VCE Specialist 3/4 Question Thread!

[lzxnl]

Did you learn this outside of school...? Where/how did you learn to make all of those connections?
How do you have the time to be learning extra stuff this year.
I actually have mad respect for you right now lol.

It never occurred to you that people said Fx=change in 1/2 mv^2 and acceleration just happens to have a d(1/2 v^2)/dx somewhere?
Nah, I looked up somewhere for a proof that kinetic energy really was 1/2 mv^2 and saw that.
And I didn't do much learning at all this year. Kinetic energy? That would have been...year ten, when I DID have the time to be learning extra stuff

[sydneyboy]

A train mass m accelerates on a straight levelled track under a constant force P. There is a resistance proportional to v^2. Let the terminal velocity = V

Using Newtown second law find a in terms of P,v,K the constant of proportionality. When a=0, v=V.

I think i can do this bit, however the hence show part throws me

Hence show a=p/(mV^2x(V^2-v^2))

Also... this one threw me when they turn around and ask me to graph out all my solutions );

Two particles represented by r(t)= 6t^2i +2t^3-18tj AND r(t)=13t-6i +3t^2-27j
After working out where they cross paths and where they collide, how do I go about graphing it all out?(inc where they cross/collide)

[brightsky]

P-kv^2 = ma
a = (P-kv^2)/m
when a=0, v=V:
0 = (P-kV^2)/m
P-kV^2 = 0
k = P/V^2
sub into original equation:
a = (P-(P/V^2)v^2)/m = (P - Pv^2/V^2)/m = P(1-v^2/V^2)/m = P(V^2-v^2)/mV^2
pretty sure this is right, although it doesn't seem to coincide with the answer you've provided...

find the Cartesian equation of the path along which each particle moves. to do so, let x = i-component and y=j-component and solve simultaneously. the Cartesian equation should be something you recognise.

[lzxnl]

A train mass m accelerates on a straight levelled track under a constant force P. There is a resistance proportional to v^2. Let the terminal velocity = V

Using Newtown second law find a in terms of P,v,K the constant of proportionality. When a=0, v=V.

I think i can do this bit, however the hence show part throws me

Hence show a=p/(mV^2x(V^2-v^2))

Also... this one threw me when they turn around and ask me to graph out all my solutions );

Two particles represented by r(t)= 6t^2i +2t^3-18tj AND r(t)=13t-6i +3t^2-27j
After working out where they cross paths and where they collide, how do I go about graphing it all out?(inc where they cross/collide)

P-kv^2 = ma
a = (P-kv^2)/m
when a=0, v=V:
0 = (P-kV^2)/m
P-kV^2 = 0
k = P/V^2
sub into original equation:
a = (P-(P/V^2)v^2)/m = (P - Pv^2/V^2)/m = P(1-v^2/V^2)/m = P(V^2-v^2)/mV^2
pretty sure this is right, although it doesn't seem to coincide with the answer you've provided...

find the Cartesian equation of the path along which each particle moves. to do so, let x = i-component and y=j-component and solve simultaneously. the Cartesian equation should be something you recognise.

I can't see anything wrong with brightsky's working. Maybe you typed the question incorrectly.

As for crossing paths and colliding, first solve r1(t) = r2(t) for colliding
then r1(s) = r2(t), i.e. different variable, as the path is independent of what you call the time variable.

[sydneyboy]

Might be a question error, sorry. Thanks for the help guys i found the collision and crossing paths the same place. (27/2 , -81/4) *** woops

[lzxnl]

Might be a question error, sorry. Thanks for the help guys i found the collision and crossing paths the same place. (27/2 , -81/4) *** woops

Are you sure that's the only solution? I haven't done the question yet though.
All collision points are at the same place. Not all path intersections are at the same time.

[zvezda]

Q13 in the essentials textbook in exercise 13E,
Im not able to get the second part of the answer (speed when next at the point of projection).
I keep getting v=u.
Help is appreciated. Thanks

[Alwin]

Q13 in the essentials textbook in exercise 13E,
Im not able to get the second part of the answer (speed when next at the point of projection).
I keep getting v=u.
Help is appreciated. Thanks

Question: A body of mass m is projected vertically upwards with speed u. Air resistance is equal to k times the square of the speed where k is a constant. Find the maximum height reached and the speed when next at the point of projection.

Hint: Remember air resistance opposes the direction of motion. When it's travelling up, F_air is down. But when its going down, the  F_air is up.

iirc not accounting for the change in direction leads to a symmetrical answer ie v=u...

hope it helps =^^=

[BasicAcid]

I'm having a bit of trouble with this question and both of its parts.

A lift cage of 200 kg starting from rest is raised vertically by a force which decreases steadily from an initial value 2100N at a rate of 5N per second until the lift again comes to rest.
a) Find the greatest speed of the lift.
b) Find how high the lift is raised before again coming to rest.

What I've done so far is figure out the relationship (equation) for a force-time function which is F(t) = 2100 - 5t
I've come up with this equation in the j direction with up being positive:
2100 - 5t - 200g = 200a

I then divided both sides by 200 to get (2100 - 5t - 200g)/200 = a and I replaced g with 9.8, a with dv/dt and antidifferentiated it on my calculator getting v = (-(5t - 140)^2)/2000 + c and as it is at rest when t = 0, c = 0.
I then differentiated the above and made it equal to zero on my calculator and solved, I got t = 28.

Now I'm sure I messed up somewhere in my process as when I subbed in t = 28 back into the velocity function, I get v = 0.

So I'm wondering where I've gone wrong and if someone could possibly run me through step by step through both parts.

Thanks in advance.

[brightsky]

yep, your method is fine.

a) P = 2100 - 5t
P - 200g = 200a
(2100-5t) - 200g = 200a
a = (28-t)/40
dv/dt = (28-t)/40
v = t(56-t)/80 + c
now the lift starts from rest, so v = 0 when t = 0:
0 = 0 + c
c = 0
so v = t(56-t)/80
velocity is positive on the interval t E [0,56]. the lift starts from rest, ascends, then stops at t = 56. so max velocity = max speed.
to find max velocity:
dv/dt = 0
t = 28
so max speed is 28(56-28)/80 = 49/5 m/s.

b) the lift stops at t = 56. we know that v = t(56-t)/80.
dx/dt = t(56-t)/80
x = t^2(84-t)/240 + c
we know that x =0 when t = 0
0 = 0 + c
c = 0
so x = t^2(84-t)/240
sub t = 56:
x = 5488/15
so the lift is raised 5488/15 m before it again comes to rest.

I think you might have stuffed up the integration...

[lzxnl]

I'm having a bit of trouble with this question and both of its parts.

A lift cage of 200 kg starting from rest is raised vertically by a force which decreases steadily from an initial value 2100N at a rate of 5N per second until the lift again comes to rest.
a) Find the greatest speed of the lift.
b) Find how high the lift is raised before again coming to rest.

What I've done so far is figure out the relationship (equation) for a force-time function which is F(t) = 2100 - 5t
I've come up with this equation in the j direction with up being positive:
2100 - 5t - 200g = 200a

I then divided both sides by 200 to get (2100 - 5t - 200g)/200 = a and I replaced g with 9.8, a with dv/dt and antidifferentiated it on my calculator getting v = (-(5t - 140)^2)/2000 + c and as it is at rest when t = 0, c = 0.
I then differentiated the above and made it equal to zero on my calculator and solved, I got t = 28.

Now I'm sure I messed up somewhere in my process as when I subbed in t = 28 back into the velocity function, I get v = 0.

So I'm wondering where I've gone wrong and if someone could possibly run me through step by step through both parts.

Thanks in advance.

Look carefully. When t=0 and v=0, c is NOT zero.

[BasicAcid]

yep, your method is fine.

a) P = 2100 - 5t
P - 200g = 200a
(2100-5t) - 200g = 200a
a = (28-t)/40
dv/dt = (28-t)/40
v = t(56-t)/80 + c
now the lift starts from rest, so v = 0 when t = 0:
0 = 0 + c
c = 0
so v = t(56-t)/80
velocity is positive on the interval t E [0,56]. the lift starts from rest, ascends, then stops at t = 56. so max velocity = max speed.
to find max velocity:
dv/dt = 0
t = 28
so max speed is 28(56-28)/80 = 49/5 m/s.

b) the lift stops at t = 56. we know that v = t(56-t)/80.
dx/dt = t(56-t)/80
x = t^2(84-t)/240 + c
we know that x =0 when t = 0
0 = 0 + c
c = 0
so x = t^2(84-t)/240
sub t = 56:
x = 5488/15
so the lift is raised 5488/15 m before it again comes to rest.

I think you might have stuffed up the integration...

Thanks a lot for that mate, I guess the people like you are the ones who'll be getting the 40+s on spesh haha.
Yep I did, look at nliu1995's post.

Look carefully. When t=0 and v=0, c is NOT zero.

Lol I wanted to smack myself in the head with the iPad after reading this post.
Thanks dude.

[Homer]

Zac connects two plastic crates with a piece of rope, fills the crates with toys and pulls them up and down the hallway. He exerts a pulling force of P newtons acting at an angle of   to the horizontal. The first crate has a mass of 4 kg and the second has a mass of 5 kg. The coefficient of friction between the hallway floor and the crates is 0.5.

Zac exerts a pulling force of 43.55 N and the system of crates accelerates at   down the hallway. Find the tension in the rope. Express your answer correct to 1 decimal place.

ANSS: 27N how?

[sydneyboy]

A Skier mass 80kg slides down a friction less slope at an angle of 15 degrees to the horizontal. A drag force, D is proportional to v. The terminal speed is 20ms.

express a in terms of v and k
find the inital acceleration
evaluate k to find v in terms of t
solve the differential equation to find the time till 19ms

ive got a= gsin(theta) - (kv)/m

then i have no idea how to get and in t into the same eq

[Conic]

For the first crate the net force is the horzontal pulling force, minus the friction force, minus the tensile force, so:

(P=pulling force, Fr=friction, T=tension)

Fr=µR, where R is the normal reaction force, so we need to find R. In this case R is equal to the weight force - the horizontal component of the pulling force.



Now we can find the friction force:



Now substitute this into the original equation:







*Note: googled the question to find the acceleration and angle (0.5 and 30).