VCE Specialist 3/4 Question Thread!

[bts]

2kg mass inital speed is 5m/s slides to rest along waxed floor over 10m. The coefficient of friction is?

please and thanks

[brightsky]

u = 5, v = 0, s = 10, a = ?
v^2 = u^2 + 2as
0 = 5^2 + 2a(10)
0 = 25 + 20 a
a = -25 / 20 = - 5/4

Let F be the friction force.

- F = 2(-5/4)
F = 5/2

Since the mass is moving, F = u N.

u N = 5/2
u (2 g) = 5/2
u = 5/(4 g)

[IndefatigableLover]

Easy question but why does:

simplify out to ? On a CAS-FREE exam, could I just leave my answer in the first form?

[Zealous]

Easy question but why does:

simplify out to ? On a CAS-FREE exam, could I just leave my answer in the first form?

 

[IndefatigableLover]

 

Very nice! Thanks Zealous!

[M_BONG]

What are the conditions in proving something is collinear (as opposed to being parallel?)

Also, does it what matter which points we prove?

Eg. A question states that to show that points O, D and B are collinear.

I tried to show OD is equal to kDB k being a constant

But answer showed that OD is equal to kOB.

Am I wrong? Also, I got a negative value for "k". Does that mean my points are collinear?

Thanks!

[kinslayer]

What are the conditions in proving something is collinear (as opposed to being parallel?)

Also, does it what matter which points we prove?

Eg. A question states that to show that points O, D and B are collinear.

I tried to show OD is equal to kDB k being a constant

But answer showed that OD is equal to kOB.

Am I wrong? Also, I got a negative value for "k". Does that mean my points are collinear?

Thanks!

You're not wrong, the two conditions are equivalent.

If you write DB = OB - OD, then your condition is equivalent to saying that OD = k(OB - OD). You can solve for OD to get the same condition as in the solution, the constant just looks a bit different.

There aren't any restrictions on the constant, except that you might want to assume that k is nonzero to keep things interesting. Negative values are fine.

[allstar]

any help is greatly appreciated

an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.

[BLACKCATT]

Can someone clarify why the distance of a plotted solution on an argand diagram is r^(1/n)? Isn't the radius the same from any given point connected to the origin? E.g why is the radius of 2cis(2pi/3) and 2cis(-2pi/3) different to 2cis(0)

[lzxnl]

Can someone clarify why the distance of a plotted solution on an argand diagram is r^(1/n)? Isn't the radius the same from any given point connected to the origin? E.g why is the radius of 2cis(2pi/3) and 2cis(-2pi/3) different to 2cis(0)

The magnitudes of the three numbers given above are the same.
The magnitude of r cis t is simply |r|

[M_BONG]

Hey, can someone help me in understanding this diagram please?
From my interpretation, P is the pushing force in the diagram.

So why is F (friction) moving in the same direction as P?

Cheers!

[keltingmeith]

Hey, can someone help me in understanding this diagram please?
From my interpretation, P is the pushing force in the diagram.

So why is F (friction) moving in the same direction as P?

Cheers!

There's also the 5kg block pulling it to the right, which is why friction is in the same direction as P.

In truth, they haven't given you enough information - there's no way to tell if P is greater than or less than 5g, and you need to know this to figure out which way friction is working. I'm assuming this is a third party paper?

[psyxwar]

Question 2
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.

The solutions say that D is smaller than 4.9N (the value of D for when it is equal to maximal friction). Just a couple of questions about this:

- Is it better to write D is between 0 and 4.9N, not inclusive of end points?
- Can someone explain the point of moving? I assumed the answer would be R+\{4.9}, as I thought it'd only be on the point of moving when D is equal to maximal friction. Is the solution wrong or am I just misunderstanding what 'point of moving' means? I'm aware that friction is variable and Fr =< mju(N), but I wouldn't have thought an object was considered to be on the point of moving unless D was equal to maximal friction.

[M_BONG]

There's also the 5kg block pulling it to the right, which is why friction is in the same direction as P.

Yeah, it's an Insight exam 1 2010

I am sort of getting the gist of this. Can you explain a little bit more?
When will friction act in the same direction as the pushing force/thrust (whatever you call it)?

I always thought friction acts in the opposite direction of motion.

Also, really petty point. I was doing an exam today, I wrote 1.5 instead of 3/2 and I realise VCAA takes marks off for non-exact answers. Would they take a mark off if I write 1.5 instead of 3/2 since 1.5 is technically non-exact?

Thanks (sorry for two questions at once)/

[psyxwar]

Yeah, it's an Insight exam 1 2010

I am sort of getting the gist of this. Can you explain a little bit more?
When will friction act in the same direction as the pushing force/thrust (whatever you call it)?

I always thought friction acts in the opposite direction of motion.

Also, really petty point. I was doing an exam today, I wrote 1.5 instead of 3/2 and I realise VCAA takes marks off for non-exact answers. Would they take a mark off if I write 1.5 instead of 3/2 since 1.5 is technically non-exact?

Thanks (sorry for two questions at once)/

1.5 is an exact answer though, it's just in decimal form. 1.5 and 3/2 are exactly equivalent, unlike say, pi and 3.14 or g and 9.8 because those are approximations.