Physics Paper4

[ruby92]

o/n 2010 variant 41 question 1 c (i) and (ii)
Same paper variant 43 question 3 b (i) and (ii)

[Sue T]

hope im answering the ryt one :
paper 41 - 1ci) you have a gravitational field due to the earth, attracting any mass within its field to the left hand side of ur ppr , and you have another (smaller) gravitational field due to the moon attracting masses to your right hand side of th paper - (its smaller cuz the moon is smaller- simple) now at some point wont the resultant force exerted on unit mass, somewhere between the two planets, be equal and opposite? so thats why the grav field strength is zero
ii) grav. field strength = GM/r² where M is the mass of the thing causing the field
so if the at a certain distance r from the centre of the earth - the grav field strengths of both planets are equal we form this eqn:

GM_earth     =     GM_moon
 _____           ____________
 r²                  (60R_E-r)²

cancel the Gs and put in the masses - eventually you'll get ans answer in terms of R_E

[Sue T]

paper 43 - 3bi) for cylinder X they gave you an eqn :  x= -4.0Cos(220t)
now the phase diff for cylinder Y is 2pie/3
so the eqn for cylinder Y becomes :  x= -4.0Cos(220t + 2pie/3) {everything else like freq and amplitude are same}
now the instant theyre talkin bout is wen x= -4 (you can see from th fig that cylinder X is in its lowest position where x=-4) so putting it in th eqn we get:
-4 = -4Cos(220t)
therefore 220t=0
put that in ur eqn for cylinder Y - you get its displacement at that instant :
x=-4.0Cos( 0 + 2pie/3 )
you get x=2
so draw it 2cm above AB
the next one is th same - th arrow direction is diff bcuz it has already done half a cycle (pie) and is continuing downwards ---> (as in 4pie/3 is larger than pie)
hope its clear - i got stuck here too a couple of days ago

[basoom16]

hello can u help me with physics paper 42 o/n 2009
1b why is the radius taken as 2.86x10^7 insted of 2.22x10^7
thank you

[ruby92]

hello can u help me with physics paper 42 o/n 2009
1b why is the radius taken as 2.86x10^7 insted of 2.22x10^7
thank you

In the equation F=GMm/r^2
r is taken from the center of mass. So r=Radius of earth+Distance of orbit from earths surface.

[ruby92]

hope im answering the ryt one :
paper 41 - 1ci) you have a gravitational field due to the earth, attracting any mass within its field to the left hand side of ur ppr , and you have another (smaller) gravitational field due to the moon attracting masses to your right hand side of th paper - (its smaller cuz the moon is smaller- simple) now at some point wont the resultant force exerted on unit mass, somewhere between the two planets, be equal and opposite? so thats why the grav field strength is zero
ii) grav. field strength = GM/r² where M is the mass of the thing causing the field
so if the at a certain distance r from the centre of the earth - the grav field strengths of both planets are equal we form this eqn:

GM_earth     =     GM_moon
 _____           ____________
 r²                  (60R_E-r)²

cancel the Gs and put in the masses - eventually you'll get ans answer in terms of R_E

Thank you

[basoom16]

Thank you

can u plz have a look at the question. its actaully 2 equations used. can u plz see it
Thanks

[ruby92]

Yes GMm/r^2 =mw^r
but i dont understand what your doubt is then?

[basoom16]

ah ok stupid me din calcualte properly. THANK U SOO MUCH

[ruby92]

Your welcome and i'm sorry i'm very good at explaining.

[ruby92]

M/j 2005 question 7 c.

[basoom16]

lol how does the value become 1/10

[Sue T]

manganese decays to form iron ryt? so if
iron/manganese = 9
it means out of 10 molecules, 9 are iron and 1 is manganese
so its like 1/10 of manganese has decayed
so u use the eqn with 1/10 and the decay constant of manganese

[ruby92]

M/j 2004 7c,b , why is h squared?
M/j 2005 5(b)
o/n 2005 for question 3b (i) why is work done negative, isnt work dont on the system?
M/j 2006 4(c), Q7 (a) and (b)
O/n2006 7b
M/j 2007 Q3 d
o/n 2008 3c(ii)
M/j 2009 9b, 11 biii

Any help would be awesome!

[Sue T]

M/j 2004 7c,b , why is h squared?

so 2007 >> E=(1/2)mv²   and lambda=h/p where p=mv
so             mv²=2E
                v=sqroot(2E/m)
                                                                                p=m x sqroot(2E/m)  (muliply numerator and denominator by sqroot m)                                             p=m x sqroot( 2E m) / (sqrootm)²
                                                                                p= sqroot(2Em)
substitute p back into lambda=h/p
you get wat they're askin

part c) get E by substituting everything else in then use E=Ve where V is th pd and e is the charge on th electron
h is squared cuz amm if u want to make E the subject of the formula you have 2 square everything else if u still dont get it tell me ill attach a working