pHysics papers help???

[niassu]

http://www.xtremepapers.me/CIE/Cambridg ... _qp_62.pdf
1-aph scales plz ???
and
http://www.xtremepapers.me/CIE/Cambridg ... _qp_32.pdf
Q2 how the max KE is at height 0.1 not 0.2

 

[ChrisChin]

By graph scales I assume you mean P6-2 Q3(c)
Appropriate scales would mean (for the 2mm x 2mm squares given) horizontal axis = 2 cm : 2 ohm, vert. axis = 1 cm : 2 V

For P3-2 Q2(b)i
Height is measured as a difference between max height and min height, i.e., 0.2m - 0.1m = 0.1m. Or in other words, the difference in PE is due to a change of 0.1m in height.

 

[niassu]

By graph scales I assume you mean P6-2 Q3(c)
Appropriate scales would mean (for the 2mm x 2mm squares given) horizontal axis = 2 cm : 2 ohm, vert. axis = 1 cm : 2 V

For P3-2 Q2(b)i
Height is measured as a difference between max height and min height, i.e., 0.2m - 0.1m = 0.1m. Or in other words, the difference in PE is due to a change of 0.1m in height.

thnx but in 2nd question why did u say difference in height sorry but really donot get it i'll tell u what i understand is the the P.E=mgh and h is the height the bob is above the ground so i think when it is vertical the lenght above ground would be 0.4-0.2
i know i am annoying but mark schem says ur answer 'the height is 0.1 '
sorry again

 

[ChrisChin]

PE --> KE, right?

Max KE is attained when bob is at minimum height, (i.e., bob is below support,) 0.1m below max height.

At this point, change in KE = change in PE = mgh.

At any point, (assuming no loss in energy from system) total energy = KE + PE.

Since KE=0 at max height, and KE is max at min height,

.: at max height, KE + PE = 0 + mgh[sub]max[/sub], and
at min height, KE + PE = 1/2 mv[sup]2[/sup] + 0

Note that we treat min height as 0 => actual max height = 0.30m, actual min height = 0.20m (both measured from ground)

.: change in PE = m x g x (change in height)
.: max KE = 0.15 x 10 x 0.10 = 0.15 J

 

[ChrisChin]

If you want to work it out using actual heights, you will still get the correct answer, though you do not show understanding that

change in PE = m x g x change in height (essentially what this question is testing for.)

i.e., change in PE = initial PE - final PE
= (0.15 x 10 x 0.30) - (0.15 x 10 x 0.20)
= 0.45 - 0.30
= 0.15 J

If you try to simplify the first line, you will also get change in PE = m x g x change in height, as shown below:

change in PE = initial PE - final PE
= (m x g x h1) - (m x g x h2)
by factorizing,
= m x g x (h1 - h2)
= mg x (change in height)

 

[niassu]

If you want to work it out using actual heights, you will still get the correct answer, though you do not show understanding that

change in PE = m x g x change in height (essentially what this question is testing for.)

i.e., change in PE = initial PE - final PE
= (0.15 x 10 x 0.30) - (0.15 x 10 x 0.20)
= 0.45 - 0.30
= 0.15 J

If you try to simplify the first line, you will also get change in PE = m x g x change in height, as shown below:

change in PE = initial PE - final PE
= (m x g x h1) - (m x g x h2)
by factorizing,
= m x g x (h1 - h2)
= mg x (change in height)

thnx alot no words can explain how much u 've helped thnx



plz sorry having naother question Q 3(c)
http://www.xtremepapers.me/CIE/Cambridg ... _qp_61.pdf

 

[ChrisChin]

Q3(c) : If you read the question carefully, you will discover that you are only asked for the value of resistance R based on the values in the table in 3(b)i when I = 0.5 x Io.

0.5 x Io = 0.5 x 0.30A (or whatever answer you got for (a))
= ~0.15A

From the table, R = 10.1 ohm when I = 0.15A.

Since you're asked to estimate (guess) the value of R, go ahead and give the answer as 10 ohm (the accuracy of your answer being only to 1sf -- cos you're only guesstimating!)

 

[niassu]

Q3(c) : If you read the question carefully, you will discover that you are only asked for the value of resistance R based on the values in the table in 3(b)i when I = 0.5 x Io.

0.5 x Io = 0.5 x 0.30A (or whatever answer you got for (a))
= ~0.15A

From the table, R = 10.1 ohm when I = 0.15A.

Since you're asked to estimate (guess) the value of R, go ahead and give the answer as 10 ohm (the accuracy of your answer being only to 1sf -- cos you're only guesstimating!)

thnx u mean the for estimation we should use 1s.f.or 2.s.f as i got the answer to 10.1
:sorry: :sorry: also in this paper i got the estimation to 1.65 but in markscheme was 1.6
http://www.xtremepapers.me/CIE/Cambridg ... _qp_62.pdf
this amy indicate an erroe i my graph mmm and yes my scales was
X-axis :1 cm =1.0 ohm
Y-axis :2cm=0.50 volt
thnx in advance

 

[CaptainDanger]

Isn't it already answered? You PM me about it?

 

[niassu]

Isn't it already answered? You PM me about it?

ya he answered me before u

 

[ChrisChin]

You should state your answer as 10 ohm instead of 10.1 ohm,

because Io > 0.30V, (based on given diagram in (a))
.: 0.5 x Io > 0.15V

but because Io is also *almost* 0.15V, you are justified in saying R ~ 10 ohm... i.e., 10.1 ohm is too accurate for a justifiable guesstimate!

However, yeah, you have some error in your graph. However, this is something examiners expect, and as long as your answer is close, you get the mark. Mark scheme for follow-on questions will take into account this error and allow the difference in answer.

Not that any of this is that important... It's only a difference of 1 mark. :grin:

 

[CaptainDanger]

Isn't it already answered? You PM me about it?

ya he answered me before u

Good.

 

[niassu]

answerr this plz

 

[CaptainDanger]

answer what?

 

[ChrisChin]

I did. Already. Your answer is fine. It's within acceptable limits from mark scheme.

 

[haochen]