Author Topic: HSC Physics Marathon[NEW SYLLABUS] (Read 2166 times) Tweet Share

DrDusk · « **Reply #30 on:** October 26, 2019, 09:42:37 pm »

Quote from: THSCStudyOnly on October 26, 2019, 08:19:16 pm

My solution, I think it may be easier to see, but bit longer to do
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/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 this attempt!

$\text{Just a few points I must make. You went from }$
$\dfrac{2K}{m} = \dfrac{r^2q^2B^2}{m^2}\hspace{2mm}\text{to} \\ B^2 \geq \dfrac{2Km}{qr} \\ \text{However you have not justified from going = to >} \\ \text{And also I think the qr should be}\hspace{2mm}q^2r^2 \\ \text{Your definitely on the right track but your just missing one observation that allows you to justify the change from = to >} \\ \text{Draw a picture of the electrons path and think about what must happen if its circular path does not intsersect the plate}$

Well done on getting this far! You've almost got it

DrDusk · « **Reply #31 on:** November 27, 2019, 09:55:39 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!

Coolmate · « **Reply #32 on:** November 27, 2019, 10:55:58 pm »

Quote from: DrDusk on November 27, 2019, 09:55:39 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!

This is very exciting, I can't wait to complete some questions!

Coolmate

mani.s_ · « **Reply #33 on:** December 04, 2019, 10:30:22 pm »

Quote from: DrDusk on September 06, 2019, 01:12:16 am

Question 1:

This one is quite easy. Make sure to set your working and equations out clearly.

What does it mean by far the ball has dropped when it strikes the hill???

DrDusk · « **Reply #34 on:** December 04, 2019, 10:32:34 pm »

Quote from: mani.s_ on December 04, 2019, 10:30:22 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.

mani.s_ · « **Reply #35 on:** December 04, 2019, 10:34:30 pm »

Quote from: DrDusk on December 04, 2019, 10:32:34 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?

DrDusk · « **Reply #36 on:** December 04, 2019, 10:39:30 pm »

Quote from: mani.s_ on December 04, 2019, 10:34:30 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)

mani.s_ · « **Reply #37 on:** December 04, 2019, 10:42:01 pm »

Quote from: DrDusk on December 04, 2019, 10:39:30 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!!!

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