AS Physics P1 MCQs Preparation Thread.

[Paparazzi478]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf
Question 13
can you show me the working?

 

[Jaf]

Q9 - this question we are supposed to use projectiles...
use v^2 = u^2 + 2as for each of the s values and add em up, you get 19...and the closest answer is 20 so D it is.

Wait... what? How? :O Elaborate.

 

[xyz!]

Q18- we subtract the pressure bcos the pressure at the surface is 100kPa and not 0 kPa...so the total pressure has to be 450-100 kPa

 

[TCHS]

Can someone pls help me with question 6 from the following paper:

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

The answer is C. I just need to know the working. Thanks!

 

[TCHS]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

Question 9 - how to do it?

 

[geek101]

need help in may june 2004

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf
Q9
Q23
Q25
Q26

make a change to no. 9 and try this:
first you calculate the time
using s= ut + 0.5at^2
use s = 1.25 you get t = 0.5 seconds...
now cuz teh time is same for both the components...
you use speed = d/t
use d = 10 and t = 0.5 you get 20 m/s

 

[geek101]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

Question 9 - how to do it?

first you calculate the time
using s= ut + 0.5at^2
use s = 1.25 you get t = 0.5 seconds...
now cuz teh time is same for both the components...
you use speed = d/t
use d = 10 and t = 0.5 you get 20 m/s

 

[hussamh10]

How do we do this? What I did was to find the resultant force down the slope(taking the friction force into account) then use that force to find the acceleration, and then use the formula v^2= 2as where s is 7m. I got 9.5m/s, which is not even in the options!!View attachment 11874

F irst of all fin the loss in mgh which is
2x3x9.81=58.86
then Work done against friction which is =5x7=35
So Ke
58.86-35=Half x2 x v^2
v=4.88
Option A

 

[Nirma]

Can somebody plz help me with N10/12/q22. And q8. I would be grateful to u.

Also, N09 q12 (pulley one )

 

[geek101]

Can somebody plz help me with N10/12/q22. And q8. I would be grateful to u.

ok here we want an arrangement highly sensitive for small masses and having a low sensitivity for smaller masses. Remembering, F = kx
F is constant, so an increase in K will mean a low extension. K and x are inversely proportional.
instantly reject D. It cant be B either makes no sense, there isnt a different sensitivity for large and small masses. But looking at A and C one of these is the answer. Just remember that the rigid box will stop any further extension, and will also reduce the sensitivity. In A, if a large mass is hung, the spring with the lower k produces a greater extension and is stopped by the box >> lower sensitivity to large masses. and also provides a high sensitivity to small masses, because the lower k spring wont touch the box, so no reduction in sensitivity, and so the answer is A. hope that helps...

 

[shuffle]

You can post your doubts about Physics P1 here, which will be cleared. Post any related material, notes etc. for help!

I have a problem in the May June 2008 paper 1. Question Number 10. How to solve this type of momentum problems??

 

[fathimaash]

For distance x from rest,

s = ut + 0.5at^2
x = 0.5a * t1^2

For distance h from rest,

h + x = 0.5a * t2^2

Place x from equation 1 into equation 2,

h + 0.5a * t1^2 = 0.5 * a * t2^2
0.5a(t1-t2)^2 = h
a = 2h/(t1-t2)^2

im sorry is da answer D bcoz accordin to ur da answr is C but MS its D

 

[Peter Check]

Why is the answer D? Should it not be B??? The tensio is the same throughout right so it will be 30N, and hence the torque will be 30 X 0.15=4.5!

 

[Peter Check]

I am getting A for this one? But that not right, why?

 

[iKhaled]

need help in may june 2005

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf
Q13
Q16
Q21
Q29

 

[USMAN Sheikh]

help needed pls in q3 and 4

 

[thunderingthunder]

help needed pls in q3 and 4

For question 4,
Speed=Distance/time,
So,40/2.50=16 m/s.
For uncertainty. Δs/s=( Δd/d)+( Δt/t)

Δs/16=(0.1/40)+(0.05/2.50)
Δs=0.36.Errors are always written correct to 1 sig. figure therefore error is ±0.4
hence,speed is 16 ±0.4 m/s.The answer is C.
Hope this helps

 

[USMAN Sheikh]

For question 4,
Speed=Distance/time,
So,40/2.50=16 m/s.
For uncertainty.Δs/s=(Δd/d)+(Δt/t)

Δs/16=(0.1/40)+(0.05/2.50)
Δs=0.36.Errors are always written correct to 1 sig. figure therefore error is ±0.4
hence,speed is 16±0.4 m/s.The answer is C.
Hope this helps

thankscan u pls answer my other uestion too

 

[thunderingthunder]

thankscan u pls answer my other uestion too

Estimate time to complete 100m=10sec
Average speed=100m/10sec=10m/s
Average k.e=1/2(80)(10^2)=4x10^3J
Therefore the answer is B

 

[thunderingthunder]

Hey guys can anyone please explain Q 14?why is the answer C?? every time I calculate it I get 0.67!
please help me..