VCE Specialist 3/4 Question Thread!

[Special At Specialist]

^^ Thanks. I get it now.

[soccerboi]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010specmath2-w.pdf

With MC Q21, If i assign acceleration as positive in the downward direction, i get :

Mg-T=M(7/5)
T-5g=5(-7/5)

Why does this not get me the value of M?

Instead when i solve using simultaneous eqns with :
Mg-T=M(7/5)
T-5g=5(7/5)
I get the correct answer. So why does the second eqn have positive acceleration when i have assigned the downwards direction as positive. Shouldn't it be negative acceleration since the 5kg mass is moving up?

Also, with MCQ22, can someone explain to me how to start it off?
http://www.itute.com/wp-content/uploads/2010-vcaa-specialist-mathematics-exam-2-solutions.pdf
I don't understand the first line of Itute solutions for it. a=F(x)/m
Thanks

[BubbleWrapMan]

If the downward direction is positive then 5g should be positive and T should be negative

[polar]

F = ma is newton's second law, F(x) is a function for a variable force



they equate two expressions for acceleration in the first line

[soccerboi]

Dot products are only applied to vectors in i,j,k form right? so if vectors are written in a,b,c notation, we can't use dot product? Just need clarification, thanks.

[b^3]

You can't apply the dot product to as you would with components, if you don't have the components, but you can still 'dot' them, you just end up with the term and may have to work with that, or if you dot a vector with itself you get the magnitude which you may need to manipulate to show something.
e.g.

[BubbleWrapMan]

That's correct, unless you know the dot products of a, b and c.

Because the dot product distributed over brackets: a.(b+c) = a.b + a.c,

and we know the dot products of the unit vectors: i.j = j.k = k.i = 0 and i.i = j.j = k.k = 1,

that's why we can use the dot product in that way, but if we don't know the dot products of the vectors then we can't use the dot product in that way.

[soccerboi]

With connected mass problems, at the instant the string breaks, does tension exist?

Also, with questions involving a person jumping out of a plane, at the instant the parachute opens, does the magnitude of the velocity immediately change? e.g velocity of man falling is 80km/hr, then parachute opens, his velocity is immediately 50km/hr or will it decrease from 80 to 50 after a certain amount of time?

One more: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010specmath2-w.pdf
With Q 2e) Where in the question did it indicate that acceleration is constant, so that the constant acceleration formula can be used?

Many thanks for all the help

[kensan]

Hey could someone help me with MC question 11 from the VCAA 2011 exam please?
 
           

The magnitude of the sum of these three forces is equal to?

 Don't really get how to do these force questions, thanks!

[Lasercookie]

Hey could someone help me with MC question 11 from the VCAA 2011 exam please?
 
           

The magnitude of the sum of these three forces is equal to?

 Don't really get how to do these force questions, thanks!

Just think of them as vectors really, the forces is really just giving the maths some context I guess. Add them up and find the magnitude. You'll be doing vector addition, so you'd be taking into account the direction of the forces that way.





edit: Then just go through each option and look for the one that has a magnitude of 1, that happens to be E.

[Jenny_2108]

With connected mass problems, at the instant the string breaks, does tension exist?

Also, with questions involving a person jumping out of a plane, at the instant the parachute opens, does the magnitude of the velocity immediately change? e.g velocity of man falling is 80km/hr, then parachute opens, his velocity is immediately 50km/hr or will it decrease from 80 to 50 after a certain amount of time?

One more: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010specmath2-w.pdf
With Q 2e) Where in the question did it indicate that acceleration is constant, so that the constant acceleration formula can be used?

Many thanks for all the help

1. Yes. Tension doesn't exist after that
2. I think velocity changes because as parachute opens, the air resistance increases, thus body deccelerates => affects velocity
3. Use a=dv/dt then integrate v from t=0 to t=3

[rife168]

2010 MC Q22:


Can someone help me with this question? I had a vague idea what I was doing and got it correct, but what is a methodical way of approaching this question? It's mostly the bit that's confusing me a bit.

The answer is D



EDIT: Don't worry, I got it!
Good old F=ma...













Edit II:
Thanks oglow100, just saw your reply.

[Biceps]

2010 MC Q22:
(Image removed from quote.)

Can someone help me with this question? I had a vague idea what I was doing and got it correct, but what is a methodical way of approaching this question? It's mostly the bit that's confusing me a bit.

The answer is D

Ok you use definite integrals on both sides were F(x)=ma a=vdv/dx you make v0 and x0 both lower limits for the integrals and v1 and x1 are upper limits.

[kensan]

Could someone help with with question 5b please? I don't really get how to show that in this scenario, thank you.
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf

[polar]