AS Physics P1 MCQs Preparation Thread.

[ashoo123]

10)
Look firstly we'll find accn from newtons 2nd law
f=ma
60=30a a= deceleration
a= -2ms-2 (negative sign indicates that its declaration)

Now we'll use, a = (Vf - Vi)/t to get Vf
-2 = (Vf - 3)/0.50

Vf = 2m/s = C

Thanku!!

 

[Hanona]

can someone please help me with q 14 o/n 09 pleaseeeee

 

[Thought blocker]

can someone please help me with q 14 o/n 09 pleaseeeee

Link ?

 

[Hanona]

Link ?

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

14)
The horizontal velocity will remain the same.
The vertical velocity will be zero at that time.
So the kinetic energy will be due to horizontal velocity alone which is v*cos*45 = v √2/2
Kinetic energy at max point = 1/2 * m * (v √2/2)² = [1/2 m v² ]* 1/2 (the thing in square brackets is equal to kinetic energy at the beginning) so k(max)=0.5K(initial)
Got it ?

 

[Hanona]

OHYES !!! I did get it! thank you soooo very much! you saved my brain from collapsing

 

[Hanona]

i have a last request , and I promise I wont disturb again!
do you have an good notes for phsyics practicals?! my exam is on the 5th of june! T_T

 

[Thought blocker]

i have a last request , and I promise I wont disturb again!
do you have an good notes for phsyics practicals?! my exam is on the 5th of june! T_T

Click this red words
Notes by Jassim
Notes by Suchal

 

[Hanona]

Click this red words
Notes by Jassim
Notes by Suchal

thank YOUUUU

 

[ziremm]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf

q21,8,3,5 please

 

[Hanona]

q 3 -- at the maximum height the horizontal component is 0 and the only component acting is the vertical one which when you resolve gives you vsin(x) (this is the final velocity) now considering the equation v^2=u^2+2gs you substitute and rearrange ...
q -5 you add up all the PERCENTAGE uncertainities so 4 pie is a constant , l has 4% and T has 1% but remember there is a square here so now it is 2 x 1% = 2% now you add them up to get 6 %
q 8 - by the time it takes over the goods train they should have travelled the same distace so s1=s2 ..... 10t = 1/2 x 1/2 x t^2 ... solve and youll get 40 s
q-21----total pressure at base = pressure of water and oil.... 17.5 x 10 ^ 6 = 9.81 x (x) x 830 + 9.81 x 1000 x (2000-x ) solve it and youll get D as the answer,,,,

 

[ziremm]

q 3 -- at the maximum height the horizontal component is 0 and the only component acting is the vertical one which when you resolve gives you vsin(x) (this is the final velocity) now considering the equation v^2=u^2+2gs you substitute and rearrange ...
q -5 you add up all the PERCENTAGE uncertainities so 4 pie is a constant , l has 4% and T has 1% but remember there is a square here so now it is 2 x 1% = 2% now you add them up to get 6 %
q 8 - by the time it takes over the goods train they should have travelled the same distace so s1=s2 ..... 10t = 1/2 x 1/2 x t^2 ... solve and youll get 40 s
q-21----total pressure at base = pressure of water and oil.... 17.5 x 10 ^ 6 = 9.81 x (x) x 830 + 9.81 x 1000 x (2000-x ) solve it and youll get D as the answer,,,,

thank you very much

 

[ashoo123]

Hey could someone pls help with these questions... q 9, 15, 17 winter 2009 http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_ms_11.pdf

thnkx in advance!

 

[ziremm]

Hey could someone pls help with these questions... q 9, 15, 17 winter 2009 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_11.pdf

thnkx in advance!

q15
total momentum is zero.
momentum of 2kg trolley is 2*2=4
so 1*(-4)=-4 resultant momentum will be zero
4+(-4)=0
Kinetic eng of 2kg: 1/2 * 2* 2^2 = 4J
KE of 1kg= 1/2*1*4^2= 8J
8+4=12 J

 

[Thought blocker]

Hey could someone pls help with these questions... q 9, 15, 17 winter 2009 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_11.pdf

thnkx in advance!

9)
Here Pi = Pf
Pi = 2mu - mu = mu
now check in all optons, and see where does Pi = Pf
A C D has Pi =Pf now we are asked for Elastic collision, so A

17)
Theory based answer. Read the chapter - Matters.

 

[ashoo123]

q15
total momentum is zero.
momentum of 2kg trolley is 2*2=4
so 1*(-4)=-4 resultant momentum will be zero
4+(-4)=0
Kinetic eng of 2kg: 1/2 * 2* 2^2 = 4J
KE of 1kg= 1/2*1*4^2= 8J
8+4=12 J

Ohh! thanks for tht .. I was trying the resultant momentum equals zero thing but i messed up the symbols...thts y i couldnt come up with a reasonable ans

 

[ashoo123]

9)
Here Pi = Pf
Pi = 2mu - mu = mu
now check in all optons, and see where does Pi = Pf
A C D has Pi =Pf now we are asked for Elastic collision, so A

17)
Theory based answer. Read the chapter - Matters.

Umm... I did the Pi=Pf nd was confused between A and C...now momentum nd collisions isnt really my best area but in an elastic collision cant both objects move in the same direction?

 

[Thought blocker]

Umm... I did the Pi=Pf nd was confused between A and C...now momentum nd collisions isnt really my best area but in an elastic collision cant both objects move in the same direction?

No! Coz its the part of inelastic collision.
Book parhlo app

 

[ashoo123]

No! Coz its the part of inelastic collision.
Book parhlo app

Ohh ok! nd I barely understand hindi/urdu

 

[Thought blocker]

Ohh ok! nd I barely understand hindi/urdu

Read the books