M1 - Edexcel, 21st May - ULTIMATE THREAD

lol its easier than that. cos if you go through that angle when resolving, sin if you don't.

Reply 62

I just use cos all the time, seems easier.

But you just need to work out 90-the angle first hand to use cos al the time.

Reply 63

rahulsood

lol its easier than that. cos if you go through that angle when resolving, sin if you don't.

You lost me :s-smilie:

I've been sticking with sohcahtoa, didn't think it got any easier than that! (I'm not really cut out to be a mathematician...)

Reply 64

when your resolving vertically for example, if you turn through the angle, it's cos. if not, then it's sin.
It's the easiest way, you don't even need to think.

Reply 65

Ah right, I get you :smile:

Thank you!

Reply 66

OggYx

I have such variation in my mocks. I just did 2 today, got 77 in one and 100 in another.
Meh.
Most hated questions: Pulleys - ESPECIALLY if it asks you the distance or time particle A takes to stop when the other particle hits the ground. Hate them.

Reply 67

A GUIDE TO RESOLVING FORCES

Take this diagram for example...

Given the plane is smooth and the particle is held in equilibrium...

We're going to resolve it to find the normal, so we're going to resolve it in the direction of the normal... So, in

\nwarrow

that kind of direction.

Right, so R is parallel to the direction we're resolving in so that can stay as it is...

The force acting horizontally on the particle is NOT parallel...so let's look at that.

We want to resolve it so that it's PARALLEL to R, so we need to move it clockwise to get there. Now, there are two methods.

First method is to use sin of the angle that is measured from perpendicular to the direction you're resolving in... which would make it

\sin 30

.
Second method is to use cos of the angle that is measured from parallel to the direction you're resolving in... which would make it

\cos 60

.

They both give the same answers so it's up to use which you find easier.
Personally, I just use cos all the time which saves me from having to remember when to use sin or not.

Okay, so I'm using the cos rule here, and resolving the 15N, I get

15 \cos 60

. But because it's acting AWAY from R, it's negative.

Next I'm going to have to resolve the 12gN. Make it parallel with R, which would give me

12g \cos 30

. Again because it's acting AWAY from R, it's negative.

Right, so we get that

R - (15 \cos 60) - (12g \cos 30) = 0

R = (15 \cos 60) + (12g \cos 30)

R = 109.3N

Little big, I grant you, but I didn't really work out the figures before I drew it out. I just stuck some arbitrary numbers in for the sake of example.

Hope this helps!

Reply 68

Legend. Rep. Love you forever :smile:

Reply 69

Inquilaab

Has anyone got the Jan 2008 paper?

Reply 70

Junker

rahulsood

your teacher must really suck.

Nah, she's a good teacher, it's just that for this certain aspect of M1 her explanation always confuses me.
EvenStevens, thanks for your guides it has helped me a bit. Can anyone help me on these topics because these are the only ones I struggle with now:
Connected particles
Pulleys on slopes, especially when you have to use a suvat equation and when you use F=ma, is it T=ma when the string from the pulley is ON the slope or T-mg=ma just like the hanging weight?

Reply 71

Jordan_U

vectors just keeps jumping out of my head, all the past papers I do I just seem to be a bit :confused:

on vectors.... and I want to do well in M1, so i don't rely on C1 and C2 considering the upper sixth at my school failed M1 and are now doing S1, M1, C3, C4 and I think C1 and C2

Reply 72

where do you get past papers from? :frown:

Reply 73

By connected particles, do you mean particles connected by string? Or when 2 particles merge into one?

Reply 74

Junker

Yeah, connected by a string like a car and a trailer.

Reply 75

Gimothy

Inquilaab

Has anyone got the Jan 2008 paper?

I have a hard copy. I don't really want to type all the questions out, but if you tell me a topic ill find the relevant question for you.

Reply 76

cant you scan it please?

Reply 77

Alright, I'm going to use the example from June 2005...

First of all, lets draw all the forces on there...

Question a) Find the acceleration of both the lorry and the car.

Well we know that

F=ma

and we've got aaall the forces on there, so lets resolve

\rightarrow

.

nB. Remember F = ma uses the MASS of the particles, not the weight. The mass of the particles is what they weigh without gravity acting upon them. So if something has mass of 900, it has a weight of 900g. Remember this. I've known it to completely mess up people's calculations as they put 900g in their F=ma calculations and got something completely different.

1500 - 600 - 300 = (1600+900) \times a

Therefore

600 = 2500a

Which makes

a = 0.24 ms^{-2}

.

Easy enough?

Right, part b) Tension in the towbar. Little trickier...

Again, we have to use F=ma for our resolving because it's moving and it's not in equilibrium...

You can resolve for T using the properties of the car or the lorry, you'll wind up with the same answer either way... but I'm going to go with the car because it just looks easier to me ^_^

So let's resolve for T with the car, in a

\rightarrow

sort of direction...

It may help to re-redraw the diagram focussing on the particle you're resolving from and what you're trying to find, to help you visualise it better...

The circle is the car and I forgot to draw on the 900gN. Apologies.

F = ma \Rightarrow T \cos15 - 300 = 900 \times 0.24

T \cos15 - 300 = 216

T \cos 15 = 516

T = \frac{516}{cos15} = 534.2N

Awesome!

Parts c) and d) to follow...

Reply 78