# Thread: Cambridge HSC MX1 Textbook Marathon/Q&A

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

Originally Posted by davidgoes4wce
$Cambridge Exercise 3E Q4 (a)$

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

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

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

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

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

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.

2. ## 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!

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

Originally Posted by hedgehog_7
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.

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

Originally Posted by InteGrand
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?

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

Originally Posted by hedgehog_7
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.

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

Originally Posted by hedgehog_7
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.

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

Originally Posted by InteGrand
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?

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

Originally Posted by hedgehog_7
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.

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

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

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

ew imperial

yuck

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

Originally Posted by si2136
How common is Volumes in the HSC Exams? I rarely see them o.O

http://i.imgur.com/kzIHB2w.jpg
Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.

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

Originally Posted by InteGrand
Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.
Yes, the volume. And are feet and foot the same (dumb question)?

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

Originally Posted by si2136
Yes, the volume. And are feet and foot the same (dumb question)?
Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.

This means viewing everything with origin A, the equation of the inner parabola is y2 = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).

We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).

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

Originally Posted by InteGrand
Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.

This means viewing everything with origin A, the equation of the inner parabola is y2 = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).

We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).
Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge

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

Originally Posted by si2136
Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge
I doubt they'd ask it in the HSC without providing a conversion rate though, so don't worry.

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

Originally Posted by InteGrand
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.
Is there something wrong with the solution for that question because it shows the speed to be $v=5 \sqrt{2} \ and \ 3 \sqrt{2}$

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

Originally Posted by davidgoes4wce
Is there something wrong with the solution for that question because it shows the speed to be $v=5 \sqrt{2} \ and \ 3 \sqrt{2}$
Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).

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

Originally Posted by InteGrand
Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).
I fixed up your mistake and redid the solution and was able to get the speeds.

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

Hey guys,

Sorry I'm a bit new to using this. :P

Could I get some help with Exercise 10I question 10???

There are two round tables, one oak and one mahogany, each with five seats. In how many
ways may a group of ten people be seated?

Thanks

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

There are (5-1)! ways to arrange the 5 people in the first table then another (5-1)! ways to arrange in the other table. But for the first table, any 5 of the 10 people can be chosen and the remaining 5 chosen for the 2nd table.
So the answer is 10C5*4!*4! = 145152

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

Cambridge EX 10 D Q 19

$Two dice are rolled. A three appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than seven.$

Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).

Am I missing something?

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

Originally Posted by davidgoes4wce
Cambridge EX 10 D Q 19

$\\ Two dice are rolled. A three appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than seven.$

Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).

Am I missing something?
$The favourable outcomes are as follows: (3,5) , (3,6), (5,3), (6,3) (Thus there are four favourable outcomes.) The number of possible outcomes (where a three appears on at least one of the dice) is 6^2 - 5^2 = 11.$
$\\ Hence the probability is \frac{4}{11}.$

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

Can understand it a bit better now, my thinking initially was 4/36 but can see why its 4/11, since the total possible combinations with a 3 are: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3) and (6,3).

Only 4 of those have combined value above 8 or more.

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

Ex 3A Q 9 (h)
$A particle moves according to x=10 \ cos \ \frac{\pi \ t}{12}, in units of metres and seconds$

$Use the graph and table of values to find when the particle is more than 15 metres from its initial position$

$I got the following graph :$

Should the solution be between the range?
8 < t < 16, 32 < t < 40, 54 < t< 60 ?

8 < t < 16

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

Originally Posted by davidgoes4wce
Ex 3A Q 9 (h)
$A particle moves according to x=10 \ cos \ \frac{\pi \ t}{12}, in units of metres and seconds$

$Use the graph and table of values to find when the particle is more than 15 metres from its initial position$

$I got the following graph :$

Should the solution be between the range?
8 < t < 16, 32 < t < 40, 54 < t< 60 ?

8 < t < 16
Since they only wanted you to consider when you could use the graph AND the table of values...

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