[Pinkhead]

sorry it took a while, i've been busy revising for chemistry unit 1 :P

[Jukeboxing]

(Original post by nm786)
sorry if i am asking so many questions.
But i really need to understand this,
why do you need to find the magnitude of R? can't you just use "F=ma"

If we are to find the mass m, a scalar quantity, then we cant leave R in vector form.

Mass cannot be represented in terms of i and j, only vector quantities can be represented in terms of i and j.

Scalar quantity - Magnitude Only
Vector quantity - Magnitude and direction

[Mouth]

I'm terrible at vectors. They throw me every time
Only one more day to practice and I've been doing history all night and also have politics to revise tomorrow as well.

[Cleoleo]

this is gonna sound really stupid, but what exactly is impulse? O_o
I use it but never understood what it was exactly

[Pinkhead]

(Original post by Cleoleo)
this is gonna sound really stupid, but what exactly is impulse? O_o
I use it but never understood what it was exactly

I was taught that it's a change in momentum of an object.
It's also the force x time.
So i guess you could say that its the change in momentum when a force is applied for a certain amount of time?
lol, probably completely wrong, but I thought i'd give it a shot.

[ak395]

(Original post by Mouth)
I'm terrible at vectors. They throw me every time
Only one more day to practice and I've been doing history all night and also have politics to revise tomorrow as well.

I found vectors really hard at first too but after doing all the qs in the book, using the solutionbank to help I think I understand them waaaaaaay better now. Mymaths helped a lot as well as it makes it all very clear and clears up misconceptions. Now when I get a vector q i'm like yaaaaaaaaaaaay!!! vectors!!!!

[ak395]

(Original post by Pinkhead)
I was taught that it's a change in momentum of an object.
It's also the force x time.
So i guess you could say that its the change in momentum when a force is applied for a certain amount of time?
lol, probably completely wrong, but I thought i'd give it a shot.

I agree with u it is change of momentum cuz ur effectively doing
m(v-u)= mv-mu=Momentum after-momentum b4

[99llewellyn99]

(Original post by Pinkhead)
sorry it took a while, i've been busy revising for chemistry unit 1 :P

good luck then for tomorrow

[nm786]

(Original post by Pinkhead)
Attachment 147996

a) R is parallel to the vector i-2j so if we draw out i-2j, we can see that the angle between the force and the component j =

b) If the force is parallel to the vector i-2j, that's the same as saying

so
so if we equate the i components: (1)
if we equate the j components: (2)

substitute (1) into (2) and you get
work that out and youll get the answer

c) using
replace q with the value they give you (q=1) and you should get p=-2

now you can put these back into the resultant force:

Find the magnitude of this force, so , which gives

now use f=ma so

thank you very much.

(Original post by Jukeboxing)
If we are to find the mass m, a scalar quantity, then we cant leave R in vector form.

Mass cannot be represented in terms of i and j, only vector quantities can be represented in terms of i and j.

Scalar quantity - Magnitude Only
Vector quantity - Magnitude and direction

alright, then, that makes sense now,

[Spicey]

(Original post by ak395)
I found vectors really hard at first too but after doing all the qs in the book, using the solutionbank to help I think I understand them waaaaaaay better now. Mymaths helped a lot as well as it makes it all very clear and clears up misconceptions. Now when I get a vector q i'm like yaaaaaaaaaaaay!!! vectors!!!!

Haha same with me, except in FP1 with parabolas and rectangular hyperbolas. I found them so difficult a few weeks ago, but when you work really hard to understand them, you actually kind of enjoy seeing them in the paper because its kind of a chance for you to show off how much you've worked

[nm786]

(Original post by Pinkhead)
sorry it took a while, i've been busy revising for chemistry unit 1 :P

good luck for tomorrow, even though you don't need it

[wam-bam]

(Original post by Pinkhead)
Attachment 147996

Hi can someone explain question 4 (especially part a) on paper june 02, i really don't understand how to do it!?!?!

thanks in advance


A box of mass 6 kg lies on a rough plane inclined at an angle of 30 degrees to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude P newtons, as shown in Fig. 2. The line of action of the force is in the same vertical plane as a line of greatest slope of the plane. The coefficient of friction between the box and the plane is 0.4. The box is modelled as a particle.

Given that the box is in limiting equilibrium and on the point of moving up the plane,
find,

(a) the normal reaction exerted on the box by the plane, (4 marks)

(b) the value of P. (3 marks)

The horizontal force is removed.

(c) Show that the box will now start to move down the plane. (5 marks)


heres a link to the digram with the question and also the mark scheme

http://www.thestudentroom.co.uk/atta...4&d=1149284890

http://www.thomas-reddington.com/upl..._june_2002.pdf

[AmrinderRai]

So have any of your guys done solomon papers for M1, or are going to do them?
Is there much benefit or point in doing them ?

[gdunne42]

(Original post by wam-bam)
Hi can someone explain question 4 (especially part a) on paper june 02, i really don't understand how to do it!?!?!

thanks in advance


A box of mass 6 kg lies on a rough plane inclined at an angle of 30 degrees to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude P newtons, as shown in Fig. 2. The line of action of the force is in the same vertical plane as a line of greatest slope of the plane. The coefficient of friction between the box and the plane is 0.4. The box is modelled as a particle.

Given that the box is in limiting equilibrium and on the point of moving up the plane,
find,

(a) the normal reaction exerted on the box by the plane, (4 marks)

(b) the value of P. (3 marks)

The horizontal force is removed.

(c) Show that the box will now start to move down the plane. (5 marks)


heres a link to the digram with the question and also the mark scheme

http://www.thestudentroom.co.uk/atta...4&d=1149284890

http://www.thomas-reddington.com/upl..._june_2002.pdf

I'll assume you know how to resolve forces on an inclined plane
You could resolve the forces parallel and perpendicular to the plane and form simultaneous equations that you solve for P and R
BUT
It's less fiddly if you resolve vertically for part (a) and (b) because then you can ignore P
In the down direction you have weight and a component of friction
6g and 0.4Rsin(30)
In the up direction you have a component of the reaction
RCos(30)

6g + 0.4Rsin(30) = RCos(30) - solve for R

You can then resolve horizontally to find P

[Alexlol28]

(Original post by arnab)
any one think there will be a big vector question in wednesday paper?

To be honest with you, i don't think so, no, in the last two past papers the last question was a vector one.
I reckon the vector question will be pretty simple, worth 6 marks, asking to find speed the the position vector at a certain time.
I have a feeling that the last question is either going to be on pulleys or on inclined planes, AND definitely something including the string breaking and finding the time until it stops

[Alexlol28]

Reading through the posts I have seen that many people are struggling with vectors.
- To find speed: add the squares of velocity and then take the square root
- r = ro + vt, where ro is its position vector a t =0
- v = u + at, using values in i and j notation
- To find the distance AB between to vectors, you do rb - ra, where ra and rb are the position vectors of A and B at a certain time
- Due north, means equate i components to 0, to resolve for t
- Due west, means equate j components to 0, to resolve for t

For practice on vectors, do the last question from June 09, June 11 and Jan 11

Hope that helps

[ak395]

I'm really stuck on a q - can anybody help me plzzz....
question 6c:
http://www.edexcel.com/migrationdocu...e_20110119.pdf

Answer:
30+ F =120sinα OR 30 − F = 120sinα
So F = 42N acting up the plane.

My question is that if u do 30 − F = 120sinα won't f be negative so f will act down the plane?

[Waqar.]

Would I be able to work out my vector answers using column vectors? I think it makes calculations simpler and I always leave my asnwer in the form they want (i's and j's) but im not sure if i'd lose method marks if the person marking doesn't quite understand my method?

[Alexlol28]

(Original post by ak395)
I'm really stuck on a q - can anybody help me plzzz....
question 6c:
http://www.edexcel.com/migrationdocu...e_20110119.pdf

Answer:
30+ F =120sinα OR 30 − F = 120sinα
So F = 42N acting up the plane.

My question is that if u do 30 − F = 120sinα won't f be negative so f will act down the plane?

By writing 30 − F = 120sinα, (i would write 30 − 120sinα = F), your F would be negative, but that is because force is a vector, it has direction, as you are solving upwards and F acts downward, this is why your F is negative.

[Alexlol28]

(Original post by Waqar.)
Would I be able to work out my vector answers using column vectors? I think it makes calculations simpler and I always leave my asnwer in the form they want (i's and j's) but im not sure if i'd lose method marks if the person marking doesn't quite understand my method?

Solving using column vectors is fine, I do that as well, as long as you put it back into i and j notation, its fine. What i also tend to do is to put a key, so lets say i have (xi + yj), i will write (xi + yj) gives ......(and write it as a column vector)