VCE Specialist 3/4 Question Thread!

[soccerboi]

Can someone please help me with this one?:

A box of mass 5kg rests on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.1 . A boy applies a horizontal dragging force of D newtons to the box in an attempt to move it.

a.   Find the values of D if the box is not at the point of moving across the floor.

[b^3]

For the box to not be at the point of moving across the floor (I'm assuming its going to stay stationary), then we need the force to be less than the frictional force at which the box would be at the point of moving. As when the force applied to the box is less than this, it won't move and the friction force will be equal to the force we are applying.

So we need



Forces in the vertical direction.

i.e.

So that means (assuming D is always in that direction).
So (I think you missed the value of mu)


NOTE: If I've interpretted the question wrong and it just wants the value of which its not at the point of moving, but includes when  it actually moves, then it will be not equal to the value found, but I'm pretty sure it asked what I did above (but just in case anyway).

[soccerboi]

Thanks b^3!
And yeh i missed the mu value, should be 0.1.

So if D > mu N, it moves to the right, and if D< mu N, it stays where it is? What if D=mu N, does it just stay?

[b^3]

Thanks b^3!
And yeh i missed the mu value, should be 0.1.

So if D > mu N, it moves to the right, and if D< mu N, it stays where it is? What if D=mu N, does it just stay?

Basically yeh.

if then the box will move to the right (for the diagram we drew, well it will move in the direction that D is in, its just that I drew D to the right)
if then the box will be 'on the point of moving' but won't move.
if then it will stay where it is as you aren't applying enough force to overcome the friction force opposing it.

[soccerboi]

A ball of mass 1/5 kg is projected vertically upwards at a speed 10 ms−1 from the top of a 20 m high cliff. Air
resistance is negligible. Correct to two decimal places, the ball will hit the foot of the cliff with a momentum, in kg.ms−1, of
A. 3.42
B. 3.96
C. 4.44
D. 17.10
E. 22.12


My question about this problem is attached below. Thanks

Also, the solutions said that by using calculus you can get v=root 492, but what did they mean? How do you find v by using calculus?

[b^3]

Remember that u is the initial velocity where as v is the final velocity.
So the initial velocity will be 10 m/s up as this is where we are taking time 0 as when it is being projected vertically upwards.

If we were interested in when the ball got to it's maximum heigth, then v would be 0 m/s. But we want when it hits the foot of the cliff.

Now with (sometimes instead of s they will use x), s here isn't distance, its displacement. I.e. Where you end up in reference to the original point. So when the ball hits the foot of the cliff it will be 20 m below our starting reference point, i.e. s=-20 m.

If you were asked to find the distance it falls from the maximum heigth then it would be the 20 m plus the extra distance.

So remember, u is initial velocity, v is final velocity and 's' (I'd rather use x but anyways) is displacement.

Hope that helps

[Bhootnike]

a particle moves such that its pos. vector is given by: r = sin(2t)i +2cos(t)j
show that the distance of the particle, is given by

so i started by rooting and squaring r.






is it right to take out cost as a factor. if not.. whats the next step!>?

[b^3]

Yeh its fine, we're only taking it out as a factor, so it's all good.

[Bhootnike]

ok im also stuck on the next bit :\
it asks for, 'HENCE, find the maximum distance of the particle from the origin and state where this occurs'

these questions always stump me..
i could integrate it, might take a while though.
theres a way to do using sin graphs ... right? or something like that?
could someone please explain

[b^3]

Think about it this way, if you make as big as you can then you are making as big as you can right?

So now to look at .
Think about what values can be.
The smallest we can get out is , the largest is
Now if we look at whats the smallest and biggest numbers we can get out?
They will be and right?

So now to make as large as possible, we need to make as small a possible.
So the smallest value of is so we have
Max distance= (then the units).

If you get a its a similar deal.

Anyway hope that makes sense, probably over explained it a little.

EDIT: fixed it up, was missing bits of sentences, guess thats what happens when you put latex in the middle of sentences...

[Bhootnike]

oo ok, that made sense, except when you said whats the smallest and biggest numbers we can get out for sin^4 (t),
i dont get how its 0 and 1 ,  - i wouldve said -1 and 1.
im not in the maths mood i think at the moment, so please excuse my crappy maths atm

[Conic]

oo ok, that made sense, except when you said whats the smallest and biggest numbers we can get out for sin^4 (t),
i dont get how its 0 and 1 ,  - i wouldve said -1 and 1.
im not in the maths mood i think at the moment, so please excuse my crappy maths atm

Since it's sin^4 the minimum is 0^4 (as -1^4=1)

[Bhootnike]

i see!
thanks guys.

another question:

given that ,
solve the equation giving your answers in the form cis(A), where

I Don't even know where to start.

[BubbleWrapMan]

Let cos(x) = 1/2

[oneoneoneone]

There are 4 intersections, But at when the first particle is in the second quadrant, the second particle is in the fourth quadrant and vice versa. So their paths cross at 4 points, but the actualy particles only meet when theyre at the same spot at the same time, which is only in the first and 3rd quadrants.