ALL CIE PHYSICS DOUBTS HERE !!!

[physichemaths]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_1.pdf
how to do question 40?? how we know the ans is C?

[EMO123]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_1.pdf
how to do question 40?? how we know the ans is C?

on taking ratio you an find that
take the ratio of proton:Neutron
For H= 1:1 i.e 1
for He= 2:2 i.e 1
for Li= 3:4 i.e 0.75
for Be= 4:5 i.e 0.8
so the least ratio has the lowest speed that is C

[JACKRABBIT]

k, im not sure about this method but i got the answer somewhow

the p.d across the 5 ohm resistor in the 1st part of the circuit = 5/15 * 2V = 2/3

THe p.d across the 2 5 ohms resistors in the second part of the circuit = 10/15 * 2 = 4/3

Hence the total potential "DIFFERENCE" = 4/3 - 2/3 = 2/3 V

------

If u apply this formula to find the p.d across, say the all three resistors in the 1st branch
The voltage across the point at the bottom and the negative terminal = 0V
the voltage across the point at the top and the negative terminal = 2V
Difference = 2-0 = 2V -> which is the P.d across the 3 resistors in the first branch..
   

Aiight thx man much appreciated...Now i have a few more...and these are like really the hard ones.....anyway...again a full detailed explanation would be awesome cos i dont take maths and that kinda makes it hard to understand all these equations and stuff

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w10_qp_12.pdf

Qns. 7, 8(my teacher couldnt solve it) ,9,15,34

AAAAND  finally..

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w07_qp_1.pdf

Qns, 23 and 25


   

[username]

hey guys
exam on wednesday
and loads of doubts
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 7,8,9,11,15,22,29,40
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 9,13,34,37

[Ghost Of Highbury]

Aiight thx man much appreciated...Now i have a few more...and these are like really the hard ones.....anyway...again a full detailed explanation would be awesome cos i dont take maths and that kinda makes it hard to understand all these equations and stuff

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w10_qp_12.pdf

Qns. 7, 8(my teacher couldnt solve it) ,9,15,34

AAAAND  finally..

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w07_qp_1.pdf

Qns, 23 and 25


   

8. XY - 40m - 12s
    XZ - 80m - 18s
80 = 18u + 0.5a*18^2
40 = 12u + 0.5a*12^2

Solve the simultaneous equations, by eliminating u and calculating a.

9. There's a little bit of guess work involved here.
Given it's inelastic, the rebound velocity won't be v. Say, 40% of energy is lost, say around 0.6v
Change = (-0.6v*2m) - 2mv = -3.2mv
SO the change is almost 3mv.
(Even if u take 0.7v, u won't get 4mv+)

15. Steady speed v, therefore weight is balanced by retarding force
mg = kv
v = mg/k
KE = 0.5mv^2 = 0.5m*(mg/k)^2 = m/2 * m^2g^2 /k^2 = m^3g^2 /2k^2

34. When x = 0, the voltage measured is across the whole of the wire, so it has to be V.
    when x decreases, and the pointer is moved at the right end of the part with resistivity p, the voltage is divided into 2 parts (p and 2p+3p)
   So the voltage across the length x is now, not V, but a lil less than V because some VOltage is lost in the remaining length of the wire.
Answer: B (note the graph is not a straight line because as x decreases, the resistance doesn't decrease proportionately)

---
23. frequency = 1/(time taken for 1 complete oscillation = 50m (check x-axis))
It covers 8m in 1 second (speed)
so 50m in 50/8 s = 6.25
That's its time period
Hence, frequency = 1/t = 1/6.25 = 0.16Hz
Speed = 2pi*a*f = 2pi*2*0.16 = 2.01m/s
KE = 0.5mv^2 = 4mJ
25. dsin45 = 3*lambda
d = 3lambda/sin 45 = 3sqrt(2)*lambda
maximum order of diffraction = dsin90 = n*lambda
n = d/lambda
n = (3sqrt(2)*lambda)/lambda = 3sqrt(2) = 4.24264.... ~ 4

[Ghost Of Highbury]

hey guys
exam on wednesday
and loads of doubts
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 7,8,9,11,15,22,29,40
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 9,13,34,37

qp11
7.  I have trouble in this too. How does horizontal component of velocity decrease to zero? Isn't it constant in a projectile?
8. It goes up 5m, comes down 3m. Displacement = 5-3 = 2m
9. v^2 = u^2 +2as
   s = v^2-u^2 / 2a
   v = 0, so 0 = u^2+2as -> 2a = -u^2/s or -u^2/x

Now, the new u = U+0.2u = 1.2u
V = 0
so s = 0-1.44u^2 /2a
s = -1.44u^2/(-u^2/x) = 1.44x
11. M1V1 -M2V2 = 0
    M1V1 = M2V2
V1/V2 = M2/M1
B
15. It can be either A or D (because they are closed vector triangles)
It's D, because the direction of the arrow representing the force by the cable is correct.
22. The length of the wire doesnt affect the extension. The load and cross-sectional area do. Here both are the same. So extensions of
     both the wires are same (1.5 and 2), that's y total e = 1.5 + 2 = 3.5mm

29. It's upward because electrons are attracted towards +ve plate. And ofcourse, it's not a part of the circle because the path is not exactly an arc.
40. Count th4e no. of protons in Argon, it's 18. Only option with 18 protons for argon is C. So C it is

qp12

9. There's a little bit of guess work involved here.
Given it's inelastic, the rebound velocity won't be v. Say, 40% of energy is lost, say around 0.6v
Change = (-0.6v*2m) - 2mv = -3.2mv
SO the change is almost 3mv.
(Even if u take 0.7v, u won't get 4mv+)

34. When x = 0, the voltage measured is across the whole of the wire, so it has to be V.
    when x decreases, and the pointer is moved at the right end of the part with resistivity p, the voltage is divided into 2 parts (p and 2p+3p)
   So the voltage across the length x is now, not V, but a lil less than V because some VOltage is lost in the remaining length of the wire.
Answer: B (note the graph is not a straight line because as x decreases, the resistance doesn't decrease proportionately)

13. Ofcourse it's D, gravitational pull is constant (mg). Mass is constant and a is a constant.
37. Check 1/R = 1/R1 + 1/R2 + 1/R3
And see when u get value of R as 50.

[SkyPilotage]

Air resistance is NOT zero So it is NOT a projectile...

[Ghost Of Highbury]

Doubt
How is it 15?
(1/300000)/(450*10^-9) = 7.4 orders of diffraction
Maxima on both sides = 7.4*2 = 14.8
Meaning, myou would see 7 on both sides, so it should add up to 14 right
Why is the answer 15?

And Can someone also explain question 27_w09_qp11

@Sky-pilotage - Ofcourse, it is a projectile? SO what if there's air resistance?

[Ghost Of Highbury]

Okay, i read it in a book. It is still a projectile, but external forces act on the horizontal component.

[SkyPilotage]

Doubt
How is it 15?
(1/300000)/(450*10^-9) = 7.4 orders of diffraction
Maxima on both sides = 7.4*2 = 14.8
Meaning, myou would see 7 on both sides, so it should add up to 14 right
Why is the answer 15?

And Can someone also explain question 27_w09_qp11

@Sky-pilotage - Ofcourse, it is a projectile? SO what if there's air resistance?

It Is projected..Correct..But It will NOT show uniform Projectile motion....
Therefore Its not a regular projjectille So there is no constand vertical acceleration nor constant horizontal velocity...
A Projectile is an object upon which ONLY the force of Gravity acts on it...

[Ghost Of Highbury]

It Is projected..Correct..But It will NOT show uniform Projectile motion....
Therefore Its not a regular projjectille So there is no constand vertical acceleration nor constant horizontal velocity...
A Projectile is an object upon which ONLY the force of Gravity acts on it...

Agreed. Could you help me with question 27_w09_qp11 and q26.

[SkyPilotage]

Agreed. Could you help me with question 27_w09_qp11 and q26.

Question 26:

You will find the number of orders on each side...So to get the maximium Possible of orders on one side You will consider sinX to be sin90....
THerefore n(4.5 x 10^-7)=(1 / (300/0.001) ) ( sin90)               *P.S:-we need to get d which is the seperation of 2 slits.
You will get n as 7.4.. Which means that there are 7 orders on each side including the zero ordder..Hence 15 orders..

Question 27:

The rule is F/charge..They want the field strength acting on charge q only..SO its F/q...

[Ghost Of Highbury]

Question 26:

You will find the number of orders on each side...So to get the maximium Possible of orders on one side You will consider sinX to be sin90....
THerefore n(4.5 x 10^-7)=(1 / (300/0.001) ) ( sin90)               *P.S:-we need to get d which is the seperation of 2 slits.
You will get n as 7.4.. Which means that there are 7 orders on each side including the zero ordder..Hence 15 orders..

Question 27:

The rule is F/charge..They want the field strength acting on charge q only..SO its F/q...

Ah! The zero order! Damn, thanks a lot man.

[wstrawberries]

I need help with question 9 and 33. If anyone could spare the time?
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s06_qp_1.pdf

[SkyPilotage]

I need help with question 9 and 33. If anyone could spare the time?
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s06_qp_1.pdf

Question 9:-( I just need a confirmation on this )
Since the mass is going up and down..
At A its halfway going up..
At B its at its highest point...
At C its halfway going down..
And finally At D its at the bottom which is the lowest point of its motion..

Question 33:
Find the Pd of the resistors accross XQY or XPY..
The voltage across the 2 parallel wires is the same i.e 12 V...
To find the voltage of the 500 ohm resistor its 500/(500+1000) x12 = 4 V
ANd the voltage across the 2000 ohm resistor is 2000/(2000+1000) = 8 V
-->Therefore The potential DIFFERENCE is 8-4 volts = 4volts..
* you could also go using XQY and find the voltages accross the 1000 oh resistors and you will find the top one to be 4 and the bottom on eot be 8 so THe P.d between XandY is 4 volts...

@Ghost I forgot it too
Glad to help