Ed to find E) then I used it to find the v (using kinematic equations) coming into the second set of plates. After this I found the acceleration in the second set of plates (using same thing I did for the above ones). Then I used s=ut+1/2at^2 to plug those values in, but I used negative acceleration (because a=qE/m gave me a negative value which I forgot to times by negative again because the electron in moving down). My bad.
I hope this teaches you to always picture the scenario in your head when dealing with situations like this. I would recommend drawing the diagram and labeling the vectors for acceleration. This will reduce the chance of you making that silly mistake again
Post by: THSCStudyOnly on October 04, 2019, 05:29:47 pm
Post by: DrDusk on October 13, 2019, 04:20:20 pm
This is definitely a kind of question you can be asked because the Math itself is simple but requires you to make an observation that may be difficult to spot =)
Definitely more of a challenging uniform circular motion/Magnetism question!
Post by: fun_jirachi on October 13, 2019, 07:19:10 pm
My answers
Hope these are right! :)
i) Into the page. (Used left-hand grip rule)
ii)
)
i) Into the page. (Used left-hand grip rule)
ii)
Post by: DrDusk on October 13, 2019, 07:22:07 pm
Wowie well done! Your smashing them. However you may want to re-consider your part (i) ;)My answersHope these are right! :)
i) Into the page. (Used left-hand grip rule)
ii)
Post by: frog0101 on October 21, 2019, 05:22:45 pm
Question 3 answer.
They wont always provide a diagram. Sometimes there really is a lot of info so you gotta learn to stay calm and work your way through it.
A question like this can definitely appear in a trial or hsc exam.
I thought that since the electron is being accelerated horizontally by the electron gun,
Not sure that the way I am visualising this is correct (horizontal acceleration and then vertical acceleration), but wouldn't only the horizontal parallel plates accelerate the electron in the vertical direction?
Thanks
Post by: THSCStudyOnly on October 26, 2019, 08:19:16 pm
Spoiler
i) Out of page (using rhpr but flipped hand because electrons)
ii) Using mass spectomrtry equation
i.e. qvB=mv2/r
r = mv/qB
rearranging v = rqB/m .............. (1)
Now Consider:
K = (1/2)mv2
rearrange to make v subject: v = sqrt(2K/m) ............... (2)
Equating (1) and (2)
sqrt(2K/m) = rqB/m
2K/m = (r2q2B2)/m2
Just rearrange to make B2 subject
B2 > 2Km/q2r2
B > sqrt((2Km/q2r2))
(Now: q=e and r=d)
B > sqrt((2Km/e2d2))
There may be lots of errors because it is too messy to read with this crap formatting.
ii) Using mass spectomrtry equation
i.e. qvB=mv2/r
r = mv/qB
rearranging v = rqB/m .............. (1)
Now Consider:
K = (1/2)mv2
rearrange to make v subject: v = sqrt(2K/m) ............... (2)
Equating (1) and (2)
sqrt(2K/m) = rqB/m
2K/m = (r2q2B2)/m2
Just rearrange to make B2 subject
B2 > 2Km/q2r2
B > sqrt((2Km/q2r2))
(Now: q=e and r=d)
B > sqrt((2Km/e2d2))
There may be lots of errors because it is too messy to read with this crap formatting.
Post by: DrDusk on October 26, 2019, 09:42:37 pm
My solution, I think it may be easier to see, but bit longer to do :P ;DWell done on this attempt!Spoileri) Out of page (using rhpr but flipped hand because electrons)
ii) Using mass spectomrtry equation
i.e. qvB=mv2/r
r = mv/qB
rearranging v = rqB/m .............. (1)
Now Consider:
K = (1/2)mv2
rearrange to make v subject: v = sqrt(2K/m) ............... (2)
Equating (1) and (2)
sqrt(2K/m) = rqB/m
2K/m = (r2q2B2)/m2
Just rearrange to make B2 subject
B2 > 2Km/qr
B > sqrt((2Km/qr))
(Now: q=e and r=d)
B > sqrt((2Km/e2d2))
There may be lots of errors because it is too messy to read with this crap formatting.
Well done on getting this far! You've almost got it
Post by: DrDusk on November 27, 2019, 09:55:39 pm
Now that the HSC has happened we know what level of questions to expect.
With my finals finishing in the next 2/3 weeks. I will be releasing a whole papers worth of questions!
Coincidentally a question that appeared in the HSC was one of the questions that I uploaded here! ;)
Post by: Coolmate on November 27, 2019, 10:55:58 pm
UPDATE:
Now that the HSC has happened we know what level of questions to expect.
With my finals finishing in the next 2/3 weeks. I will be releasing a whole papers worth of questions!
Coincidentally a question that appeared in the HSC was one of the questions that I uploaded here! ;)
Thanks DrDusk! :D
This is very exciting, I can't wait to complete some questions! ;D
Coolmate 8)
Post by: mani.s_ on December 04, 2019, 10:30:22 pm
Question 1:What does it mean by far the ball has dropped when it strikes the hill???
This one is quite easy. Make sure to set your working and equations out clearly.
Post by: DrDusk on December 04, 2019, 10:32:34 pm
What does it mean by far the ball has dropped when it strikes the hill???It means what is it's 'y' displacement from where it was launched.
Post by: mani.s_ on December 04, 2019, 10:34:30 pm
It means what is it's 'y' displacement from where it was launched.So its height at the marked point? or are you talking about its range for the launch position?
Post by: DrDusk on December 04, 2019, 10:39:30 pm
So its height at the marked point? or are you talking about its range for the launch position?It's how far it as fallen so (the height of the building) - (it's height at that point)
Post by: mani.s_ on December 04, 2019, 10:42:01 pm
It's how far it as fallen so (the height of the building) - (it's height at that point)ohhhh ok. Thank you so much!!!
Post by: mani.s_ on February 25, 2020, 10:06:10 am
Question 3:Hi, for this question I was wondering why
This is a good question. You will need to think!
A very good question for the new syllabus!
Let me add just in case, you may make the assumption that the effect of gravity on the electron is negligible
Also typo. The time spent between the plates should be 2.6 times 10^{-9} in the 2nd last line!
Post by: JonTheToot on January 22, 2021, 06:24:34 pm
From the wording of their question, the correct answer is B as electrons above the wire will always be pushed upwards depending on their velocity and the strength of the magnetic field. There is absolutely no way that the cathode rays (electrons) can deviate downwards from their starting position unless the current direction is reversed. I can only arrive at their solution if I use mangled English rather than the exact wording in their question.
The exam question asks "Which direction will this spot move towards if the resistance is increased?" Now the effect on the electrons is they will not deviate upwards by as much as if the current had remained at a higher amps (less current = reduced magnetic field intensity), but as I stated above they will never deviate downwards which is the given answer. So in effect what the examiners from NESA/BOSTES/BOS/(whatever they change their name to next week) should have asked is "What is the CHANGE in direction that the spot will move if the resistance is increased?" That's one lost mark for all the hard working students who knew their Module 6 but sadly could not read the minds of the boofheads at NESA.
This is in effect how the old style CR televisions worked, though the magnetic fields were supplied by 2 pairs of solenoid coils arranged to provide a horizontal and vertical field.