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OCR Physics A - G481 Mechanics - [13/01/09][@1400]

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Reply 120
Avatar for neomilan
neomilan
OP
17
dave2k
definition?? No derivation:p:

People who hardly revised for this are VERY lucky now that they give you equations...


that what i mean, i editted my post 30 seconds after anyway!
Reply 121
Avatar for InsaneBob
InsaneBob
InsaneBob
Has anyone got a list of all the correct definitions we need to know or knows a place to find them? Because I dont know what words we may be askedto define and also some books have differen definitions.

Anyone?
thanks good advice...
If you have the textbook i'd imagine it's all the ones on the sides of the pages where it says Key Definition.
InsaneBob
Anyone?


Download the syllabus from OCR and it should all be there
Definitions glossary attached :smile:
RES7_phys_glossary.doc361.0KB
Reply 126
Avatar for neomilan
neomilan
OP
17
ShadowBronze
Definitions glossary attached :smile:


Helpful! But some of the long definitions they have given are not necessary i.e. they have said
displacement = The distance travelled in a particular direction, measured in metres (m), e.g. 3 m; a vector quantity.

Whereas in the exam, Rate of Change of velocity would suffice

You know what i get stuck on: defining a joule, watt and a newton
I always get really confused since i can never remember the perfect definitions!
any tips?

Joule: 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force.
Newton: 1 N is the force which gives a mass of 1 kg an acceleration of 1ms^-2
neomilan
Helpful! But some of the long definitions they have given are not necessary i.e. they have said
displacement = The distance travelled in a particular direction, measured in metres (m), e.g. 3 m; a vector quantity.

Whereas in the exam, Rate of Change of velocity would suffice

You know what i get stuck on: defining a joule, watt and a newton
I always get really confused since i can never remember the perfect definitions!
any tips?

Joule: 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force.
Newton: 1 N is the force which gives a mass of 1 kg an acceleration of 1ms^-2



Work Done (or energy transfered) = force x distance
J = N x m

1 Joule of energy is transferred when a force of 1N moves an object 1 meter

F = ma
N = kg ms^-2

1 newton is the force required which gives a mass of 1kg an acceleration of 1ms-2
Spongebob*No*Pants
According to the spcimen data sheet for 2008 Physics A

it says:

"The information in this sheet is for the use of candidates following
GCE Physics A H158 and H558.
Clean copies of this booklet must be available in the examination room, and must
be given up to the invigilator at the end of the examination.
Copies of this sheet may be used for teaching"

These forumlas will be given in the exam!

F cos (theta)
F Sin (theta)

a= delta V/ delta T

the 4 suvat equations

F=ma

moment = fx

Torque = fd

press. = F/A

Dens. = M/V

K.E = 1/2 mv^2

P.E = mgh

efficiency

stress

strain

and young modulus


This is a bad thing!

It means we will get deep thinking applying questions which are much harder than simple recall ones such as "what is youngs modulus etc"
neomilan
Helpful! But some of the long definitions they have given are not necessary i.e. they have said
displacement = The distance travelled in a particular direction, measured in metres (m), e.g. 3 m; a vector quantity.

Whereas in the exam, Rate of Change of velocity would suffice

You know what i get stuck on: defining a joule, watt and a newton
I always get really confused since i can never remember the perfect definitions!
any tips?

Joule: 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force.
Newton: 1 N is the force which gives a mass of 1 kg an acceleration of 1ms^-2


That is the correct definition for displacement. Rate of change of velocity is the def for acceleration.
Robbie10538
This is a bad thing!

It means we will get deep thinking applying questions which are much harder than simple recall ones such as "what is youngs modulus etc"


Yeah, That's what I'm scared of too!
Reply 131
Avatar for DrVas
DrVas
Robbie10538
This is a bad thing!

It means we will get deep thinking applying questions which are much harder than simple recall ones such as "what is youngs modulus etc"



How is that hard?
DrVas
How is that hard?


You missunderstood my post

It means we will get deep thinking applying questions which are much harder than simple recall ones such as "what is youngs modulus etc"


:smile:
Reply 133
Avatar for neomilan
neomilan
OP
17
annaroberts20
That is the correct definition for displacement. Rate of change of velocity is the def for acceleration.


woops! yeah i completely read that wrong!!!!
I meant to say that displacement is simply velocity x time and not the long crap they've told us
Um no - you'll get displacement = vel x time marked wrong.
Reply 135
Avatar for neomilan
neomilan
OP
17
teachercol
Um no - you'll get displacement = vel x time marked wrong.


why? that is correct and every mark scheme has awarded the marks for that!
Displacement = velocity x time
displacement could also be: distance travelled in a particular direction
velocity = rate of change of displacement
Reply 136
Avatar for neomilan
neomilan
OP
17
can someone help me with questions like these, i can never do them!
i just want to know how i should do questions like this where there are 3 forces acting!
thanks:

a) W = mg
W = 60 x 9.81 = 588.6

b)
R must equal the force the goes opposite to R (Lets cass this C (for component)). The ONLY force that effects C is W (because P is at a right angle)
The component of W in the direction of C is determined using a vector triangle with W as the hypothenuse
588.6cos(55) = C = R
therefore R = 337.61N

c) Just another vector diangram
588.6cos(35) = 482.15N

d)No resultant force; hence no acceleration; hence constant velocity. The force down the slope from the component of weight is canceled out by the force up the slope from friction.
Reply 138
Avatar for neomilan
neomilan
OP
17
Robbie10538
a) W = mg
W = 60 x 9.81 = 588.6

b)
R must equal the force the goes opposite to R (Lets cass this C (for component)). The ONLY force that effects C is W (because P is at a right angle)
The component of W in the direction of C is determined using a vector triangle with W as the hypothenuse
588.6cos(55) = C = R
therefore R = 337.61N

c) Just another vector diangram
588.6cos(35) = 482.15N

d)No resultant force; hence no acceleration; hence constant velocity. The force down the slope from the component of weight is canceled out by the force up the slope from friction.


i was asking how to do questions like that, but thanks for answering it anyway!
i always get stuck on those questions with 3 forces
Reply 139
Avatar for DrVas
DrVas
neomilan
i was asking how to do questions like that, but thanks for answering it anyway!
i always get stuck on those questions with 3 forces


That question is nothing!

Can you answer this one?

hardq.JPG20.2KB

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