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Last 3 days for Physics P4- Post your doubts

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change in potential energy (height) is what we are looking for and not the total potential energy after so there is no need to add. you could work out total potential energy before then total after and find difference, in that case the final potential energy is f(r+4cm)
 
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Right Thanks baby :)
Others can post here too, so we know about difficult problems which may help us ;)
I believe P4 will be more about theory this yr. I see a change.
 
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I am seeing these years, they are asking more on theory, not sure. I saw the 2006, 2008, 2009 pattern
 
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I saw some 2008, 2009 questions repeat in 2010-2012. I am not sure i need to check them I will let u know. Chemistry P4 went to take numbers from 2004 . I couldn't memorize the answer :( And I messed up. Oh well!
 
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_41.pdf

Please help with some of my questions,thank you in advance!!

Q1 b(ii) and (iii)

Q2 a.

Q3 b(ii)

Q4 (c)


Pleeeeeeeease!! :'(

ok here are the solutions:
1(b)(ii):
you should use the formula F=m(w^2)x and F=(GMm)/(x^2) then u should equate both equations:
m(w^2)x = (GMm)/x^2, the "m" will cancel out and then u'll multiply by x^2 on both sides of the equation so u'll b left with this:
GM=(w^2)(x^3).....but g= GM/(R^2) hence GM=g*(R^2)
therefore u get gR^2 = (x^3)(w^2)

1(iii)
u just use the formula given that is: gR^2 = (x^3)(w^2)
but first u have to find "w"...which is (2pi)/T and then u just substitute the values ur given into equation.

2(a)
here u use pV=nRT and n=N/Avogadro's constant
so u'll have ur equation as:
[(2.5*10^5) * 4.5*10^-3] = (N/Avogadro's constant) * 8.31 *290
this will give u: N= 6.02*10^23 * 0.4668
=2.8*10^23

3(b)(ii)
u use the formula:
Q= m*c*change in temperature, so
c= Q/(m*change in temp)
the change in temp is 3.7/60 = 0.061667...
hence c= 54/(960*10^-3 * 0.061667...)
= 912

4(c)
u should use K.E = 0.5mv^2
where v = xw
but w = 2pi*f (f=frequency)
=2pi*833.333...
=1666.667pi
hence v= 0.2*10^-3 * 1666.667pi
= 1.05
therefore kinetic energy= 0.5 * 2.5*10^-3 * (1.05)^2 = 11.37*10-3

hope u understood!:)
 
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can someone help me with november 2003 number 6(c)(ii) and november 2011-43 number 3(c)(ii)
 
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Since you guys kind of know the trends that cambridge uses, could you pls tell me which past papers i should do?
Obviously there isn't enough time to do them all :(
 
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Since you guys kind of know the trends that cambridge uses, could you pls tell me which past papers i should do?
Obviously there isn't enough time to do them all :(
There is time to do all. Do them chapter wise and you can do all :) I am going to do them all today and tomorrow My personal challenge. I believe you should do 2007-2010 and 2003 but we can never know, so do the maximum ;)
 
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There is time to do all. Do them chapter wise and you can do all :) I am going to do them all today and tomorrow My personal challenge. I believe you should do 2007-2010 and 2003 but we can never know, so do the maximum ;)

Lol i like doing whole past papers so that i can time myself and get a feeling for the exam, but guess everyone has their own method of studying... And i like to deeply analys what i get wrong :p

Could you also pls tell me, i'm writing paper 42 right... So should i just do the paper 42 past papers or should i do papers 41 and 43 too, or doesn't it really matter?

Thanks for the advice though :)
 
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You are the best. Thank you so freaking much!! I totally get it!! Amazing explanation,God bless you!! <3


ok here are the solutions:
1(b)(ii):
you should use the formula F=m(w^2)x and F=(GMm)/(x^2) then u should equate both equations:
m(w^2)x = (GMm)/x^2, the "m" will cancel out and then u'll multiply by x^2 on both sides of the equation so u'll b left with this:
GM=(w^2)(x^3).....but g= GM/(R^2) hence GM=g*(R^2)
therefore u get gR^2 = (x^3)(w^2)

1(iii)
u just use the formula given that is: gR^2 = (x^3)(w^2)
but first u have to find "w"...which is (2pi)/T and then u just substitute the values ur given into equation.

2(a)
here u use pV=nRT and n=N/Avogadro's constant
so u'll have ur equation as:
[(2.5*10^5) * 4.5*10^-3] = (N/Avogadro's constant) * 8.31 *290
this will give u: N= 6.02*10^23 * 0.4668
=2.8*10^23

3(b)(ii)
u use the formula:
Q= m*c*change in temperature, so
c= Q/(m*change in temp)
the change in temp is 3.7/60 = 0.061667...
hence c= 54/(960*10^-3 * 0.061667...)
= 912

4(c)
u should use K.E = 0.5mv^2
where v = xw
but w = 2pi*f (f=frequency)
=2pi*833.333...
=1666.667pi
hence v= 0.2*10^-3 * 1666.667pi
= 1.05
therefore kinetic energy= 0.5 * 2.5*10^-3 * (1.05)^2 = 11.37*10-3

hope u understood!:)
 
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Lol i like doing whole past papers so that i can time myself and get a feeling for the exam, but guess everyone has their own method of studying... And i like to deeply analys what i get wrong :p

Could you also pls tell me, i'm writing paper 42 right... So should i just do the paper 42 past papers or should i do papers 41 and 43 too, or doesn't it really matter?

Thanks for the advice though :)
I'm doing 42 too and I have started doing papers some hours ago :p Not sleeping today I guess :D I think you should do 2009-2011 well all variances. Then 2002-2008 pick the numbers which did not repeat from 2009-2012 ;) I am going for that route. So many things I still don't know.
 
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I'm doing 42 too and I have started doing papers some hours ago :p Not sleeping today I guess :D I think you should do 2009-2011 well all variances. Then 2002-2008 pick the numbers which did not repeat from 2009-2012 ;) I am going for that route. So many things I still don't know.

Lol you do realize thats virtually all the papers, and they are 2 hours each, you will be up till you write the exams man!
Haha i must congratulate you on such dedication, gl ;)

I just did o/n 2007 and think i got an E or something hey... If you note any trend please post it over here :)
Like which year you saw repeats

***Edit
I just marked the 2007 paper and got a C not an E, yay :D
Now to pray i do well in the real exam

Thanks
 
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