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AS Physics P1 MCQs Preparation Thread.

Oliveme

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waleedsmz

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Oliveme said:
Please answer the questions above. Third time. :(
Click to expand...

I'm not really sure about 9 and 13...
14: At the maximum height , there is only the horizontal velocity acting on the particle ( Vcos45 ). So...


3.jpg​

27: The electric field of a particle ( Q ) is the force done BY the particle on a unit charge ( divided by q ). Although they basically have the same force of attraction between them, each charge has a different electric field. The electric field of ( q ) is F/Q ... However, the electric field of (Q) is F/q --> So the answer is D.

29: Force = qE
= 1.6 x 10^-19 x 3 x 10^7
Notice how we completely ignore the number of protons ( the charge ) of the alpha particle since the electric field is already given. In other words, the force by an alpha particle on a unit charge is given. We multiply the charge of the electron and the force per unit charge of the alpha particle.
So the answer is A.

Hope that helped. : )
 
L

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Oliveme said:
Please answer the questions above. Third time. :(
Click to expand...
For 9, the collision is elastic so K.E is conserved. Find the total K.E of the system before, and then see which collision after has the same K.E. You could use fake values of m and u to make the calculation easier. I would do it now but I have to go now. :|
 
Oliveme

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waleedsmz said:
I'm not really sure about 9 and 13...
14: At the maximum height , there is only the horizontal velocity acting on the particle ( Vcos45 ). So...



27: The electric field of a particle ( Q ) is the force done BY the particle on a unit charge ( divided by q ). Although they basically have the same force of attraction between them, each charge has a different electric field. The electric field of ( q ) is F/Q ... However, the electric field of (Q) is F/q --> So the answer is D.

29: Force = qE
= 1.6 x 10^-19 x 3 x 10^7
Notice how we completely ignore the number of protons ( the charge ) of the alpha particle since the electric field is already given. In other words, the force by an alpha particle on a unit charge is given. We multiply the charge of the electron and the force per unit charge of the alpha particle.
So the answer is A.

Hope that helped. : )
Click to expand...
Thank you very much. That did help a lot! :)
 
umarashraf

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Oliveme said:
Can you please answer these question? thank you very much.

Question 27
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_ms_1.pdf
Click to expand...

distance between 2 maximas received by the receiver was 15mm... maxima is heard at nodes only.. and distance between 2 nodes is half a wavelength...

so s=f(wavelength).. f=3*10^8/0.015*2.. it comes C as answer...
 
Oliveme

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umarashraf said:
distance between 2 maximas received by the receiver was 15mm... maxima is heard at nodes only.. and distance between 2 nodes is half a wavelength...

so s=f(wavelength).. f=3*10^8/0.015*2.. it comes C as answer...
Click to expand...
i thought maxima was head at anti nodes and low sound at nodes? thank you very much for your answer. :)
 
Oliveme

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hmlahori said:
Please people do help i've posted these questions several times but no replies:( . If someone can explain any one of the questions if not all please do. Thanks.

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
http://www.xtremepapers.com/papers/...A and AS Level/Physics (9702)/9702_w04_er.pdf
Please can someone explain Q18 i get B but answer is A. and Q20 and for Q 33 i get B 2I but correct answer is C 4I. How?

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_11.pdf
Q10 Why cant answer be A?
Q18,26,28???
Q29 How is a stationary wave formed in Q since both ends are open?
Q30??
Q33 Why cant answer be C?
Q36???
Click to expand...
2004
18) work done by an external force = (k.e. + p.e.)+ work done against friction. here the work done against friction is the heat dissipated. so work done by force = 9000 N x 40m and potential energy p.e. is mgh = 2000 kg x 10 x 12m = 24 0000. all the kinetic energy at the beginning is transferred to p.e. at the top. so k.e. = p.e.
------------> (9000 x 40) - (24 0000 + 24 0000) = 120 000 joules that is 120 kJ which is the WD against friction or heat. :)

33) one method of doing this could be to assume false values for V, d and l.
V/I = rho x L/A. so for the first batch we can assume V = 20, l = 4 and d = 3 --------> 20/I = 4/(pi x (3/2)^2) = 35.34 amps.
for the nest batch, we take 2V = 40, l = 8 and d = 6 -------------> 40/I = 8/(pi x (6/2)^2) = 141.372 amps.
so 141.372 is 4 x 35.34 :)

Sorry I can't help with 20 as I'm stuck there as well. hope that helped.
 
omg

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Oliveme said:
that's okay. can you help me with questions 9 and 13? if you don't mind. no one seems to be able to answer them. thank you. :)
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_ms_11.pdf
Click to expand...
for q.9,
momentum before the collision = 2mu + (-mu) => mu
now as the collison is elastic, momentum will b conserved. it will b same after the collison ,
find momentum of all 4 cases given, the one havin momentum equal to mu is the ans, and dats A!
 
Oliveme

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hmlahori said:
Please people do help i've posted these questions several times but no replies:( . If someone can explain any one of the questions if not all please do. Thanks.

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
http://www.xtremepapers.com/papers/...A and AS Level/Physics (9702)/9702_w04_er.pdf
Please can someone explain Q18 i get B but answer is A. and Q20 and for Q 33 i get B 2I but correct answer is C 4I. How?

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_11.pdf
Q10 Why cant answer be A?
Q18,26,28???
Q29 How is a stationary wave formed in Q since both ends are open?
Q30??
Q33 Why cant answer be C?
Q36???
Click to expand...
this might help for 20. http://www.xtremepapers.com/communi...-preparation-thread.17044/page-64#post-332401 :)
 
umarashraf

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Oliveme said:
that's okay. can you help me with questions 9 and 13? if you don't mind. no one seems to be able to answer them. thank you. :)
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_ms_11.pdf
Click to expand...

13 first... since applied force is constant, the torque (F*d) will be proportional to d.... torque for 100mm is 3.. so for 150mm , it will be 4.5Nm... so we have to look for the tensions now..torque is the product of force and moment arm.. and the only force present here is the tension in the belt so 3=F*.1 F=30N
for the other wheel, 4.5=F*.15 F=30... so total force is equal to 30N to 30N = 60N...
 
umarashraf

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10.. Why cant the answer be A... look , its simple. newton was the one to discover the idea of force... and he defined the force in his second law of motion... the eq F=ma is a derived formula from the basic eq of force... F=mv-mu/t here mass is considered as constant and v-u/t is replaced by a... so A is actually the derived one from a basic definition...
 
omg

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umarashraf said:
10.. Why cant the answer be A... look , its simple. newton was the one to discover the idea of force... and he defined the force in his second law of motion... the eq F=ma is a derived formula from the basic eq of force... F=mv-mu/t here mass is considered as constant and v-u/t is replaced by a... so A is actually the derived one from a basic definition...
Click to expand...
Excuse me sir, which qs are u explanin? :p
 
umarashraf

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omg said:
Excuse me sir, which qs are u explanin? :p
Click to expand...
hmlahori said: ↑
Please people do help i've posted these questions several times but no replies:( . If someone can explain any one of the questions if not all please do. Thanks.

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
http://www.xtremepapers.com/papers/...A and AS Level/Physics (9702)/9702_w04_er.pdf
Please can someone explain Q18 i get B but answer is A. and Q20 and for Q 33 i get B 2I but correct answer is C 4I. How?

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_11.pdf
Q10 Why cant answer be A?
Q18,26,28???
Q29 How is a stationary wave formed in Q since both ends are open?
Q30??
Q33 Why cant answer be C?
Q36???​
 
umarashraf

umarashraf

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Oliveme said:
yes, please. the question 9 I just asked. I still don't understand it. thank you.
Click to expand...

since the collision is elastic, total kinetic energy will be conserved... Ek before is 0.5(2m)(u^2) + 0.5(m)(u^2).. this will give u 1.5mu^2....
now check the answer one by one.. and look for Ek= 1.5
luckily it is the first option... 0.5(2m)(u/3)^2 + 0.5(m)(5u/3)^2.... this is also equal to 1.5mu^2...

elastic collisions have their Ek conserved... Remember!!!!!
 
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