AS Physics P1 MCQs Preparation Thread.

[ShaanSiddiq090909]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 4

 

[ahmed abdulla]

Can u plz tell the year..i need to see the diagram..is there any?

on 2008 q 11

 

[hope4thebest]

i did this 1 by elimination method
see by using more layers of spring increases extension.
and using smaller spring constant also increase extension.f=kx
so f/k=x
so smaller spring constant increase the extension
and y.m=stress/x/l
so e=stress*l/x
s0x=stress*l/e
so smaller young modulus increases extension
so the correct ans is B
Read the question carefully.they said which change will nt have the desired effect that means less x

Thanks brother

 

[magnesium]

on 2008 q 11

sory ..i didn't get it bt em copy pasting the solution posted in the other thread
. for the 2Kg box...because it is accelerated downward so the weight is greater than the tension, so W-T=ma...so ( 2 x 9.81) - T = 2a , then for the 8 Kg box...the tension is grater than the friction because the 2 kg box is pulling it downwards so, T-F=ma, so T - 6= 8a...add both equations..so [((2 x 9.81)-T= 2a) + (T - 6)= 8a] so T will cancel each other and it will be like this: (2 x 9.81) - 6= 10a so a = 1.4

 

[Jasmine 12]

8. In an elastic collision, the relative speed of approach = the relative speed of separation.
9. Use F=m(v-u)/t.
12. All you gotta do is locate distances and determine forces.
13. Vertical component of the velocity of a body in projectile motion is 0 at the highest point, so the K.E. can come only from the horizontal component, which is constant throughout.
14. K.E. of the trolleys when they're moving = the E.P.E. stored in the string initially.
15. Power output/Power input=Efficiency. You have Po and Efficiency; Pi will be x.

Thanks!
Can u tell q18. http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf

 

[ShaanSiddiq090909]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 13 and 14 pleaaaseeee!!!

 

[ahmed abdulla]

sory ..i didn't get it bt em copy pasting the solution posted in the other thread
. for the 2Kg box...because it is accelerated downward so the weight is greater than the tension, so W-T=ma...so ( 2 x 9.81) - T = 2a , then for the 8 Kg box...the tension is grater than the friction because the 2 kg box is pulling it downwards so, T-F=ma, so T - 6= 8a...add both equations..so [((2 x 9.81)-T= 2a) + (T - 6)= 8a] so T will cancel each other and it will be like this: (2 x 9.81) - 6= 10a so a = 1.4

what i was doing .. solving the simultanous eqn wrong
any way thanks

what about this ?​

​

An electric power cable consists of six copper wires c surrounding a steel core s​

1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a​

resistance of 100 Ω.​

What is the approximate resistance of a 1.0 km length of the power cable?​

1.6 Ω​

​

o/n2008 q32​

 

[ShaanSiddiq090909]

what i was doing .. solving the simultanous eqn wrong
any way thanks
what about this ?

An electric power cable consists of six copper wires c surrounding a steel core s
1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a
resistance of 100 Ω.
What is the approximate resistance of a 1.0 km length of the power cable?
1.6 Ω

o/n2008 q32

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 13 14 and 15 pleaaaseeee!!!
JAZAKALLAH!

 

[ahmed abdulla]

q

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 13 14 and 15 pleaaaseeee!!!
JAZAKALLAH!

q13 use torque=tension * radius

 

[Lyfroker]

q#11, 13, 17, 19 & 32
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_1.pdf

 

[ahmed abdulla]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 13 14 and 15 pleaaaseeee!!!
JAZAKALLAH!

q14

​

​

at highest point, its vertical velocity is zero,​

and horizontal speed remains same,​

let initial velocity is V, then K.E. is 1/2mV*2​

at highest point horizontal velocity is V / root2​

so K.E. = 1/2mV*2 x 1/2​

ie .05E​

 

[ahmed abdulla]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 13 14 and 15 pleaaaseeee!!!
JAZAKALLAH!

i think u didnt understand
here is better expaination


vertical velocity= u cos 45
horizontal velocity = u sin 45
so, initial K.E. : the vertical and horizontal components of kinetic energy = 1/2 m ( u cos 45 ) + 1/2 m ( u sin 45 ) ......... ( 1/2 m v^2 formula )

so, sin 45 = 0.71
....cos 45 = 0.71 ( check them in the calculator )

so they will give the same kinetic energy for both velocities ( horizontally and vertically )

so at the highest point, (theta) = 90 ( for the vertical velocity ) , horizontal velocity doesn't change
so final kinetic energy = 1/2 m ( u cos 90 ) + 1/2 m ( u sin 45 )..................... ( horizontal velocity is constant ) ( cos 90 is 0 )
final kinetic energy = 1/2 m ( u sin 45 ) + 0 only which is half of the initial kinetic energy.

 

[TaffsAsLevel]

13. All you gotta do is choose a pivot and work out the moments on either side of it...
14. mgh=1/2m(v^2). mgh/2 = 1/2m(x^2). Eliminate m, g, and 1/2. Cross multiply.
15. Find the angle to the horizontal of the slope. Then find the component of the weight parallel to the slope. Then use F=ma to find the acceleration. Then use a suvat. You can also do it using 1/2m(v^2)=mgh.
34. Resistance = ql/A. A=lxl, so Resistance=ql/lxl. l and l cancel out; R=q/l. l^3=V, so l = cube root of V = V^(1/3), so R=q/V^(1/3).

Hi please solve june 2010 paper 11 question 33

 

[ahmed abdulla]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Question no 13 14 and 15 pleaaaseeee!!!
JAZAKALLAH!

use total momentum before collision=total momentum after collision
therefore, 2*2=1*v
therefore speed of 1 kg trolley is 4..
now u have mass and velocity of both trolleys , right ?
find kinetic energy of both ... and simply add them !
hope that helps !

 

[Raghad Dia]

When you are told that an object is falling freely under the effect of gravity, does this mean that there is no drag??

 

[Your-Blood]

When you are told that an object is falling freely under the effect of gravity, does this mean that there is no drag??

Yeah. Acceleration is considered to be constant 9.81ms^-2

 

[Raghad Dia]

May june 2010 varyant 11 #22... How do we solve such questions??:s

 

[magnesium]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_1.pdfQ`15

 

[Tkp]

Thanks brother

welcome

 

[nerds and geeks]

can anyone help me with june o6 ques 36