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The Official Edexcel M1 June 2005 Revision Thread!

Surprised nobody has created this yet with the exam on Tuesday! This looks like its going to be my worst unit unless something changes between now and Tuesday. Anyway, I think we should use this thread to discuss the exam and post any M1 questions.

My first question is not a past exam question but about resolving, etc. Could someone please explain how you resolve perpendicular and parallel to a plane? I keep thinking I get the sin and cos the right way round and when I do I miss out a force in one of the equations! Help. :frown:

Reply 1

username22878

2

It's cos when it's across (adjacent to the angle) and sin when it's up (opposite to the angle).

Reply 2

foxylady

does anyone have a copy of the sylabus? my revision is going so badly! ive just started today and realised that im so bad at mechanics!! :frown:

oh and do we get a formula sheet?

Reply 3

Christophicus

18

lol, for m1! no.
now for m5 it's acceptable, cos you can't learn all the formulae for moments of inertia. See page 19 of 40.

http://www.edexcel.org.uk/VirtualContent/83441/Formulae_book_Revised_Specification.pdf

Reply 4

Chris87

OP

1

ljfrugn

It's cos when it's across (adjacent to the angle) and sin when it's up (opposite to the angle).

Yeah I can do it easily when its just a horizontal surface but say when its raised at an angle it gets confusing. Could someone post a diagram with a general example to make things look a bit easier? Thanks.

Reply 5

username22878

2

Do you mean, like, when you have to find the component of weight down an inclined plane and stuff?

Reply 6

DaKe

I've got a question....

Its from the M1 mock paper...question 7 (a)

A small parcel of mass 2kg moves on a rough plane inclined at an angle of 30 degrees to the horizontal. The parcel is pulled up a line of greatest slope of the plane by a means of a light rope which is attached to it. The rope makes an angle of 30 degrees with the plane. The coefficient of friction is 0.4.
Given that the tension in the rope is 24N,

(a) Find to 2sf the acceleration of the parcel.

I've done it three times in various ways but still keep getting -0.2157.... which isn;t right according to some answers i already have. Apparently it should be 4.5ms-2???

I;m sure that i'm resolving it correctly...so i don't know what it could be?????

N e help is much appreciated....

Thanks

Reply 7

Audionut

Dake I did the question and I got it right, see the attachment below.

M1Question7.PNG16.5KB

Reply 8

ciara

13

I'm just revising now, am resitting cos i only got 73 last year and that was after a lot of revision. Am really really terrible! I know all the stuff but it takes me so long for each question to work out what to do that I run out of time :frown:

.

Cxx

Reply 9

goku999

3

Its all about looking at the whole diagram and where the forces act. If it is at an angle then the particle will experience both horizontal and vertical movement.
Its the ability to apply the correct trigonometric equations/formula on that force applied to that particle.
It takes sum practice and remember this saying:

Sex On Holiday

Sin = Opposite/Hypotenuese

Comes After Having

Cos = Adjacent/Hypotenuese

Tonnes Of Alcohol

Tan = Opposite/Adjacent

Though the tan is not needed when taking components on the plane

Gud Luk and remember it.

Reply 10

DaKe

PoseidonX

Dake I did the question and I got it right, see the attachment below.

kool thanks for your help....

i c where i went wrong now. I was resolving R=ma which is totally wrong cos R isn't moving is it!!!!!

Thanks

Dake

Reply 11

KAISER_MOLE

7

One thing i find is that they like to keep using the angle 36.87...degrees
i.e arctan=0.75 arcsin=0.6 and arccos=0.8 ...so if you know these values before the exam, it is likely to save you 10 seconds, hoorah!

:biggrin:

Reply 12

Malik

12

I always prefer to resolve horizontally and vertically rather then parallel and down the plane. Does anyone know if it really matters what way you use, for example is there a question which cant be solved with only horizontal and vertical.

Thanks

Reply 13

mafakuza

Does any one have the mark scheme for november 2004 ?

Im stuck on a few questions.

Here is one of the questions im stuck on:

"A tent peg is driven into soft ground by a blow from a hammer. The tent peg has mass 0.2kg and the hammer 3kg. The hammer strikes the peg vertically.

Immediatley before the impact the speed of the hammer is 16ms^_1 . It is assumed that immediatley after the impact the hammer and peg move together verticaly downwards.

(a) find the common speed of the peg and hammer immediatley after the impact.

Untill the peg and hammer come to rest the resistance exerted by the ground is assumed to be constant and of magnitude R newtons. The hammer and peg are brought to rest 0.05 s after the impact.

(b) find to 3 s.f the value of R."

I have done part (a) and i get the speed to be 15.

for part (b) i have worked out the aceleration of the peg and hammer using a constant acceleration equaiton and i get the aceleration to be (-0.75).
Im prety sure i have to apply this to "F=ma" I know the mass and acceleration but how do i work out the resultant force.

Reply 14

foxylady

I thought I'd mastered pulley questions now, and maybe its just me getting tired but can anyone help me out with this?

Two particles A and B are connected by a light inextensible string which passes over a smooth fixed pulley. A has mass 4kg and descends 2.8m in the first 2 seconds after the system is released from rest, the hanging parts of the string being verticle. Find the mass of B and the tension in the string.

Reply 15

goku999

3

mala2k

I always prefer to resolve horizontally and vertically rather then parallel and down the plane. Does anyone know if it really matters what way you use, for example is there a question which cant be solved with only horizontal and vertical.

Thanks

The way you choose to resolve depends on the number of unknowns u have and whether there are more forces parallel and perpendicuar to the plane or whether there are forces that are not.

Reply 16

ciara

13

foxylady

I thought I'd mastered pulley questions now, and maybe its just me getting tired but can anyone help me out with this?

Two particles A and B are connected by a light inextensible string which passes over a smooth fixed pulley. A has mass 4kg and descends 2.8m in the first 2 seconds after the system is released from rest, the hanging parts of the string being verticle. Find the mass of B and the tension in the string.

I'm tired too but thought I could do with some practice..

s = ut+1/2at^2
2.8 = 0.5 x a x 2^2
2.8 = 2a
a = 1.4ms^-2

R(verticallyforA): F=ma
4g - T = 4 x 1.4
T = 4g - 5.6
T = 33.6N

R(verticallyforB): F=ma
T - mg = 1.4m
33.6 - 9.8m = 1.4m
33.6 = 11.2m
m = 3kg

Sorry if that's wrong..
Cxx

Reply 17

mizfissy815

12

foxylady -
I thought I'd mastered pulley questions now, and maybe its just me getting tired but can anyone help me out with this?

Two particles A and B are connected by a light inextensible string which passes over a smooth fixed pulley. A has mass 4kg and descends 2.8m in the first 2 seconds after the system is released from rest, the hanging parts of the string being verticle. Find the mass of B and the tension in the string.

First use
S= ut +1/2 at²
2.8= 0 + 1/2a(2)²
a= 2.8/2
a= 1.4 m/s²

Now draw the force diagram and you'll get 2 equations...add them up and T (Tension) is cancelled out.
x= unknown mass of B

T - xg = xa
4g - T = 4a
^{_____________}
4g - xg = 4a +xa

Sub in value of g=9.8 and a= 1.4

(4*9.8) - 9.8x = (4*1.4) + 1.4x
39.2 - 9.8x = 5.6 + 1.4x
39.2-5.6 = 1.4x+9.8x
33.6= 11.2x
x= 33.6/ 11.2
x= 3 kg

Since... 4g-T = 4a
T= 4g - 4a
T= ( 4*9.8) -(4*1.4)
T= 39.2- 5.6
T= 33.6 N
---------

Can someone plz tell me if this is right?

Reply 18

mizfissy815

12

I see Ciara has already beat me to it. :wink:

Our answers are the same, thats a good sign.

Reply 19

Malik

12

goku999

The way you choose to resolve depends on the number of unknowns u have and whether there are more forces parallel and perpendicuar to the plane or whether there are forces that are not.

So in some cases the only way to solve a question would be to do it a certain way. One method may not work on certain situations correct?

The Official Edexcel M1 June 2005 Revision Thread!

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