CM_Tutor leaving (1 Viewer)

[CM_Tutor]

for the snow for a week's holiday this weekend.

So, the last time I'll be around before going will probably be Sat AM, and I won't be back until late on Sun 18th.

If there are any things people want to ask about before I go, please post here.

Also, if things come up while I'm away that you want me to look at, you can leave me a message here.

 

[:: ck ::]

ooOo ~! have fun CM =)

do u reckon u cud post up some challenge questions for pretty much any topic that will be gud revision for trials n stuff ~! if it isnt too much trouble

 

[Grey Council]

have an awesome time!
or should I say: Break a leg!




On a side note, I was doing some circle geo revision last night. I have a collection of past trial questions (I think you've seen the collection, unless i'm mistaken, I think I sent it to you. I think), and I was doing that collection. Thing is, they're so damn hard. So I stopped doing them, and started doing the fitzpatrick 4u chapter on harder circle geo. Whaddya know, thats damn hard too.


Is the fitzpatrick chapter worth doing? Or should I stick to doing the past 4u trials question 7/8 circle geo?
Or is there an even better way to revise this topic?
*takes a breath*
whoa, that was quite a mouthful. :-\

 

[CrashOveride]

A stone of mass m is proijected with velocity 30m/s from a point at the foot of an inclined plane making an angle of 30degrees to the horizontal. The path of the projectile can be assumed to be in the vertical plane containing the line of the greatest slope of the inclined plane (btw, whats this about?) If angle of projectile is > 30 degrees, for wahts values of & will the stone strike the inclined plane:

a) horizontally b) at right angles

 

[Rorix]

They mean that the mass is above the inclined plane.

 

[CrashOveride]

The nozzle of a water hose is at a point O on the horizontal ground. The water comes out of the nozzle with spped U m/s. Neglecting the air-resistance, prove that the water can reach the wall at a distance d from O, if U² > gd, where g is accel. due to gravity. If U² = 4gd, also prove that the max height that can be reached by the jet on the wall is given by 15d / 8.

I interpreted the question as having an angle first but then took it as 0. I get something close to their answer but i think my approach was flawed. If someone could to teh 1st part i think i could do the 2nd part. Kind regards.

 

[BlackJack]

1 way to do it:
Starting from the equations of motion: Assume this nozzle is angled at Θ,
y=U sinΘt - 0.5gt²
x=U cosΘt

The minimum conditions required for the water to reach the wall is for the stream to hit its exact foot. Or, the water could hit it above the foot. Hence, we get y ≥ 0 when x = d.

By substitution, from eq. for x we get:

d = U cosΘ t,
t is the time required for the water to reach the wall.

express as t = d / (U cosΘ ) and sub into the other eq., with y = 0:

0 ≤ U sinΘt - 0.5gt²
0.5gt ≤ U sinΘ
U sinΘ ≥ 0.5gt
≥ 0.5gd/(U cosΘ )

=> U²sinΘcosΘ ≥ 0.5gd

Note sinΘcosΘ = sin2Θ / 2, which has maximum of 1/2, the minimum guaranteed U is the given answer.

 

[Estel]

lol! Got a heart attack reading the title...

 

[CrashOveride]

A stone is projected with an initial velocity v1 in a vertical plane at an angle ¥ to the horizontal and hits ground at A. another stone launched with initial veloc. v₂ at same angle but v₂ > v₁. Let B(x2, y2) and C(x1,y1) be two points on the respective flight paths at the same time t. Show that gradient of segment BC is independant of time t. <B>Done</B>

When 2nd stone just clears a wall of hieght h, the first stone hits the ground at A. If the wall stands at point D on the level ground, prove that AD = hcot¥. <B>Done</B>

Further show that tan(-Ø) = tan - [gt / (v2 cos¥)], where Ø is the angle made by the downward flight of the faster stone with the horizontal and T is the time of flights for the slower stone. Hence show v₂(tan¥ + tanØ) = 2v₁tan¥

EDIT: Ok i can show that tanØ = tan¥ - (gt/(V₂cos¥)) which is kind of close, just 't' is not the time of flight of the slower stone, its just a time 't'. Hrmm

 

[Grey Council]

lol! Got a heart attack reading the title...

heehee!
same same.



although for some stupid reason I never end up doing the questions he posts, with a few exceptions.

@_@

I will, I've got them all saved up in a file, lol.

 

[flyin']

Almost thought BoS was going to lose a very helpful member, fortunately that isn't the case.

 

[CrashOveride]

If anyone feels like doing some revision on harder projectile motion questions, plz have a go at the ones i posted. Im stuck :'(

 

[CrashOveride]

Some more. It's not going too well.

Missile is fired from O with initial velocity U at an angle ¥ with the horizontal. Prove that it describes a parabola of focal length U²cos²¥ / (2g). <B>Done.</B>

Also prove that any point P(x,y) within and on the circle x² + y² = v⁴/g² is in danger of being hit by the missile.
I think here the v is suppose to be U. I was made aware of the fact we could show max range of projectile was U²/g but i wasnt totally clear on how we can translate this for the whole scenario. I thought perhaps we could solve the eqn. of the flight path and the circle simultaneously because all contact would be made on the circle. So the result of this simultaneous equation should be the original circle ??

EDIT: I started doing this and it appeared to get messy]

And also for the earlier water hose question, i had got up to the line U² > gd/(sin2Θ). I already asked BlackJack but just to confirm could you tell me why we can just take the max value for sin here?

EDIT: Because we're allowed to build the question how we want because its general? Because my line here would give us a general case right, for all angles? Hmmm thanks in advance.

 

[CM_Tutor]

do u reckon u cud post up some challenge questions for pretty much any topic that will be gud revision for trials n stuff ~! if it isnt too much trouble

Will try to post some questions before I go. (Have you tried the induction Q to do with Fibonacci and the golgen number I posted in Extn 1 - there's some good stuff in that one. )

 

[CM_Tutor]

On a side note, I was doing some circle geo revision last night. I have a collection of past trial questions (I think you've seen the collection, unless i'm mistaken, I think I sent it to you. I think), and I was doing that collection. Thing is, they're so damn hard. So I stopped doing them, and started doing the fitzpatrick 4u chapter on harder circle geo. Whaddya know, thats damn hard too.


Is the fitzpatrick chapter worth doing? Or should I stick to doing the past 4u trials question 7/8 circle geo?
Or is there an even better way to revise this topic?
*takes a breath*
whoa, that was quite a mouthful. :-\

Some of Fitzpatrick's 4u exercise on circle geometry is pretty nasty, and it's also undirected (ie. it is prove ..., with no guidance by breaking up into sub-parts).

If you don't have trouble with the more straight-forward questions, and you know all of the theorems, etc, then I'm afraid past trial practice is probably best.

 

[CM_Tutor]

A stone of mass m is proijected with velocity 30m/s from a point at the foot of an inclined plane making an angle of 30degrees to the horizontal. The path of the projectile can be assumed to be in the vertical plane containing the line of the greatest slope of the inclined plane (btw, whats this about?) If angle of projectile is > 30 degrees, for wahts values of & will the stone strike the inclined plane:

a) horizontally b) at right angles

Taking the angle of projection as @, and firing from the origin, stone has equations of motion of:

x = 30tcos@ and y = -5t² + 30tsin@

(a) To strike horizontally, stone must be moving in x direction only. That is, y' = 0
-10t + 30sin@ = 0
t = 3sin@

So, strikes slope at t = 3sin@, and at that time tan 30 = y / x. From this, I get @ = tan^-1(2&radic;3 / 3) = 49.1° (3sf)

(b) To strike perpendicular to the surface, must land at 60° to the vertical, and it must do so at a time t > 3sin@

It follows that tan 60° = -y' / x' = (10t - 30sin@) / 30cos@
So, find t, then require that point to lie on the slope.

 

[CM_Tutor]

The nozzle of a water hose is at a point O on the horizontal ground. The water comes out of the nozzle with spped U m/s. Neglecting the air-resistance, prove that the water can reach the wall at a distance d from O, if U² > gd, where g is accel. due to gravity. If U² = 4gd, also prove that the max height that can be reached by the jet on the wall is given by 15d / 8.

I interpreted the question as having an angle first but then took it as 0. I get something close to their answer but i think my approach was flawed. If someone could to teh 1st part i think i could do the 2nd part. Kind regards.

The nozzle definitely fires water at an angle, and nike33 and I discussed a related problem in http://www.boredofstudies.org/community/showthread.php?t=29880&page=1

 

[CrashOveride]

Ah ok, makes perfect sense now

If you have any more spare time, could you try the other two i posted. Then you should get some sleep before the long journey

 

[CrashOveride]

(b) To strike perpendicular to the surface, must land at 60° to the vertical, and it must do so at a time t > 3sin@

It follows that tan 60° = -y' / x' = (10t - 30sin@) / 30cos@
So, find t, then require that point to lie on the slope.

You mean to say 30° to the vertical

I got t = 3sqrt(3)cos@ + 3sin@ ...(1)
Now this tells us its 30 degrees to teh vertical anywhere, so to require it that this is the case when it strikes the plane (i.e. perpendicular to the plane)

Sub the result in (1) into:

tan30 = -5t² + 30tsin@ / [30tcos@]
When i do that i guess this messy thing which i try simplifying a few times trying to get everything into tans.

My last 'dead-end' line was: tan²@(270 - 45sqrt(3)) + tan@(90sec@ - 180) + (270 - 105sqrt(3)) = 0

 

[BlackJack]

(b) So it is, the final angle is 30&deg; to the vertical...
tan 60&deg; = -y'/x' = (10t - 30sin@) / 30cos@
10t = 30 sqrt(3) cos@ + 30 sin@
t = 3sqrt(3)cos@ + 3sin@ ...tick.

Ah,
tan30 = (y)/(x) = (30tsin@ - 5t²) / (30tcos@) <- this line.
= tan@ - t / (6cos@)
sub => = tan@ - (3sqrt(3)cos@ + 3sin@) / (6cos@)... etc. Should be fine now.

edit: welll, yeah I meant minus.