VCE Specialist 3/4 Question Thread!

[Mattjbr2]

This is for Q:12 on page 555 of the Cambridge book (attached)

Let T(12-5)=T1, T(8-12)=T2 and down=positive

5kg: mg-T1=ma ∴ T1=49+5a
12kg: T1-mg-T2=ma ∴ T1-T2=118+12a
8kg: T2-mg=ma ∴ T2=78.4+8a

Solving by CAS (menu371): a=-5.9, T1=78.5, T2=31.2

But that's obviously not correct since the 5kg block can't possibly accelerate upwards in this scenario. I have the worked solutions, but exactly what part of my logic is wrong?

[Eric11267]

This is for Q:12 on page 555 of the Cambridge book (attached)

Let T(12-5)=T1, T(8-12)=T2 and down=positive

5kg: mg-T1=ma ∴ T1=49+5a
12kg: T1-mg-T2=ma ∴ T1-T2=118+12a
8kg: T2-mg=ma ∴ T2=78.4+8a

Solving by CAS (menu371): a=-5.9, T1=78.5, T2=31.2

But that's obviously not correct since the 5kg block can't possibly accelerate upwards in this scenario. I have the worked solutions, but exactly what part of my logic is wrong?

For the 8kg and 12kg blocks you have their mass as a force acting against tension, but the only relevant forces acting on these blocks are tension forces.
So you should have T1-T2=ma

[Syndicate]

This is for Q:12 on page 555 of the Cambridge book (attached)

Let T(12-5)=T1, T(8-12)=T2 and down=positive

5kg: mg-T1=ma ∴ T1=49+5a
12kg: T1-mg-T2=ma ∴ T1-T2=118+12a
8kg: T2-mg=ma ∴ T2=78.4+8a

Solving by CAS (menu371): a=-5.9, T1=78.5, T2=31.2

But that's obviously not correct since the 5kg block can't possibly accelerate upwards in this scenario. I have the worked solutions, but exactly what part of my logic is wrong?

I would separate this into two different tensions T1 (between the 5 kg and 12kg blocks) and T2 (between the 12kg and 8 kg blocks), and write down three linear questions for each block.

12a = T1-T2 (1)
5a = 5g - T1 (2)
8a = T2 (3)

Sub (3) into (1)
12a = T1-8a
Therefore 20a = T1 (4)

sub (4) into (2)
5a = 49 - 20a

Therefore a = 49/25 m/s^2
T1=196/5 N
T2 = 392/25 N

EDIT: I got beaten to it, but I'll leave my solution up anyways (just in case someone might need to refer to my working out).

[Mattjbr2]

For the 8kg and 12kg blocks you have their mass as a force acting against tension, but the only relevant forces acting on these blocks are tension forces.
So you should have T1-T2=ma

My bad! Thanks

[Willba99]

Hello lads and lasses,
is there a way to go from parametric equations to cartesian on casio??

[peanut]

A few questions from Northern Hemisphere VCAA:
Question 1) I tried making a triangle of forces using a right angled triangle (reaction force as hypotenuse, and P and 5g as the other sides) but couldn't solve
Question 4) I found the conjugate pair and thus the quadratic factor, but couldn't proceed from there.
Question 11) I used the formula shown, but got a definite integral I couldn't solve.

[Shadowxo]

A few questions from Northern Hemisphere VCAA:
Question 1) I tried making a triangle of forces using a right angled triangle (reaction force as hypotenuse, and P and 5g as the other sides) but couldn't solve
Question 4) I found the conjugate pair and thus the quadratic factor, but couldn't proceed from there.
Question 11) I used the formula shown, but got a definite integral I couldn't solve.

1. Try resolving the forces parallel to and perpendicular to the plane. Then you should be able to get an equation involving P and mg for the parallel component which you can then solve for P.
4. I'm a bit rusty on this, but try multiplying that quadratic factor by a linear factor eg z-x-yi (ie the third factor is z=x+yi) and expanding it out to find the relevant values. Since a and b are real constants, y should end up being 0 and you should be able to solve for x by knowing the constant term in the cubic equation you're given is -4. Once you expand, simply equate coefficients and solve for a and b. If you're still stuck, let me know and I'll write out a solution.
11.

See if you can solve from there

Edit: Misread your third question so changed my solution

[Eric11267]

A few questions from Northern Hemisphere VCAA:
Question 1) I tried making a triangle of forces using a right angled triangle (reaction force as hypotenuse, and P and 5g as the other sides) but couldn't solve
Question 4) I found the conjugate pair and thus the quadratic factor, but couldn't proceed from there.
Question 11) I used the formula shown, but got a definite integral I couldn't solve.

1.You should either resolve P parallel to the plane, or resolve the normal force parallel to P
let x be the component of P up the plane
x=P/cos30
5gsin30=P/cos30
P=5g/root(3)
4.Okay so the quadratic factor is z²-2z+2
For these questions I like to use short division
Since the cubic has -4 as the last term and the quadratic has 2 as its last term, I know that the remaining root must be (z-2)
(z²-2z+2)(z-2)=z³-4z²+6z-4
This gives us:
a+b=-4
b²-a=6           
This gives a=-5,b=1 or a=-2,b=-2
11.dx/dθ=a-acosθ
dy/dθ=asinθ
Putting these into the formula gives you
a sqrt(2-2cosθ) in the integral
But you're still left with a square root on the inside, so you need to use the half-angle formula to change it into sin² or cos²
Using the fact that cosθ=1-2sin²θ/2, you can complete the question

[gnaf]

do we need to know how to do q10 of vcaa 2008 exam 1 since modulus is still in the course?

[mtDNA]

do we need to know how to do q10 of vcaa 2008 exam 1 since modulus is still in the course?

I would assume it’s still apart of the course (when I did that paper I didn’t see anything wrong with it, so my judgement is based on the fact that it didn’t look foreign at all)

[peanut]

Thanks shadowxo and Eric11267. Here a few more questions from exam 2 that I am struggling with

[RuiAce]

Thanks shadowxo and Eric11267. Here a few more questions from exam 2 that I am struggling with

[Eric11267]

Thanks shadowxo and Eric11267. Here a few more questions from exam 2 that I am struggling with

For question 3, the limiting value of P occurs when dP/dt=0. So you just solve the provided equation for P, giving you 0.894 as the largest solution

[LifeisaConstantStruggle]

Thanks shadowxo and Eric11267. Here a few more questions from exam 2 that I am struggling with

for question 13) you can test with all choices and eliminate the incorrect ones:

A) a+c=b+d is OA+OC=OB+OD
If you visualise your sum of vectors (by moving the vector OC to start from point A and the vector OD to start from point B) you'll realise that they reach the same point, so this would be the correct answer.
B) (a-c).(b-d)=(OA-OC).(OB-OD)=(CO-OA).(DO-OB)=(CA).(DB)
By multiplying it you are evaluating the product of the magnitude of both vectors (CA and DB) and the cosine of the angle between these vectors. If cos(angle)=0 they are perpendicular. Are they?
C) Same approach as A
D) Same approach as B
E) Same approach as B
Might be a slight error but A is the correct answer

[Willba99]

For u to exist (the vector resolute of a in the direction b), does the mod of a have to be smaller than the mod of b?