Post Your AS LEVEL PHYSICS PAPER 2 DOUBTS

[Rakayz]

umm can u explain m/j 2008 p2 question 2..the whole of it?

 

[Rakayz]

ur welcum nd best of luck for ur examz

thanks! same to u

 

[leosco1995]

Thanks a lot for the quick answers. One more if you don't mind:

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q6 (b) (i) and (ii)

You don't need to tell me the whole experiment, just how it would be done.

 

[applepie1996]

umm can u explain m/j 2008 p2 question 2..the whole of it?

ok

 

[Saad Sarfraz]

Okay so please help me on this sum 09 paper 21 Q5B..

 

[applepie1996]

umm can u explain m/j 2008 p2 question 2..the whole of it?

2)a)i) the constant k=F/e since the graph is extension over the load u have gotto take the reciprocal of the gradient so to find the gradient just take a co-ordinate i'm takin (20,2.5) gradient=y co-ordinate/x co-ordinate convert 2.5 cm to metres =0.025m
gradient =0.025/20 = 1/800 k=reciprocal of gradient so k=800N/m

ii)strain energy = area under the graph
=1/2 x 3.5/100 x 28 =0.49 J

b)i) momentum b4 collision= momentum after collision so in this case momentum b4 collision is zero
0=2400 x V - 800 x v i'm subtracting as both the trolleys will move in opposite direction so 800v=2400V
so v/V=2400/800 = 3/1
ii)0.49 is the total kinetic energy wid which dey movedthe trolley wid mass 2.4kg moves wid a speed of (1/3)v and the trolley wod mass 0.8kg moves wid a speed of v
0.49=(1/2 x 2.4 x (1/3v)^2) + (1/2 x 0.8 x v^2) solve for v and u get 0.96m/s

 

[applepie1996]

Thanks a lot for the quick answers. One more if you don't mind:

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q6 (b) (i) and (ii)

You don't need to tell me the whole experiment, just how it would be done.

okay i'll check it out now and tell and sorry for da delay

 

[applepie1996]

Okay so please help me on this sum 09 paper 21 Q5B..

ok i'll check it out and tell u

 

[Rakayz]

2)a)i) the constant k=F/e since the graph is extension over the load u have gotto take the reciprocal of the gradient so to find the gradient just take a co-ordinate i'm takin (20,2.5) gradient=y co-ordinate/x co-ordinate convert 2.5 cm to metres =0.025m
gradient =0.025/20 = 1/800 k=reciprocal of gradient so k=800N/m

ii)strain energy = area under the graph
=1/2 x 3.5/100 x 28 =0.49 J

b)i) momentum b4 collision= momentum after collision so in this case momentum b4 collision is zero
0=2400 x V - 800 x v i'm subtracting as both the trolleys will move in opposite direction so 800v=2400V
so v/V=2400/800 = 3/1
ii)0.49 is the total kinetic energy wid which dey movedthe trolley wid mass 2.4kg moves wid a speed of (1/3)v and the trolley wod mass 0.8kg moves wid a speed of v
0.49=(1/2 x 2.4 x (1/3v)^2) + (1/2 x 0.8 x v^2) solve for v and u get 0.96m/s

thank u soo much!!!!

 

[applepie1996]

Thanks a lot for the quick answers. One more if you don't mind:

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q6 (b) (i) and (ii)

You don't need to tell me the whole experiment, just how it would be done.

ok for diffraction of transverse waves
use a laser screen and a slit and pass the laser through the slits as the light passes through the slit it is diffracted so it spreads out onto the screen showing diffraction of transverse waves
for diffraction of longitudinal waves instead of using a screen and a laser use a c.r.o to detect the sound wave and a loudspeaker as a source of longitudinal wave

 

[applepie1996]

Okay so please help me on this sum 09 paper 21 Q5B..

ok so phase difference is 128-100=28cm this is because s1 is 100cm from M and s2 is 128cm from M u form a triangle and find the hypotenuse which is 128cm so now u find lambda = speed/frequency lambda for frequency of 1000hz=330/1000=0.33m=33cm
lambda for frequency 4000hz=330/4000=0.0825m=8.25cm
thts all i have cleared out just wait aliitle and i'll figure out the other part of the question

 

[USMAN Sheikh]

that your new graph should be about quarter the size of your old one

mate i did not get that could u pls sketch te old one and new one of both parts in any formats even image in paint will be encouraged pls yar me nt getting it *worrrid* :'( :'(

 

[USMAN Sheikh]

and one more question pls Q4 part b plsss be quick mate

 

[Silent Hunter]

2)a)i) the constant k=F/e since the graph is extension over the load u have gotto take the reciprocal of the gradient so to find the gradient just take a co-ordinate i'm takin (20,2.5) gradient=y co-ordinate/x co-ordinate convert 2.5 cm to metres =0.025m
gradient =0.025/20 = 1/800 k=reciprocal of gradient so k=800N/m

ii)strain energy = area under the graph
=1/2 x 3.5/100 x 28 =0.49 J

b)i) momentum b4 collision= momentum after collision so in this case momentum b4 collision is zero
0=2400 x V - 800 x v i'm subtracting as both the trolleys will move in opposite direction so 800v=2400V
so v/V=2400/800 = 3/1
ii)0.49 is the total kinetic energy wid which dey movedthe trolley wid mass 2.4kg moves wid a speed of (1/3)v and the trolley wod mass 0.8kg moves wid a speed of v
0.49=(1/2 x 2.4 x (1/3v)^2) + (1/2 x 0.8 x v^2) solve for v and u get 0.96m/s

did not understand the highlighted part.... like we know the ratio of 800g:2400g is 3:1 ... after that what to do to get the speeds in terms of 'v' of them like you have calculated?

 

[Saad Sarfraz]

ok so phase difference is 128-100=28cm this is because s1 is 100cm from M and s2 is 128cm from M u form a triangle and find the hypotenuse which is 128cm so now u find lambda = speed/frequency lambda for frequency of 1000hz=330/1000=0.33m=33cm
lambda for frequency 4000hz=330/4000=0.0825m=8.25cm
thts all i have cleared out just wait aliitle and i'll figure out the other part of the question

Thanx man but u should knw that i also did it till this part couldn't do the rest of it

 

[applepie1996]

Thanx man but u should knw that i also did it till this part couldn't do the rest of it

kay i'll do da rest and tell u tommorow

 

[freakybandi]

s09 variant 1 qs 5b....plz help
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf

 

[vivo990]

s09 variant 1 qs 5b....plz help
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

listen,
First u calculate S2M (Pythagoras)
√(80²+100²)= 128 cm
then path difference is 28.0 cm (128-100)
Calculate lambda at 1.0 KHz which is (330/[1.0 x 10³]) = 0.330 m = 33 cm
then claculate lambda at 4.0 KHz which is (330/[4.0 x 10³]) = 0.0825 m = 8.25 cm
so lambda (wavelength) has decreased from 33.0 cm to 8.25 cm
u know if the path difference is half a wavelength then it is destructive interference [zero intensity].
if u calculated the path difference in terms of lambda at both extremes u'll get
28/33= .850 lambda & 28/8.25= 3.93 lambda
so the path difference (p) ,when frequency increased might be; .850<p<3.39
within this range how many X.5 are there (where X is any number) ? Ans: two
p= 1.50, or 2.50 Lambda

 

[Shouly]

http://xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s08_qp_2.pdf

PLEAAASE 6B how do you calculate the POWER IN EACH PARTTTTTTTT.

 

[lalaleeza]

Paper 23 2010 october....help me for question 3 pleeeeeeaaaaase