VCE Specialist 3/4 Question Thread!

[Zlatan]

thanks for that. Could you also help with these questions b)ii and c

[silverpixeli]

bi)
the relevant circle theorem is that angle AOC is twice angle ABC (z) and since angle AOY is half angle AOC,

angle AOY = z

bii)
is angle AOY is z, what's length AY? it's a right angle triangle, so

sin(z)=AY/AO (soh-cah-toa)
AY = AO*sin(z)

AY is half of AC since it's an equilateral triangle and Y is halfway along the base

AC =2AOsin(z)

c)
we know that x + y + x = pi (180 degrees in a triangle)
thus z = pi - (x+y)
RHS = sin(z)
= sin( pi -(x+y) )
= sin( (x+y) ) by symmetry
= LHS

fairly confident in those answers, haven't studied circle rules since the start of 2013 though so i might have missed something

[allstar]

at what angle does the kite hit the ground?

please

P.S the ans is 24.1 degrees

i've attached what i did.. but its not the ans

[Phy124]

At the component is , so you should have rather than for that side of the triangle as it is going from to for a change of .

Then you end up with

[eagles]

Hello, can I have some help with Q.21?
I understand that to find the initial speed, a calculation of |r(0)| is required.
I'm unsure why the answer for Q.21 is C.

Thanks.

[theshunpo]

Hello, can I have some help with Q.21?
I understand that to find the initial speed, a calculation of |r(0)| is required.
I'm unsure why the answer for Q.21 is C.

Thanks.

First you have to derive the position vector to get the velocity vector (as this is what represents the direction of motion). You then have to sub into the velocity vector and find it's magnitude in order to find the unit vector.

Spoiler

[Zlatan]

It seems straight forward, but i can't get an answer for this for some reason.

[psyxwar]

Take the dot product of the two vectors, and solve the quadratic for p using the quadratic formula

[psyxwar]

Is there a mistake in the solutions?

THe part where is says 20g= Tcos (theta), shouldn't it be 20g = Tsin(theta) especially when they are resolving in the vertical direction?

uhh do you have the full question?

[M_BONG]

uhh do you have the full question?

Yeah that's the full question. First part is just label the forces acting on particle on diagram. By looking at that are answers wrong?

[psyxwar]

Yeah that's the full question. First part is just label the forces acting on particle on diagram. By looking at that are answers wrong?

I mean, where did the sin theta come from and all that? Actually just tell me which paper this is from and I'll look at it.

[psyxwar]

You have to infer yourself it's sin theta.... by separating the tension force in vertical and horizontal components yourself.

TSSM 2007 Exam 1.

EDIT: my bad. Realise I didn't attach the first part of the question properly. Awks

their answer is right, I can tell you that. I guess they defined theta differently to me but what I did:

cos(A)=1/4, taking A to be the angle between T and F
Drawing a triangle: we have tan(A)=sqrt(15)
tan(A)=20g/F=sqrt15
=>F/20g=1/sqrt(15)
=> F=20g/sqrt(15)=20g(sqrt(15))/15=4sqrt(15)/3

[M_BONG]

their answer is right, I can tell you that. I guess they defined theta differently to me but what I did:

cos(A)=1/4, taking A to be the angle between T and F
Drawing a triangle: we have tan(A)=sqrt(15)
tan(A)=20g/F=sqrt15
=>F/20g=1/sqrt(15)
=> F=20g/sqrt(15)=20g(sqrt(15))/15=4sqrt(15)/3

Oh, I realise what I did wrong, I assumed theta to be at the wrong place (between the 2m and 0.5m) so it appeared to me they had sin and cos the wrong way around. But yeah, I understand why  theta should be at the other angle.  Damn, sneaky question.

Thanks for your help (lol sorry for derping up with the attachments)!
 

[lucas.vang]

For questions that say show that these three vectors are linearly independent

Is showing that they aren't linearly dependent sufficient?

like m(vector a) + n(vector b) = (vector c)

then simultaneous solve?

Or
 m(vector a) + n(vector b) + p (vector c) = 0i + 0j + 0k

 should show that all components (m,n.p)  equal zero?

[allstar]

for part a) can someone please explain the solutions?

Why are we finding the perpendidular and the parallel parts and then the hypotenuse? between the two?