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Physics: Post your doubts here!

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Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand
AoA,
Q3:
We know that when a liquid turns into gas, work is done by the liquid
In this case, they are asking about work done on the system, so this must mean - (work done by the system)
(i)
W = - p(deltaV)
W = - 1.03 × 10^5 [(2.96 × 10^-2) - (1.87 × 10^-5 )]
W = - 3050 J

(ii)
Heating we know it is given Q = 4.05 × 10^4 J

(iii) U = Q + W
U = 4.05 × 10^4 + (- 3050)
U = 3750 J
I dont think the calculation is much tricky :)
And, I dont get their approach of penalising 2 s.f in (b) but not in (c)
What i think is because in (b) they have given all data values to 3 s.f so they want the answer to atleast 3 s.f
In (c) they have not given any values, so they accept s.f below 3,
This is what i think altough im not sure
:unsure:
 
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Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand
AoA,
Q4:
Speed is max when displacement is zero,
Consider the ball thrown in the air and down again case (which you must have studied)
At max. height the ball has max. gravitational potential energy (P.E) and zero kinetic energy (K.E) meaning zero speed
Similarly, when the ball falls and touches the ground,
P.E is zero as the height is zero and by law of conservation of energy the whole P.E at top is converted into max. K.E at bottom, so speed is max at ground (at zero displacement)

Same applies in S.H.M:
At amplitude, speed is zero (K.E is zero) because the object has max P.E as it is at max height and at momentary rest.
At zero displacement, speed has increased to maximum value as whole of P.E at amplitude has been converted into K.E
That is why speed is taken max at displacement = 0
:)
 
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Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand
AoA,
Q5
When an electron is moving fast in a magnetic field, it experiences the magnetic force for a shorter duration of time thus is deflected less!
When it is passing slowly (with smaller value of speed), it experiences the magnetic force for a longer duration of time thus is deflected more!

Imagine you have to pass a patch of quicksand;
If you run through you will go through easily and will not be much drenched in the sand
If you walk slowly, you will be going down and down in the sand ;)

^ Just came across my mind, hope it helps :D

Q6(b)
You just have to derive the expression for magnetic flux linkage (Φ) :
We know that
Φ = BAsinθ
For N number of lines,
Φ = NBAsinθ
Done
Small derivation so worth 2 marks only
:):p:)
 
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
q5 part c
ms says plastic is an insulator / not a conductor / has no free electrons B1
charges do not move (on an insulator) B1
either so no single value for the potential
or charge cannot be considered to be at centre
wats the purpose of this point that charges do not move (on an insulator)
Just defining the term 'insulator'
Thats all, otherwise i agree there is no need
;)
 

XPFMember

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AoA,
Q5
When an electron is moving fast in a magnetic field, it experiences the magnetic force for a shorter duration of time thus is deflected less!
When it is passing slowly (with smaller value of speed), it experiences the magnetic force for a longer duration of time thus is deflected more!

Imagine you have to pass a patch of quicksand;
If you run through you will go through easily and will not be much drenched in the sand
If you walk slowly, you will be going down and down in the sand ;)

^ Just came across my mind, hope it helps :D

Q6(b)
You just have to derive the expression for magnetic flux linkage (Φ) :
We know that
Φ = BAsinθ
For N number of lines,
Φ = NBAsinθ
Done
Small derivation so worth 2 marks only
:):p:)
waalaikumassalam wr wb
haha...nice one :D

ok and for ques 6..i wanna know what's the difference b/w flux and flux linkage..see ms...it says smthng like that...

and jazakAlllahhhhhhhhhhhhh for the help...
many duassss 4 u
 

Jaf

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Wavelength of the waves [from the graph] = 50m
Velocity of the wave = 8m/s
[f]Frequency = Velocity/Wavelength = 8/50 = 0.16 Hz
[a]Amplitude (or max displacement) of the waves [from the graph] = 2m
Maximum velocity= 2(pi)af = 2 x pi x 2 x 0.16 = (16/25)pi
Maximum KE of the object = (1/2)m(v^2) = (1/2) x (2x(10^-3)) x ((16/25)pi)^2 = 0.00404 J. Approximately 4mJ.
 
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tha
Wavelength of the waves [from the graph] = 50m
Velocity of the wave = 8m/s
[f]Frequency = Velocity/Wavelength = 8/50 = 0.16 Hz
[a]Amplitude (or max displacement) of the waves [from the graph] = 2m
Maximum velocity= 2(pi)af = 2 x pi x 2 x 0.16 = (16/25)pi
Maximum KE of the object = (1/2)m(v^2) = (1/2) x (2x(10^-3)) x ((16/25)pi)^2 = 0.00404 J. Approximately 4mJ.
thanx very much...if u may jaf im spending more than 1 hour to solve q1 b in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf, we all know that cosx=adj/hyp....so it must be 25/cos(35) to get R but the mark scheme says 25cos(35)... ill blow pls say how???
 

Jaf

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^I don't get why they're complicating things and finding 'R' in the marking scheme lol. This is how I got the answer (sorry my handwriting is kinda messy :p) :
IMG_1872.JPG
 

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...some more doubts in June 2006


· Q:2 b (ii) how to do?? :s
· Q:4 c can you show, plz?
· Q:6 a what I understand is the field should be circle..right? I don’t get this :s
· Q:6 c what values do I use? :s
· Q:7
 
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