PHY PAPER 2 QUESTIONS...URGENT HELP

[new2sys]

Hello Everyone,

ummmm I need urgent help with Physics Paper 2

I have a lot of questions...f ne 1 just cud answer most of them i wud really vey much appreciate it...thanx

A)
o/n 2006 :
Question # 1 a ii) i wrote power: work done per unit time
mark scheme says that the "ratio idea should be clear"??? What does that mean?

1 c i) Why can't we use the formula P = F v in this?

1 c ii) I don't get this??

Question 2 b ii?? I dont get it? :/

Question 3 d) Why is a third particle emitted in the direction of alpha particle? Why not in direction of astatine nucleus?

Question 4 a) What's the right definition of interference...I dont get the mark schemes definition
b) ii I don't get this at all

Question 5: The whole question is out my scope of understanding unfortunately

Question 7 part a)What should I write?

__________________________________________________________________________________________
May/June 2006
Question 1 c ii?? Please explain !

I don't get all of Question 3
Neither do I get Question 5
Nor do I get Question 7
__________________________________________________________________________________________________

October/November 2005

Question 2 b?
Question 3 c?
I don't get all of Question 4
Question 5 d ii
Question 8 b i and ii
_________________________________________________________________________________________
May/ June 2005:
Question 1: How do I guess the density of air and the mass of a protractor?
Question 2: Please explain part b
All of question 3...don't really get it
Question 4 part c
Question 5 b iii and c?

Question 7 part c
_____________________________________________________________________________________________________________
October/November 2004:
Question 1 part a iii) and b ii)

Question 6 part b and c

Question 7 part a) I am confused what the mark scheme says :?
___________________________________________________________________________________________________________
May/June 2004

Question 1 part b
Question 3 part b ii
Question 6 part b i and ii
Question 7 part c
Question 8 part c
_______________________________________________________________________________________________
October/November 2003
Question 1 b ii number 2

Question 2 b
Question 6 part c
All of Question 7
____________________________________________________________________________________
May/ June 2003
Question 2 b iii
Question 5 part b
Question 3 part b
:? :? :? :? :? :?

I know these are a lot of questions!

But please try to answer as many as you guyz can
Jazakallahu Khairun
Sincerely,
new2sys

 

[new2sys]

ummm and also for paper 2

o/n 2007: question 2,3,4, 5 part c, 6, and 7

 

[Azazel]

holy cow, those ARE a lot of questions.

W06:

Q1.a) ii) Work done per UNIT time implies it's a ratio, your answer is correct.
c) i) You need to find the RATE at which kinetic energy is supplied.

Kinetic Energy = 0.5 mv^2
Kinetic Energy / time (i.e kinetic energy per unit time) = 0.5mv^2/t

You can't use P = Fv because it's ACCELERATING, it's not moving at a constant speed!

c) ii) The acceleration cannot be constant because as the speed of the car increases, the amount of air resistance acting on the car increases too. Since the power supplied is constant and the resistive force is increasing, the acceleration will decrease.


Q2. b) ii) Again, same logic. As time increases, the air resistance acting on the trolley increases. Thus, the acceleration is decreasing.

Q3. d) /Not sure/

Q4. a) I have no clue how to answer your question. Read the book for the definition.

b) d sin theta = n * lambda

theta = 90 deg, lambda = wavelength = 644nm = 644 * 10^-9 m, d = 1/n where n is the lines per metre.

550 lines : 1 mm
n lines : 1000mm (i.e 1 metre)

n = 550,000 lines per metre

sin 90 / 550,000 = n * 644x10^-9
n = 2.8
Therefore, 2 complete orders.

Q5. This is really basic stuff. Again, read from the book.

Q7. The mark scheme is really clear in this answer.

 

[Azazel]

June 03:

2 b) iii) The answer calculated in (ii) is in perfect conditions i.e no friction with tires or air resistance. So in real life, these factors need to be taken into account. More power is needed to overcome these forces.

3 b) i) Use the formula for arc length. s = r(theta)

Theta is in radians, so convert 6.5 degrees to radians and you get (6.5/180)*pi = 0.113 rad

s = (d/2) x 0.113 = 1.5 * 0.113 = 1.7

Strain = change in length / length

ii) Stress = Force / Area

iii) Young Modulus = Stress / Strain

iv) If the elastic limit is exceeded, the wire will be permanently deformed.

Come on, ^ was an extremely easy question.



5 b) i) The variation of V and I in the resistor will be a straight line passing through origin.

ii) The lamp works at a p.d of 6V with a resistance of 150 ohms. The resistor has a resistance of 200 ohms. The total resistance is 350. 40mA is the current. E = IR1 + IR2 = I(R1 + R2) = 40x10^-3 (200+150) = 0.04 (350) = 14V

Alternatively, when you've drawn a line passing through origin, you can see when I = 40 mA, V = 8V. Add this V to the lamp's V.

 

[new2sys]

@ Azazel: yeah these R A LOT of questions! thanks a lot for trying to answering them! they really helped ummm but i dont get 1 thing in ur explanation 2 w 06 questions.....how is increasing the speed related to increase in air resistance? (i dont c ne connection) (referring to ur explanation
"he acceleration cannot be constant because as the speed of the car increases, the amount of air resistance acting on the car increases too. Since the power supplied is constant and the resistive force is increasing, the acceleration will decrease.")

 

[Azazel]

W03:

1 b ii) 2) The 6th, 7th and 8th dots all have a constant distance between them as the acceleration is zero. So find the distance between, say, the 7th and 8th dot. Common sense: if the distance b/w the 7th and 8th dot is, say, D. Then add 2D to the 9th dot. The 9th dot represents 0.9 seconds. Adding 2D to it would represent the 11th dot i.e 1.1s

2 b) In a liquid, because a upthrust will be acting on its weight countering it and if the gravitation field strength (g) is of a different value e.g on the moon. Weight = mg

6 c) Check the marking scheme, it's quite clear.

7. a) i) P = VI, plug in P = 1200 and V = 240

ii) P = I^2 R, V = IR, plug in ANY formula.

b) i) potential difference across the CABLES: V = IR, I = 5A, R = 4ohms.

ii) The main supply needs to make sure 240V reaches the heater. Hence, we add the potential diff. lost across the cables to 240V.

iii) Power lost in the CABLES, P = VI,

c) The power input is (1200 + power dissipated)

 

[Azazel]

@ Azazel: yeah these R A LOT of questions! thanks a lot for trying to answering them! they really helped ummm but i dont get 1 thing in ur explanation 2 w 06 questions.....how is increasing the speed related to increase in air resistance? (i dont c ne connection)

Hmph, I'm sure you've done terminal velocity?

If you recall correctly, if an object is dropped from a height from rest, it'll accelerate with 'g' before the acceleration gradually decreases to 0 at which point the object begins to travel at a constant speed called terminal velocity. This decrease in acceleration is due to increased air resistance. It's the reaction of the environment on the driving force, you may say.

 

[new2sys]

oh yeah~!!! thnx 4 reminding !

 

[Azazel]

June 04:

1b. You need to complete the diagram by drawing a parallelogram and then measuring the distance of the resultant. Use a scale like 1cm to 1N force. This will give you a parallelogram with sides 6cm and 8cm. The resultant which passes from one corner of the parallelogram to the other is calculated using a ruler and then converted to N. For example, if you get the resultant to be 10cm, then it'll be a 10N resultant force.

3b ii) The gradient of the graphic = y/x = cm / s^2 i.e clearly acceleration.

y = mx + c
y = d, x = t^2
The only equation of motion that fits this is s = ut + 0.5at^2
since u = 0, we can cut that out. s = 0.5at^2

Compare this to y = mx
y = s
m = (0.5a)
x = t^2

0.5a = gradient
and you can find the value of a.

6b i) ii) This question has been done in the other thread, please open that up.

7c) the answer to b) was 0.137
This is common sense again, a bulb is small. You can't fit a 14cm wire in it, it's too long.

8c) The voltmeter has a resistance of 7800 ohms, so the overall resistance in parallel with the resistor is now: 1/R = 1/7800 + 1/7800, R = 3900 ohms

V = 3900/(3900+Rt) * 1.5

V = 3900/(7800) * 1.5 = 0.75V

 

[x.daydreamer]

Could anyone solve this one with diagram =)
Oct/Nov/2009 Variant 22.
Q3--alll =(
And OCT/NOV/2009 VARIANT 21.Q1

 

[new2sys]

@

Could anyone solve this one with diagram =)
Oct/Nov/2009 Variant 22.
Q3--alll =(
And OCT/NOV/2009 VARIANT 21.Q1

3 a) i Horizontal velocity remains same through out projectile motion so horizontal velocity= 4 m/s
ii) Use kinematic equation for vertical motion:
s=u t - 1/2 gt^2 (there is - instead of + b/c accleration due to gravity is negative 9.8)

u get time from this which comes out to be about .63 s

then use v= u - gt (1nce again its v= u - gt instead of the usual equation v = u + at b/c g is negative)

v= -6.19 (ignore the negative sign b/c it just tells direction)

For part b) it's addition of vectors (i really suck at it!!! but i will try my best )

I drew a diagram for this in microsoft 2007 i dont know how to post it here b/c this forum wont let me upload an attachment.......


......

(working on rest.....

 

[sizbeauty]

hey can sm1 plz tell me tht in nov09/variant 22,q4(partc2(ii))
frm where did v get 0.0362?
2) ES = ½ × 1.8 × 10-2 ( 0.0362 – 0.0212)
= 0.077 J


plz sm1 help me!!!!

 

[Azazel]

hey can sm1 plz tell me tht in nov09/variant 22,q4(partc2(ii))
frm where did v get 0.0362?
2) ES = ½ × 1.8 × 10-2 ( 0.0362 – 0.0212)
= 0.077 J


plz sm1 help me!!!!

You need to use the total extension, not the difference in extensions.

The total extension of 3.8N is (16.3 - 14.2) = 2.1m
The total extension of 3.8N + F N is (17.8 - 14.2) = 3.6m

Converting them to m, 0.021m and 0.036m