AS Physics P1 MCQs Preparation Thread.

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lavanyamane said:
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Q.14, 15, 22, 26, 28, pretty please? :)
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For Q15
initial momentum is zero cuz both trolleys are at rest.When they are released let velocity of 1 kg trolley be v
Using conservation of momentum
0=2*2-1*v which will give u v=4 m/s.Now
Energy of spring=Kinetic energy of each trolley
E=1/2*2*2^2 + 1/2*1*4^2 givin u D answer
For Q22
you have to use the formula
E=Fl/eA where E is young modulus,l is length,e is extension and A is area.First calculate area then put the values in the formula.Bring the formula in the form so that e is divided by l cuz he has asked by what percent string contracts.Dont forget to multiply it by 100.
For Q26
First find line spacing
d=10^-3 divided by 300
Now use the formula d sintheta=n lambda
Put 90 in place of theta and find n.It will be something like 7.4 i dun know.But take n as 7 dont round it to 8.Now he has asked for maximum orders so 2n+1=15
For Q28
Since drop is stationary so that means some force must be equal to its weight.Since weight is acting downwards this force must be acting upwards.The upper plate is +ve so that means the drop must be negative for it to be attracted to upper plate.If it was positive it wont have been stationary.W=mg To find charge use the formula F=qE where q=F/E Now substitute F with W since they both are equal mg=qE which will give charge to mass ratio as g/E
 
umarashraf

umarashraf

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MysteRyGiRl said:
guyz help me too plz
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q25... c=f(lambda)
for the given length of loop, which is equal to half the wavelength, we have, speed=f(1/2lambda)
for complete wave now, speed wont change, so wavelength is doubles as before, so inorder to maintain constant speed, frequency has to be doubled to maintain the constant product...
 
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MysteRyGiRl said:
guyz help me too plz
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q27... one way to look is that no displacement is covered in the direction of the force... so no work is done... secondly, force and displacement has an angle of 90 in between.. so fdcosx will again be zero... thirdly, both P and Q are at the same potential.. so moment of charges at same potential does no work...
 
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MysteRyGiRl said:
guyz help me too plz
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q33.. make it a numerical my assuming each resistor to be of 2 ohms and battery to supply emf of 6V... now total effective resistance will be 3 ohms... total current will now be p=IV 12=i6 i=2A... now look the diagram again, P will have 2A current while Q R will have 1A each... now by applyin P=I^2 R... P=1^2 * (2) you will have @ as answer... i didnt have any shorter way to solve this one...:p
 
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Oliveme said:
Please help me with these questions ----------> 9, 14, 22, and 25. This is probably the 5th time I'm posting. Thank you very much. May Allah Bless you.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_ms_12.pdf
Click to expand...

22. Mate option B and D will produce the extension on 1/2 ratio for every weight.. so forget about them... consider A.. spring with high K is difficult to be expanded.... by placing a high k spring down will enable it to work both ways around.. when a heavy mass is to be suspended, rigid box will not allow the lower k spring in the box to move.. this will produces little extension... so low sensitivity with higher mass... when lighter mass is to be suspended, it will produce by rapid extension by release of the box.. option C cannot work with this phenomenon....
 
umarashraf

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Oliveme said:
Please help me with these questions ----------> 9, 14, 22, and 25. This is probably the 5th time I'm posting. Thank you very much. May Allah Bless you.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_ms_12.pdf
Click to expand...

14. total distance travelled will be 3m.. 0.5/sin30... work done by the will be used to overcome frictional work done and the resistance by the inclined plane... resistive W.D = 150 * 3 = 450J
work done against slope= mgsin30 = 200sin30=100J.
and now word in lifting it up = mgh =300J.. so total is 750J
 
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umarashraf

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Oliveme said:
Please help me with these questions ----------> 9, 14, 22, and 25. This is probably the 5th time I'm posting. Thank you very much. May Allah Bless you.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_ms_12.pdf
Click to expand...

for the 9th.. i think we cant prove it numerically but one think is for sure that initial momentum is 2mv... and if collision were perfectly elastic ones, max. momentum change would have been 2mv-(-2mv) =4mv... now it is given that collision was inelastic, it momentum change should have been less than 4mv and greater than 2mv.. so we have only one option to satisfy this... option C...
 
umarashraf

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omg said:
umarashraf when u get tym solve ma qs too =p
15, 26 and 34!
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34 first... r is directly propotional to length/cross sectional area... for X.. R= L/(pie)(D/2)^2 ----. R=4L/pie D^2
for Y.. R=2L/(pie)D^2 ----- R= 2L/(pieD^2)
simplify it and you will have the A as answer...
 
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omg said:
umarashraf when u get tym solve ma qs too =p
15, 26 and 34!
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15... work done is equal to change in kinetic energy.. that is 8-4=4J in first case.. we know that W.D=Fs.. so F=4/s
by the second statement, let the final Ek be equal to x... so.. x-4=Fs ... x-4=2(4/s)(2s) this will give u x-4=16
and then x=20J...
 
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umarashraf

umarashraf

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bloody_mary said:
can someone please explain question 17 and 31
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_12.pdf

thanks
Click to expand...

17.. efficiency=output/input * 100 if input is 100 output is 8... so 8/100 * 100 =8...

31... electron would be attracted by the positive charge, so B and D are fundamentally wrong... now since the electron is moving against the electric field, it will be deflected out of the field by the negative charge... line of field represents the direction of the field ...
 
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