AS Physics P1 MCQs Preparation Thread.

[JalalKaiser]

it says 'when the sample contracts it follows the same force extension pattern;hence definitely not plastic'

X is the limit of proportionality and Y is the elastic limit.

i guess it is because the qs says 'when the sample contracts it follows the same force-extension curve as when it was being stretched'
it means the behaviour is ELASTIC not plastic!!

Ahhhhh, okay. Thank you, both of you!
Jazakumullah Khair.

 

[dawood235]

people taking bio u can also get help u can see some difficult questions here and practice them !!! u dont need to run through papers!!! people not taking BIO (including me ) lets blast through the past papers and rock our examinations!! ok here i go for october 2002 but i dont have it's marking scheme!

?

 

[A.ELWY 7]

AHH! thanks a lot! i didn't notice that before..

welcome...we post year by year there..happy to help

 

[JD REBORN]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf
In Q7 the acceleration is zero cuz the gradient at Q is zero?Am I right?But i donot uunderstand the displacement?

 

[DARK DRAGON]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
q14 plz answer is D

 

[sagar65265]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
q14 plz answer is D

For question 14, you need to realize that the momentum initially was equal to zero, so the momentum after the spring is released is also equal to zero.

Thus: Momentum of 2 kg trolley = mv = 2 * 2 = 4 kgms^-1 and since total momentum should be zero, 4 - 1kg * v = 0, so the velocity of the 1 kg trolley = 4 ms^-1.

We can also assume that all the energy stored in the spring is transferred to the carts, so the KE of the 2 kg Trolley = 0.5 * 2 * 2 * 2 = 4 J
KE of the 1 kg Trolley = 0.5 * 1 * 4 * 4 = 8 J
Therefore the total energy transferred from the spring to the trolleys is equal to 4 + 8 = 12 Joules of energy = D

Good Luck for your exams!

 

[DARK DRAGON]

For question 14, you need to realize that the momentum initially was equal to zero, so the momentum after the spring is released is also equal to zero.

Thus: Momentum of 2 kg trolley = mv = 2 * 2 = 4 kgms^-1 and since total momentum should be zero, 4 - 1kg * v = 0, so the velocity of the 1 kg trolley = 4 ms^-1.

We can also assume that all the energy stored in the spring is transferred to the carts, so the KE of the 2 kg Trolley = 0.5 * 2 * 2 * 2 = 4 J
KE of the 1 kg Trolley = 0.5 * 1 * 4 * 4 = 8 J
Therefore the total energy transferred from the spring to the trolleys is equal to 4 + 8 = 12 Joules of energy = D

Good Luck for your exams!

thx i wasn't thinking of momentum

good luck to you too!!

 

[sagar65265]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
In Q7 the acceleration is zero cuz the gradient at Q is zero?Am I right?But i donot uunderstand the displacement?

This question is tricky in that at first sight the amplitude, which is equal to the maximum displacement, may appear to be reached after 2.5 seconds, but in reality,
on a displacement time graph, at the maximum displacement you can see that the gradient of the graph is zero, i.e. the velocity is zero so the answer for the right hand side is in reality 5 and 15 - these numbers correspond to 2 different times when the displacement is maximum, i.e. when the amplitude is zero. On the right hand column, we need to select the acceleration at point Q. The acceleration at Q is equal to the gradient of the velocity time graph which is flat, parallel to the x-axis and is thus equal to zero. This means that the answer is D.

Good luck for your exams!

 

[Oliveme]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
Please help me with these questions. 8, 9, 12, 14, 15, 22 and 25. Thank you very much.

 

[Rashmit Arora]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Q.4 - Shouldn't it be D?

 

[beauty princess]

F=ma
so, F =(m+3m)a
F=4ma
hence a=F/4m
block x in exerting a force on block y , so Fx=3m(F/4m)
so Fx=3/4F!

HOPE IT HELPED

thank u!!

 

[omg]

thank u!!

anytime

 

[omg]

q 13 plsss anybody??????

 

[DARK DRAGON]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Q.4 - Shouldn't it be D?

is the answer c? if it is this is because the least number of significant figures given in the data is 0.1 (1 sf) so the uncertinty must also be to 1 sf, so we round 0.36 to 0.4. same with the speed given to 3 sf

 

[A.ELWY 7]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Q.4 - Shouldn't it be D?

no it is C..as the uncertanity must be rounded to 1 SF.....in the question when u see the 2 uncertanities the least one have 1 sf so rounded and ull get 0.4..wish i helped

 

[omg]

no it is C..as the uncertanity must be rounded to 1 SF.....in the question when u see the 2 uncertanities the least one have 1 sf so rounded and ull get 0.4..wish i helped

will u help me wid the qs i just posted??

 

[DARK DRAGON]

q 13 plsss anybody??????

this a hard one i will try my best to explain

first u need to know that at the highest points the vertical component is 0 so we only need to consider the horizontal component (vcos45) sub this in as velocity in 1/2mv sqr and u will get 1/2mV*2 x 1/2 this is half the original KE
hope u get it

 

[omg]

this a hard one i will try my best to explain

first u need to know that at the highest points the vertical component is 0 so we only need to consider the horizontal component (vcos45) sub this in as velocity in 1/2mv sqr and u will get1/2mV*2 x 1/2 this is half the original KE
hope u get it[/quot
im still nt clear :/

 

[DARK DRAGON]

which part

 

[A.ELWY 7]

will u help me wid the qs i just posted??

im sry i cant at the moment have to go...but i promise i will!!