AS Physics P1 MCQs Preparation Thread.

[Peter Check]

Was your acceleration -1.7?

No it was 6 point something...

 

[fatima 007]

Help needed :
J03 q17,22,28,30
Nov02 q7,10,18,24,32,36
Nov03 q4,5,7,22,28,31

 

[fatima 007]

June02 marking scheme and question paper please?

 

[fatima 007]

Yes please! Its getting too confusing this way. :/
How about we make a separate thread for the random MCQ doubts and leave this one for the year by year solving..?

http://www.xtremepapers.com/community/threads/as-physics-p1-mcqs-yearly-only.17330/

 

[fatima 007]

I second that idea.

http://www.xtremepapers.com/community/threads/as-physics-p1-mcqs-yearly-only.17330/

 

[fatima 007]

that would be good too but the problem is we will have to make group of 3-4 people or else server gets bussy i reckon we can do that for doubts which are really difficult hard to capture from here

http://www.xtremepapers.com/community/threads/as-physics-p1-mcqs-yearly-only.17330/

 

[fatima 007]

Agreed. :|

http://www.xtremepapers.com/community/threads/as-physics-p1-mcqs-yearly-only.17330/

 

[smzimran]

Question no 8 pls i have gone for C but its A why ????

Its quite simple:
When an object falls under gravity for quite sometime, it eventually gains constant 'terminal velocity'. That means 'g' drops to zero or almost zero in order for speed to be constant!
C is wrong as it shows no indication of g reducing to zero and terminal velocity being reached.
A is correct!

 

[USMAN Sheikh]

Question no 8 pls i have gone for C but its A why ????

no one is reply me pls the above ques and q31 and 33 of this year pls waiting for help pls mates

 

[USMAN Sheikh]

Its quite simple:
When an object falls under gravity for quite sometime, it eventually gains constant 'terminal velocity'. That means 'g' drops to zero or almost zero in order for speed to be constant!
C is wrong as it shows no indication of g reducing to zero and terminal velocity being reached.
A is correct!

thanks alot mate can u pls ans my other questions too

 

[fatemakhan]

How do we do this? What I did was to find the resultant force down the slope(taking the friction force into account) then use that force to find the acceleration, and then use the formula v^2= 2as where s is 7m. I got 9.5m/s, which is not even in the options!!View attachment 11874

i did the same...
2*a*s=v^2-u^2
i got 4.8785 (A) is that correct?
u must hav made some silly mistake...did u get the resultant force 3.4ms^-2?? And acceleration 1.7?

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
Q16 Q24

Q16:
input energy = F.s
output energy = mgh = mg (s sin@) = mgs sin@

efficiency = output / input
= mgs sin@ / Fs
= mg sin@ / F

So, D is correct

Q24: wave properties (check the text book)

 

[fatemakhan]

Q16:
input energy = F.s
output energy = mgh = mg (s sin@) = mgs sin@

efficiency = output / input
= mgs sin@ / Fs
= mg sin@ / F

So, D is correct

Q24: wave properties (check the text book)

nae bhae input shuld be mgh or 1/2 m v^2 and output shuld be F*s...i thnk...?

 

[smzimran]

no one is reply me pls the above ques and q31 and 33 of this year pls waiting for help pls mates

Q31:
R = pL / A [p is resistivity]
V/I = pL / A
p = AV / IL
Since resistivity is same due to same material
A1 V / I1 L = A2 V / I2 L
Since L and V are same the formula becomes
A1 / I1 = A2 / I2
Area of cross section = pie * r^2 = (pie * d^2) / 4

So the formula becomes
{ pie * (d1)^2 } / 4 I1 = { pie * (d2)^2 } / 4 I2

cancelling pie and 1/4 on both sides

(d1)^2 / I1 = (d2)^2 / I2
I1 / I2 = ( d1 / d2 ) ^2
= ( 2 / 1 ) ^2
= 4

D is the answer!

Q33:
The formula is
V = I (R + r)

C is the answer!

 

[smzimran]

nae bhae input shuld be mgh or 1/2 m v^2 and output shuld be F*s...i thnk...?

No, always remember in these cases, output is the workdone in lifting the load through a height h
And workdone by object is the input which is F.s where F is the applied force by the object!

 

[A.ELWY 7]

really impressed by the explanation to Q27...i was trying so hard since O level to figure out some imagination of this sort for these questions...thnx n best of luck

welcome...any time

 

[A.ELWY 7]

How do we do this? What I did was to find the resultant force down the slope(taking the friction force into account) then use that force to find the acceleration, and then use the formula v^2= 2as where s is 7m. I got 9.5m/s, which is not even in the options!!View attachment 11874

oooohhh u just couldn't imagine how was this question a torture for me...but i finally solved it, see:
first when resolving forces...the forces acting down the inclined plane is mg sin(theta)...to get this theta..u will use the opp/hyp=sin(theta), so sin^-1(3/7)=25.4 is the angle needed...then we have the mass which is 2kg x 9.81 sin(25.4)=8.4N...this is the f0rce down the plane...the resultant force is the 8.4 - the friction force which is 5 = 3.4...then u need to get the acceleration using F=ma...so 3.4/2 = 1.7 ms^-2..and finally using the equation v^2=u^2+2as...so 2 x 1.7 x 7=23.8...v^2=23.8 and so V=4.87 and so the answer is A

 

[Ahmed Ali Akbar]

can someone please explain to me question 19


isn't work done in a gas = p * V ?
so there is loss of energy overall[/quot if u have mark scheme of this paper do post it...PLEASE..

 

[USMAN Sheikh]

Q31:

Q33:
The formula is
V = I (R + r)

C is the answer!

cant iD???

 

[A.ELWY 7]

can someone please explain to me question 19


isn't work done in a gas = p * V ?
so there is loss of energy overall

i think it is A the answer because yes work is done when there is pressure in the volume...but he said air resistance is negligible...and all the other options do have work done across them so the only one left is A