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Author Topic: ALL CIE PHYSICS DOUBTS HERE !!!  (Read 172024 times)

Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #315 on: November 13, 2010, 09:24:04 pm »
Thankx :)
and S09 Q29 if you may please? :)
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Offline TJ-56

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #316 on: November 13, 2010, 09:25:34 pm »
Quote from: Dania on November 13, 2010, 09:21:29 pm
Why is the answer A?

Momentum before = momentum after so:
mv - mv = 2mV ( the negative sign is given due to the 2 trolleys moving in opposite directions ) so
60m - 40m = 2mV (V is the speed of the 2 together, and the masses are added because they are taken as 1 trolley now)
so simplified would give 10 cm/s
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Offline Hypernova

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #317 on: November 13, 2010, 09:29:08 pm »
Quote from: $!$RatJumper$!$ on November 13, 2010, 08:56:27 pm
Help with S09 Q29 and S08 Q26 please :)

S09 P1 Q29

The electron has kinetic energy = 1/2 m v2

We know that the electron travels a distance x, then stops. This means that the it lost its kinetic energy since velocity changed from something to zero.

The loss in Ke should equal the work done on the force.

F x X = 1/2 x m x v2          E = F/e
                                   F = Ee  so...
EeX = 1/2 x m x v2
  X = m x v2
        2Ee
« Last Edit: November 13, 2010, 09:32:44 pm by Hypernova »
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Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #318 on: November 13, 2010, 09:31:06 pm »
Quote from: Hypernova on November 13, 2010, 09:29:08 pm
S09 P1 Q29

The electron has kinetic energy = 1/2 m v2

We know that the electron travels a distance x, then stops. This means that the it lost its kinetic energy since velocity changed from something to zero.

The loss in Ke should equal the work done on the force.

F x X = 1/2 x m x v2          E = F/e
                              F = Ee  so...
EeX = 1/2 x m x v2
  X = m x v2
        2Ee

+rep :)
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Offline Dania

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #319 on: November 13, 2010, 09:57:11 pm »
Quote from: TJ-56 on November 13, 2010, 09:25:34 pm
Momentum before = momentum after so:
mv - mv = 2mV ( the negative sign is given due to the 2 trolleys moving in opposite directions ) so
60m - 40m = 2mV (V is the speed of the 2 together, and the masses are added because they are taken as 1 trolley now)
so simplified would give 10 cm/s


Thank You!
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:)

Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #320 on: November 13, 2010, 10:30:59 pm »
W08 P1 Q 10, 27, 31, 36, 37 please :)

Sorry for all my questions! :(
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Offline TJ-56

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #321 on: November 13, 2010, 10:49:27 pm »
Quote from: $!$RatJumper$!$ on November 13, 2010, 10:30:59 pm
W08 P1 Q 10, 27, 31, 36, 37 please :)

Sorry for all my questions! :(

Q10

Relative speed of approach = relative speed of separation
u1 + u2 = (-)v1 + v2
so:
u1+ u2 = v2 - v1 rearranged (v1 i given a negative sign due to its direction)

Q27

Between 2 maxima is 15 mm , so wavelength = 2 * 15 * 10-3  (remember its a stationary wave)
so speed c = 3 * 108
c=f * lambda
and f = c/wavelength
we have the 2 values required, so the frequency is calculated as 1 * 1010

Q31
Wire p has twice the diameter, not twice the area, its quadruple the area because the diameter is squared,
current I=V/R V is constant, so its a ratio of resistances, which is 4 to 1

Q36
Looked confusing to me at first, but turned out to be rly easy:
V=IR
V=7.5 (1.5 volts are lost to internal resistance)
R=15
Go for I to give 7.5/15
which is 0.5A

Q37

If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.

Hope that helped.
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Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #322 on: November 13, 2010, 11:00:14 pm »
Quote from: TJ-56 on November 13, 2010, 10:49:27 pm
Q10

Relative speed of approach = relative speed of separation
u1 + u2 = (-)v1 + v2
so:
u1+ u2 = v2 - v1 rearranged (v1 i given a negative sign due to its direction)

Q27

Between 2 maxima is 15 mm , so wavelength = 2 * 15 * 10-3  (remember its a stationary wave)
so speed c = 3 * 108
c=f * lambda
and f = c/wavelength
we have the 2 values required, so the frequency is calculated as 1 * 1010

Q31
Wire p has twice the diameter, not twice the area, its quadruple the area because the diameter is squared,
current I=V/R V is constant, so its a ratio of resistances, which is 4 to 1

Q36
Looked confusing to me at first, but turned out to be rly easy:
V=IR
V=7.5 (1.5 volts are lost to internal resistance)
R=15
Go for I to give 7.5/15
which is 0.5A

Q37

If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.

Hope that helped.


Thank you that did help a lot :)
I just have a few questions

For Q10, why cant we say u1 + -u2 = v1 + v2
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Offline TJ-56

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #323 on: November 13, 2010, 11:10:08 pm »
Quote from: $!$RatJumper$!$ on November 13, 2010, 11:00:14 pm
Thank you that did help a lot :)
I just have a few questions

For Q10, why cant we say u1 + -u2 = v1 + v2

Well the original equation is
u1 - u2 = - (v1 - v2)  Which is derived by using both the principle of conservation of momentum and energy, where it is assumed all directions are the same (for example both to the left before and both with velocities to the left after)
since u1 and u2 are in different direction , its now u1 -  -u2, which is u1 + u2
and on the right hand side, -(v1 -v2) which is v2 - v1
Thats all there is to it i guess, hope that cleared it up.

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Offline Hypernova

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #324 on: November 13, 2010, 11:14:25 pm »
Quote from: TJ-56 on November 13, 2010, 10:49:27 pm

Q37

If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.

Hope that helped.


Q37)

The potentiometer!!

In this circuit,

E2 = E1 x resistance of XT
            resistance of XY + R


since the galvanometer is reading a null result, we know that the pd across XT is equal to E2.
The galvanometer is basically an ammeter. 0 amps means 0 p.d which means that the p.d across XT exactly nullifies the emf produced by E2

So we know the length of XT. All we need is E2 and divide it by the length to find its pd per length.
« Last Edit: November 13, 2010, 11:20:22 pm by Hypernova »
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Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #325 on: November 13, 2010, 11:16:46 pm »
Quote from: Hypernova on November 13, 2010, 11:14:25 pm
Q37)

The potentiometer!!

In this circuit,

E2 = E1 x resistance of XT
            resistance of XY + R


since the galvanometer is reading a null result, we know that the pd across XT is equal to E2.
The galvanometer is basically an ammeter. 0 amps means 0 p.d which means that the p.d across XT exactly nullifies the emf produced by E2

So we know the length of XT. All we need is E2 and divide it by the length to find its pd per length.

Thank you that helped :)

Quote from: TJ-56 on November 13, 2010, 11:10:08 pm
Well the original equation is
u1 - u2 = - (v1 - v2)  Which is derived by using both the principle of conservation of momentum and energy, where it is assumed all directions are the same (for example both to the left before and both with velocities to the left after)
since u1 and u2 are in different direction , its now u1 -  -u2, which is u1 + u2
and on the right hand side, -(v1 -v2) which is v2 - v1
Thats all there is to it i guess, hope that cleared it up.


Many thanks that helped a lot :)

+rep both
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Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #326 on: November 13, 2010, 11:26:00 pm »
S08 Q 24, 29, 32, 34 please
:)
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Offline thecandydoll

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #327 on: November 14, 2010, 03:20:01 am »
W09 VARIANT 11 P1 PHYSICS.
Q15
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Offline thecandydoll

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #328 on: November 14, 2010, 03:34:43 am »
may/june 2009
20/21

I dont get it :(
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Offline thecandydoll

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #329 on: November 14, 2010, 03:59:25 am »
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