VCE Specialist 3/4 Question Thread!

[ahat]

(Image removed from quote.)

(VCAA 2011 MC Q2)

The asymptotes will be perpendicular if the hyperbola is rectangular, i.e. if the values of a and b are the same. You can then instantly rule out all of them except the last option. Notice the coefficient of 2 in front of the x² term, so when you complete the square and divide both sides by 4, the value of a will be different to b so the hyperbola will not be rectangular. By the way, you can check this by:
1) CAS: Graphs > Menu > 3 > 2 >  6 > 1 and then analyze the hyperbola asymptotes by menu > 6 > 8 > 6
2) CAS: use the complete square function for the x terms and then the y terms: Calculator > 3 > 5 > input: (ax² + bx, x)

Hey guys, I have a question.
A girl is sitting on a swing while her brother exerts a horizontal force that holds her stationary and the rope of the swing makes an angle of 45^o to the vertical. In this case, would the normal reaction force be included in the vertical component of the rope's tension? Do we only ignore the normal reaction force for a free-air particle? Thanks.

[lzxnl]

OK, so what would cause this "normal reaction" force? The air? I don't think there would be one here. There only seem to be three forces that I can see in this scenario: the boy's push, the girl's weight force and the tension in the swing.

[brightsky]

yeah we only consider normal reaction forces when we have an object resting on something. normal reaction force is basically the force exerted by something on the object to prevent it from falling through it. for example, if we have a book resting on a table, obviously the book has a weight force. but this can't be the only force or else the book will fall through the table. the normal reaction force prevents this from happening. in this case, nothing is resting on anything.

[ahat]

OK, so what would cause this "normal reaction" force? The air? I don't think there would be one here. There only seem to be three forces that I can see in this scenario: the boy's push, the girl's weight force and the tension in the swing.

yeah we only consider normal reaction forces when we have an object resting on something.

I was assuming the seat of the swing would cause the normal reaction force? Thanks though guys.

[lzxnl]

We're considering the seat and the girl as one object. The seat pushes up, but the girl pushes down in accordance with Newton 3

[ahat]

We're considering the seat and the girl as one object. The seat pushes up, but the girl pushes down in accordance with Newton 3

I see, thanks

[satya]

Hey guys,

for 2009 exam 2, Q5g) How come you can't do 1200 = integral from 0 to T v(equation) dt. To work out the time to fall to 1200m ?

I am pretty sure we have to use Pythagoras for that and use constant acceleration formula as it Is travelling at constant 2m/s

[ahat]

Hey, if I wanted to show that two particles didn't collide would it be sufficient to equate the components of each particle's velocity position vectors?
For example, when I did this, I found that for the i components: sin(t) = -1/5 and for the j components sin(t) = 2/5.
First of all is this correct, second of all, enough for two marks?

Thanks.

[brightsky]

I think you mean position vector. velocity vector is irrelevant. and yes, to show that particles don't collide, equate components of position vectors and show that there is no solution for t.

[duhherro]

btw guys, does point of moving down mean acceleration = 0? I'm a bit confused on this term though. Does that just mean the object has YET to move?

[eddybaha]

btw guys, does point of moving down mean acceleration = 0? I'm a bit confused on this term though. Does that just mean the object has YET to move?

yes the acceleration is 0. on the point of moving down the slope means if there is a slight change in force down the slope it will cause it to accelerate.

[~T]

btw guys, does point of moving down mean acceleration = 0? I'm a bit confused on this term though. Does that just mean the object has YET to move?

To your first assumption, yes it is means a=0. It just means that, as eddybaha stated, the object will accelerate if there is any net positive change in force down the plane.

Most importantly, however, it will give you the direction of the frictional force and also lets you know that the maximum frictional force is acting (ie, F = uN).

If it is on the point of moving down the plane, then friction must be acting up (try visualise that it's holding the object in place, preventing it from going down).

If it is on the point of moving up the plane, then friction must be acting down the plane (this will only occur with a tension force or something else acting up on the object, of course)

[Jaswinder]

how do we find asyptotes of graphs like on the casio classpad? also is there a way of getting something like this in cartesian form through classpad? thanks

[lzxnl]

Asymptotes? Get your calculator to divide it (expand function? I don't use classpad but it should have a function like that) and then you can read off the asymptote.

[duhherro]

cheers guys! I swear once i remembered seeing a question on a paper with same wording but they had bla bla bla = ma    -.-