Cambridge HSC MX1 Textbook Marathon/Q&A (1 Viewer)

InteGrand · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Oops, my bad on the wording. I meant what to use as n and r.

Basically, two clumsy attempts whilst being tired weren't correct.

30C3.(3/5)^3.(2/5)^27
+30C4.(3/5)^4.(2/5)^26
+30C5.(3/5)^5.(2/5)^25 which I realised was obviously wrong but

5C3(3/5)^3 (2/5)^2 + 5C4 (3/5)^4 (2/5)^1 + 5C5 (3/5)^5 (2/5)^0. still didn't work

For the first two days chosen to be fine, the probability is: (prob. a day is fine)^2 = (12/30)^2

This Q. is like a coin toss Q. basically.

(We are assuming for this Q. that the weather of the days are independent of each other (which is probably not a valid assumption, but oh well).)

leehuan · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Oh right I was staring at part b) though I did get stuck on both

The answers were a) 0.0124, b) 0.7102 (at 4d.p.)
_______________________________

But I'm feeling really dumb right now. I still can't see anything. Might need a more specific sketchout...

For a) If I assume the same day can get chosen twice I used (2/5)^2(3/5)^3 but that just gives 0.03456.

I don't know what I'm doing

InteGrand · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Oh right I was staring at part b) though I did get stuck on both

The answers were a) 0.0124, b) 0.7102 (at 4d.p.)
_______________________________

But I'm feeling really dumb right now. I still can't see anything. Might need a more specific sketchout...

For a) If I assume the same day can get chosen twice I used (2/5)^2(3/5)^3 but that just gives 0.03456.

I don't know what I'm doing

I found how they got their answer to part a): they did (12/30)^5 (but this should be the probability that all days are fine, not the probability that the first two days are fine).

davidgoes4wce · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Oops, my bad on the wording. I meant what to use as n and r.

Basically, two clumsy attempts whilst being tired weren't correct.

30C3.(3/5)^3.(2/5)^27
+30C4.(3/5)^4.(2/5)^26
+30C5.(3/5)^5.(2/5)^25 which I realised was obviously wrong but

5C3(3/5)^3 (2/5)^2 + 5C4 (3/5)^4 (2/5)^1 + 5C5 (3/5)^5 (2/5)^0. still didn't work

$Cambridge Ex 10J Q18. During winter it rains on average 18 out of 30 days. Five winter days are selected at random. Find, correct to four decimal places, the probability that:$

$a) The first 2 days chosen will be fine and the remainder wet$

$b) More rainy days than fine days have been chosen$

$I reckon part a) and b) of the answers is wrong. Turns out to be 0.03456, obviously conflicting with the answers cambridge provided$

$a) P(2 fine then 3 wet) = (12/30)^{2}.(18/30)^{3}= 0.03456$

$(b)$

davidgoes4wce · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Reckon they did a mistake in the book for both parts a) and b)

InteGrand · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
$Cambridge Ex 10J Q18. During winter it rains on average 18 out of 30 days. Five winter days are selected at random. Find, correct to four decimal places, the probability that:$

$a) The first 2 days chosen will be fine and the remainder wet$

$b) More rainy days than fine days have been chosen$

$I reckon part a) and b) of the answers is wrong. Turns out to be 0.03456, obviously conflicting with the answers cambridge provided$

$a) P(2 wet then 3 fine) = (12/30)^{2}.(18/30)^{3}= 0.03456$

$(b)$

In leehuan's version of the question (the one he typed up some posts ago), part a) just asked for the probability that the first two days are wet, which is different to your part a), which also requires the remaining days to be fine. Is leehuan's edition of the textbook different to yours?

leehuan · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Facepalm... I didn't type a bit of the question...

InteGrand · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Facepalm... I didn't type a bit of the question...

Ah OK, that would explain the difference. Though if davidgoes4wce's version of the Q. is correct, then yes, the answers had a mistake, they did the probability all days were fine. This mistake could be added to the textbook mistakes thread if anyone wants to add it in.

leehuan · Apr 16, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
Ah OK, that would explain the difference. Though if davidgoes4wce's version of the Q. is correct, then yes, the answers had a mistake, they did the probability all days were fine. This mistake could be added to the textbook mistakes thread if anyone wants to add it in.

@david Please.

This disheartened my morning lol.

davidgoes4wce · Apr 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

$Cambridge Exercise 3E Q4 (a)$

$$ A stone is dropped from a lookout 500 metres above the valley floor. Take g $ =10 m/s^{2}, $ ignore air resistance, take downwards as positive, and use the lookout as the origin of displacement- the equation of motion is then $ \ddot{x}=10$

$(a) Replace $ \ \ddot{x} \ by \frac{d}{dx} (\frac{1}{2} v^2) $ and show that $ v^{2}=20x. $ Hence find the impact speed.$

$$ (b) \ Explain why, during the fall, $ v=\sqrt{20x} $ rather than $ v=-\sqrt{20x}$

$(c) Integrate to find the displacement-time function, and find how long it takes to fall.$

$Ans a) 100 m/s b) Downwards is positive, so while the stone is falling, its velocity is positive c) $ x=5t^{2}, $ 10 seconds$

Just had a bit of trouble with part (a)

davidgoes4wce · Apr 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
$Cambridge Exercise 3E Q4 (a)$

$$ A stone is dropped from a lookout 500 metres above the valley floor. Take g $ =10 m/s^{2}, $ ignore air resistance, take downwards as positive, and use the lookout as the origin of displacement- the equation of motion is then $ \ddot{x}=10$

$(a) Replace $ \ \ddot{x} \ by \frac{d}{dx} (\frac{1}{2} v^2) $ and show that $ v^{2}=20x. $ Hence find the impact speed.$

$$ (b) \ Explain why, during the fall, $ v=\sqrt{20x} $ rather than $ v=-\sqrt{20x}$

$(c) Integrate to find the displacement-time function, and find how long it takes to fall.$

$Ans a) 100 m/s b) Downwards is positive, so while the stone is falling, its velocity is positive c) $ x=5t^{2}, $ 10 seconds$

Just had a bit of trouble with part (a)

I have nothing else to do on a Monday night and have no social life, so I decided to answer my own question.

hedgehog_7 · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

The amplitude of a particle moving in simple harmonic motion is 5 metres, and its acceleration when 2 metres from its mean position is 4 m/s^2. Find the speed of the particle at the mean position and when it is 4 m from the mean position.

do i form an equation of: x double dot = -n^2 (x - 2) ??

which becomes 4 = - n^2 (x-2) but then i dont know how to solve for n

or is the equation just 4 = -n^2 x 2

but then how do i get rid of the negative to get a value for n.

yeah i know really easy question, i just havent grasped the concept of shifting away from the centre. any reply will help heaps thanks!

InteGrand · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

hedgehog_7 said:
The amplitude of a particle moving in simple harmonic motion is 5 metres, and its acceleration when 2 metres from its mean position is 4 m/s^2. Find the speed of the particle at the mean position and when it is 4 m from the mean position.

do i form an equation of: x double dot = -n^2 (x - 2) ??

which becomes 4 = - n^2 (x-2) but then i dont know how to solve for n

or is the equation just 4 = -n^2 x 2

but then how do i get rid of the negative to get a value for n.

yeah i know really easy question, i just havent grasped the concept of shifting away from the centre. any reply will help heaps thanks!

We can set the mean position to be x = 0.

So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).

(If we have centre of motion (mean position) x₀ instead, the formula becomes v^2 = n^2 (A^2 - (x - x₀)^2).)

We already know A = 5 (given), so v^2 = n^2 (25 - x^2).

Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x₀ have opposite signs at all times.)

So -4 = -n^2 * 4, so n = 1.

So v^2 = 25 - x^2.

Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.

hedgehog_7 · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
We can set the mean position to be x = 0.

So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).

(If we have centre of motion (mean position) x₀ instead, the formula becomes v^2 = n^2 (A^2 - (x - x₀)^2).)

We already know A = 5 (given), so v^2 = n^2 (25 - x^2).

Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x₀ have opposite signs at all times.)

So -4 = -n^2 * 4, so n = 1.

So v^2 = 25 - x^2.

Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.

question, why is x double dot equal to -4 and not just 4? because isnt the acceleration just 4m/s^2. the problem i had with that was then since

x double dot = - n^2 x

is the x (the distance for the mean position) can it be equal to 2 or -2 ? since it can be above or below the mean position?? can i take it as x = -2, so when i sub it into x double dot, i can actually solve for n?

InteGrand · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

hedgehog_7 said:
question, why is x double dot equal to -4 and not just 4?

As I said, in simple harmonic motion, the acceleration and displacement from centre of motion are always opposite signs (this is because the acceleration is always trying to push the object back to its centre of motion, as you can visualise by imagining a pendulum for instance). So when x = 4 (displacement from centre of motion), the acceleration must be negative.

InteGrand · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

hedgehog_7 said:
question, why is x double dot equal to -4 and not just 4? because isnt the acceleration just 4m/s^2. the problem i had with that was then since

x double dot = - n^2 x

is the x (the distance for the mean position) can it be equal to 2 or -2 ? since it can be above or below the mean position?? can i take it as x = -2, so when i sub it into x double dot, i can actually solve for n?

If you take x to be negative, the acceleration you plug in needs to be positive.

And when they said the acceleration is 4 m s^-2, that is a bit misleading, since what they really were referring to was the magnitude of the acceleration. But as I said, the acceleration and displacement have opposite signs, so you can deduce everything from that to solve for n.

hedgehog_7 · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
If you take x to be negative, the acceleration you plug in needs to be positive.

And when they said the acceleration is 4 m s^-2, that is a bit misleading, since what they really were referring to was the magnitude of the acceleration. But as I said, the acceleration and displacement have opposite signs, so you can deduce everything from that to solve for n.

oh ok thanks for clarifying, so when it states that the displacement from the mean position is 2, i CAN take it as x = 2 or x = -2? considering above and below respecitvely?

InteGrand · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

hedgehog_7 said:
oh ok thanks for clarifying, so when it states that the displacement from the mean position is 2, i CAN take it as x = 2 or x = -2? considering above and below respecitvely?

Yeah. You only need to take one of them to find n.

si2136 · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

How common is Volumes in the HSC Exams? I rarely see them

Halp Please

http://i.imgur.com/kzIHB2w.jpg

Paradoxica · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

ew imperial

yuck

Cambridge HSC MX1 Textbook Marathon/Q&A (1 Viewer)

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