VCE Specialist 3/4 Question Thread!

[StupidProdigy]

How do they get to that final answer (the second step shown is 1/Z). Thanks

[The Mathemagician]

Whenever you have a complex denominator you can multiply both the numerator and the denominator by the complex conjugate, which is a method to get a real denominator.

For example if you have you can multiply by (which is essentially multiplying by 1, so it won't change the number) to get . For that particular question you will also need to use the identity .

[Escobar]

how do you do this question?

[StupidProdigy]

With the question attached, why do they ask for m/n for dependence? I can't get my head around why they put these two as a ratio..like how does that still allow for dependence..?
And I found a bit about this question on an old thread but didn't get all the info I was after. What I found was that you can simply find m and n by adding vectors a and b and equating them to the c vector, which literally takes stuff all working. Why is it suggested we try and solve it by letting c=p(a)+q(b)?? Thanks!!

[grannysmith]

2006 Exam 2 Q5 a. ii.

Any help would be appreciated

Edit: DW, figured it out. Forgot about loci.

[The Mathemagician]

how do you do this question?
(Image removed from quote.)

I would probably use Lami's Theorem for this question as it's the quickest method and the particle is in equilibrium (sum of forces on it is 0). First draw out the forces (the force at P, tension force and weight force). Then you would need to calculate the angles in between each force (you will need to use trigonometry for this)

Solution

Force Diagram

First calculate the angle a. From the information given, so a=60 degrees. Then angle p must be 150 degrees. Angle t is 90 degrees, thus angle w must be 120 degrees. The weight force, W is equal to 0.5 kg wt because the mass of the object is 0.5 kg. Lami's Theorem states that so which is 0.3 when rounded to one decimal place.

With the question attached, why do they ask for m/n for dependence? I can't get my head around why they put these two as a ratio..like how does that still allow for dependence..?
And I found a bit about this question on an old thread but didn't get all the info I was after. What I found was that you can simply find m and n by adding vectors a and b and equating them to the c vector, which literally takes stuff all working. Why is it suggested we try and solve it by letting c=p(a)+q(b)?? Thanks!!

Beacuse that's the definition of linear dependence - c is linearly dependent on a and b if you can find constants p, q such that . If you have a linearly dependent vector, any scalar multiple of it is also linearly dependent.

To give an example with numbers, the vector is linearly dependent on the vectors and as . But any scalar multiple of will also be linearly dependent, for example . But a vector that is not a scalar multiple of the first one will be linearly independent of the other two, such as . This is why you are asked for a ratio.

Also, while solving might work since the j components of a and b add to give 0, what would happen if they didn't add up to give 0? I'm not sure if they're this strict at VCE level, but saying something straight away like "For linear independence, " is mathematically incorrect, without explaining how you got there.

[Escobar]

thanks bro, can you repost the force diagram (in the solution)?

[I am a unicorn]

Hello
I'm having trouble with this multiple choice question... no idea how to do it...
Any help appreciated!
Thanks

P.s answer is C

[Splash-Tackle-Flail]

Hello
I'm having trouble with this multiple choice question... no idea how to do it...
Any help appreciated!
Thanks

P.s answer is C

The key thing to notice in this question is that the acceleration must be constant. So what I would do (there may be a faster way idk) is rearrange each option into relation between acceleration and time, and if the answer is a constant, then the option is true (and can hence be eliminated).

I'll show you with options A, B, and C, because once you realise C doesn't have constant acceleration you can circle it and move one (unless you really want to try each option/have the time to )

So A: diffrentiating twice gives , where a is the second derivative of x. You could also simply note that the equation with displacement in it is a second degree polynomial so the second derivative will give a constant.

and B: , can be rearranged to get (ignoring negative value cause the acceleration will either be a constant or not so it really doesn't matter). Then use and flip the equation and integrate to get t=4root(x-2), which can be rearranged for x to get , which, again is a second degree polynomial that will give a constant acceleration if derived twice.

And C, rearrange for x, and differentiate twice. You will notice t will still be a variable in the acceleration formula, and thus acceleration is not constant for C, making it the correct answer.

[StupidProdigy]

To show that an object on a horizontal plane with one force acting horizontal on it will not move, do I just show that friction force>applied force? thankya

[keltingmeith]

To show that an object on a horizontal plane with one force acting horizontal on it will not move, do I just show that friction force>applied force? thankya

Yup, that should do it.

[cosine]

Help

[knightrider]

How would you do this question attached?

Prove  the following identity.

[natdogg]

How would you do this question attached?

Prove  the following identity.

[knightrider]

Thanks natdogg