VCE Specialist 3/4 Question Thread!

[e^1]

Hey everyone

I got some troubles with some questions from MQ again. I'm not sure how to approach the first question -- it's about projectile motion.

Secondly, I know what the triangle of forces does, but not sure how to draw one. If anyone could give some tips about it, that would be helpful.

Thank you!

[SocialRhubarb]

Question 1a.):

Spoiler

Okay, for the first q uestion:

We know that the distance from A to the hoop is the same as the distance from the hoop to the target. This means that since there are no retarding forces along the horizontal axis, the time taken to travel from A to the hoop at B is the same as the time taken for the ball to get from the hoop to the target.
Let's consider just the vertical component of the ball's velocity. Let the initial y-velocity be , and the time taken to travel from A to the hoop at B be .



The vertical component of velocity at B will be .

The y displacement at B is given by .

The y displacement at C is given by , since we know that the time taken to get from A to B is the same as the time taken to get from B to C.

Two times the first equation minus the second equation gives:







However, was the time taken to get from A to the hoop. The time taken to get from A to the target is , and .

Question 1b.)

Spoiler

The horizontal component of velocity is simply distance over time, which is

We can work out the vertical component of velocity by going back to our answers in part A and rearranging to find .





Four times equation 1 minus equation 2:





The angle of projection can then be found using

Question 1c.)

Spoiler

Speed of projection is . This is the easy part, and we just sub in the values we worked out before.

Triangle of forces: add all 3 vectors together head to tail, interior angles of the triangle can be found by either extending the vectors or recognising alternate angles between  parallel lines.

[satya]

Okay, two more questions from me.

1)

Spoiler

2) Suppose we were asked in a question (with an appropriate context), "Find, to the nearest minute, how long, after it is brought

Hope this makes sense, thanksindoors, it takes the water to reach a temperature of 25 deg C".  Would we round our answer to the nearest minute, or would we find the nearest minute at which the water has gone beyond the 25 deg C?? (so if the rate of change was COOLING, would we round to a minute where the temperature has gone equal or BELOW 25?) 

for part a i got 6617 years?? correct or nah??

[Jeggz]

I'm really struggling with dynamics and i keep getting every single question wrong..so any help would be greatly appreciated

1) What force is necessary to accelerate a train of mass 200 tonnes at against a resistance of 20000N?

2) A puck of mass 0.1 kg is sliding in straight line on an ice-rink. The co-efficient of friction between the puck and the ice is 0.025. Find the resistive force owing to friction.

Thanks!

[Professor Polonsky]

(I haven't done dynamics yet, so I hope I'm not stuffing anything up. This is all from my general physics knowledge.)

1) So we know that . We are given m=200,000 and a=0.2.

Substituting our values in, we get

Now, this is done against a resisting force of 20 000 N. Letting be our accelerating force, we have



Hope this helps

[SocialRhubarb]

Question 1.)




Question 2.)


If you're really having trouble, drawing force diagrams might help, even if making them feels really lame.

[barydos]

1) i found it pretty tricky with words but is the answer somewhere around 6636 years?

2) i think equal or more but not less but im not 100% sure

for part a i got 6617 years?? correct or nah??

Wow completely forgot about this, sorry.

Umm, I didn't actually get to find out what the answer was LOL
but yeah I got 6617 years as well using the points  t = 0, carbon remaining = 1
and t = 5760, carbon remaining = 0.451

[~T]

Just doing the revision chapter 11 questions... Extended response question 15b)

For a), you get

For b), I thought you would let y equal the depth (as a constant... sort of) and calculate the volume as the area of the cross section multiplied by the length:




The answer is:


Any help would be appreciated - why my answer is wrong, and/or how to get the correct solution. Thanks

[sin0001]

From memory, wouldn't the area of the cross-section be just equal to: int.(function)dx, with an upper limit of 'y' and a lower limit of '0' (Instead of integrating 'y - function')
I could be wrong as I don't remember the exact question

[abcdqd]

you can integrate with respect to y to get the cross-sectional area at a certain y value, so get x in terms of y first

take one half of the cross section

then volume will equal 2*that area*length

[~T]

Thanks guys, far more logical way right there. Let's blame tiredness! I now also realise my method would have worked if I had used the correct endpoint, which definitely wasn't ... sigh

[Professor Polonsky]

180, surely? And I don't think the course is bigger than Methods.

[Stick]

Oh crap calculator active is 2 hours? I did not know that hahahaha.

But still, 3 hours to cover everything we've done?
It feels like they could examine me for 6+ hours and still not cover everything I've learnt so far this year.
And we only just started dynamic lol.

There's a high degree of overlap in some of the topics so that's one way they go about it.

[lzxnl]

Oh crap calculator active is 2 hours? I did not know that hahahaha.

But still, 3 hours to cover everything we've done?
It feels like they could examine me for 6+ hours and still not cover everything I've learnt so far this year.
And we only just started dynamic lol.

Tell me, what HAVE we done this year?
Ellipses, hyperbolas, vectors, dot product, linear dependence, vector proofs, inverse trig functions, reciprocal trig functions, basic complex number operations, polar form, de Moivre, solving polynomials over complex numbers, reviewing basic Methods calculus, differentiating inverse trig functions, implicit differentiation, finding volumes, integrating the results of our differentiation, integrating by substitution, simple first-order differential equations of the form dy/dx = f(y) or f(x), Euler's method, direction fields, constant acceleration formulas, velocity in terms of time, acceleration in terms of time and velocity and displacement, differentiating vector functions, what the derivative vector actually means, net forces, Newton's second law, variable forces, resolving forces, forces on inclined planes, friction, connected bodies and systems...

Please tell me if I've missed anything.

Is it THAT hard to fit three hours worth of exams on most of this? Some of these only need one MC question, like direction fields and basic complex number operations.

[lzxnl]

Mad respect for typing all of that out.
Well I guess you've convinced me haha.

What prompted me to ask in the first place was because I just spent like two hours on variable forces (which involved a lot of calculus I seem to be extremely rusty on haha).

Well, ideally when you're done with the course you should be able to list what you've learned throughout the year.
I can't do that for physics and chemistry though as I didn't learn those in a systematic order like I did with maths.

When you're familiar with variable forces they take...a minute or two to do? It's just using F = ma and then the acceleration expressions.
Funny, using F = ma = m*d(1/2 v^2)/dx, you can show that integrating F dx = integrating m*d(i2 v^2)/dx * dx = change in 1/2 mv^2, which is kinetic energy. Hey look it's physics in spesh.