VCE Specialist 3/4 Question Thread!

[sin0001]

I doubt you'd get 4/4 because there's always 1 mark for getting the right answer. You wouldn't get that mark. There's a chance that you can get consequential marks but I'm not sure how many you'd get with consequential marks.

Ahh, thought so!
Gave myself 3/4; minus 1 for the answer mark

[e^1]

Hey everyone, I have a few questions from a trial exam.


For the first question, I used Lami's theorem (ie. ) to get T₁=T₂, but the answer is C (using the fact that ). I know why C is right, but could A be wrong because I generalised the angles?


Also, I have trouble with how to do the following 2 questions; I'm not sure how to them correctly. Should add that for question 3, the vector between the tugboat and the oil rig is


Thank you if you can help!

[brightsky]

1) T1 and T2 are vectors not magnitudes; the question states explicitly that T1 and T2 are tension FORCES. so T1 and T2 can't be the same, because they're pointing in different directions. instead use the fact that the sum of all FORCES for object in equilibrium is 0 vector.
2) m1gsin(t1) - T = 0 and T - m2gsin(t2) = 0
so we have m1g sin(t1) = m2gsin(t2)
m1/m2 = gsin(t2)/gsin(t1) = sin(t2)/sin(t1)
3) distance at time t = sqrt((t+1/t)^2 + (t-1/t)^2) from the position vector
differentiate and then let = 0 to find t.

[lzxnl]

I disagree with both of the above.

If we let the position vector of the tugboat be OT, and the position vector of the rig be OR = 5j

Then we want to minimise the distance RT which is mod(OT-OR) or mod((t+1/t)i+(t-1/t-5)j)
It's easier to minimise the square of the distance as that is still a minimum for the distance. Minimise (t+1/t)^2+(t-1/t-t)^2 however you like.

[e^1]

I disagree with both of the above.

If we let the position vector of the tugboat be OT, and the position vector of the rig be OR = 5j

Then we want to minimise the distance RT which is mod(OT-OR) or mod((t+1/t)i+(t-1/t-5)j)
It's easier to minimise the square of the distance as that is still a minimum for the distance. Minimise (t+1/t)^2+(t-1/t-t)^2 however you like.

Haha thanks for the clarification, I knew initially that he didn't write it right but I didn't want to tell him off. Again, thank you nliu and brightsky

Just to add on to this post, when trying to find a solutions of a polynomial (in this case ), is it fine to do the working out below:






From the exam answer sheet, they did it differently in that they expanded and used that to divide in order to find the answer.

Thanks!

EDIT: I just made this up while I was doing trial exam 1, but yeah thanks heaps nliu , appreciate it alot

[lzxnl]

LOL I use that method all the time.
Although to be picky, in your working you could just have said "equating constant terms" to avoid expanding out the product of the quadratic and the linear term
It's not a wrong method; it is perfectly valid and certainly appropriate for the course, so I wouldn't expect you to lose any marks for that working. Provided you don't mess it up of course.

[barydos]

Spoiler

I always stuff up tension questions (spesh & physics both), so I would like some help with this question.
Thanks

[b^3]

Firstly, ask yourself, which direction are the masses going to accelerate in? Since they're both hanging vertically (and there is no external torque on the peg), the heavier mass will accelerate downwards and the lighter mass will accelerate upwards. So if we look at the 3kg mass we have the weight force acting downwards, and the tension force acting upwards. The sum of these  (taking into account direction) will give the net force, i.e. . Since the object is accelerating downwards, the weight force is larger, so for the 3 kg mass we will take down as positive. So we have


Now for the 1 kg mass, we have the weight force acting downwards and the tension force acting upwards, and since it is accelerating upwards we will take up as positive.




The problem that a few people have is that they put the the wrong way around, as they don't define directions as positive, and just take both up as positive. This actually results in a different system. By making sure you have the larger "mg" or "T" first, then you keep your acceleration positive, and account for directions. (You could also flip both, but that would be redundant).

Easiest way to solve these is normally solve for or and then substitute into the other equation.
Solving:

[barydos]

Firstly, ask yourself, which direction are the masses going to accelerate in? Since they're both hanging vertically (and there is no external torque on the peg), the heavier mass will accelerate downwards and the lighter mass will accelerate upwards. So if we look at the 3kg mass we have the weight force acting downwards, and the tension force acting upwards. The sum of these  (taking into account direction) will give the net force, i.e. . Since the object is accelerating downwards, the weight force is larger, so for the 3 kg mass we will take down as positive. So we have


Now for the 1 kg mass, we have the weight force acting downwards and the tension force acting upwards, and since it is accelerating upwards we will take up as positive.




The problem that a few people have is that they put the the wrong way around, as they don't define directions as positive, and just take both up as positive. This actually results in a different system. By making sure you have the larger "mg" or "T" first, then you keep your acceleration positive, and account for directions. (You could also flip both, but that would be redundant).

Easiest way to solve these is normally solve for or and then substitute into the other equation.
Solving:

Great explanation. I understand now, thanks a lot! I appreciate it

[saba.ay]

Please and thank you.

[abcdqd]

it works like any other function of x. eg. if you had f(x) = x^2, f(3) requires replacing the x with 3
here, you replace the x with sqrt2, and evaluate the integral

[saba.ay]

That's what I did but it didn't work out as being the right answer.

 I put it into the calculator wrong.. 

[barydos]

"The position vector of a particle at time t, is given by:
r(t) = i + j,

Find the angle between the vector r(t) at time t = 1 and the vector -i - j."

So after plotting the two vectors out, you can see that there are two possible angles to choose from (duh), but how do we know which one? It doesn't specify obtuse/reflex angle.

[Hancock]

Wouldn't you use the dot product identity to work it out? And if it was me, I'd take the obtuse angle over the reflex one.

[barydos]

Wouldn't you use the dot product identity to work it out? And if it was me, I'd take the obtuse angle over the reflex one.

This was an exam 1 question and the answer was , so using dot product was a bit tough ( and don't go nicely together .. or do they?). Either way, r(1) and - i - j were nice numbers whose angles were also nice, so I thought it was a bit quicker to just plot the vectors onto an graph and add the angles visually.

Any reason why you'd take the obtuse other than the reflex?